Man Running on a train

1. Apr 11, 2006

yogi

A train travels at a high speed v wrt the earth. A runner on the train sprints toward the back of the train at velocity v wrt the train.

Clocks on the train should run slow compared to earth clocks. The runner's clock should run slow compared to the train clock. Therefore the runner's clock should run doubly slow compared to the earth clock. But the runner is not moving wrt the earth!

2. Apr 11, 2006

masudr

The runner's clock ticks slowly compared to the train's clock, for an observer on the train. The train's clock ticks slowly compared to the runner's clock, for the runner.

Similar considerations applying to the train's clock and the earth's clock will show why the following
does not follow. The relative velocity between the earth and the sprinter are, as you say, 0. Hence the clocks will tick at the same rate.

Last edited: Apr 11, 2006
3. Apr 11, 2006

robphy

In addition, note that "time dilation factors" are not multiplicative.

Specifically, assuming "spatial motion only in the (say) x-direction",
since rapidities are additive, the $\gamma$-factor (i.e. $\cosh\theta$) is not multiplicative since
$$\cosh(\theta_C-\theta_A)\neq \cosh(\theta_C-\theta_B) \cosh(\theta_B-\theta_A)$$.
Instead, it is the doppler factors $k=\exp\theta$ that are multiplicative since
$$\exp(\theta_C-\theta_A) = \exp(\theta_C-\theta_B) \exp(\theta_B-\theta_A)$$.

4. Apr 11, 2006

Staff: Mentor

Isn't this just the same mistake people always make with the twins paradox? Ie, taking the trip only halfway? To really compare the runner's clock with the stationary one on the train, without having to deal with the simultenaity issues, the runner has to turn around and run back to the stationary clock. Then he is unambiguously moving (has moved) wrt both the train and the ground.

The whole point of the twins paradox is to have the two twins standing next to each other at the end of the trip, and really being different ages. Not doing that confuses people into thinking time dilation is just a signal-delay trick.

: Just to clarify, those quesitons are not rhetorical - I'm an engineer, not a physicist and I'm trying to make sure my own understanding of the concepts is correct.

Last edited: Apr 11, 2006
5. Apr 11, 2006

robphy

I don't think this is really a twin paradox problem.
It's more of a relative velocity prolem.

Focussing on the issue of relating time dilation factors in different frames, I think the original problem can be reformulated so that the "man on the train" [effectively "running in place"] is "passive". Regard the floor inside the train to be frictionless. Consider the situation when the everything is initially at rest. Then the train suddenly accelerates for an instant to the right... Now, study the problem. The train moves to right wrt the earth. The man moves to the left wrt the train with the same speed. A spacetime diagram will clarify that the earth and the man have parallel worldlines, and hence, a relative velocity of zero.

By the way, in the usual twin paradox (with symmetrical inbound and outbound speeds) the return speed referred to is measured by the earth. In the problem posed, the "return speed" is measured by the "outbound vehicle".

6. Apr 11, 2006

yogi

Russ- I didn't see it as a twin round trip analogy - I did see it as a direct parallel to statements made by Einstein in part 4 of the 1905 paper involving the one way transport of a clock that has been synchronized with a distant clock that remains at rest.

In 1918 Einstein wrote another paper where he rejected all the traditional SR "twin paradox" solutions and concluded that the round trip twin journey could not be properly analysed w/o GR.

Anyway, Wheeler poses the paradox on p 168 in the second edition of Spacetime Physics but doesn't solve it - and I am not sure the hints he gives will resolve the problem in a manner that is consistent with the one way transport of previously synchronized clocks which Einstein derives from the observational relationships. In other words, if one way transport leaves the clocks out of sync when they are compared, time dilation is actual - and it cannot be argued that there is in reality no time dilation because the observations are only apparent (a condition demanded by reciprocity).

Last edited: Apr 11, 2006
7. Apr 11, 2006

pervect

Staff Emeritus
There are many ways of working out this problem, but everyone will agree that when the train clock next meets the walking clock, the walking clock will read more time than the train clock.

If we let the train encircle the earth, one of the key steps in eliminating the apparent pardox is noting that we cannot synchronize all the clocks on the train using the Einstein convention. We can start at one clock and start synchronizing clocks with this particular clock in one direction (say clockwise) using the Einstein convention. However,when we attempt to complete the circle, we will find that the first and last clocks are not, according to the Einstein convention, synchronized.

Note that this implies that the train is not in an inertial frame - though if we were thinking, we already knew that as it has to accelerate to be moving in a circle.

This is therefore a problem in SR with generalized coordinates. Peadagogically this is pretty much equivalent to GR in sophistication of methods, though because the problem does not actually make use of Einstein's field equations it could still be classed as a "SR" problem.

Last edited: Apr 11, 2006
8. Apr 11, 2006

robphy

(I just read the problem in Spacetime Physics.)
The problem reduces to a problem in Minkowski vector algebra, eventually leading to the following relation:

the time-dilation factor between the (r)unner and (e)arth is
$$\cosh\theta_{re}=\cosh\theta_{rt}\cosh\theta_{te}+\sinh\theta_{rt}\sinh\theta_{te}$$, where $$\theta_{re}=\theta_{r}-\theta_{e}$$. As I said above, the time-dilation factors are not multiplicative... Instead, it's this expression... Note when the relative velocity of the runner and train $$\tanh\theta_{rt}$$ is opposite to the relative velocity of the train and earth $$\tanh\theta_{te}$$, that is to say,
$$\tanh\theta_{rt}=-\tanh\theta_{te}$$, then $$\theta_{rt}=-\theta_{te}$$.. or $$0=\theta_{rt}+\theta_{te}=\theta_{r}-\theta_t+\theta_t-\theta_e=\theta_{re}$$

I don't have time to write out the whole calculation but it starts like this:

Write the unit 4-velocity vector of the train in terms of the earth's unit 4-velocity:
$$\tilde t=\gamma_{te}(\hat e + \tilde v_{te})=\cosh\theta_{te}\hat e + \sinh\theta_{te}\hat e_{\perp}$$
the runner with respect to the train:
$$\tilde r=\gamma_{rt}(\hat t + \tilde v_{rt})=\cosh\theta_{rt}\hat t + \sinh\theta_{rt}\hat t_{\perp}$$
and now try to write the runner with respect to the earth by substitution of the first into the second and identifying with the following
$$\tilde r=\cosh\theta_{re}\hat e + \sinh\theta_{re}\hat e_{\perp}$$

It's essentially a change of basis problem... (under the assumption that all motion is in (say) the x-direction.)

The calculation is more complicated thinking in terms of $$\gamma$$'s and $$\beta$$'s... and even worse with relative velocities $$v_{rt}$$, etc... (Rapidity is a good thing... it suggests analogues of trigonometric methods from Euclidean geometry.)

Last edited: Apr 11, 2006
9. Apr 11, 2006

yogi

pervect - I think the intent of the author was that the earth be regarded as a SR inertial frame, and therefore the train becomes a valid inertial frame (after the train has accelerated to a uniform velocity wrt the earth). After the man begins running along the top of the train at a constant velocity, the man is also a good inertial frame.

10. Apr 11, 2006

yogi

robphy - what is the English translation of your post #8.

11. Apr 11, 2006

JesseM

Compared how, exactly? If two clocks start out synchronized in a single location and move apart at constant velocity, different frames will just disagree about which clock is behind the other at any given moment. Just analyze this problem from the point of view of two different inertial reference frames using the Lorentz transformation, you'll see that each frame observes the normal laws of SR being obeyed (for example, the rate of a clock's ticking is given by the normal time dilation formula which is based on its velocity), and all frames agree about what a clock reads at the moment it passes some other object (like another clock at rest in a different frame).

It seems like a lot of your problems with SR come from the assumption that there must be an objective truth about which of two clocks is "really" running slower. But the essence of SR is that there's no need for such an assumption, you can analyze any problem from the point of view of different frames which disagree on this, and the laws of physics will work exactly the same in each frame so there's no way to pick out one frame's view as more "real" than any other.

12. Apr 12, 2006

pervect

Staff Emeritus
If the train is not accelerating (i.e. it's moving in a straight line) then you basically have the classic twin paradox with no turn-around. The man thinks the train's clock is slow, any individual on the train thinks the man's clocks are slow.

Once the running man passes a clock on the train, he never passes the same clock again.

If the train is moving in a circle, it must be accelerating, and the man will pass the same clock more than once. This is really the only case that I thought was interesting, the standard twin paradox has been discussed extensively.

In the case where the train is accelerating, the easiest approach is to totally ignore the non-inertial coordiante system of the train, and work purely in some inertial coordinate system.

Last edited: Apr 12, 2006
13. Apr 12, 2006

robphy

Sorry. It's the composition of two boosts in the x-direction.

\begin{align*} \tilde r &= L_{re}\tilde e &=& L_{rt}\tilde t &=& L_{rt}\left( L_{te} \tilde e \right)\\ \left(\begin{array}{c} \cosh\theta_{re}\\ \sinh\theta_{re} \end{array}\right) &= \left(\begin{array}{cc} \cosh\theta_{re} & \sinh\theta_{re} \\ \sinh\theta_{re} & \cosh\theta_{re} \\ \end{array}\right) \left(\begin{array}{c} 1\\ 0 \end{array}\right) &=& \color{red}{ \left(\begin{array}{cc} \cosh\theta_{rt} & \sinh\theta_{rt} \\ \sinh\theta_{rt} & \cosh\theta_{rt} \\ \end{array}\right) \left(\begin{array}{c} \cosh\theta_{te}\\ \sinh\theta_{te} \end{array}\right) } &=& \left(\begin{array}{cc} \cosh\theta_{rt} & \sinh\theta_{rt} \\ \sinh\theta_{rt} & \cosh\theta_{rt} \\ \end{array}\right) \left[ \left(\begin{array}{cc} \cosh\theta_{te} & \sinh\theta_{te} \\ \sinh\theta_{te} & \cosh\theta_{te} \\ \end{array}\right) \left(\begin{array}{c} 1\\ 0 \end{array}\right) \right] \end{align*}

I'm trying to describe "the runner's 4-velocity in the earth's frame".... using the given information for "the runner's 4-velocity in the train's frame" and "the train's 4-velocity in the earth's frame".

The expression is red (after you matrix-multiply that out) tells you how to express the time-dilation factor between the runner and the earth in terms of those factors for the runner-train and the train-earth. As I said above, it's not the product of the time-dilation factors (it's not the product of $\cosh(\Delta\theta)$'s).

If rapidities look foreign, this might make things look more familiar:
$$\left(\begin{array}{cc} \cosh\theta_{rt} & \sinh\theta_{rt} \\ \sinh\theta_{rt} & \cosh\theta_{rt} \\ \end{array}\right) = \cosh\theta_{rt} \left(\begin{array}{cc} 1 & \tanh\theta_{rt} \\ \tanh\theta_{rt} & 1 \\ \end{array}\right) = \gamma_{rt} \left(\begin{array}{cc} 1 & \beta_{rt} \\ \beta_{rt} & 1 \\ \end{array}\right) = \frac{1}{\sqrt{1-\beta_{rt}^2}} \left(\begin{array}{cc} 1 & \beta_{rt} \\ \beta_{rt} & 1 \\ \end{array}\right)$$

Last edited: Apr 12, 2006
14. Apr 12, 2006

nazgjunk

Relativity is weird... I don't know much about it (yet, i'm planning on doing physics at uni), but I have one question: What happens when the train, or the runner suddenly stands still? As said, the train clock, from the runners view, ticks slower than the runner's clock. From the train's viewpoint however, the runner's clock however ticks slower than the train clock. I now wonder what happens when the runner suddenly stands still (decelerates to zero compared to the train).

15. Apr 12, 2006

yogi

But Einstein tells you there is an objective reality to the one way trip of a clock - we keep going over this. When two spaced apart clocks are synchronized, and one is put in motion (brief acceleration followed by a coasting period) the clock put in motion will read less when it arrives at the second clock - so as to the first part of the analysis the train clock will read less than the earth clock, and the time lapsed on the earth clock will be greater than the time lapsed on the train clock. Its not reciprocal, its not apparent, it is an actual physical difference in the location of the clock hands when they are brought together at some later time. Now when the runner begins his sprint toward the caboose, its the same as if the train negatively accelerated to a halt - the runners watch much tick at the same rate as the earth clock since he has zero velocity wrt the earth when he is moving at v toward the caboose. We know this much in order to avoid internal inconsistencies

16. Apr 12, 2006

JesseM

In a one-way trip, he only says there's an objective reality about what the two clocks read when they meet, but not an objective reality of which was ticking faster as they approached each other. You can analyze the situation from the perspective of any frame and they will all get the same answer as to what the two clocks read when they meet, so there's no contradiction with the idea that each frame's perspective is equally valid. Likewise, in the round trip case, every inertial frame also makes the same prediction about what the clocks read at the moment they reunite, which in this case means they all agree about the ratio between the average rate of ticking of each clock over the whole trip--but they still disagree about the ratio between the rates of ticking of each clock at particular moments during the trip. For example, in the twin paradox, one frame may say that the travelling twin's clock was running faster than the earth twin's clock during the outbound leg and slower during the inbound leg, while another may say the opposite, even though they both agree the average rate of ticking of both clocks from the beginning to the end.

So, the twin paradox still does not provide any basis for preferring one inertial frame's analysis to another (or preferring one view of which clock was ticking faster at a particular moment), according to Einstein. And it's a pretty simple matter to see that in the case of two objects moving at constant velocity without accelerating, different frames will disagree about which clock is ticking slower. Since Einstein would say no frame's perspective is more "objectively true" than any other, obviously he would not say there is any objective truth about which clock is ticking slower in this situation.
We do keep going over this. Don't you remember you already agreed that you can analyze this situation from the point of view of any inertial frame you wish, including frames where the clocks were not synchronized when they were at rest relative to each other, and where the clock that was behind when they met was actually ticking faster? You also admitted that you have no clear idea of how to pick which frame represents the "truth" in any situation, in post #187 of this thread. And regardless of what you think, surely you would agree that Einstein would say that each inertial frame's view of this situation is equally valid, since they each observe the same laws of physics and they each get the same answer as to what the clocks read when they meet?
The view of each frame is completely reciprocal, including each frame's different opinion of how fast each clock was ticking at a particular moment. Again, you're free to believe that physicists are wrong about this somehow, but if you want to answer questions by mainstream physicists like the one by Wheeler in Spacetime Physics, you have to realize this is how they all think about things, and be comfortable thinking this way yourself.
And since all inertial frames agree on what two clocks read at the moment they unite, it's completely illogical to use this as an argument for preferring one frame's view of a situation over any other's.
I don't know what you mean here--if he runs towards the caboose, he shares the same rest frame as the earth, and the earth-frame is the one that sees the train moving at high speed, not halted.
Exactly, and in his frame, the train's clock is ticking slower than his. On the other hand, in the train's frame, both the runner's clock and the earth's clocks are the ones that are slowed down, by the same amount. Both frames agree that the runner's clock is ticking at the same rate as the earth's clocks, but one frame sees them running faster than train's clocks and one sees them running slower. Once you accept that there need not be an objective truth about which of two clocks is running slower, there's no paradox here--just because clock B is running slower than clock A in frame 1, and clock C is running slower than clock B in frame 2, that doesn't imply that clock C needs to be running slower than clock A, in any frame. It may be that clock A and clock C are running at the same rate in every frame, as in this example.

Consider the following analogous "paradox" in galilean relativity:
If you agree there is nothing paradoxical here because there is no objective truth about whose speed is greater than whose, that speed is purely frame-dependent, what's wrong with treating the rate of ticking of clocks in the same way?

Last edited: Apr 12, 2006
17. Apr 12, 2006

yogi

Jesse:To sum up - I fully agree there are many ways to analyse the situation - just as you have gone through the numbers on a previous thread. But the interest in these issues is deeper. Some persons can find a very simple resolution to the classic clock or twin paradox - Example: SR doesn't apply where accelerated frames are involved - thr turn around twin feels the acceleration, therefore we know which one is not inertial - and the time dilation at the end of the round trip is real for whatever reason - but whatever the reason, it doesn't violate SR because SR isn't applicable to the situation where the traveling twin switches frames. This line of reasoning yields an answer that comports with what we all believe would be the age difference if we actually made the experiment. If you are satisfied there is nothing to be gained by attempting to explore the cause - that is fine.

When Einstein first published his 1905 paper he offered no explanation of why the moving clock recorded less time. His intuition was correct - but the world had to wait many years before actual time differences could be measured. In the meantime, many physicists questioned the cause - in 1918 Einstein responded to the many questions regarding this so called paradox by inserting some physics - he proposed that acceleration was really responsible for the tiime difference in the twin scenario, and that the accelertion would take the form of a pseudo gravitational field. Does this really explain things? Well, again, it gives the right answer - but we know the twin paradox can also be structured so there is no turn around force - the outbound twin can sling-shot around a distance planet in which case the entire journey can conceivably be imagined to take place in a single reference frame, or he can hand off his clock data to an inbound triplet.

While it is true that inertial frames are equivalent for the formulation of physical laws, They are not idential in every respect. Take a look at Rindler at page 30 and 31 in connection with the twin thing: ...."Granted that the motions of A and B are not symmetric, yet it could be maintained they are symmetric most of the time...the three asymmetric activities can be confined to arbitrarily short times....how is it then that such a large asymmetic effect can arise, and, morover, one that is proportional to the symmetric portions of motion? " The reason is that accelerations, however brief, have immediate and finite effects on A but not B."

Continued below

18. Apr 12, 2006

yogi

More from Rindler: "During the period T1 (A - the outbound traveler) finds he ...has transferred himself to a frame in which the distance between earth and object is halved."

Now - one of two things appear to be happening - on the outbound leg A accelerates to a new frame where the distance is halved - so you can say, A's clock doesn't really run slow during the trip - it just recorded less time during its voyage to the object point because the trip length was less in the new frame

But one could also say, in the new frame, A's clock ran slower so the trip appeared to take less time.

Which is reality.

Is this a distinction w/o a difference?

19. Apr 13, 2006

yogi

Either way, there is a difference between the two inertial frames - in the case of the man running on the train, the initial acceleration of the train as it leaves the station places the man and the train in a new inertial frame (after the acceleration is over) which has different properties than the earth frame from whence it was accelerated. The issue is whether the different properties are a consequence of time running slower in the moving train frame (relative to the earth), and therefore there will be a difference in the train clock and the earth clock reading when they are subsequently compared - or if the earth clock and the moving train clock beat at the same frequency, but the total length of the trip in the train frame is contracted, ergo the clocks accumlate a different number of seconds even though they ran at the same frequency.

We know from GPS satellite data that two clocks, both in an inertial environment, do not run at the same frequency after one has been accelerated to a velocity wrt to the other. Whatever your feeling about why the train clock and the earth clock are out of sync when they are compared, the experiments would seem to support the notion that the moving frame is different from a temporal perspective.

20. Apr 13, 2006

masudr

Proper time is something that is calculated/measured along a worldline.

$$\tau = \int_P d\tau = \int_P \sqrt{g_{\mu\nu}dx^\mu dx^\nu}.$$
*put a minus in there if you choose a different metric

The twin paradox is also explained quite well in http://en.wikipedia.org/wiki/Proper_time#Example_1:_The_twin_.22paradox.22 with example figures. The common misconception is that SR cannot deal with acceleration -- it can, it just cannot deal with accelerating reference frames. I could happily be watching the twins do their "paradoxing" from my inertial reference frame, and SR correctly predicts that the twin executing the motion will be the one who ages less.

21. Apr 13, 2006

robphy

Indeed, no matter how "tries" to interpret it... time-dilation for this guy, length-contraction for that guy, sudden switching of frames, etc.. etc.. etc.. the bottom line is this interpretation that http://www.google.com/search?q="proper+time"" is the spacetime-arclength. In moving to more general situations (e.g. general relativity [generally, curved spacetime on a general 4-manifold]), every other explanation will fail for one reason or another.

If SR is interpreted in its modern form as "a flat [i.e. zero-curvature] spacetime on R4", SR can deal with accelerating reference frames. Certainly the mathematics (calculus and differential geometry in flat spacetime) and its physics interpretation may be more difficult. Nevertheless, in whatever coordinate system you choose, the square-interval of any spacetime segment is an invariant. The issue is akin to using polar coordinates or (say) http://mathworld.wolfram.com/ParabolicCoordinates.html in Euclidean space.

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22. Apr 13, 2006

masudr

I'm not sure how my knowledge of Lorentz transformations acting on four-vectors and tensors could be adapted to accelerating reference frames.

I have:

$$T^{\left[i_1',i_2',...i_p'\right]}_{\left[j_1',j_2',...j_q'\right]} = \Lambda^{i_1'}{}_{i_1}\Lambda^{i_2'}{}_{i_2}...\Lambda^{i_p'}{}_{i_p} \Lambda_{j_1'}{}^{j_1}\Lambda_{j_2'}{}^{j_2}...\Lambda_{j_q'}{}^{j_q} T^{\left[i_1,i_2,...i_p\right]}_{\left[j_1,j_2,...j_q\right]},$$

where

$$\Lambda^\alpha{}_\beta = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\mbox{ and }\eta^{\alpha\beta} =\eta_{\alpha\beta} = \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$

In my opinion, this more or less sums up all of special relativity (with a few more definitions [which I assume] about proper time, four-momentum, four-force etc). How can this be adapted to accelerating reference frames (if at all)?

Last edited: Apr 13, 2006
23. Apr 13, 2006

Staff: Mentor

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24. Apr 13, 2006

masudr

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25. Apr 13, 2006

pervect

Staff Emeritus
MTW's "Gravitation" has a good and fairly long treatment of the uniformly accelerated observer as well, if you can get a hold of it.

A brief overview of the process goes like this

Phase 1: Compute the motion of a uniformly accelerated obsever

Answer: let $u$ be the 4-velocity of the accelerated observer, and $a = du/d\tau$ be the 4-acceleration of the accelerated observer.

To say that the acceleration is uniform is to say that $a^j a_j$ is constant, i.e. that $\eta_{ij} a^i a^j$ is constant.

Because $u^i u_i = -1$, which is a fixed magnitude, the 4-acceleration will always be orthogonal to the 4-velocity, i.e. $a^i u_i = 0$.

This yields the solution

$$x(\tau) = \frac{cosh(g \tau)}{g}$$

$$t(\tau) = \frac{sinh(g \tau)}{g}$$

which can be verified by calculation of u and a.

Phase 2: Compute a coordinate system for the accelerated observer.
Phase 2a: compute a tetrad carried by the uniformly accelerated observer

$$(e_{0'})^u = [\mathrm{cosh} g\tau, \mathrm{sinh} g\tau, 0, 0]$$
$$(e_{1'})^u = [\mathrm{sinh} g\tau, \mathrm{cosh} g\tau, 0, 0 ]$$
$$(e_{2'})^u = [0, 0, 1, 0]$$
$$(e_{3'})^u = [0,0,0,1]$$

Phase 2b Compute the coordinate system of the accelerated observer. We can do this implicitly, by letting $x^i$ be an inertial coordinate system, and $\chi^i$ be the coordinate system of the accelerated observer. Using the previous results

$$x^0 = (\frac{1}{g} + \chi^1) sinh (g \chi^0)$$
$$x^1 = (\frac{1}{g} + \chi^1) cosh(g \chi^0)$$
$$x^2 = \chi^2$$
$$x^3 = \chi^3$$

Note that many coordinate systems are possible for an accelerated observer, this is just one of the standard ones.

Radar coordinates are also useful, you can see some of George's past posts for how they work.

Phase 4. Compute $ds^2 = \eta_{ij} dx^i dx^j$ in terms of the new variables.

$$ds^2 = -(1 + g \chi^1)^2 (d\chi^0)^2 + (d\chi^1)^2 + (d\chi^2)^2 + (d\chi^3)^2$$
This gives you the metric $g_{ij}$ associated with the new coordinates $\chi^i$.
Then you can just use the $\chi^i$ as coordinates along with the new metric $g_{ij}$.