I On the Relativity of Lengths and Times

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Summary
In this section of his 1905 paper, Einstein sets out to demonstrate that clocks that are synchronized in one frame of reference are not synchronized in another frame.
I'm struggling to understand Section 2 of Einstein's 1905 paper, "The Electrodynamics of Moving Bodies." This is my source for the paper: http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

In section 1 Einstein states that by definition light travels at the same speed, c, in every direction. He then proposes a means of synchronizing clocks at a distance:

Let a ray of light start at the “A time” tA from A towards B, let it at the “B time” tB be reflected at B in the direction of A, and arrive again at A at the “A time” t'A. In accordance with definition the two clocks synchronize if tB − tA = t'A − tB.

In section 2 he introduces a "rigid rod" and places a clock at each end of the rod:

We imagine further that at the two ends A and B of the rod, clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are therefore “synchronous in the stationary system.”

He then places the rod in motion so as to demonstrate that the synchronization of the clocks depends on frame of reference:

We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in section 1 for the synchronization of two clocks. Let a ray of light depart from A at the time tA, let it be reflected at B at the time tB, and reach A again at the time t'A. Taking into consideration the principle of the constancy of the velocity of light we find that tB − tA = rAB/c − v and t'A − tB = rAB/c + v where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

First off I assume by "the two clocks" Einstein means the clocks placed at the two ends of the rod, A and B (not the clocks in the stationary system). Observers co-moving with the rod find the clocks out of sync. But if the clocks are placed at each end of the rod and therefore moving with the rod, and if the observers are also moving with the rod, then the observers will not think the rod or the clocks are moving. This is because, from your own point of view, you're never in motion. In fact this is Galileo's principle of relativity. So why would "observers moving with the moving rod" find that the two clocks are no longer synchronized? Perhaps the answer is that the two clocks are not actually attached to the rod. This way the observers co-moving with the rod approach the clock at B while receding from the clock at A. As the light ray travels from A to B, the velocity v of the observers makes it seem as if the light ray's speed is c - v. As the light ray travels back from B to A, the velocity of the observers makes it seem as if the light ray's speed is c + v. Of course, as we know from Lorentz, the speed of light is always measured at exactly c. Assuming Einstein is right and this is because the speed of light is a law of nature and therefore frame-invariant, the motion of the observers is causing them to dilate in time so that they see the clock at B displaying an earlier time than what a stationary observer would perceive.

At this point I think I've got it. But then I realize the time dilation of the observers will cause them to see exactly the same earlier time displayed on clock A. So the clocks are once again synchronized!

Can anyone explain what I'm missing here?
 

Ibix

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I have to say I find this a bit unclear too. I wonder if there's some linguistic peculiarity of German that doesn't translate well.

As always with this kind of thing, look at the maths (which, incidentally, should say ##t_B − t_A = r_{AB}/(c − v)## and ##t'_A − t_B = r_{AB}/(c + v)## - the brackets you missed are rather important). Einstein is simply pointing out that if the clocks are attached to the ends of the rod and synchronised according to the rest frame of the rod (i.e., on the basis that the two transit times are equal) they cannot also be synchronised in the frame where the rod is moving, because the transit times are manifestly not equal due to the motion of the rod. This stands in contrast to the Newtonian/Galilean expectation that the transit times would be equal in both frames because the speed of light would be frame dependent.
 
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clocks are placed which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be.
Note, this means that they are not normal clocks.

So why would "observers moving with the moving rod" find that the two clocks are no longer synchronized? Perhaps the answer is that the two clocks are not actually attached to the rod.
No, they are attached to the rod, but they are not normal clocks. They do not keep the correct time in their own frame.

Note, something similar to this is done in the GPS system. Clocks onboard the satellites are rigged to keep time in the earth centered inertial frame, not their own frame. They are not normal clocks.

These unusual clocks are not synchronized in their own frame. They also do not run at the correct rate in their own frame, but I believe that is not shown at this part of the paper.
 

Orodruin

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I think it is also worth pointing out that reading the original research papers is generally not a good way to learn a new subject. In particular, regarding something such as relativity, which has had over 100 years of didactical development, a modern textbook would be a much better source for learning it. Once you know the subject, going back to read the original texts can be interesting from a curiosity point of view, but the original papers are typically not a good learning material.
 
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@pheurton I heartily second the comment above by @Orodruin . It is much easier to understand Einstein's paper after you know relativity than it is to learn relativity from that paper. The seminal papers in a field are of great historical importance, but are usually not the best way to learn the topic because the community takes the ideas in those seminal papers and refines them and improves them considerably over time. A great scientist like Einstein has the first word on their theory, but not the last word.
 

vanhees71

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In this case, great progress came through Minkowski's mathematical analysis, which makes the theory much clearer. Though the original papers may sometimes indeed not be the best way to start learning about a topic, I think Minkowski's papers have a lot to offer.
 
Thanks to Dale for providing exactly the piece of the puzzle I was missing. I did not understand the significance of Einstein's emphasis on synchronizing the clocks of the moving frame with the clocks of the stationary frame. Incidentally, I began my study of special relativity with An Illustrated Guide to Relativity by Tatsu Takeuchi, an excellent introductory book. Next up is Minkowski's "Space and Time." Wish me luck!
 

Janus

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Thanks to Dale for providing exactly the piece of the puzzle I was missing. I did not understand the significance of Einstein's emphasis on synchronizing the clocks of the moving frame with the clocks of the stationary frame. Incidentally, I began my study of special relativity with An Illustrated Guide to Relativity by Tatsu Takeuchi, an excellent introductory book. Next up is Minkowski's "Space and Time." Wish me luck!
I set up a visual aid for this concept a while back, that used more than just two clocks in each frame.
Imagine two rows of clocks passing each other with some relative velocity. All the clocks in the lower row have been synchronized to each other according to the rest frame of the lower row.
We have also arranged its so that the clocks in the upper row are spaced apart and tick at a rate such that as measured from the frame of the bottom row, they are in sync with the lower row. Any time a lower and upper row pass each other, they read the same:
clock_sync1.gif

(here we just show that duration between the clocks reading 12:00 and 2:00)

Now we view things from the rest frame of the upper row.
Now, in the above animation, the upper row is in relative motion, so the lower row is measuring a length contracted distance between the upper row of clocks, and is also measuring a time dilation of the upper clocks.
In its own rest frame, the upper row measures the distance between it clocks as being the proper length (further apart than as measured by the lower row) and It's clocks ticking at a non time dilated rate.
At the same time, it measures the distance between the clocks in the lower row as being length contracted, and the clocks ticking at a time dilated rate.

In order to maintain the fact that when lower and upper clocks pass each other they read the same, the clocks in neither row can be in sync as measured from the upper row rest frame, and you get this according to the upper row:

clock_snyc2.gif


No clock in either row reads the same as another clock in its own row, but every time a lower and upper clock pass each other, they read the same.
 

Hans de Vries

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In section 1 Einstein states that by definition light travels at the same speed, c, in every direction. He then proposes a means of synchronizing clocks at a distance:
Time and distance just depend on how we define our coordinate system (t,x,y,z).

This is the simplest it gets:

Question 1: Why is c constant?
Answer 1: We define our coordinate system (t,x,y,z) in such a way that light travels at the same speed, c, in every direction.

Question 2: What method do we use to define this coordinate system?
Answer 2: If two clocks are at the same distance then they should show equal time.

Question 3: How do we know that two clocks are at the same distance?
Answer 3: Clocks further away look smaller. If clocks seem to have the same size we assume they are at the same distance.

Question 4: So if two clocks both cover 10 degrees of our field of vision then we assume they are equidistant?
Answer 4: Exactly. If a clock covers only 1 degrees then we assume that it is 10 times farther away in distance as well as 10 times further away in time.

Question 5: Is it that simple?
Answer 5: Yes, this completely defines the coordinate system.

Question 6: But why do reference frames change if we are moving?
Answer 6: After we accelerate things in front of us look bigger while things behind us look smaller.

Question 7: So the viewing angle changes? Does this completely define the transformed coordinate system?
Answer 7: Yes, from the different viewing angles we assume different distances and time.

Question 8: Okay. but why, and how, do the viewing angles change?
Answer 8: First simple vector addition: Draw a light-ray hitting you from an angle, add your velocity and you have a new angle. Then take into account that you are Lorentz contracted to obtain the angle as you observe it.

All of relativity follows from the relativistic wave equations. This includes the classical wave equation and Maxwell's equations that both comply with Special Relativity, as well as the equations of relativistic Quantum Field Theory.


See the following chapters from my book:

Lorentz contraction from the classical wave equation.
This requires going through quit some mathematics which is due to Liénard & Wiechert around 1900.

Time dilation from the classical wave equation.
Relatively simple, basically using the bouncing photon clock.

Non simultaneity from the classical wave equation.
This chapter contains many illustrations demonstrating the above Q&A (Starting for instance at section 4.3 on page 5). It goes into great depth on why everything looks the same, including physics, if we boost our world from one reference frame to another. If your goal is to obtain a deep insight in special relativity going beyond being able to handle the mathematical calculations, then you can do so using the illustrations in this chapter.
 
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So why would "observers moving with the moving rod" find that the two clocks are no longer synchronized?
Einstein's example can be summarized as follows:
  • given the constancy of the speed of light in all reference frames, clocks cannot be simultaneously synchronized in both a stationary and a moving reference frames
Since the speed of light is constant in the stationary reference frame, the time intervals tB-tA and tA'-tB are different. The reason for that is that the rigid rod with clocks A and B at its ends is in motion and therefore a light pulse emitted from A to B travels longer than a light pulse emitted from B to A (again, from the stationary perspective).

Inside the moving reference frame, however, the speed of light is constant as well. Therefore, the time intervals tB-tA and tA'-tB in this reference frame should be identical, as the clocks are at rest in this frame.

So, we have an example where two time intervals are different in two different reference frames due to the constancy of the speed of light.

If the clocks in the moving reference frame are synchronized with the clocks in the stationary reference frame (as in Einstein's example), then they won't measure identical time intervals in the moving reference frame, as they should. Thus, observers moving with the moving rod would find that the two clocks are not synchronous.

The key to understanding this example is to keep reminding yourself that the speed of light is constant in all reference frames. This is the unintuitive part.
 
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If two clocks are at the same distance then they should show equal time.
Also both clocks need to be at rest relative to the observer. (Given your approach I would also expect "at rest" to be defined in terms of a direct observable; the most obvious one would be zero Doppler shift of light signals.)
 

Orodruin

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given the constancy of the speed of light in all reference frames, clocks cannot be simultaneously synchronized in both a stationary and a moving reference frames
It should definitely not be summarized like this as this formulation implies the existence of a reference frame that is ”stationary” contrary to the relativity principle. It would be more appropriate to refer to two inertial frames with a non-zero relative velocity. Referring to stationary and moving reference frames goes against both galilean and special relativity.
 

Hans de Vries

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Also both clocks need to be at rest relative to the observer. (Given your approach I would also expect "at rest" to be defined in terms of a direct observable; the most obvious one would be zero Doppler shift of light signals.)
Yes that's correct! Good remark, that was indeed assumed.
 

pervect

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Time and distance just depend on how we define our coordinate system (t,x,y,z).

This is the simplest it gets:

Question 1: Why is c constant?
Answer 1: We define our coordinate system (t,x,y,z) in such a way that light travels at the same speed, c, in every direction.

Question 2: What method do we use to define this coordinate system?
Answer 2: If two clocks are at the same distance then they should show equal time.

Question 3: How do we know that two clocks are at the same distance?
Answer 3: Clocks further away look smaller. If clocks seem to have the same size we assume they are at the same distance.

Question 4: So if two clocks both cover 10 degrees of our field of vision then we assume they are equidistant?
Answer 4: Exactly. If a clock covers only 1 degrees then we assume that it is 10 times farther away in distance as well as 10 times further away in time.

Question 5: Is it that simple?
Answer 5: Yes, this completely defines the coordinate system.
Point #3 , comparing angular sizes, works for clocks that are stationary, but I think things need to get a bit more complicated for moving clocks, which are constantly changing their angular size with time as they move.

Attempting to fix this leadds out to questions such as "what size is the clock, which is changing it's angular size, _now_". And that gets into the issue of _now_, which is more complex than it seems due to the the issue of the relativity of simultaneity.

For moving clocks, we "see" Terrell rotation <<wiki>>, not length contraction.

So I don't think we can make things quite as simple as you suggest, we really need to talk about the relativity of simultaneity at some point. And that's often the point where people get confused about special relativity.
 

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