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

the weight rises half the distance the monkey climbs up

Can you explain why Ben?

By conservation of potential energy - so not too overactive! - the weight should fall by the same amount the monkey climbs

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

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.

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);

1. 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?
   
   
2. 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?
   
   
3. 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?

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.

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

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.

I think that the overactive monkey gets pulled over the top of the frictionless pulley by the weight on the other side.

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!

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

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

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.

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

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.

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

We think that there is a simpler way of doing this. We don't know what, but we will keep thinking!

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.

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

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".

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

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..."

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

Hiya!
Ooops!

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 $\gamma$ negative).
Graham

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.

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).

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.

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

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

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

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?

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).

$r\text{'}\text{'}-r\theta {\text{'}}^2$

The transverse acceleration is:

$1/rd/\mathrm{dt}(r^2\theta \text{'})$

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:

$Fr=d/\mathrm{dt}({\mathrm{mr}}^2\theta \text{'})$
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

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:

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

but, if the system is frictionless, how could the monkey pull himself up in the first place.

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.

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:

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 bodius is given by:

$F\le \nu R$

where $R$ is the normal reaction force and $\nu$ is the constant of proportionality, dependent on the material which both bodies are made out of. As $R$ and $F$ are both forces, $\nu$ is a dimensionless constant, so cannot be used to get the unit second either.
Yours,

Michael

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

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 -mc² sqrt(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 -mc² sqrt(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

How about considering an electron accelerated through a potential difference?

Total energy before = m_o c² according to Einstein, so total energy after :

E = m₀ c² + 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

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

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.

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: c² dm/dt=v.d/dt(mv)

Now the trick is to multiply both sides by 2m and integrate, we obtain

c² m² =m² v² +C

and since v=0, m=m₀ , we have C=m₀ ² c² , where m₀ is the rest mass and m is the relativistic mass.

m² (c² -v² )=m₀ ² c²

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

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):

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

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

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.

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

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

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.

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

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) - c² m)/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 ) -c² M

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 )-c² M
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

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

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 v =/= c, 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

Michael,

$p\lambda =h$

and so $p=h\lambda =mc$
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

I'll write more later, I have to go right now.

Brad

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

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 x 7.09 l/c not 4 x 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 x 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

Everything follows from here. In particular it is useful to note that this equations imply that the interval $c^2(\mathrm{dt}{)}^2-(\mathrm{dx}{)}^2$ is the same in all frames, which gives quick answers to many questions.

Yes, it should be 4 x 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

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

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?

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

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 x 7.09l/c.

Brad

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

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 = m² c⁴ + p² c²

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

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?

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 g mv, 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,p_x ,p_y ,p_z ). A property of these objects is that "norm" is preserved: x₀ ² - x₁ ² - x₂ ² - x₃ ² 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

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

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

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.

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

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^-51 f. f may be around 10¹³ for visible light, making the mass of a photon of visible light about 5*10^-38 kg. Of course it has zero rest mass, but this is different.

Yours,

Michael

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.

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 x v₂ /c² )

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 argument:

take E = gamma m v

m is again the rest mass. Now gamma = 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

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

Aha - Sean, I've just realised that if your equation read:

v₁ = (v₂ + v₃ )/(1 + v₂ x 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 = 10⁶ c 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

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

Sean

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

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 xtime 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

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

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

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 m₁ c² and m₂ c² , 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 = m₁ c² + m₂ c²

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 x 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 pi ). 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

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?

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

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

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