By Mike Pearson (Gmp26) on Wednesday, February 17, 1999 - 03:51 pm:
No-one likes to be the first one to create
a new topic, so I thought I'd get things underway.
There's a famous puzzle popularised by Lewis Carroll, that goes
something like this...
Imagine a rope over a frictionless pulley. On one side hangs an
overactive monkey exactly balanced on the other side by a weight.
What happens when the monkey climbs the rope?
Mike Pearson
By Ben Plommer on Monday, March 1, 1999 - 06:53 pm:
the weight rises half the distance the monkey climbs up
By Mike Pearson (Admin) on Monday, March 8, 1999 - 12:11 pm:
By Richard Marriott on Tuesday, March 30, 1999 - 06:58 pm:
By conservation of potential energy - so not too overactive! - the weight should fall by the same amount the monkey climbs
By Mike Pearson (Gmp26) on Wednesday, March 31, 1999 - 10:04 am:
Richard,
I'm not sure this reasoning works!
Is Conservation of Potential Energy a sound principle to
use? Can it not be converted to kinetic energy or heat? Also, the
monkey may add a certain amount of kinetic/potential energy by
climbing the rope.
Still, I think looking at Kinetic + Potential energy may get
us somewhere, provided you can quantify the contribution the monkey
makes from energy extracted from its last banana! (That may be
possible!)
Mike
By marriott on Saturday, April 3, 1999 - 01:35 pm:
yeah, my mistake. I meant mechanical energy with no Kinetic energy involved. Still wooly thinking on mu part. How about saying that what we have is an impulse caused by the monkey which will mean that both monkey and weight will both rise. There has been a net input of mechanical energy into the system of 'monkey+weight' reflected by the relative positions of both being at a higher potential energy level.
By Daniel Goldberg on Thursday, April 8, 1999 - 11:53 am:
Daniel Goldberg wrote in with some good
advice on solving this problem. Thanks Daniel!
This situation must be looked at carefully, not being of a
'standard' type taught at school. Experience has shown that one
must try to distance oneself from intuitive notions about actions
and desires (such as 'pull' and 'climb') and stick rigidly to
accepted principles (for example, Newton's 3 laws, F=ma and the
principles of conservation of energy and momentum), carefully
applied. It also helps to think of analogous but perhaps simpler
situations. Consider, for example, the following scenarios (in
which all ropes are light and all pulleys smooth);
- A man stands on the ground and pulls on a rope attached, via a
pulley, to an object of considerably lower mass than he. What
happens? Can we solve such a problem by consideration of the
principle of conservation of energy? Can we solve such a problem by
consideration of the principle of conservation of momentum?
- Two people of identical mass are standing on skateboards (ie on
effectively frictionless surfaces) on horizontal ground. They are
attached by a rope which is just taut. They both simultaneously
begin to pull on the rope, each pulling as hard as the other. What
happens?
- The two people in scenario 2 go back to their initial positions. This time, only one of them pulls on the rope. How does this alter the subsequent motion?
Once you are aware that scenario 3 contains not only a trick question but also a 'trick statement' then you will be able to apply your answers to 2 and 3 to the monkey problem and very quickly find the solution.
Note that there is also an important difference between scenarios 2 and 3 and the monkey problem, as set.
By Ben Plommer (P731) on Saturday, May 1, 1999 - 05:53 pm:
The weight would rise half the distance the monkey climbs because this would keep the monkey at the same height as the weight, and as the monkey is the same weight as the weight, this is the equalibrium position.
By Mike Pearson (Gmp26) on Tuesday, May 4, 1999 - 09:13 am:
Hi Ben! Thanks for explaining your
reasoning. I think you probably have the right idea. It is possible
to use a symmetry argument to work out what is going on here -
especially in the light of Dan's comments earlier. The only thing
that bothers me is your last phrase. Given that the monkey and the
weight weigh the same, (and if we assume a negligible weight of
rope), then ANY final resting position is stable. So it doesn't
make sense to talk about THE equilibrium position.
Mike
By Anonymous on Monday, May 24, 1999 - 03:26 pm:
I think we have to be careful about what we mean by 'climbs'.
When we claim that the weight rises half the distance that the
monkey climbs, I think we mean relative to the rope, which is
moving. Thus the monkey and the weight move at the same rate, and
eventually meet at the pulley.
I think the monkey exerts a force on the rope, which pulls the
weight up. The tension in the rope ensures that the force pulling
both the monkey and the weight up is the same.
Since the same force is applied to both monkey and weight, they
rise at rates determined by their relative masses according to
F=ma. Since the masses (m) are the same, and the forces (F) are the
same, so the accelerations (a) will be the same.
By megan on Saturday, July 31, 1999 - 10:55 pm:
I think that the overactive monkey gets pulled over the top of the frictionless pulley by the weight on the other side.
By Graham Lee (P1021) on Tuesday, August 24, 1999 - 10:47 pm:
To use Newton's 3rd Law, any action must have an equal and
opposite reaction.
Therefore, if the monkey pulls himself up on the rope, and there is
no friction to counter his action, the rope must move down. There
is no change of momentum in the monkey, so the weight moves up the
distance that the monkey "climbed", although the monkey does not
move. Only when the weight (which I assume is not a point source of
mass) jams the pulley will the monkey be able to move... unless he
lets go!
By Graham Lee (P1021) on Friday, September 3, 1999 - 07:02 pm:
Actually, when I said the monkey wouldn't move, I was lying.
Let's assume he can climb at some constant velocity v. As he
climbs, the rope accelerates beneath him... or does it? I've just
changed my mind. I will now tell you why.
If the monkey pulls down on the rope (in an effort to climb), he
exerts a force mg on the rope, and the rope exerts a force mg on
him (if he ways mass m, and if the monkey is male. I suppose only a
male monkey would be stoopid enough to get into this situation, as
half of the population would agree:-). The force mg on the rope is
exactly countered by the mass on the other end, but there is
nothing stopping the monkey from moving, ergo he does. The monkey
can climb the rope, and as long as he doesn't let go the rope and
mass won't move.
GL
By Michael Doré (P904) on Friday, September 3, 1999 - 11:46 pm:
The rope does accelerate: the force on the rope must be greater
than mg otherwise the monkey would not accelerate (I think).
However if the monkey starts with an upward speed of v then as you
say by making very skilful arm movements (maintaining contact with
the rope continuously) it would be possible for the monkey to
ascend with velocity v and keep the weight in the same
position.
Michael
By Graham Lee (P1021) on Monday, September 6, 1999 - 08:09 pm:
All I know is that the monkey cannot pull down on something with
a force greater than mg newtons, unless he also has a supply of
bananas carried on an antigravity plate (so that they don't weigh
anything until he eats them).
Anyway, on another note, I have now decided that the monkey
accelerates the entire system by pulling, as it was never specified
that there was a gravitational field (or indeed any force field) in
operation on the system.
By Michael Doré (P904) on Monday, September 6, 1999 - 09:52 pm:
Why can't the monkey pull down on something with a force larger
than mg? Imagine it had a bar magnet in its hand and that the rope
was made of metal. Now there would be an force exerted on the rope
by the monkey and it could be as large as you would like. All
contact forces are basically electromagnetic forces in disguise
anyway so the force on the rope isn't limited by the weight.
By the way in the question it does say that there was a "weight" on
the end of the rope so this implies that the whole system is in a
uniform g-field.
Michael
By Graham Lee (P1021) on Wednesday, September 8, 1999 - 07:17 pm:
I'm afraid to tell you that the EM force from the monkey on the
rope IS limited by the weight of the monkey.
The force between the monkey and the rope is not the weight of the
monkey directly, as you pointed out but is instead friction between
his(her?) hands/feet and the rope. His hands are connected to the
rest of his body, which is all trying to accelerate towards the
mutual CoG of himself and the Earth (we assume this is the Earth we
are talking about) with an acceleration g. Therefore the force on
the rope (which is stuck to the monkey) is mg newtons downward.
By Michael Doré (P904) on Wednesday, September 8, 1999 - 09:03 pm:
But when the monkey pulls it is accelerating upwards (say with
acceleration a). So now the acceleration difference between his
actual acceleration and his "preferred" acceleration (g) is g + a
not g. So the overall force is m(g + a). If the force on the rope
was mg then by Newton's 3rd law the force on the monkey would be mg
so the monkey wouldn't be able to get started!
Michael
By las on Friday, December 17, 1999 - 09:49 am:
We think that there is a simpler way of doing this. We don't know what, but we will keep thinking!
By Graham Lee (P1021) on Thursday, December 23, 1999 - 06:40 pm:
Hello! Me again!
The monkey can most definitely NOT accelerate the system, as there
is no external force acting. He pulls down on the rope, the rope
pulls up on him. I think.
P.S. does anyone know if these Lewis Carrol (Charles Dodgson)
puzzles are published anywhere, and give me ISBN number etc?
Cheers,
Graham.
By Michael Doré (P904) on Thursday, December 23, 1999 - 08:28 pm:
The monkey definitely DOES accelerate the system by climbing.
Resolving forces on the monkey upward:
T - mg = ma
In order to start climbing upward a>0 where a is the upward
acceleration. So T - mg > 0 and T > mg.
Now as the rope is light and inextensible, the resultant upward
force on the weight is T - mg as the weight has the same mass as
the monkey. Now T>mg so the upward resultant force on the weight
is positive, therefore the monkey accelerates the weight upward by
climbing.
It is true that if you consider the centre of mass of the ENTIRE
system including the rest of the Earth there is no acceleration as
there is no external force. But this is because the Earth moves
very slightly downwards to compensate for the monkey and weight's
ascent.
Michael
By Graham Lee (P1021) on Monday, December 27, 1999 - 07:40 pm:
A good point Michael - I keep forgetting that the Earth is a
non-inertial reference point. Just remember not to listen to any
physics teachers until you get to Uni, and you'll go a lot
further.
An example of this is when I got a physics lecturer at a uni I went
to completely flummoxed by saying that a field was "The space in
which fundamental particles emitted by an object can have a
measurable effect when absorbed by another object".
By p904 on Wednesday, December 29, 1999 - 06:30 pm:
Hi Graham,
Well that's certainly a very original definition of a field but I
guess it works. Would you say that what you described has to be a
field? (I'm not really sure.)
I'd better not comment on "don't listen to physics teachers"...
just in case my school are reading this site.
Happy New Year/Millenium
Michael
By Graham Lee (P1021) on Thursday, January 6, 2000 - 02:27 pm:
If you believe QG theory, then yes, a field is where the
fundamental particles can be measured to have an effect.
If you think about it, all objects emit the particles relating to
the fields that they exhibit (such as gravitons, photons, gluons,
W+/W- and Z0). If a particular
particle (!) did not find anything to interact with inside what we
consider to be the field, it would keep going to infinity (unless
the fundamentals have a half-life, which I think defeats the
object). It might be absorbed outside the field, but the momentum
change due to one particle is very small as to be
immeasurable.
On this subject (I know we weren't, but...) can anyone rationalise
the graviton theory for me? As far as I can see, an object emits
gravitons uniformly (if it is a point mass) in all directions,
which I am happy with because the object itself does not move, so
momentum is conserved. But when a graviton hits something else,
something strange happens.
A graviton travelling with momentum p(c) in the positive-x
direction hits and is absorbed by a body, let's say it's at rest.
Now this body has a momentum P(v) in the negative-x direction! Is
the graviton meant to have a negative inertial mass (in which case,
a body emitting a graviton would gain energy and become
heavier!)?
I realise this is a bit deep, but consider it awhile.
GL.
"The time has come the walrus said to think of many things. Of
chocolate bars and jelly sweets, of crips and onion rings..."
By Michael Doré (P904) on Thursday, January 6, 2000 - 08:05 pm:
Hi!
I see what you're saying about QG theory (which I didn't know
actually existed yet). What about a photon (e.g. a gamma ray)
interacting with matter depositing energy? Is this directly
associated with an electric field?
For your question, I really shouldn't attempt an answer as I almost
certainly don't know what I'm talking about... The bosons an entity
emits causing a field are due to the uncertainty principle. (They
are virtual particles whose energy is below the uncertainty limit.)
These interact as you described and transfer momentum. (For reasons
that have never been fully clear.) So there is no real problem with
the mass being negative and the momentum in the opposite direction.
After all during a quantum fluctuation, one of the virtual
particles has negative mass. (And this is what gives rise to
Hawking radiation from black holes.)
So now I think I'll give somebody who has a clue the chance to
reply...
Michael
By Graham Lee (P1021) on Monday, January 31, 2000 - 09:44 pm:
Hiya!
Ooops!
I accidentally said that bosons could go to infinity, didn't I?
Well, of course bosons are created from non-existent energy, and
therefore must obey the uncertainty principle DEDt>h/2p, implying that they can only exist for a time
inversely proportional to their mass before they must interact,
either with another particle or with the one that created them
(which in certain cases can mean that they go back in time -
cool).
Gravitons cannot be bosons, as they could never go to infinity
without having infinitesimal (i.e. 0) energy. But then, neither can
they be fermions without seriously affecting the quantum numbers
(e.g. spin, but don't think of a spinning particle) of their
emitter.
Unfortunately objects that have negative energy tend to be short
lived and must be followed by a burst of positive energy bigger
than their negative counterpart. I don't think though that true
negative mass from quantum fluctuations is what comes from black
holes, but matter-antimatter pairs (an antimatter particle is a
matter particle with g negative).
Graham
By Alex Barnard (Agb21) on Tuesday, February 1, 2000 - 12:34 pm:
Urm... some problems in the above
reply!
The photon is a spin 0 particle (and hence it is also a boson -
this is something called the Spin-Statistics connection). It is the
force carrier for the EM force and this is very well known to have
infinite range. The 'reason' for the infinite range is that the
photon has 0 rest mass and hence it can be 'virtually created' with
any energy larger than 0. Hence by an uncertainty principle they
can last for arbitrarily long periods of time.
The graviton can be shown to be a spin 2 particle (and hence it IS
a boson). It is also massless and hence has an infinite range.
Being a particle of spin 2 it is represented by a tensorial field
(rather than a scalar or vector which do for spins 0 and 1) and it
is due to some rather subtle properties of the tensor field that it
only interacts attractively.
AlexB.
By Graham Lee (P1021) on Thursday, February 3, 2000 - 02:28 pm:
If the graviton is spin 2, how do spin conservation equations work during graviton interactions? Unless there are an infinite number of them floating around, which is plausible but not practical (as it requires a uniform homogenous gravitational force throughout the universe, which would mean that any particle in the universe would be in freefall when compared with any other particle in the universe).
By Julian Steed on Wednesday, April 5, 2000 - 10:06 am:
To return to reality...
Surely the point is that looking at the system as a whole, it
starts in an equilibrium. The monkey is not in reality a point
mass, though the weight may as well be and it is tempting to think
of the monkey as such. The monkey is really both a mass in a
gravitational field (and so possessed of potential energy) and a
store of "biological energy" in that he can convert stored food
into mechanical energy and thereby do work. That is why biological
systems like monkeys and people are able to jump up in the air
without needing antigravity plates loaded with bananas.
Now we know the system starts in equilibrium with the weight forces
balanced. Think about what happens next - the monkey takes one hand
off the rope (nothing significant happens); the monkey raises that
arm in the air (nothing significant happens); the monkey attaches
that free hand to the rope higher up (nothing hapens); now the
monkey tries to pull himself up, a bit like doing a one-handed chin
up. His weight is exactly balanced by the weight on the other side,
so he can get started on this. In doing the pull up he converts
stored biological energy into mechanical energy, which in turn, if
the other weight was anchored, would be converted into increased
potential energy. The monkey would necessarily lose some mass in
the process. But the instant the monkey does work he converts some
mass to energy and as soon as that happens the constant weight of
the weight on the other side exceeds the weight of the
monkey.
Result - over he goes.
By Michael Doré (P904) on Monday, April 10, 2000 - 04:01 pm:
Hi, I think that you are nearly right, but you say that the
monkey converts some mass into energy. In fact the mass of the
monkey is constant. (Well, okay, he does lose a tiny bit of mass
due to E = mc2, where E is the gain in GPE and m is the
loss in mass. But this would be enough to create a force imbalance
that would pull him up by 1cm in approximately 491 days!) The
reason the monkey gets pulled up is simple: during the climb the
tension in the rope exceeds the monkey's weight. This also ensures
that the point weight on the other side gets pulled up.
I've learnt a bit more mechanics in the meantime, and the concept
of angular momentum can help with this problem.
The angular momentum of a set of objects, about an axis is the
moment of the linear momentum of the objects. Now angular momentum
is always conserved about an axis if the axis doesn't exert a
torque on the system. Because of the symmetry of the situation
there is clearly no torue exerted by the axis. So the total angular
momentum of the monkey and the weight is zero. The perpendicular
distance to the monkey is the same as that to the weight, therefore
it follows that the velocities are the same (so that the moment of
their momenta can cancel). Therefore the monkey and the weight
travel with the same speed at any point - so the distance the
monkey climbs relative to the rope is double the distance the
weight rises.
Michael
By Brad Rodgers (P1930) on Monday, April 10, 2000 - 10:07 pm:
I am not sure on this, and in the most theoretical standpoint I
believe that the last post is correct; however, it seems to me that
as the monkey must first move his arm downwards and then move his
body upwards-it would not be a simultaneous occurance(in the
newtonian sense). Therefore, as he exerted a force on the rope, he
would pull the rope a distance, causing the wieght to go up, and
himself to go down. If he were in a reasonably strong gravitational
field, he would then be pulled down, while the weight would be
pulled up and this action would continue in greater exageration.
So, I think that the strength of the gravitational field might play
a part.
Brad
By Michael Doré (P904) on Monday, April 10, 2000 - 10:29 pm:
No the field strength cannot make a difference to what
eventually happens - it can only affect when it happens or to what
extent it happens. You can show this using dimensional analysis.
According to you there should be a critical field strength, for
which an event occurs. This is measured in m/s2. But if
you look at the quantities you know about the system (the masses,
the lengths of the objects etc) there is absolutely no way you can
get the unit s from anywhere. Therefore there cannot be a crictical
field strength, so the field strength doesn't affect what
eventually happens. I'm assuming there is no friction.
Also, you say that the monkey could pull on the rope causing
himself to go downwards and the weight to go upwards. This can
never happen. (Assuming the rope is weightless and the system
frictionless of course.) If it did happen then the angular momentum
of the system would no longer be zero. If you like I could show you
the derivations of the laws of angular motion - they look very much
like those of linear motion. For example you have
F = ma
replaced by
L = Is
where s is the angular rather than linear acceleration and L is the
resultant torque on the system from the centre of mass or the
instantaneous centre of rotation. It's quite an interesting
area.
Yours,
Michael
By Neil Morrison (P1462) on Tuesday, April 11, 2000 - 08:04 pm:
Michael-
If you were using the idea of angular momentum, surely you could
get the unit s:
If L is angular momentum,
L = Iw and I = mr2 depending on what objects you are
using. v=wr, and so L = mvr = msr/t. substitution of v is
unneccessary depending on various circumstances.
In any case, the monkey is not moving in a radial path relative to
the pulley, so how can you be using angular momentum?
By Michael Doré (P904) on Tuesday, April 11, 2000 - 09:48 pm:
Hi Neil,
The angular momentum about a point is simply the sum of the moments
of the linear momenta of all the particles in the object. There is
no need for the path to be radial. You can see this result by
looking at the kinematics of a particle moving in a 2-D
plane.
Here are two standard results: (which can be verified by
representing the particle's position as a complex number in polar
co-ordinates then differentiating).
For a particle with polar co-ordinates (r,q) the acceleration torwards the centre is:
r''-rq'2
The transverse acceleration is:
1/r d/dt(r2q')
It's the latter we're interested in. For a particle, if m is its
mass and F is the force on it, r is its distance from the origin
you get:
Fr = d/dt(mr2q')
Add this equation to itself for every particle present in the
object we're consdering and you get:
Total vector moment = d/dt(total angular momentum)
(analogous to total force = d/dt(linear momentum) for linear
mechanics)
In our case the total moment of the forces about the top of the
pulley is zero fairly obviously (the moment due to the weights
cancel and the reaction doesn't have a moment), so the angular
momentum of the rope, combined with the two weights is constant. As
the rope is light it will not have any angular momentum so we just
need worry about the monkey and the weight. Suppose the radius of
the pulley is r, the speed of the monkey u and the speed of the
weight v then:
rv - ru = constant
(The sign convention is to reflect that the moments of the speeds
act in different senses. This is quite easy to see from the
diagram.)
So v - u = const
But initially v - u = 0 so v - u = 0 always, so v = u always, and
their speeds are identical.
The point I was making about the dimensions was that the field
strength cannot affect whether the monkey gets started or not. (One
thing I forgot to mention was that I assumed that the monkey is
always strong enough to pull with a force exceeding his/her own
weight. I don't think this was what Brad was getting at though, as
this is rather trivial.)
Say the situation was such that beyond a certain field strength,
the monkey simply can't get started. The point is that then you
could work out the critical field strength in m/s2. Now
this should be a function of the conditions of the setup. So we
could reasonably expect this to be dependent on the mass of the
monkey, or the radius of the wheel. But there is no way we can get
m/s2. The angular momentum is not a function of the
setup, as this will depend on the field strength and the force with
which the monkey feels like pulling.
Yours,
Michael
By Michael Doré (P904) on Tuesday, April 11, 2000 - 10:28 pm:
Just a couple of things to add.
Three messages up - I spoke of the angular momentum about an axis,
and the torque exerted on the axis. This might have been slightly
confusing as I was referring to an axis perpendicular to the plane
of the wheel at the top. (It's easiest to look at the diagram given
in the very first message in this topic.) Even easier is to
conserve angular momentum about a point rather than an axis because
this always works. So read my above message as: there is no
external moment on the system of ropes/weights about the point the
rope touches the pulley, so the angular momentum of the
rope/weights are constant. Sorry about this.
Also, above I forgot to mention that the internal forces within the
object have a total moment of zero. This follows from Newton's 3rd
law. This is why when you add the equation:
Fr = d/dt(mr2q')
to itself for all the particles you get:
Total external moment on system = d/dt(angular momentum)
about any point.
Yours,
Michael
By Brad Rodgers (P1930) on Wednesday, April 12, 2000 - 01:27 am:
Ok, I was envisioning that it was just the pulley that was
frictionless and not the whole system. With this assumption, you've
shown quite well that the weight will travel up(with respect to an
inertial observer not moving with respect to the pulley) half the
distance the monkey climbs the rope.
Brad
By Brad Rodgers (P1930) on Wednesday, April 12, 2000 - 01:29 am:
but, if the system is frictionless, how could the monkey pull himself up in the first place.
By Dan Goodman (Dfmg2) on Wednesday, April 12, 2000 - 02:57 am:
This isn't too related to the rest of the
conversation, but here is a computer program that actually has a
monkey on a rope!
http://www.hypermatter.demon.co.uk/trapeze.zip
[Read on before downloading - The
Editor]
Well, OK, it's on a trapeze, but I couldn't pass up the opportunity
to post it here. I couldn't get it to run on my computer, it says
you need a pentium 450 or more to run it. Oh well.
By Michael Doré (P904) on Wednesday, April 12, 2000 - 12:52 pm:
Hi Brad,
Yet another confusing statement on my part. When I said that the
system was frictionless I did mean externally, i.e. the pulley
doesn’t exert friction on the system (the rope, monkey and
weight). (Interesting things happen if you drop this requirement.
Anyone got any ideas? Persumably the monkey and the weight will
start to swing from side to side.) Anyway the conclusion still
holds good if there is internal friction. (As there has to be.)
When you add the equation:
Fr = d/dt(mr2q’) to
itself for all the particles, then if an equation contains an
internal frictional moment L then by Newton’s third law,
another equation must have –L, so all internal friction
cancels when you add the equations. You get the equation:
Total external moment = d/dt(angular momentum)
from which the conclusion, that at any point their speeds are
equal, holds.
The fact that there is internal friction also doesn’t affect
the dimensions argument. According to experiment the friction
between two bodies is given by:
F £ uR
where R is the normal reaction force and u is the constant of proportionality, dependent on
the material which both bodies are made out of. As R and F are both
forces, u is a dimensionless constant,
so cannot be used to get the unit second either.
Yours,
Michael
By Michael Doré (P904) on Wednesday, April 12, 2000 - 01:18 pm:
Dan,
I got that program to work on my computer (100MHz I think). It is
certainly worth having a look at if possible - the way the wire
holding the trapeze curves at different stages is very convincing,
(and probably very hard to calculate!)
Thanks,
Michael
By Michael Doré (P904) on Wednesday, April 12, 2000 - 01:47 pm:
OK, I don't know too much about special relativity, but I'd like
to know what is wrong with this argument (my attempted derivation
of E=mc2).
Suppose we have a particle of rest mass m, which is being pushed by
a constant force F, starting from rest. The equation of motion for
the particle should be:
m dv/dt / sqrt(1-v2/c2) = F
where the sqrt(1-v2/c2) is to account for the
mass increase, due to its speed.
If you write dv/dt as v dv/dx and integrate both sides you
get:
mc2-mc2sqrt(1-v2/c2) =
Fx
Now Fx is the work done. So by energy conservation:
Fx = increase in kinetic energy + increase in mass energy
So increase in mass energy
= mc2-mc2sqrt(1-v2/c2)
- 1/2mv2/sqrt(1-v2/c2)
But if E=mc2 is true this should equal:
(m/sqrt(1-v2/c2) - m)c2 which I
don't believe it does. Help!
Michael
By Neil Morrison (P1462) on Wednesday, April 12, 2000 - 07:36 pm:
How about considering an electron accelerated through a
potential difference?
Total energy before = moc2 according to
Einstein, so total energy after :
E = m0c2 + qV = mc2
so m-m0 =qV/c2
This is a nice simple way of expressing the change in mass.
Michael- if you're using relative rules (or trying to prove them)
should you really be using (1/2)mv2 at all?
On the subject of relativity, suppose we're travelling on a
spaceship at something like 0.99c. I'm standing at one end, and
you're at the other, which are l metres apart (when the ship is not
moving). How I shine a torch towards you, and you raise your hand
when you see it go on, and when I see that you've done this, I
switch the torch off. When you see that its off you lower your
hand. If I start my stopwatch when I switch the torch on, and stop
it when I see your hand go down, what will it read?
Regards,
Neil M
By Michael Doré (P904) on Wednesday, April 12, 2000 - 08:22 pm:
I think you're perhaps right about my use of 1/2mv2.
I will need to think through the assumptions special relativity
makes (which I'm not entirely clear on) a bit more carefully.
As for your puzzle, I make the answer 4l/c, providing of course all
human reactions are instantaneous. Or have I missed
something?
Yours,
Michael
By Brad Rodgers (P1930) on Wednesday, April 12, 2000 - 09:33 pm:
I see what you are saying; I am now in full agreement with your
mathematics on the Monkey on a Rope problem.
As far as the answer to Neil's puzzle, it depends on whether the
spaceship undergoes any acceleration. Assuming that it is
inertial(and all human actionsare instantaneous), I get 2l/c, as
the light would only need to travel the distnce of l once to get
from you to me and again to get from me to you. But I think I may
be misreading the puzzle.
By Kerwin Hui (P1312) on Thursday, April 13, 2000 - 03:21 am:
I think we can deduce the formula
m=m0/sqrt(1-(v/c)2) from E=mc2,
thus 'proving' E=mc2.
Assuming E=mc2.
From: Power=F.v
We have: dE/dt=F.v
Newton's second law gives: F=d/dt(mv)
substituting everything gives: c2dm/dt=v.d/dt(mv)
Now the trick is to multiply both sides by 2m and integrate, we
obtain
c2m2=m2v2+C
and since v=0, m=m0, we have
C=m02c2, where m0 is
the rest mass and m is the relativistic mass.
m2(c2-v2)=m02c
2
A little bit of rearranging gives the relativistic correction of
mass, which is observed experimentally.
Michael, if you are to use F=rate of change of momentum, surely you
must put the m inside d/dt to give
F=d/dt(mv2/sqrt(1-(v/c)2), as m is not
constant?!
Kerwin
By Michael Doré (P904) on Thursday, April 13, 2000 - 12:32 pm:
Ah yes you're absolutely right - I always make that mistake. F =
ma applies to isolated systems but here we've got mass being
transferred from the source of the force to the rocket. Thanks for
pointing it out. I usually try to avoid F = d/dt(mv) because it
assumes that the mass being gained/lost ends up/starts off with a
velocity of zero.
Neil's point was also very valid - there is no need to take into
account KE, because E = mc2 is designed to take into
account the entire energy.
So my first line should have read:
F = d/dt(mv/sqrt(1-v2/c2))
Multiply both sides by dx/dt, you get:
dE/dt = v d/dt(mv/sqrt(1-v2/c2))
Make the substitution u = v/sqrt(1-v2/c2) and
integrate from u = 0 to u = V/sqrt(1-V2/c2)
(when v = V):
DE = mc
sqrt(c2/(1-V2/c2)) -
mc2 = c2Dm as
required.
Anyway, your argument also works very nicely - it is basically what
I was trying to do except in reverse. What made you think of
starting with E=mc2?
Here is a question that has always puzzled me on special
relativity. We have a rocket, firing out fuel, gaining momentum,
approaching the speed of light, apparently gaining mass. But how
can the rocket gain mass? It will gain some mass by the
1/sqrt(1-v^2/c^2) factor, but then it must lose the same amount of
mass by virtue of the chemical/nuclear energy it uses in its fuel.
In other words the rocket cannot be gaining energy as it
accelerates because there is no external source. Therefore how can
it gain mass?
Thanks,
Michael
PS. What do you think of Neil's problem?
By Neil Morrison (P1462) on Thursday, April 13, 2000 - 08:47 pm:
I think my problem is less simple than you first thought, but I
haven't tried to answer it, I just stated it! Ok, the rocket isn't
losing any mass because its not using fuel so its not accelerating
either. That makes it easier. What's difficult is that light will
move relative to the ship: Its overall speed cannot be more than c
right? But the ship is moving 0.99c, so the light quanta cannot
move faster than 0.01c relative to the ship on the way forward, but
they can on the way back. What I wonder is, how fast can they move
back? If they move at c backwards, then relative to the ship they
are going at -1.99c. So is this possible, or can relative speeds
not exceed c either?
Anyway, there's time and length dilation to consider. I'll try and
solve it myself, and I'll think about your problem.
Regards
Neil M
By Michael Doré (P904) on Thursday, April 13, 2000 - 09:24 pm:
Hi Neil,
All the effects you are describing must surely cancel. As long as
the spacecraft is going at a constant speed (or is in an inertial
frame) then its speed doesn't matter at all. Simply switch the
frame of reference to one which is also moving at 0.99c, and the
laws of physics still work. This is the first assumption of special
relativity. (The other being that the relative speed of light in a
vacuum is always c, no matter what your speed.)
The 0.99c bit was irrelevant - you will always be moving at 0.99c
relative to something. We might as well be talking about your
school lab (its frame of reference is inertial, nearly). Remember
there is no such thing as ether...
Yours,
Michael
By Neil Morrison (P1462) on Friday, April 14, 2000 - 07:07 pm:
To simplify my question:
A has the torch, B waves his hand.
The direction of A to B is the way the ship is moving, at
0.99c.
Surely light can only travel along this distance at c overall, and
therefore at 0.01c relative to the ship. But in the opposite
direction (B to A), light
can travel at -1.99c relative to the ship. Now the ship has length
L. We will assume that reactions are instant, and
that the raising and lowering of B's hands is an instant
occurance.
The time taken for B to see that the torch is on is L/0.01c
Time for A to see B's raised hand is L/1.99c
Time for B to see that the torch is off is L/0.01c again
Time for A to see hand down is L/1.99c again.
Now the sum of these times will be the value on the
stopwatch.
Time = 2L/0.01c + 2L/1.99c
= 400L/1.99c
I mentioned that time/length dilation may effect our measurements,
but it would only
effect the measurements if we were in a different frame of
reference from the ship. As
we are inside it, we experience the same velocity, so our
measurements will be unaltered: our sense of time
and distance will be the same. If we measured the length of the
ship at rest, and also at 0.99c (while we were on it) then
our answers would be the same. An observer at rest would measure
the ship to be much shorter than usual.
So if none of this dilation affects us, why is the answer not 4L/c
? Because the velocity of light relative to us changes. A can see
what B is doing much earlier than usual,
and B only sees what A is doing later than usual. My initial idea
when setting the question was that the answer would be 4l/c,
because the light changes both ways, so I though the
change might balance out, but obviously this isn't that
simple.
The only dodgy bit that worries me is the assumption that the
light's overall speed limitation of c does not affect its relative
speed limit of 2c:
to explain: if two ships are travelling in opposite directions at
0.99c, then one's relative speed to the other is 1.98c. I am not
sure about this bit at all: maybe the light only
goes at c from B to A (relative to the ship) instead of 1.99c, but
the first way seem more sensible.
By Kerwin Hui (P1312) on Friday, April 14, 2000 - 07:13 pm:
Be careful, Neil. Time dilation will affect the reading on the
stop-clock, and we cannot use the absolute frame of reference in
the calculation!
Michael: I think I read the proof in one of the Freyman lecture
books(volume II?!)
Kerwin
By Michael Doré (P904) on Friday, April 14, 2000 - 08:25 pm:
Hi Neil,
I'm not entirely clear on this myself, but I think you may be
mixing up two different frames of reference. First of all you claim
that the speed of light is c relative to Earth (this is what
someone on Earth says) then you say that the lengths aren't dilated
because we are not in a different frame of reference to the
ship” (this is what someone on the spacecraft says). We need
to pick which frame we are working from and then stick to it
– as long as you don’t change your frame of reference
everything makes sense, I think.
OK, suppose we are on the spacecraft. According to us, we are
still. As we are still, the light will of course appear travel at
c, relative to us. Therefore when the light travels from A to B to
A to B to A, the time will quite clearly be 4L/c, and this is what
the stopwatch will read. (Remember there is never any doubt that
this is the answer – the principle of relativity says that
the results of all experiments are independent of the speed they
have place at, as long as it's constant).
Now we are on Earth, and we are using our telescopes to view the
spacecraft, which has glass walls and floors. As you say, the light
beam will travel forward with a speed of 0.01c relative to the
spacecraft, and back with a relative speed of 1.99c. But now there
is time and length dilation to consider – because we are
measuring from Earth. In fact you can actually work out what the
time and length dilation must be in order to keep the stopwatch
answer as 4L/c.
I think Einstein worked it out with thought experiments similar to
this – though all his thought experiments would have been
useless if his assumption of the principle of relativity was wrong.
This assumption is based entirely on intuition as far as I know (if
I'm wrong then please tell me). The other assumption used is that
the speed of light is independent of your speed, and by implication
the speed of the light source. This was predicted by Maxwell's
theory of electromagnetism, and verified by the Michelson-Morley
experiment, about 10 years before Einstein published special
relativity in 1905. He claimed never to have heard of this
experiment.
Anyway, here is another very famous thought experiment Einstein
did, which relates to some more of the interesting points Neil made
earlier:
We have a spacecraft, moving at 0.99c along AB; ends A and B. A and
B are light detectors – they emit a bright flash of light as
soon as they detect light. A lamp is placed on the mid-point of AB.
The lamp is switched on. In what order do the flashes appear to
someone in the spacecraft, at the centre? What about to the person
on Earth, a very long way away, looking through his telescope? What
about someone who is travelling at 0.99c relative to the spacecraft
in the other direction? This experiment questions the concept of
simultaneity – which to most people is more sacred than the
principle of relativity – not to Einstein though, and
experiments have proved him right.
Yours,
Michael
By Brad Rodgers (P1930) on Saturday, April 15, 2000 - 03:16 am:
Yes, the answer is 4l, now that I have read the problem
correctly. We mustn't consider any time dialation as we are in an
inertial vehicle and all inertial velocities are to be treated on
equal footing, with the same physical rules applying for those
inside. This dates back to Galileo. Anyway, even tough this process
would take 40l/c+40lx.99/c time for an external observer, an
internal observer would see himself at rest.
I'll work on your problem Michael; currently I can just think that
for anything planning to travel very fast, It probably would need
to be pwered by something external-such as solar energy. Therefore,
whatever energy would be lostwould be being gained at the same
rate, but I don't think this helps solve the problem very well.
By Brad Rodgers (P1930) on Saturday, April 15, 2000 - 03:31 am:
Eureka!
You'll have to excuse me if I am wrong as this is my first time to
use Calculus on anything but "change x+1 to x+2" problems, but it
seems to me that with a very small bit of differentiation, you find
that dE/dm=sqrt[(c-v)/c5]. This is positive until c-v=0.
I am not sure that this is right though, for the reason stated
above.
Brad
By Michael Doré (P904) on Saturday, April 15, 2000 - 07:48 am:
Hi Brad,
If E = mc2 then actually dE/dm = c2 - in
other words if you increase the mass, the energy always increases
linearly. This can be shown going back to the definition of a
derivative.
If you have a function f(x) then its first derivative at x is
defined by:
(f(x+d)-f(x))/d
for very small d.
Can you see why this is going to the the gradient of f(x)? Also can
you see why, if x is a time and f(x) is a particle's distance then
this will give its speed?
Anyway, back to our example.
E(m) = mc2 (I write E(m) because the energy is a
function of mass)
So the derivative of E(m) with respect to mass is:
(E(m+d) - E(m))/d = (c2(m+d) - c2m)/d =
c2
So E'(m) = c2
Could you just run me through how you got to 40l/c + 40lx0.99/c? As
we are viewing from Earth, we now must take into account length
dilation...
I wouldn't worry about the problem on external energy source. I was
really just making the point that many books incorrectly say that
if a rocket accelerates itself through its rockets then its overall
mass will increase. I don’t think it does – it only
increases if the source is external (e.g. solar energy as you
mentioned) I think it is still clear why c cannot be obtained, by
normal means.
Let fuel rest mass of the rocket be m and the total rest mass of
the rocket be M.
At speed V:
Total mass >= (M-m)/sqrt(1-v2/c2)
(not more of m can be burnt up)
Change in mass >= (M-m)/sqrt(1-v2/c2) -
M
(As the remaining rest mass is at least M-m.)
So total energy required >=
c2(M-m)/sqrt(1-v2/c2) –
c2M
But the total energy that can be supplied to the shuttle cannot be
more than mc2 where m is the mass of the fuel. So:
mc2 >=
c2(M-m)/sqrt(1-v2/c2)-c2M
m [sqrt(1-v2/c2)+1] >= M -
Msqrt(1-v2/c2)
And if v = c ever then it simplifies:
m >= M
Therefore the rest mass of the fuel is bigger than or equal to the
rest mass of the rocket and the fuel. This is a contradiction. (The
non-fuel part would need negative mass!) Therefore speed c cannot
be obtained.
If you have the energy being supplied externally then it gets even
simpler. Let M be the rest mass.
Mass at v = M/sqrt(1-v2/c2)
Energy required = c2
[M/sqrt(1-v2/c2)-M]
which tends to infinity as v tends to c.
A much more interesting problem is Einstein’s thought
experiment that I outlined in my last message. The question was,
what order will someone see the flashes of the light detectors
inside the spacecraft, and on Earth viewing through a telescope?
The answer has stunning consequences for our model of the world, as
it forces us to drop what most of us think of as an obvious
assumption.
Yours,
Michael
By Neil Morrison (P1462) on Saturday, April 15, 2000 - 12:59 pm:
Thanks Michael, I knew that length and time dilation wouldn't
effect us on the spaceship, but I didn't know if light would still
be going at c in our frame, so thats why I suspected my answer was
dodgy! It all comes from that thing about school physics... they
tell you how to work out some things, but not where to do it, and
how to solve questions such as the one we have been
answering.
What I think is the most interesting part of all this, is that
light (and e.m radiation) can travel at c, whereas or everything
else it is a limit which cannot be reached. Going by the equations
for length and time dilation and so on, time will stand still for
light particles, and their relativistic mass will be 0/0 (as their
rest mass is zero), which is of course undefined. So is a light
beam everywhere at once? And thus suppose you could ride a light
beam as is travelled from some distant star to earth.. wouldn't the
whole journey take zero time for you?
Neil M
By Michael Doré (P904) on Saturday, April 15, 2000 - 04:03 pm:
Hi Neil,
You are quite right that a beam of light is always in the first
instant of its creation. Of course it is important to point out
that this is from its perspective, whatever that means. But things
don't really have to make sense from this frame – you are
only guaranteed to get sensible experimental results in situations
where it is possible to actually do experiments. This seems to be
an underlying theme in modern physics.
I think you can also work out the relativistic mass of a photon of
frequency f, but I am very unsure about this so I would be grateful
if someone who knows can either confirm it or correct me. If you
equate the energy associated with a photon E = hf with its mass
energy you get:
mc2 = hf
m = hf/c2
The real problem with this logic is that E = mc2 is
derived for v¹c, for example by
Kerwin's argument (in reverse) or the way I did it using F =
d/dt(momentum). There is no particular reason it should hold for a
photon.
One thing is certain – light does have mass (albeit not rest
mass). This is shown with an experiment in which light turns a
wheel, showing it has momentum and therefore inertia.
Have you heard of tachyons? I think these are conjectured
particles, which actually start their existence at a speed faster
than light. This does not contradict special relativity – all
this says is that it is impossible for an object to make the
transition between moving slower than light and faster than light,
or vice versa. Of course, there is no experimental evidence I know
of for the existence of tachyons.
Anyway it turns out that any particle that had imaginary rest mass
would fit the bill for being a tachyon. They would be quite useful
for communication. There was some speculation a while back about
whether a neutrino (an incredibly weakly interacting particle)
might be a tachyon, but eventually I think experiments showed they
had a mass of about 2eV /c2, so travelled at very high
speeds <c.
Anyway, I'll describe a bit more of the experiment I outlined
earlier, as this is central to special relativity.
To recap: we have a spacecraft; its ends are A and B. End A and End
B both have light detectors, which are labelled A and B for
convenience. Both emit a flash of light when they detect a photon.
We have a lamp at the midpoint of A and B. The spacecraft moves at
0.99c relative to the Earth, in the direction AB. The lamp is
switched on. What does someone in the centre of the spacecraft
see?
Well that's easy. By the principle of relativity, the spacecraft
might as well not be moving. Therefore the light will take the same
amount of time to reach either detector, the detectors flash at the
same time, the person in the centre detects the flashes at the same
time. As far as he is concerned, the flashes occurred
simultaneously.
Now we have the person on Earth, watching through his telescope.
Because Earth is such a long way away, we can consider it
equidistant to the detectors. So we don't have to worry about
correcting for the time it takes for the flash to reach
him…
Yet he sees A flash before B. Why? Well from his perspective the
light travelling backwards towards A has a relative speed of 1.99c
to the spacecraft. The light travelling forward has a relative
speed of 0.01c. They have the same distance to go (the length
dilation is the same in both cases, therefore the lamp is still
equidistant from A and B). So light takes 199 times longer to get
to A than it does to B. Therefore, lamp A flashes BEFORE lamp B.
Therefore he sees A flash before B.
So the consequence of this experiment is that while A and B may
have simultaneously flashed according to the person in the
spacecraft, A happened before B for someone on Earth. For someone
travelling even faster than the spacecraft, B would have flashed
before A. Therefore in Einstein’s universe, we can no longer
think of things as simultaneous with one another. Simultaneity
depends on your frame of reference.
You may now be able to try the following. The spacecraft is again
going at 0.99c. At each end A and B there are two clocks. The
person in the centre of the spacecraft uses his computer to adjust
them until he is satisfied that they are synchronised. The person
on Earth looks on (again effectively equidistant from A and B). Are
the clocks synchronised? If not, which one is ahead, and by what
factor does it move faster?
Yours,
Michael
By Kerwin Hui (P1312) on Saturday, April 15, 2000 - 04:32 pm:
Michael,
If we use de Broglie, then the E=mc2 bit seems to work,
i.e.,
pl=h
and so p=h/l=mc
and the result follows.
An interesting question: Suppose we have a mirror in vacuum. We
shine a light beam through the glass and a very high energy
electron is fired above the glass, both events occur
simultaneously. Since the value for c in the glass is much less(by
a factor of about 40%, depending on the type), we have the electron
travelling faster than the speed of light. What happens to the
image charge? What would we observe?
Kerwin
By Brad Rodgers (P1930) on Saturday, April 15, 2000 - 07:20 pm:
But, if we know that m'=m/l, then
E=mc2/l. Therefore
dE=dmc2/l. And
thus, dE/dm=c2/l. from
that we know that dm/dE=l/c2.
I think that this should be right. As long as c doesn't equal v
then your inrease in mass over energy should be positive. However,
I now doubt this result as you proved that dm/dE=1/c2;
so therefore sqrt(1-(v/c)2)=1, and v=0. This has me
confused, and I think I have done something wrong but I don't know
what.
I'll write more later, I have to go right now.
Brad
By Brad Rodgers (P1930) on Saturday, April 15, 2000 - 08:53 pm:
I get the result of 40l/c +40lx.99/c by first assuming that the
movement of the light is from side to side of the vehicle, not from
front to back. By this it is possible to rule out any sort of
Lorentz contraction in our expiriment. we then know that as the
process would take 4l/c for an internal observer, this process
would be slowed 10x. so it would take 40l/c. But, it would also
take 40l/c*.99c/c time for the light telling that this occured to
reach the observer on earth.
Even if the movement was from front to back, I believe it would
still be the same as light must always travel at c and thus would
take longer to get to the front, but less time to get to the back.
This is, of course, if there is no gravitational interference
whatsoever-e.i. the spaceship emits no gravitational field.
Brad
By Michael Doré (P904) on Saturday, April 15, 2000 - 11:17 pm:
Brad –Aha I see what you've done. You've differentiated
the energy with respect to rest mass while keeping the
velocity constant. This is not wrong – it's just a different
result to mine, conveying different information. My result says
that the derivative of energy with respect to the relativistic mass
is c2. Your argument says that for any particular v, the
derivative of energy with respect to rest mass is:
c2/sqrt(1-v2/c2).
I'm still not sure about your calculation of the time taken
according to someone on Earth. I think you have the right idea for
the side to side case, but shouldn't it be 4×7.09 l/c not
4×10 l/c? Also let's not bother about the time light takes to
get to Earth. By putting in that factor, you are assuming that the
Earth is directly along the same line that the light beams travel.
If it is almost anywhere else, and far enough away then all points
on the spacecraft are virtually equidistant. If you're not
convinced, draw a triangle and use the cosine rule. We use this
technique a lot in A-Level physics in working out interference
patterns. We assume the source of light is so far away that you can
neglect differences in the distance the light travelled from the
source and hence you can neglect phase differences. It just makes
the maths ridiculously complicated if you don't.
So anyway, for the side to side case, it is 4×7.09 l/c. Now
you seem to think that you get the same answer if the light beam
travels forwards and backwards rather than side to side. I'm not
sure about this. You are certainly right that the one going back
will hit the wall before the one going forward (this relates to my
last message on simultaneity). But I don't think that the time
differences cancel. An analogy is helpful here.
You are in a river with two ends A and B. You have to swim from A
to B and back to A, and you want to minimise your total time. Is it
better for there to be a steady current in one direction, or for
there to be no current? Thinking about it, it is clear that it is
better for there not to be a current. (Consider what happens when
the current is really strong. You get from A to B in less than a
second, but then hours and hours to get back.)
What's more I think we now have length dilation to consider.
Kerwin – Thanks for your derivation using the De Broglie
equation. I will have to have another think about this, as it is
not clear why E = mc2 should carry through… It's
like trying to integrate to a limit that doesn't exist.
As for your problem, I'm not quite sure what an image charge is. Is
it a disturbance in the charge density of the atoms of the glass,
by virtue of the presence of the electron? I remember something
similar to this in a STEP Physics paper I was practising not long
ago. If so, here is my immediate, unreasoned, intuitive (i.e.
wrong) response:
The disturbance travels at the same speed as the electron. There is
no contradiction here as the disturbance is not a self-propagating
wave, so there is no reason it should go below c/n, n the
refractive index.
If I'm well out, please give me another chance – I'd really
like to solve this.
Yours,
Michael
By Sean Hartnoll (Sah40) on Sunday, April 16, 2000 - 02:19 am:
A general point that I think will help the
discussion:
The best way to do these sort of relativity questions is to use the
Lorentz Transformations. For one dimensional problems these
are
x' = g (x - Vt)
t' = g (t - Vx/c2)
with g =
1/sqrt(1-V2/c2).
Where the frame S' is moving at velocity V with respect to S.
Everything follows from here. In particular it is useful to note
that this equations imply that the interval c2
(dt)2 - (dx)2 is the same in all frames,
which gives quick answers to many questions.
By Brad Rodgers (P1930) on Sunday, April 16, 2000 - 03:12 am:
Yes, it should be 4×7.09l/c- I forgot to square the v/c
ratio. I now see why I recieved v=0 in my results. But we might as
well, with this being true, simplify to dm/dE=1/c2. So,
I now agree with you that a body not recieving any external energy
would remain constant in mass.
For the light path being from front to back, I think that it would
take 4l/(1-.992)c, but this is taking into consideration
length contraction, which I do not think is needed. This has
puzzled me for some time: if the equations of time dialation give
the equations of length contraction, wouldn't the length
contraction change the result of the time dialation which would
change the results for length contraction, ad infinitum. I think
time dialation and length contraction are just two ways of saying
the same thing. However, I am not sure about this and in fact, I
nearly doubt it as I have nothing of the like before. Also, has
there ever been any evidence for length contraction?
Brad
By Michael Doré (P904) on Sunday, April 16, 2000 - 04:57 am:
Hi Brad,
I don't think that the length dilation will change the time
dilation. The reason that the time dilation causes length dilation
is to keep the speed of light constant. If the dilation factors are
the same then the speed of light will still be c so no further
corrections are required.
I still think length dilation does need to be considered in our
problem. We are measuring from Earth now, so the distance the light
has to travel is shortened from our perspective. The distance it
has to travel each way is:
L sqrt(1-v2/c2)
First of all the relative speed is v-c then v+c. So the time taken
is:
L sqrt(1-v2/c2) [1/(c-v) + 1/(c+v)]
(From Earth's point of view).
For the original question we'd need to double this because Neil
wanted the light travelling from A to B to A to B to A. Have I made
a silly mistake? If not it simplifies to:
4L/sqrt(c2-v2) which makes sense when v =
0.
Sean – I'll have a think about how these equations can help
in general. I certainly recognise the second from the extension of
Pythagoras, with a factor of i inserted into the time.
Thanks,
Michael
By Neil Morrison (P1462) on Sunday, April 16, 2000 - 12:23 pm:
Kerwin,
surely the de Broglie equation of wavelength was derived using
E=mc2 in the first place.
Michael,
I agree that light has momentum, but if it has no rest mass, how
can it have any mass at any velocity? its "mass" at a velocity v
will be m0/(1-v2/c2)1/2
so for light, which travels at c all the time (??) we get m = 0/0
as before. Since this is undefined, I cannot draw any conclusions
as to whether light has mass or not.
I think the idea that light has rest mass is stupid anyway, because
if light always moves at c, it will never be at rest.
Michael-
In these light/inertia experiments, the light will lose momentum
right? So then it will not be travelling at c anymore, or is there
something I'm missing?
By Sean Hartnoll (Sah40) on Sunday, April 16, 2000 - 05:48 pm:
Light DOES NOT have mass. It DOES have
momentum. And it ALWAYS travels at c, although c may be different
in different media.
The general expression for energy in special relativity is
E2 = m2 c4 + p2
c2
Where the well-known E=mc2 comes about by setting p=0,
i.e. the rest case, this is why m is called the rest mass.
Light has no mass but does has momentum hence the energy for light
is
E = pc.
Neil,
The de Broglie relation is NOT derived from E=mc2 or
indeed anything else, it is a postulate in itself.
Brad,
There is experimental evidence for length contraction. The
Michelson-Morely experiment itself requires a length contraction to
explain the constancy of light (in fact, this is why Fitzgerald and
Poincare and Lorentz had postulate length contraction BEFORE
Einstein's 1905 paper).
Sean
By Brad Rodgers (P1930) on Sunday, April 16, 2000 - 06:31 pm:
Michael- the movement of back to front would not change the
amount of time for the process. Basically, what we have described
so far has been a light clock. And, as a moving body cannot have
two seperate time flows, there must be the same time taking place-
this is how Einstein derived his length contraction. i can show you
my own mathematics to prove this if you want. The time needed is
4×7.09l/c.
Brad
By Michael Doré (P904) on Sunday, April 16, 2000 - 07:05 pm:
Sean - what evidence is there for light having no
kinetic/relativistic mass? Mass has two properties:
1) inertia - resistance to motion. When an object's velocity
changes, an impulse is imparted to another body, provided that our
object has inertia. In our case, the light turns a wheel. Therefore
light does have inertia, so it is already half way torwards having
mass.
2) its gravitational field. Within Newtonian mechanics the field
strength at a given point is proportional to its inertia. Are you
saying that there is no gravitational field associated with a
photon? If so, is this the reason that the photon is classified as
massless?
Neil - remember momentum is a vector quantity and speed is a
scalar. Light's momentum changing doesn't imply its speed changes.
And there is no rule that says that the velocity of light is
a constant. Otherwise we would have a migration of photons from one
side of the universe to another!
The equation m = m0/sqrt(1-v2/c2) holds for
v<c. This is because it is derived using any thought experiments
in a frame of reference moving at a relative speed of v to another
frame. Once you set v = c, the thought experiment cannot be done so
any result you may get will be invalid. However, it is quite
signifcant you get 0/0. Let f(x) and g(x) be continuous functions.
If you have f(x)/g(x), where f(c) = g(c) = 0 then as x tends to c
the limit can be anything. The same would not be true if f(c) = 1
and g(c) = 0.
Brad - if you substitute v = 0.99c into my above expression I think
you get the result you gave. I don't think we are in
disagreement.
Yours,
Michael
By Michael Doré (P904) on Sunday, April 16, 2000 - 07:55 pm:
So in fact you are quite right when you say that the side-side
time is the same as the front to back time. This is obvious
thinking about it - from the point of view of the spacecraft they
are the same, so from the point of view of the Earth they are the
same, because the light travelling the two different routes can
start and stop at the same place, so must be synchronized.
Here is the mistake I was making. If the spacecraft was travelling
at v, when the light beam travelled sideways, I assumed its
sideways speed relative to Earth was c. This is wrong - its entire
speed is c. By Pythagoras: x2 + v2 =
c2 where x = sideways speed of light, v = forward speed
of light = forward speed of rocket. Therefore the sideways speed is
sqrt(c2-v2) not c. Hence the river analogy
was invalid, because the "speed in the calm water" has been reduced
from c to sqrt(c2-v2).
You can use this result to calculate the time dilation very easily
as happening by a factor of sqrt(1-v2/c2).
Therefore it follows that length also contracts by this factor (to
preserve the constancy of c). So the mass factor is
sqrt(1-v2/c2) too.
Sean - what is your definition of momentum? Your equation:
E2 = m2c4 +
p2c2
is presumably when E = total energy, m = rest mass, p =
relativistic momentum. If you assume that relativistic momentum =
mv/sqrt(1-v2/c2) then your equation is
equivalent to the one which we've been using which states:
E = mc2 where m = total (relativistic) mass
rather than rest mass.
Yours,
Michael
By Neil Morrison (P1462) on Sunday, April 16, 2000 - 09:03 pm:
Sean,
In my Physics course, de Broglie is shown by equating
mc2 with hf.
For e.m radiation, p = mc
so pc = hf and f = c/w (w is wavelength)
so w = h/p
I emphasise that this is only the way I was taught it.
Michael (did you get my mail BTW)-
I thought you said something about tachyons (?) travelling faster
than light, and you then used the equation, which you said has
v<c to say these have complex mass. Is this not a
contradiction?
c seems to be behaving like infinity for speeds and velocities, as
it is impossible to reach, and what does go that fast is causing so
many problems to investigate.
Does antimatter have positive or negative mass? If something with
negative mass hits a moving object from behind, does it make the
object slow down? Or are these situations even possible?
By Sean Hartnoll (Sah40) on Monday, April 17, 2000 - 03:22 am:
okay,
Michael -
0) I can't remember the details, but I know there are experiments
which have confirmed that the photon has no rest mass to an
accuracy of something like 10-31. It is interesting
though that it was long thought the neutrino had no mass and very
recently it has been discovered that it did. A photon with mass
would upset a lot of established physics.
1) inertia of photon: strictly F=dp/dt is the second law, so an
object can have inertia without mass.
2) light does NOT create a gravitational field. It is affected by
gravitational fields, this was a key prediction of general
relativity, but this is due to the geometry of spacetime changing,
not to a mass of the photon.
3) the relativistic defn of momentum is not strictly gmv, although this is true for a particle with
mass. The momentum of a photon is probably better defined as p =
E/c. A rigurous definition involves objects called 4-vectors, which
are 4-components objects that transform between frames according to
the Lorentz transformations, one such 4-vector is the position
4-vector (ct,x,y,z), another turns out to be the momentum 4-vector
(E/c,px,py,pz). A property of
these objects is that "norm" is preserved:
x02 - x12 -
x22 - x32 is the same
in all frames.
4) The definition you give for energy is certainly true and is
equivalent to the one I gave for particles with mass, the advantage
of the other formula is that it is true for photons also.
5) In general it is probably best to call only the rest mass "mass"
and denote it by m, a constant, the so called total or relativistic
mass is dispensible as a concept.
Neil -
0) one could use p = mc to define some kind of effective mass for a
photon but this isn't useful and is never done. For example, it is
a central part of general relativity that the gravitational mass is
equal to the inertial mass, and as I said above, the gravitational
mass is certianly zero. As someone else said earlier, don't believe
anything they tell you until Uni! (and even then...)
1) E = hv and p = h/w (w wavelength) are equivalent definitions,
related through E=pc.
2) Antimatter has positive mass. Antiparticles are not essentially
different from particles. For example, the positron, the
antiparticle of the electron is just another particle with certain
properties.
3) Tachyons. It is true that a tachyon would go backwards in time,
this is the main reason why people think they don't exist! It is
not a contradition to go faster than light, what special relativity
shows is that you cannot pass through the light barrier, so you
must be always either faster or slower than light. I don't know how
tachyons would interact with other particles, but I imagine it
would get pretty messy!
Basically, photons are not understandable without both special
relativity and quantum mechanics. They embody the craziness of both
SR and QM.
Sean
By Michael Doré (P904) on Monday, April 17, 2000 - 04:23 am:
Hi Sean,
Thanks for your reply. There is still just one thing that worries
me about all of this. What if the third cosmological model were
true? (This is the one predicted by inflation - where the mass of
the universe is on the borderline for being enough to halt the
expansion. The recent astounding observational evidence suggests
that this is wrong but never mind - a lot of theoreticians still
believe it.)
This requires the mass of the universe to be exactly right.
Now suppose a few electrons and positrons in the universe crossed
paths - they would be annihilated and photons would be produced. So
the gravitational mass of the universe would have gone down.
(Electrons and positrons both have a g-field, but photons don't.)
So how can they be sure the universe's mass is exactly right and
will always be so?
Also, even if the mass wasn't on the limit; suppose there is too
much for instance - isn't it conceivable that enough positrons and
electrons, or protons and anti-protons etc meet, so as to change
our open universe into a closed universe? I know I'm working
backwards here, but if mass conservation is disregarded, there seem
to be problems. If mass conservation is true then mustn't photons
have mass?
Neil; if you make the assumption that there exist objects that
started off with a speed >c, and that the rule: speed of light
is always constant still holds, then m =
m0/sqrt(1-v2/c2). It still can't work for v =
c, because it is impossible to conduct a thought experiment in this
frame, ever. You are quite right that c acts like infinity in some
respects. Remember the reason that special relativity is different
to Newtonian mechanics is that the speed which requires infinite
energy is not infinite itself.
Thanks,
Michael
By Michael Doré (P904) on Monday, April 17, 2000 - 01:19 pm:
So it looks as if De Broglie follows from mass
conservation, energy conservation, E = mc2 (m = total
mass) and that particles can be converted into photons (and vice
versa).
The energy particles have can be converted into light energy,
obliterating the particles (provided certain properties of the
particles can cancel). Now as the mass of the particles is also
lost, then by mass conservation the new photons must have this mass
too. As E = mc2 holds for particles, it must therefore
hold for photons too. From there Neil's argument works. So it looks
like it's all down to mass conservation (energy conservation and E
= mc2 are accepted) and that particles can be
obliterated to form photons.
However I'm prepared to believe it's not nearly that simple –
I doubt it took them 20 years to figure that out! (E = hf and E =
mc2 were known in 1905. De Broglie was put forward in
the mid twenties. However it was designed to apply to entities such
as electrons; not just photons. Maybe this was the new result.) On
the other hand, antimatter was not known about till the thirties
when Dirac came on the scene, so maybe electrons and positrons
annihilating to produce photons is a prediction of De
Broglie. This would make the argument in Neil's textbook
circular.
I think that the general problem I (and perhaps others) am having
is that I know more facts about SR, GR and QM than I can possibly
understand at the level I'm at. I still think it is a big mistake
teaching QM at A-Level / SYS – it is completely glossed over,
and this is at the expense of learning some more basic physics
properly, which would later provide a better foundation to study
20th century physics.
Yours,
Michael
By Neil Morrison (P1462) on Monday, April 17, 2000 - 06:28 pm:
Michael-
Your point about the teaching is absolutely spot on- we know
tit-bits of information, but we can't apply them to any
situation.
Also, if v>c, does something have complex/imaginary mass?
Sean-
Here I agree, the antiparticle of a certain type of particle will
have only charge or spin opposite to the original particle; mass,
strangeness etc will be the same.
By Sean Hartnoll (Sah40) on Monday, April 17, 2000 - 11:59 pm:
Michael,
Mass conservation is not true in non-Newtonian mechanics, there is
only energy conservation and momentum conservatins (well, and
angular momentum conservation etc.).
This also relates to your cosmological concerns. The gravitational
field in GR is not just created by mass, but more generally by
energy density. This is making me think that perhaps photons DO
curve space because they are just the energy quanta of the
electromagnetic field, however I was fairly sure they didn't, but
perhaps this is just an approximation everyone makes.
Michael and Neil,
More than a new relation what De Broglie gave was an
interpretation, namely that particles and waves are somehow cobined
into one object. And yes, I think a major part of his hypothesis
was to apply it to electrons.
Antimatter was not predicted until Dirac, its predition being a
major success of Dirac's theory.
It certainly seems that you would get complex numbers dealing with
tachyons, I don't know anything about how such theories are
constructed. I imagine you could make the imaginary bit cancel by
making the constant "m" (which can no longer be the rest mass
because Tachyons cannot be at rest) itself complex, so that what
you are left with is real quantities, just a random thought.
As for teaching, I agree it is certainly annoying knowing facts but
not where they come from, and hence not knowing what they really
mean. And I copletely agree with Michael when he says time should
be spent teaching classical mechanics properly, and I would extend
that to learning maths properly, the best basis for understanding,
particularly quantum mechanics, is a solid mathematical
understanding of, for example, vector spaces and matrices
(operators).
One thing I think could be taught is what is called the Old Quantum
Theory, that is the stuff that was done before Schrodinger,
Heisenbeg and Dirac formulated what is now quantum mechanics. That
is the Bohr atom, compton effect, etc. which is a useful historical
background to real quantum mechanics which is completely different.
And it would save people wasting time at Uni doing it in the first
year instead of Q.M. (which is not done properly until the second
or even third year).
Sean
By Michael Doré (P904) on Tuesday, April 18, 2000 - 09:40 am:
Hi,
For tachyons; I've just realised: Suppose you have a spacecraft
made of tachyons, going at several times the speed of light. Now it
is certainly feasible that a particle in the spacecraft could emit
a photon sideways, by the principle of relativity. Now if we let v
= forward speed of the tachyonic rocket, and x = sideways speed of
the light from Earth's perspective you get:
v2 + x2 = c2
by Pythagoras, because the total speed of light is c. v>c, so x
is imaginary. In other words, from Earth's perspective the sideways
speed of the light would be imaginary. Hmm. This also means that
time and length dilation of the spacecraft would be
imaginary!!?
Anyway, I've just realised that it would be impossible to build a
tachyonic emitter and receiver, which emit and detect very high
speed tachyons. (These are ones with low energy.)
Suppose we have a normal spacecraft, ends A and B. The midpoint of
AB is O where there is a lamp and a detector of tachyons. At A and
B there are light detectors. As soon as they detect light they emit
tachyons equally in all directions, with extremely high
velocity.
The lamp is switched on in the centre. Soon afterwards, two pulses
of tachyons arrive back at the centre. As tachyons travel fast, the
computer in the centre can time how long it took the photons to
travel from O to A and from O to B approximately. Then it can
display the time on a big screen.
Now imagine you are viewing this in the same frame as the
spacecraft. The photons obviously take the same time to travel from
O to A and O to B, and the return journey is negligible because of
the high speed of the tachyons. Therefore the two times on the big
screen should be the same.
Now view the situation from a frame moving at 0.99c relative to the
spacecraft. One time will now be bigger than the other, because the
relative speed of light to the spacecraft will have changed.
If the frame moved the other way at 0.99c, the times will be
reversed.
In other words, the result of this experiment will depend on the
frame you view it in, contradicting the principle of relativity. So
tachyonic emitters and receivers are impossible.
So that shows that instantaneous communication and special
relativity are non-compatible. But now I’m confused –
isn’t the collapse of the wave-function instantaneous in
quantum mechanics? Couldn’t this be used for instantaneous
communication. If so, how can this co-exist with special
relativity, because you could perform the experiment I have
outlined and obtain a contradiction!?
Sean; it seems to me that if photons can turn wheels (meaning they
have inertia) and they curve space (meaning they have gravitational
mass) then there isn't much of a case for them having no mass - but
I am more or less a beginner at this subject, so there may be
another property of mass I have missed out. It would of course be
absolutely tiny; m = hf/c2, where h is of the order of
5*10-34 and 1/c2 is about 10-17 in
SI units, making the mass about 5*10-51f. f may be
around 1013 for visible light, making the mass of a
photon of visible light about 5*10-38kg. Of course it
has zero rest mass, but this is different.
Yours,
Michael
By Neil Morrison (P1462) on Tuesday, April 18, 2000 - 06:22 pm:
Michael-
in your experiment, did you take into account time and length
dilation when you were in a different frame to the
spacecraft?
Also, I still don't agree that if light has zero rest mass, it can
have mass at a velocity. Unless you say that new mass is 0/0 again.
By Sean Hartnoll (Sah40) on Tuesday, April 18, 2000 - 11:58 pm:
Michael - Neil's point about your argument is correct. But there is
a more serious problem: you can't add velocities like in classical
mechanics.
When you write v2 + x2 = c2, in
fact x=c, by Einstein's original postulate, the cornerstone, if you
like, of special relativity, with or without Tachyons. Now this
doesn't imply v=0, because in fact the addition of
velocities formula should have been different to start with, for
velocities in the same direction (the case for velocities in
different directions is longer and I can't remember it offhand)
is
v1 =
(v2+v3)/(1+v1×v2/c
2)
Perhaps you could try and prove this from the Lorentz
trasnformations (use the lorentz transformations in differential
form, and remember to transform the time also).
In particular this formula shows that if you add a velocity to the
speed of light you get the speed of light again, which is what you
should expect.
Michael and Neil,
About the mass of light. When I said 10-31 I meant this
it had been established to 1 in 10-31, this is very
small.
Perhaps this argument will convince you:
The energy is
E2 = c2 m4 + p2
c2
where m is the rest mass and p the momentum.
Now, for light it is found that E = pc. So we MUST have m=0. So
light has zero rest mass. And this is what is meant by having zero
mass. And as Neil says it just doesn't make sense to say an object
with zero rest mass has mass at some other velocity. It can, and
does, have momentum though.
Another argumet:
take E = g m v
m is again the rest mass. Now g =
1/sqrt(1-v2/c2), so if v=c, this is infinite.
But the Energy is not infinite, so the only way to make energy
finite is by setting m=0. Now this gives 0/0, so what it is really
saying is that there is aproblem with the formula, but what it does
show is that if mass were nonzero, then the energy would be
infinite. The energy isn't infinite so the mass must be zero, and
we must another formula for energy, namely the one above.
Hope this helps,
Sean
By Michael Doré (P904) on Wednesday, April 19, 2000 - 10:02 am:
Hi Sean and Neil,
I'm not questioning for one second whether the rest mass is
zero or not - it must be zero otherwise SR would collapse. I'm
saying that it should have relativistic mass. There is no reason
that
m = m0/sqrt(1-v2/c2)
should work for v = c. The formula is derived for v<c - by
thought experiments. If you let v = c the thought experiments
become invalid because it is impossible to travel at c. If you
travel at c in a thought experiment you cannot expect things to
make sense as you have made an illegal assumption that you can
travel at c.
So we don't get 0/0 problems - because the formula simply doesn't
apply in this instance.
It’s like using the geometric formula:
a/(1-r) = a + ar + ar2 + ...
Let a = 0 and r = 1:
0/0 = 0 + 0 + 0 + .. = 0
Is this true? Of course not; the formula only applies for –1
< r < 1. Same here. This formula cannot give any conclusions
one way or the other as to whether light has relativistic
mass.
My point is:
1) Light turns wheels - it can exert an impulse.
2) It must curve space - otherwise would upset the cosmological
principles.
3) It has momentum
4) It has energy.
Are there any more characteristics required to have mass? (Again I
don't mean rest mass.)
Now back to the experiments I was outlining. I'm sorry if I phrased
these poorly in my last message (perfectly possible) - but I think
you've both misunderstood them. There were actually two; I'm more
interested in the second.
1) Let v = forward speed of a rocket that is made of tachyons. Now
the rocket shines a light sideways from its perspective. Now
we're on Earth; velocity 0. Let x = sideways speed of the light
relative to Earth from Earth's perspective.
We are dealing entirely from Earth's point of view. According to
Earth:
forward speed of light = v
sideways speed of light = x
total speed of light = c
Now as we are measuring everything relative to Earth, we can apply
Pythagoras to the speeds.
You can get:
x2 + v2 = c2
Making x imaginary because v>c. But according to the tachyonic
spacecraft, the sideways speed of light is c. All I was saying is
that the idea of travelling on a tachyonic spacecraft is slightly
ridiculous because their time must be dilated by an imaginary
factor! (You can see this by quoting the result that time dilation
factor = 1/sqrt(1-v2/c2). I'm just trying to
show that this result would still have to hold.)
2) More important. I'll outline it again, in case I wasn't very
easy to understand.
We have a spacecraft. Ends A and B, midpoint of AB is O. At O there
is a lamp. The aim is to time how long it takes for the light to
travel from O to A, and to time how long it takes to travel from O
to B. So to do this the lamp at O switches on giving out light
towards both ends A and B. The stopwatch at O starts. The beam
travels out to A and B. As soon as the beam gets to A, the detector
there (virtually) instantaneously communicates with O using
tachyons to note down the time on the stopwatch. This reading will
thus be the time it takes for light to travel from O to A.
Now exactly the same thing is done with B. The computer at O notes
down the time it takes light to travel from O to B. Now it compares
these two times. It's their relative value I'm interested in.
So what does it say? Well if you are travelling at the same speed
as the rocket, then according to you the light travels from O to A
and B, at exactly the same speed. Therefore the detectors pick up
the signals at exactly the same time. They communicate with O using
tachyons at exactly the same time. Therefore the two stopwatch
readings are the same.
Now imagine observer X is travelling at 0.99c in the direction of
AB. What does he see? Well speed of the beam going from O to A is c
from his perspective (as always). But the spacecraft is moving in
the same direction according to X at 0.99c. So the relative
speed of the light beam OA to the spacecraft from X's
perspective is 0.01c. By the opposite argument, the
relative speed of the beam OB to the spacecraft from X's
perspective is 1.99c. Now OA and OB have both been dilated
according to X; but by the same amount. Therefore the light has the
same distance to go in the spacecraft, but the one going from O to
A is going much slower relative to the back wall. So according to
X, the light takes longer to get to A than it does to B.
Now the tachyons communicate back the signal. As they are going at
several million times the speed of light, it doesn't matter that
their relative speeds to the spacecraft from X's perspective will
be different by 1.98c – this is irrelevant. The signal from B
will get there before the signal from A. Therefore the stopwatch
will record a lower time for the beam OB than OA. Its time will
have been dilated of course, but we want the relative times, and
there is no doubt that if the signal arrives later, the stopwatch
will record it as a larger time.
So according to X, the time the stopwatch on the spacecraft records
for the light to travel from O to A is larger than O to B.
Now imagine observer Y. She travels at 0.99c relative to the
spacecraft in the direction BA. Everything is the same but the
other way around. According to Y, the time the stopwatch on the
spacecraft records for the light to travel from O to A is lower
than O to B.
In other words if you allow instantaneous communication (or even
just communication at a speed faster than c), special relativity
falls apart. I used to be under the impression that instantaneous
communication was impossible. Tachyonic emitters and receivers
would be good enough to ruin special relativity. I’m
wondering how to get round the collapse of the wave-function
problem in QM.
Yours,
Michael
By Michael Doré (P904) on Wednesday, April 19, 2000 - 12:09 pm:
Aha - Sean, I've just realised that if your equation read:
v1 = (v2 + v3)/(1 +
v2×v3/c2)
where the speeds being added are v2 and v3,
then the problem with experiment 2) might just be averted. This
form is symmetrical on v2 and v3 so makes a
bit more sense. Is this correct now?
Now if we let the speed of the craft = v2 = 0.99c and
the speed of the tachyons relative to the craft from its
perspective = 106c for example then v1 (the
speed of the tachyons from Earth's ponit of view) will now be only
just above c. This is now also making more sense from the energy
point of view.
Therefore from Earth's perspective, the instantaneous communication
is lost, so the stopwatch readings are consistent with other
frames. Therefore the problem in 2) is gone, and tachyons can be
consistent with SR. The mistake I was making initially was assuming
that if B is moving at a very high speed according to A, and
A is moving at 0.99c according to O then B would be moving at a
very high speed according to O. In fact the speed lowers to just
over c, if you change your equation to the one I gave.
I still don't know how to get around the collapse of the
wavefunction, but at least we don't have a case of SR squabbling
with itself - just with a different branch of physics, that
superficially looks incompatible anyway.
Thanks for all your help,
Michael
By Michael Doré (P904) on Wednesday, April 19, 2000 - 05:37 pm:
OK, I've just had a go at deriving the addition velocity
equation. Apologies for not using the Lorentz transformations but I
haven't really got the hang of them yet, so I need to start off
easy. OK, we have a spacecraft ends A and B going at speed a
relative to the Earth, once again in the direction of AB. A
particle P is moving in the same direction with speed v > a. P
looks to someone in the spacecraft as if it is moving with speed b.
We need to find the speed v in terms of a and b. This will be our
addition formula.
Let O = midpoint of AB. AO = OB = L.
We are going to measure b, the apparent speed of P. To do this, we
have a detector at A and B. When P passes the detectors, they send
a signal back to O. O measures the difference in the time of
arrival of the two pulses. It then calculates b using the formula b
= 2L/difference in time of arrival.
Now we are on Earth. What time difference will the computer at O
register?
Let's measure each time from the time at which P reaches A. The
beam from A will take L sqrt(1-a2/c2)/(c-a)
to reach O (length is dilated). Remember we are on Earth. The c-a
is the relative speed of the beam to O, the destination.
Now how long will the beam from B take to arrive at O. Well first
the particle P has to arrive at B. This will take an extra 2L
sqrt(1-a2/c2)/(v-a) (v-a is the relative
speed of the particle to B). And then of course we still have the
time the beam needs to return to O from B. This is L
sqrt(1-a2/c2)/(c+a). So the total time
difference in the arrival of the beams at O, according to Earth
is:
2L sqrt(1-a2/c2)/(v-a) +
Lsqrt(1-a2/c2)/(c+a) – L
sqrt(1-a2/c2)/(c-a)
But the clock at O is dilated. So it will have measured the time
difference as only sqrt(1-a2/c2) of this. So
the time difference measured by O is given by:
2L(1-a2/c2)/(v-a) +
L(1-a2/c2)/(c+a) – L
(1-a2/c2)/(c-a)
Now from the spacecraft’s point of view, this time difference
is the time it takes for P to travel from A to B. (There is no
other way a time difference can creep in according to the
spacecraft.) Therefore this time difference is equal to 2L/b where
b is the speed of P relative to the spacecraft. Therefore:
2L/b = 2L(1-a2/c2)/(v-a) +
L(1-a2/c2)/(c+a) – L
(1-a2/c2)/(c-a)
Now the Ls cancel. This expression doesn't look at all symmetrical
on a and b – but it is! If you re-arrange it you get:
v = c2(a+b)/(c2+ab) which is what I think
Sean's equation was going to be.
Is this method sound?
Yours,
Michael
By Sean Hartnoll (Sah40) on Thursday, April 20, 2000 - 01:02 am:
Michael -
Sorry about the misprint earlier, of course your equation is right,
I hope my typo didn't cause too much trouble!
Your derivation seems okay, it certainly gave the right result. It
is good to be able to do things in terms of time dilations and
length contractions, this physics is sometimes lost under more
mathematical ways of doing it.
I'll give another derivation which has is faster, I'll set c=1
(this is standard, you can put the c's back in by dimensional
analysis).
From Lorentz (notice the suggestive symmetry now that the c's are
not there):
dx' = g (dx + Vdt)
dt' = g (dt + Vdx)
(I've taken V negative so that the result looks nicer)
so v' = dx'/dt' = (dx+Vdt)/(dt+Vdx) = (v + V)/(1+Vv)
Which is the result in three lines!
As for the wavefunction collapse, I don't know how one goes about
formulating an acausal quantum mechanics, there certainly are going
to be problems.
As for the mass of the photon, I have looked in some
electrodynamics books, there is a whole formalism developed for
dealing with a photon with mass. Many things such as Maxwell's
equaitons (and hence the inverse square law) would no longer be
true. The best current measurement of the photon mass comes from
measurements of the Earth's magnetic field, which put it at less
than 10^-51kg, much less than, say, the electron mass, although not
low enough to rule it out completely.
The gravitational field of a photon is due to its electric energy
(it carries the erngy of the electromagnetic field), not
mass.
Momentum is a more fundamental concept than velocity. I can't
really give you a straight answer to your question of what does
something need to do to have mass, because I haven't done a course
on Quantum Field Theory (and beyond!) yet, but a photon with no
mass does seem to fit into things nicer than a photon with mass,
sorry that that's not a good answer...
Sean
By Brad Rodgers (P1930) on Thursday, April 20, 2000 - 03:19 am:
I'm not sure of this, but I think that a photon does have "mass"
in the sense that m(q)=hv/c2. this is justified by the
result that hv=hv'+(mv2)/2.
Brad
By Michael Doré (P904) on Thursday, April 20, 2000 - 09:31 am:
Hi,
Thanks for your replies. So as that equation is right, that means
that tachyons would have one further amazing property – if
you travel in the other direction to them they slow down, as far as
you're concerned. If you try to catch them up by getting close to
c, then they appear to zoom away at close to infinite speed.
I think I understand the Lorentz transformations a bit better now,
having read Sean's derivation. And at last I can see why the second
one must be true. I guess the symmetry in the equations for
distance and c × time was what led Minkowski to think of 4-D
space-time, which Einstein extended later for GR.
I'm trying to find out how the collapse of the wave-function works.
According to one of my physics dictionaries, QM also predicts
quantum entanglement, which is supposed to make instantaneous
communication possible in principle. Maybe it is only instantaneous
from the frame which the particle is in, but this seems rather
arbitrary. I can see why Einstein tried to save locality as well as
reality, in QM, but eventually failed.
I am still thinking about Kerwin's puzzle above. If you have a beam
of light travelling in glass, below an electron travelling in a
vacuum, what happens to the image charge if the speed of the
electron > speed of light in glass? Anyone got any ideas?
Thanks,
Michael
By Sean Hartnoll (Sah40) on Thursday, April 20, 2000 - 09:04 pm:
Brad - in the result you quote for the
photoelectic effect the m is the mass of the electron and the
equation represents a transfer of energy, the hv and hv' terms are
the energies of the photons before and after collision and
mv2/2 is the energy of the electron. It doesn't say
anything about photon mass, just that energy can be transferred
through collision.
Sean
By Brad Rodgers (P1930) on Friday, April 21, 2000 - 12:17 am:
Oh, but does m(q)=hv/c2 still show that a photon has
mass? I didn't know what relation the two equations above had in
common, but a physics book I have showed the first result, then
said,"we then have the energy equation..." and stated the second,
so I figured that the first justified the second.
Brad
By Michael Doré (P904) on Friday, April 21, 2000 - 11:31 am:
Hi,
You are assuming that E = mc2 holds for a photon. This
is true only if the principle of conservation of mass is true, for
the following reason:
Suppose you have an electron and a positron coming together at high
speed. As speed is high they have high masses, m1 and
m2 let's say. (Several times the electron rest mass.) So
their energies are m1c2 and
m2c2, because E = mc2 definitely
holds for electrons/positrons. They annihilate each other (as
electrons and positrons do) and form photons.
By conservation of energy, the energy of photon =
m1c2 + m2c2
By conservation of mass, the mass of the photon = m1 +
m2. (The mass of the ex-electron/positron pair cannot be
transferred to anything else.)
Therefore the energy of the photon = c2 × mass of
photon.
Therefore photons satisfy E = mc2. Therefore
E = hf = mc2, m = hf/c2 etc
However - we have assumed that mass is conserved. Sean's point is
that this is not always the case in the quantum world. It certainly
seems as if photons have mass (because they certainly satisfy the
two classical properties of mass - gravitation and inertia - which
are equal for a photon as required) but the quantum world may have
another property an object with mass needs to satisfy. It is worth
emphasising again, that we are not talking about the rest
mass of the photon. If photons had rest mass, then this would mess
everything up, as they would have infinite energy. It would be even
more worrying to be treated with gamma particles, if they could
deposit an infinite amount of energy onto your cells when they were
absorbed!
Anyway, just to go back to the point about instantaneity, and
whether SR and QM contradict each other. I've found out about an
experiment that was used to investigate this called the Aspect
Experiment. The idea was first proposed by Einstein as a
thought experiment to demonstrate that the Copenhagen
interpretation was nonsense, because this experiment would have to
give absurd results, if Copenhagen were true. Eventually in the
1980s they managed to adapt his thought experiment to a real
experiment which they carried out. The experimenters were actually
trying to prove Einstein right. But unfortunately they got the
"absurd results" which Einstein said couldn't possibly occur.
The experiment concerns what Einstein described as Spooky Action at
a Distance. This is basically the passage of information faster
than light. An atom is stimulated so as to emit two photons, which
travel in opposite directions. Now the rule is that the polarities
of each photon are random, but must cancel each other (i.e. have a
phase difference of p). According to
Copenhagen, the polarities exist in a sort of "fuzzy" undetermined
quantum state until they are measured. As soon as one polarity is
measured, then both polarities collapse from being in a fuzzy state
to being fixed (because you can work out one polarity from the
other).
It actually works as follows. Each polarisation is measured as an
angle. The chance of a photon passing a certain filter at a certain
angle depends on both the photon's polarisation angle and the angle
between its polarisation and the filter. If locality is violated,
changing the angle at which to measure the polarisation of
the first photon will immediately alter the chance of the second
photon travelling through a polarising filter set at a different
angle.
Therefore it is done statistically. There is an inequality called
Bell's inequality, which should be satisfied if instantaneous
communication doesn't happen. When they did the experiment, Bell's
inequality was violated, meaning that information really did pass
between the photons faster than light. In principle they could have
been on the other side of the universe by this time.
The big puzzle now then is: how on Earth can this be consistent
with SR? In SR instantaneity depends on your frame of reference.
There is no such thing as universal instantaneity. Does viewing the
polarising filters from a different frame somehow change their
properties, so that the communication is no longer instantaneous?
If the polarising filters A and B are let's say, 1000 light years
apart, then to someone travelling at close to the speed of light in
the direction AB, it will look as if the collapse of the
wave-function happens at B 1000 years before it happens at A.
If the controls for the polarising filters are at A then to someone
going at close to the speed of light, it will look as if the signal
has travelled back in time. It will arrive at B 1000 years before
it was sent at A (by altering the polarisation of the filters). If
anyone could help me sort out these conceptual difficulties with
combining SR and QM then I would be very grateful.
Thanks,
Michael
By Neil Morrison (P1462) on Friday, April 21, 2000 - 07:47 pm:
Michael
This way of instantaneous communication is very much cheating! At
the level we're investigating, physics is all in terms of
probabilities anyway. Information doesn't pass between the photons:
I can't imagine one phoning the other one telling it what
polarisation to be! This is defined (in terms of probabilities) by
the conditions surrounding the photons existance/motion.
However, if we were to go along this line, what about gravitation.
If a large mass is suddenly called into existance somewhere, then
does the information that it is there translate to all surrounding
objects instantly? Suppose a vast mass suddenly appeared 1 light
year from the earth. Our astronomers couldn't detect it by
telescope for at least a year, but would it take this long for its
gravitational effect to kick in?
By Michael Doré (P904) on Friday, April 21, 2000 - 11:38 pm:
Yep - the information a gravitational field carries travels at
c. (This was a prediction of GR, where info can't travel
faster than c.) The EM field clearly travels at c also, because its
boson is the photon.
As for whether the polarisation experiment is cheating - to some
extent. My book says - "in effect, the non-locality would allow
someone on the other side of the universe from you to receive a set
of random numbers, determined by the angle of your polarising
filter." But now surely you can encode information in the time
delays between sending different sets of random numbers, by
removing the polarising filter at intervals. Or maybe not.
If it were impossible to send useful information then this would
probably solve the problem surrounding the combination of SR and
QM. Obviously I can't prove to you the results that my book states,
but it is a very well accepted experiment apparently.
Yours,
Michael
By Sean Hartnoll (Sah40) on Friday, April 21, 2000 - 11:55 pm:
Although I won't know the details of the
theory until next year (it's not taught until the
final year), I am completely sure that SR and QM were unified by
Dirac in the 1930s. His work was then developed to form what is now
known as Quantum Field Theory (QFT). This was extended by Feynman
and others in the 50s to include the electromagnetic field,
producing Quantum Electrodynamics (QED) which is a QFT with photons
in it, whereas Dirac's theory dealt with the QM of relativistic
electrons and had predicted the positron. This was then extended
further to Quantum Chromodynamics (QCD) by Gell-Mann and others
which formed the basis of the standard model of particle
physics.
Now, this model, IS, as far as I am aware, a coherent combination
of SR and QM, at least mathematically. I am not sure where it
stands with regard to the nonlocality of the wavefunction collapse.
I seem to remember reading somewhere that the apparent transfer of
information at speeds faster than light, cannot actually be used in
practice to transmit a force of some kind faster than light, but I
can't remember the details. But the important point is that
unifying SR and QM is not the main problem for theorists at the
moment, and I think most people accept it has been done, unifying
GR and QM is. And it is hoped that a QM theory of gravity will also
somehow resolve the issue of the wavefunction collapse, which is
thought by some to be related to gravity.
Sean
By Michael Doré (P904) on Wednesday, April 26, 2000 - 06:04 pm:
Thanks Sean. Sorry it's taken me so long to get back.
Anyway, I'm sure that QM and SR can be made consistent (even if I
don't know how). I guess the answer to the problems associated with
the non-local collapse of the wavefunction is that the collapse can
never be used to transmit useful information. Therefore, you
cannot send signals faster than light.
Thanks again,
Michael
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