# Special Relativity Basics

1. Jun 30, 2009

### JM

In his 1905 paper Einstein introduced two coordinate systems which he called K and k. It seems that both systems are inertial because there are no external forces and they are moving at the constant relative speed v.
But I'm not certain, so are they both inertial? Is it correct that either one can be considered the 'stationary' system and the other one the 'moving' system?

2. Jun 30, 2009

### George Jones

Staff Emeritus
Yes and yes.

3. Jun 30, 2009

### JM

So, with K( X,Y,Z,T) stationary the clocks of k( x,y,z,t) are moving in the + X direction at speed v. Putting X=vT in the transfforms leads to t=T/m, where m is the quantity Einstein calls beta. So t is less than T.
With k being stationary the clocks of K are moving in the -x direction at speed v. Putting x=-vt in the transforms leads to T=t/m, ie T is less than t.
Is there general agreement with this result? Is everyone OK with it?

4. Jun 30, 2009

### HallsofIvy

Staff Emeritus
Yes, an observer in the K system would take the K system as stationary, K' as moving, and observe time in the K' system, t', as slower than t in the K system. An observer in the K' system would take the K' system as stationary, K as moving, and observe time in the K system, t, as slower than t' in the K' system.

5. Jun 30, 2009

### Naty1

Yes.....It will have to do until a better theory is developed!

6. Jun 30, 2009

### JM

Thanks to you both. So can this example be generalized to say that ' Any analysis carried out with K stationary can also be carried out with k stationary, with suitable choice of k coordinates' ? ( In the example the -x coordinate was considered equivalent to the +X coordinate.)

7. Jul 4, 2009

### JM

To continue:
The 1905 paper adopted K as the stationary system for all its analysis. For the 'clock paradox' the result was t=T/m, the same as the slow clock.
Based on the above discussion, k can be chosen as stationary, and the 'clock paradox' calculated using the same concepts and math, resulting in T=t/m.
These results are compatible because the comparison is between the time of the stationary system and the time of the moving system 'as viewed from the stationary system'
The equivalence of K and k has apparently been overlooked, and consequently many writers have treated K as the only alternative, and have extensively elaborated this case.
OK?

8. Jul 4, 2009

### JesseM

What specific part are you referring to when you talk about the "clock paradox"? Is it this part from section 4 of the paper?
Is so, note that there is an asymmetry in how the problem was described--he said that the two clocks were synchronized in the K frame up until the moment clock A was moved. This is crucial to understanding why clock A is the one that is behind when the two clocks meet. If he had instead said the clocks were still at rest in the K frame, but synchronized in the k frame, and then when clock A was accelerated it came to rest in the k frame while B continued to move towards it in the k frame (i.e. B continued to be at rest in the K frame), then in that case it would be clock B that was behind when they met.

Last edited: Jul 4, 2009
9. Jul 6, 2009

### JM

Hello again, JesseM,
Yes, I refer to p.4 of '1905.
I agree with your description of the original 'clock paradox'. A and B start at rest in K, A moves and returns, and A ends up slow compared to B.
From your last sentence I gather that you recognize a valid alternative analysis of the clocks where B ends up slow compared to A.
Do you think that these two results are compatible, and that the results apply to the 'twin paradox?

10. Jul 6, 2009

### JesseM

It's not just an alternative "analysis" of the same physical situation though, it's an actually physically distinct scenario where the two clocks have initially been synchronized differently. In any given physical scenario involving two clocks that meet, there will be only one correct answer to what the clocks read at the moment they meet.
Again, the alternative physical scenario I mentioned doesn't contradict the fact that in the scenario as Einstein defined it, clock A will definitely be behind clock B, this is true regardless of what frame you analyze the scenario in. Likewise, in the twin paradox you will find that the inertial twin elapsed more time than the twin who accelerated regardless of which frame you use to analyze the voyage.

11. Jul 8, 2009

### JM

JesseM,
I have already said that I agree with Einsteins analysis.
What I am saying is that k has an equal right to consider himself to be stationary! Thus , if K sees k moving away, turning around, and returning, then k, considering himself to be stationary,sees K moving away, turning around and returning. Then by your description, k being stationary elapses more time than K who turns around. Einstein did not describe this situation.
Do you deny k the right to consider himself stationary?

12. Jul 8, 2009

### sylas

If k is not inertial, then k cannot be stationary, in any inertial frame. If k turns around, then k is not stationary, and there is no symmetry between k and K. But I don't think Einstein gave a single co-ordinate system for an observer that turned around and came back. You can't given an inertial co-ordinate system for such an observer.

13. Jul 8, 2009

### JesseM

What do you mean "moving away, turning around, and returning"? That would suggest that k and K are not moving inertially relative to each other as Einstein defined them to be. In the clock scenario, I was assuming k was the inertial frame where A and B were initially moving at the same constant speed, then after A accelerated A was at rest in k while B continued to move at the same constant speed until it met with A and they compared clock readings. Are you thinking of k as a non inertial frame whose velocity in K's frame is not constant for all eternity? In this case k would not have an equal right to consider himself stationary and use the ordinary SR time dilation equation to calculate the time elapsed on moving clocks, because the SR time dilation equation only works in inertial frames. And remember, there is an absolute truth in SR about whether a given observer is moving inertially or non-inertially--a non-inertial observer will feel G-forces as he accelerates which can be measured by an accelerometer he carries alongside himself.

14. Jul 10, 2009

### JM

Lets attend carefully to what Einstein said as quoted be JesseM in his post no.8.
The first paragraph, 'From this...', is his statement of the 'slow clock' idea. His result is the approximation of the square root in the formula t=T/m, see post 3. I have identified clock B with K and clock a with k to agree with Einsteins definitions in his par. 3, where K is stationary and B doesn't move and k and clock A move. Both A and B are inertial.

In the second par. he says 'It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.'

By 'this result' he clearly means the 'slow clock' formula he just derived in the immediately preceding par. In addition he cites the 'slow clock' formula in the third par. regarding the continuously curved line.

Thus he is saying that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight, inertial line. So he is totally ignoring any effects of accelerations, or any other effects of the curved or segmented path of the moving clock.

Since A and B are both inertial, either may consider himself to be stationary. The reasoning of the first par. is equally valid with A stationary and B moving, with the resulting time lag expressed by T=t/m.

15. Jul 10, 2009

### sylas

Good grief, no.

k is NOT the clock moving in polygons. It is a single inertial co-ordinate system, for the single leg moving from A to B. The "result that holds good" is that the clock moved from A to B ends up lagging behind the clocks at A and B which are synchronized in frame K.

Hence, when you continue to move the clock in a new direction but at the same speed, it will continue to lag further and further behind clocks that are at rest wrt A and synchronized with A. Hence, the end result is that the clock moving in a closed curve from A back to A again ends up by showing less time -- lagging behind -- the clock which remains at A throughout. THAT is what Einstein is saying if you play careful attention to his words; in terms of the twin paradox, the twin that moves in a curve is the one that ages less.

I've seen these kinds of discussions before; so be aware, I have no interest in arguing with you if you take the line that you are reading correctly and trying to explain Einstein. Einstein is quite clear and speaks for himself with admirable clarity... and you've misunderstood it. I can help explain if you are are genuinely uncertain.

Nothing in Einstein's words suggests that there's a frame "k" which applies for the clock all through its polygon or curved trip from A back to A again.

Cheers -- sylas

Last edited: Jul 10, 2009
16. Jul 10, 2009

### JesseM

But the "slow clock" formula is just meant to calculate the time elapsed on a clock which is moving in whatever inertial frame you're using. It is indeed true that if you pick different inertial frames K and k, and use them to analyze the time elapsed on a clock which moves on a polygonal line (an approximation to a continuous curve), you can use the same time dilation formula in each frame to calculate the time elapsed on each segment of the polygon--this would just be (difference in time coordinates between the beginning and end of the segment in your frame)*(time dilation factor in your frame)--and add them up to get the total time elapsed on the polygonal path, and both frames will end up predicting exactly the same time elapsed (if you like I could given an example of a simple polygonal path analyzed in different frames). But nowhere did Einstein mean to suggest that the same time dilation formula could be used in a non-inertial frame!
No, he certainly isn't! He's just saying you can break the polygonal path into segments and calculate the time elapsed on each segment the same way you would for the time between two events on the path of an inertial clock, then add up the segments to find the total time elapsed. But again, for each segment you must use an inertial frame to calculate the time elapsed (though it's not necessary to use the same inertial frame for different segments, since all frames make the same prediction for the time elapsed on a given segment).

Of course you are also free to take the worldline of the clock B that never changes its velocity and divide it into segments, calculating the time elapsed on each segment and adding them up to find the total time elapsed. But if you do this you must make sure that the end of each earlier segment corresponds to the same event as the beginning of the next segment. So you can't have one segment's end be "the event on clock B's worldline that's simultaneous with clock A accelerating in frame K", and the beginning of the next segment be "the event on clock B's worldline that's simultaneous with clock A accelerating in frame k", because according to the relativity of simultaneity these are two entirely different events which may leave an unaccounted-for section of B's worldline between the end of the first segment and the beginning of the second.

17. Jul 11, 2009

### JM

JesseM
Refering to your quote of Einstein in post 8:

In the first par. it states that the clock moved in a line from A to B lags behind B by the amount t v.v/2c.c. (c.c = c squared, etc.)

In the third par. it states that a clock moved in a closed curved line at constant velocity lags behind the statiionary clock by the amount t v.v/2 c.c.

These two formulas are identical. Therefore the conclusion is that the time lag experienced by a clock moving in a closed curve is the same as the lag experienced by a clock moving in a straight line.

One possible conclusion from this is that the curved path has no effect on the time lag. This is supported by Wikipedia-Twin Paradox. According to Langevin the acceleration ( of the curved path) is the cause of the asymmetry and not of the aging itself.

18. Jul 11, 2009

### DrGreg

Just a thought. You do understand the difference between "speed" and "velocity" don't you?

Velocity has both magnitude (= speed) and direction. A change in direction counts as a change of velocity and as acceleration. Something moving at constant speed around a closed curve is still accelerating and is not moving inertially. It would feel a "G-force" to prove the difference.

Apologies if you realise this already.
This is true for two clocks both being measured by the same inertial observer. It's not true the other way round, if an inertial observer and an accelerating observer are both measuring the same clock. (Einstein doesn't consider accelerating observers in his paper.)

19. Jul 14, 2009

### JM

Yes I understand the difference, and Einstein surely did also. The appearnce of 'velocity' instead of 'speed' is likely due to the translation.

Since clock B is stationary for both the 'straight line' path of A and the 'closed curve' path, I assume you agree with my statement you quoted. Do you also agree with my last comment in post 17?

20. Jul 14, 2009

### JesseM

The v in these formulas represents speed rather than velocity, and the formulas only work if you are using v to represent the speed as measured in an inertial frame. Do you understand that when he talks about motion in a closed curve at constant speed v, he is still talking about constant speed as measured in some inertial frame?
Yes, that's correct. If you know a clock's speed as a function of time v(t) in some inertial frame, and you want to know how much time will elapse on the clock between two coordinate times t0 and t1 in that inertial frame (which could be the times of the clock departing from and returning to an inertial clock), then the formula is just a function of speed, not acceleration:

$$\int_{t_1}^{t_0} \sqrt{1 - v(t)^2/c^2} \, dt$$

For motion at constant speed v this just reduces to $$(t_1 - t_0)\sqrt{1 - v^2/c^2}$$, and if you subtract this from the time elapsed between the same events on a clock which is motionless in this frame, i.e. $$(t_1 - t_0) - (t_1 - t_0)\sqrt{1 - v^2/c^2}$$, you get a value for the time difference which is close to Einstein's approximation of $$\frac{tv^2}{2c^2}$$ (he mentions that this is just an approximation in the paper).