By Brad Rodgers (P1930) on Monday, August 7, 2000 - 06:00 pm:
I'm trying to find a way to find the amount of elapsed time taken place upon a body going through acceleration. If we are given uniform acceleration, say v=t+C, then I think we can rewrite this as
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for T equals amount of time experienced by body and t is amount of time experienced by us.
Could someone tell me if this is right, and if so a way to integrate the RHS.
Thanks a lot,
Brad
By Simon Judes (P2636) on Tuesday, August 8, 2000 - 06:02 pm:
There is always a problem with the question: 'can SR handle
acceleration?'. The answer is yes and no. SR is perfectly capable
of dealing with accelerating objects that are seen by observers in
inertial frames. Indeed, if you differentiate the Lorentz
transformations twice, that is exactly what you will get. (I don't
think that is what you are after though).
On the other hand, SR says nothing about what accelerated observers
might see. This is the sense in which the General theory is
required to deal with acceleration. Note that there is only one
parameter in the Lorentz transformations: the relative velocity of
the two inertial frames considered. Were SR capable of dealing with
accelerated observers, there would have to be some acceleration
dependence in the transformations themselves, i.e. another
parameter.
In physical terms, what I think you are doing is trying to
approximate to an accelerated observer, by treating him/her as a
succession of inertial observers at different velocities. You then
take the limit of 'time spent at any one velocity' -> 0. This
seems like a sensible thing to do, but there is really no reason
why it should work. Remember that at each stage, your approximation
is already unphysical, because you are assuming that the step from
one velocity to the next is instantaneous. So, I doubt that the
result from integration will correspond to experiments with real
accelerated observers.
I don't think there is any way to solve the problem in the context
of SR. Say you made your approximation more physical by having
sudden, but finite accelerations between the velocity steps. You
would have solved one problem, but got nowhere, because you would
then need to know exactly what you are trying to find out, i.e. the
time-dilation formula for accelerated observers.
By Sean Hartnoll (Sah40) on Tuesday, August 8, 2000 - 06:58 pm:
As it goes, SR can sort of deal with
physics from within accelerated frames, the way it is done is to
consider the accelerated frame as a series of instantaneous
non-accelerated frames, but it is somewhat artificial and GR is
needed for a nice solution.
Brad - I don't really understand your notation, but one point is
that the acceleration v=t+C is not a valid acceleration because the
body will end up going faster than light.
The formula you want is, I think, that the proper time elapsed is
(t here is proper time)
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where Vm is the velocity four-vector.
Vm will be given by the force law.
Sean
By Simon Judes (P2636) on Tuesday, August 8, 2000 - 09:13 pm:
I'm not sure if calculations based on the approximation of a series of inertial frames actually produce the correct answers.
By Sean Hartnoll (Sah40) on Tuesday, August 8, 2000 - 10:59 pm:
Simon - It does work and is in no way an
approximation, but it only gives you instantaneous information. So
for example, it couldn't give you the trajectories of objects
(which require an integration over time), but it can give you, say,
what an accelerated observer sees as the instantaneous velocities
of other objects.
Sean
By Brad Rodgers (P1930) on Tuesday, August 8, 2000 - 11:24 pm:
I hadn't thought of the fact that my object will exceed c, but
for the moment let's just assume that t+C<c. Is there a way to
integrate my original equation(or indeed Sean's, though I don't
understand it as well-I'm just beginning to understand four-vectors
and the like)? Or is there even a point.
Why wouldn't acceleration as a series of seperate velocities work?
I am not sure I entirely understand why it wouldn't, so long as the
acceleration is uniform. I am not trying to find out what the two
observers see, just a proportion of the times they go
through.
The reason I wanted to find this out was to find a resolution to
the twin paradox, and if that integration integrates to something
even near what I think it would, I don't see much of a resolution
to the paradox, so maybe you are right, but I still don't see why.
I'll perhaps have to learn GR first though.
Also, the reason I was trying to figure the amount of time for the
accelerated observer is mainly for simplicity. This perhaps, was my
error. But, could we still intgrate that to form an idea of the
amount of time the inertial observer sees the body go through(not
the actual time traveled through)?
One last question: anyone know of a good book to learn tensors
from? I purchased D'Inverno's book, and the SR section was written
extremely understandably (though I'm still working on mastering
some of the topics). But then I got to tensors. The fastest I can
move is a page a day. So, is there a better book to learn from, or
will I just have to be content with finishing tensors in
January?
Thanks,
Brad
By Sean Hartnoll (Sah40) on Wednesday, August 9, 2000 - 04:53 pm:
Actually, D'Inverno's book has one of the
easier treatments of tensors. Don't worry if you find this
difficult, it is conceptually quite different from things you have
probably done before.
I think you will find it easier if you read the SR section on
Four-vectors thoroughly, you will not be able to move onto GR until
you understand four-vectors.
The problem with the treating acceleration instantaneously is that
you have to use different Lorentz transforms at each time instant
to transform into the accelerated frame. But it does work for a
given instant. It is not useful though for doing the integration
you want to do.
The twin paradox CAN be resolved without GR. The best way to do
this is to draw some spacetime diagrams, I think this is done in
D'Inverno, although I may be wrong.
Sean
By Brad Rodgers (P1930) on Wednesday, August 9, 2000 - 10:16 pm:
Yes D'Inverno's book does resolve the paradox through spacetime
diagrams, but the argument he uses doesn't quite make sense to me.
It seems as though the accelerated observer could draw the same
spacetime picture relative to himself and thus the paradox is as it
was before. Is the reasoning behind the resolution that in SR
acceleration is absolute and not relative.
I don't know that it really matters, but using clock hypothesis, I
was able to integrate my equation above to
T=(1-(t/2+v0/c)2)1/2
Where T is once again the proper time. Then, is the clock
hypothesis wrong?
Brad
By Sean Hartnoll (Sah40) on Wednesday, August 9, 2000 - 11:20 pm:
The clock hypothesis is not wrong.
The spacetime picture for the accelerated observer is not the same.
When there is an acceleration the observer is no longer inertial so
you have to change frames. So if you want to do the calculation for
accelreated observer you need to do it in two bits. This is why
most discussion of the twin paradox is done with a single
instantaneous acceleration, a more continuos treatment is not easy
in SR.
If you do the calculation in two bits like this, you find exact
agreement with what you would find using the other frame.
Perhaps you could explain where you got your original expression
from, I suspect it involved an invalid use of accelerating
frames.
Sean
By Simon Judes (P2636) on Thursday, August 10, 2000 - 01:10 am:
I don't think a continuous treatment is possible in SR.
Spacetime diagrams seem quite dubious too. There's always this
curvy bit where the acceleration is supposed to be taking place.
The whole point of a space time diagram is that you can just
transform the axis, and then look at the whole motion (i.e. the
same worldlines) from another observer's point of view. This is how
we get the wonderful graphical explanation of how two observers in
relative motion can both see the other foreshortened. However, this
only works when the worldlines of the observers are straight. The
curvy part seems like an ad-hoc addition that serves no purpose
other than to solve the twin paradox.
I'm not sure that the argument of an instantaneous velocity change
is at all valid either. If we won't allow acceleration, then
infinitely quick acceleration is not much better.
The only way I can think of to solve the twin paradox within the
confines of SR is by setting up an exactly equivalent problem that
doesn't involve acceleration:
First, twin 1 is passed by twin 2 moving with relative speed v.
Twin 2 goes on its way, and then meets twin 3, who is travelling in
the opposite direction, back towards twin 1. But, when twins 2 and
3 meet, they synchronise their clocks. So, twin 3 meets up with
twin 1, and they compare the readings on their clocks.
Here there is definately no problem in treating the motion in 2
parts, since it is 2 separate objects that are involved.
By Brad Rodgers (P1930) on Thursday, August 10, 2000 - 04:11 am:
Here was my logic used:
There is no doubt that the increase in proper time is equal to the
inverse of the rate at which moving clocks slow multiplied by the
increase in inertial time.
Thus dT/dt=1/b
Because the velocity changes at a rate of t, we arrive my above
equation apart from the flaw that v0+dt should be
v0+t. Now the clock hypothesis relies upon the fact that
the average velocity used in the calculations is the same as taking
into consideration all velocities over infinitesimal increments of
time . With this, it holds that my original equation integrates
to
T=(1-(t/2+v0/c)2)1/2t
(I forgot the ending t before)
I have not checked the differential of this, so I'll do so right
now, thus disproving or proving my conjecture on what the clock
hypothesis is based on(for v=t+C). It looks from where I am
currently as though the two are incompatible, though I don't see
why. I would have thought that they were saying the same thing as
each infinitesimal velocity multiplied by the amount of time that
it occured at from zero to t would equal the same distance as the
average velocity times t. But perhaps that's where I'm making my
mistake. Or maybe I'm making no mistake at all (ha!, I like to
amuse myself every once in a while). As it is, it's late (hence how
incomprehensibly I'm writing) and I have to wake up early, so I'll
write in with my finished calculation tomorrow.
Brad
By Brad Rodgers (P1930) on Thursday, August 10, 2000 - 05:02 am:
Deciding to just go ahead and do the raw algebra, I have found
that the two methods of calculation are indeed not equivalent! From
this solely, we can not conclude that even either of the methods
are true, just that one must be wrong. But, taking into mind
expiriments, one must be forced to conclude that my method stated
originally, is not true. But why then, is my method wrong? If given
info for instantaeneous rates, why can't we figure total rates? It
just doesn't seem to add up. I'll check my results in the
morning.
Brad
By Simon Judes (P2636) on Thursday, August 10, 2000 - 12:31 pm:
Brad, I think there are 2 reasons why your method didn't work.
One is that as Sean says, v=t+C doesn't make sense from a
relativistic point of view. We know that no observer can measure
the speed of any object to be > c, and clearly t+C will be >
c for sufficiently large t.
The other problem is exactly the one you highlight, i.e. that
although you have information at any particular velocity, you can't
integrate this in the normal way. This is because of the definition
of the integral as the limit of a sum. Here's an example of another
problem that can't be solved in this way. Consider a rocket
accelerating. Say we want to know how much fuel is burnt. We know
that fuel is only burnt when the rocket is accelerating. So, at
constant velocity, no fuel is burnt.
So, apparently,
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But of course, total fuel burnt at velocity v is always zero, and time spent at velocity v tends to zero in the limit. So it would appear that we can build a rocket that need no fuel at all. The reason this doesn't work no matter how accurate we make the sum, is that we are still not taking into account the bits that matter, i.e. the acceleration that occurs between the velocity steps. The formula above assumes that things that happen instantaneously will not contribute to the fuel used, which of course they do.
By Sean Hartnoll (Sah40) on Thursday, August 10, 2000 - 05:55 pm:
Simon - The last method you mentioned is
obviously the best and works. However, you really can do it with
two observers only in SR using an instantaneous acceleration.
Physically dodgy this may be, but I guarantee the spacetime
diagrams give you the right answer in whatever frame you do it. I
would write it down here but it's a bit long and requires the
diagrams.
The way you would have to do continuous acceleration is to use
different Lorentz transformations at each instant, doing this
basically gives you GR! (or at least the basics of it - see Misner,
Thorne and Wheeler, the big black book, the chapter on acceleration
in SR)
Sean
By Simon Judes (P2636) on Friday, August 11, 2000 - 12:42 am:
It kind of depends on who you ask. There is a sort of 'if it
gets the right answer then that's all we need to know' trend in
physics at the moment. So the physicists I know have no problem
with using space-time diagrams in this way. On the other hand, the
lecturer of the philosophy of physics course had other ideas! But
then they thought that all sorts of tiny details needed arguing
for, e.g. the fact observers travelling uniformly in opposite
directions measure the same speed for each other (principle of
reciprocity).
Thanks for the reference. Is the big black book good? The one I
have been using for GR is the smaller blue book by Weinberg.
Simon
By Sean Hartnoll (Sah40) on Friday, August 11, 2000 - 05:29 pm:
I think the 'trend' is hardly new, but has
been around as long as physics has. It tends to be the right sort
of attitude that allows you to break new ground and make
discoveries, because when dealing with something new, it is
inevitable that not everythign will be consistent right from the
start.
However, to SR, it is interesting that I recently flipped through a
book written as late as 1970, called The Logic of SR, the basically
contained a debate between eminent physicists as to whether the the
twin paradox was resolved and mnay seemed to think not!
But even so, I don't think there is anything unrigurous or
unphilosophical about using spacetime diagrams, they are just a
graphical representation of the mathematical situation which are
useful because they provide a visualisation.
The big black book is good as reference, I don't like the way it
does tensors, and it is far to long to try and learn from. I would
have thought the Weinberg book a little long to learn from also,
but maybe not.
Sean
By Simon Judes (P2636) on Sunday, August 13, 2000 - 05:45 pm:
There was a Prof. Dingle of somewhere or other who apparantly
maintained that:
1. the moving twin would be younger on arrival
2. SR doesn't predict this
3. so SR is wrong
I can't remember his justification, but apparantly he was quite
eminent; although his reputation suffered a downturn.
Space-time diagrams are perfectly good graphical representations of
many situations. But they only have the useful properties they do
under certain conditions.
The axes are distance and time, and so are particular to some
reference frame or another. The advantage is that if you want to
look at the situation from another inertial observer's point of
view, then you don't have to alter the diagram, you can just refer
to it from a set of axes oblique to the first one. But this only
works if the observer you want to switch to has a straight world
line, i.e. is inertial. Where the line curves, you would
effectively have to use different axes for each point, which
needless to say would remove the elegance of the diagram. This is
exactly analogous to using a different Lorentz transformation for
each instant.
I think the problem with the spacetime diagram explanation of the
twin paradox is that when you draw the lines representing the
signals from one twin to the other, you don't really know how
densely to put the signals from the moving twin to the inertial one
during the period of acceleration. You can't just place them evenly
along the line, because you don't know how the curve transforms to
the straight line that it would be when viewed from the point of
view of the accelerating observer. What you find in 'explanations'
in books is I think just a fudge, i.e. they put in as many or as
few signals as they need to to solve the twin paradox.
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 01:15 am:
Yes, Prof. Dingle comes up a lot in the
book I mentioned.
The only problems with the spacetime diagrams is that to use them
properly in this case you need a point acceleration. This is
unphysical, agreed, but if you look at it as a limiting process of
more and more instantaneous accelerations (something like you
normally treat a delta-function in physics), I think the limiting
process is valid. So the curved part of the world line becomes
smaller and smaller and hence less important.
Basically, I'm trying to say that the instantaneous acceleration
isn't a fudge, but a convenient limiting process that we use in all
sorts of situations in physics, pretty much whenever delta
functions come up.
A clear way of looking at the situation is to note that the proper
time elapsed for an observer is proportional to the length of the
world line (this is the equation I wrote a few posts ago), however,
we are in Minkowski space so length has a minus sign when it is in
one of the directions. So in fact the longer line on the spacetime
diagram is the line for which less time has elapsed. This is why
the inertial observer sees that the accelerated observer has aged
less.
Now for doing the reverse calculation (i.e. from the accelerated
observer), we need two diagrams, one for before the point
acceleration and one for after. In both this diagrams both
observers are straight lines. However, if you actually draw them
you find that the total length of the line due to the inertial
observer is longer. So more time has passed for him and there is no
paradox.
(some pictures would have helped that explanation, hopefully you
can draw them...)
Sean
By Simon Judes (P2636) on Monday, August 14, 2000 - 02:16 am:
The example was only really relevant to the case of continuous acceleration. If we accept the validity of the point acceleration model then I think the space-time diagram explanation is fine. I don't think that it is acceptable though. You say the curved part becomes smaller and therefore less important. Why should this be so? After all, as the time spent accelerating becomes smaller, the degree of acceleration has to increase for the moving twin to get back in the same amount of time. The comparison with delta functions presents another problem. Say we define it in the usual way (slightly simplified):
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I can't remember exactly what f(a,x) should be, but it's just some function that gets more and more peaked around x=0 as a -> ¥.
1. this shows clearly the last point, that even though the integrand tends towards a null function, the integral stays approximately constant. So, the diminishing time period over which acceleration occurs makes no difference when the integral is carried out.
2. We can evaluate the integral for any value of a. On the other hand, we can't work calculate the time difference between the twins' clocks for any value of the acceleration period. If you want to show that the limit of a point acceleration is going to produce approximately right results, then you need to find the equations that govern the situation when the acceleration is continuous, and then take the limiting case. That is the only way we can know that the bit we are ignoring is negligable.
You said that the proper time elapsed is proportional to the length of the world line. This is another thing that is only true in the limit of a point acceleration. Indeed that is exactly what we don't know about accelerating observers.
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 02:22 am:
I am pretty sure the last sentence is
wrong. It IS true ALWAYS that the length of the world line,
understood as
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This is why the limiting process is valid. It is true that the integral remains the same, what I was saying is that as a spacetime event it becomes more localised, sorry for the ambiguous language.
So the instant acceleration itself does not take any time by either of the proper times invloved. So it is in fact negligible. I think this resolves the problems you mention.
Sean
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 02:26 am:
... it is in fact a delta function, not just a comparison.
By Simon Judes (P2636) on Monday, August 14, 2000 - 05:31 pm:
How do you know it is always true that the
length of the world line is the proper time? If this is true for
all worldlines, then surely that's all we need. It's easy to show
it for straight worldlines, as the time axis of the spacetime
diagram is just rotated under the Lorentz transformation.
Ok, so in the point acceleration approximation, you just have a
world line divided into two parts, both of which are straight
lines. So you can calculate the proper time along it. But, consider
one of the steps towards this limit, i.e. a small period of
acceleration. How do we know that the proper time will tend towards
the value given by the instant acceleration approximation?
The analogy I am thinking of is with the length of the diagonal of
a unit square. We know it is sqr(2). It is also the limit of the
following series:
All of these lines have length = 2, even though they tend towards
the diagonal with length sqr(2).
So, if the real physical situation is some way towards the limit,
and all we know is the value at the limit, then we don't
really know what will happen.
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 07:23 pm:
It is true for all
world lines.
The series you give doesn't converge to the expected value because
of the way it's been constructed (basically to do with the
difference between discrete and continuum ideas). This does not
happen here because if we agree that the length of the world line
gives the proper time, and I will prove this in a minute, then it
is clear that the proper time associated with the acceleration is
getting smaller and smaller and tends to zero, there is no problem
with the limit.
Now, the proof, in some sense it is obvious from the 3D analogue
that sqrt(dx/dt dx/dt) dt = dx (in fact, this is exactly what is
happening). In another sense it is by definition, because all the
proper time is is a way of parametrising the world line and in fact
the integral in my previous post is invariant under
reparametrisation t -> t'(t) . Here t is the parameter along the
world line, it is called the proper time if it is such that
dXm/dt gives the 4-velocity
Vm which satisfies
Vm Vm = c2. Which leads us to the
proof, because it implies the integral becomes ò(c dt) = c ttot = c ×
total proper time.
Sean
By Simon Judes (P2636) on Monday, August 14, 2000 - 09:20 pm:
Ok, I agree your proof works. So why
bother at all with the point acceleration business?
Also, in a sense, your integral is giving something additional to
Einstein's postulates: It is ruling out any extra explicitly
acceleration-dependent phenomena. So in that sense, it is
extraneous to SR, although to be honest, I'm getting less and less
sure where the borderline lies between SR and GR.
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 09:33 pm:
The point
acceleration limit is useful because in this case we can use
spacetime diagrams, which we can't do while the particle is
accelerating over a finite time (well, we can, but we can't do them
from the point of view of the accelerated observer, which is what
we want to do to reolve the paradox).
In a certain sense it is true that acceleration is not mentioned
explicitly in Einstein's postulates. However, once you have set up
Minkowski space, you can construct 4-vectors, including
4-acceleration, and this doesn't involve any new physics. But
perhaps you are right that there is an additional postulate which
is that an accelerated observer is instantanously an inertial
observer... I'll think about it.
The boundary between SR and GR is reasonably clear, I think. In SR
the metric is the minkowski metric and in GR it is a more general
metric, that is determined by the GR field equations and which
influences motion through the geodesic equation. Everything else is
the same.
Sean
By Sean Hartnoll (Sah40) on Monday, August 14, 2000 - 09:34 pm:
whooa, sorry, I forgot to add that in GR you have the equivalence principle which tells you that you can transform into the accelerating frame and gives you the transformation to do so..
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