Einstein's definition of synchronization

[Jonathan Stanley]

At the time $t _0$, a ray of light goes out $A$, reflected at $B$ at time $t_1$, and arrives back at $A$ at time $t_2$. So Einstein provides:
$t_2-t_0 = (t_1-t_0) + (t_2 - t_1) = \frac{l _{rod}}{c - v _{rod} } + \frac{l _{rod}}{c + v _{rod} }$
Where:

• Rod with ends A and B
• $v _{rod}$ is rod's velocity in the direction of B from A
• $l _{rod}$ is the length of the rod.
• $c$ is the speed of light in a vacuum

Then we have Einstein’s definition of synchronism:

[synchronism is] the time which light requires in traveling from A to B is equivalent to the time which light requires in traveling from B to A. ​

His definition is therefore represented as $t_1 - t_0 = t_2 - t_1$, or more explicitly:

$\frac{l _{rod}}{c - v _{rod} } = \frac{l _{rod}}{c + v _{rod} }$, which is always false, except when $v _{rod} =0m/s$

We also have Einstein's equation for light's magnitude:
$c = \frac{2l _{rod}}{t_2 - t_0} = \frac{2l _{rod}}{\frac{l _{rod}}{c - v _{rod} } + \frac{l _{rod}}{c + v _{rod} }}$

which refactors to:
$c = c - \frac{v _{rod}^2}{c}$, which is always false and $c \ne c$, except when $v _{rod} =0m/s$

Thoughts?

 

[Dale]

A and B are different in the two examples. In the first example A and B are ends of a rod and in the second example A and B are spatial points in a given reference frame. In the second example v=0 by definition.

 

[Ibix]

You seem to be saying that clocks on a rod that are synchronised in the rest frame of the rod are not synchronised in other frames. This is true.

 

[Janus]

So in other words, if A and B are moving with respect to the frame you are making the measurement, then events unfold as shown in the top three images here, with the yellow arrows representing the light paths.

But if you are making the measurement from the frame that share's the same velocity as A and B, then events unfold like shown in the last image. Events that are simultaneous according to someone riding along with A and B are not simultaneous according to someone for which A and B are moving.

Only the observer who measures the relative velocity between himself and A&B(along the line separating A&B) will measure the light as taking equal times to travel from A to B or B to A. Observers who measure A and B as moving to the Right ( or left) will measure the light as taking different times to travel between the two.

 

[Jonathan Stanley]

Observers who measure A and B as moving to the Right ( or left) will measure the light as taking different times to travel between the two.

Thanks, as always, for the illustrative response. Are we in agreement that Einstein established his definition of synchronism according to a stationary "co-ordinate system, in which the Newtonian equations hold" ?

Are we in agreement that Einstein's definition of synchronism can be represented as $t_1−t_0=t_2−t_1$ according to the "stationary system" in which the Newtonian equations hold?

 

[Dale]

Thanks, as always, for the illustrative response. Are we in agreement that Einstein established his definition of synchronism according to a stationary "co-ordinate system, in which the Newtonian equations hold" ?

There is no need for us to agree or disagree, we can simply see what he actually said.

https://www.fourmilab.ch/etexts/einstein/specrel/www/

In section 1 (Definition of Simultaneity) he says "a system of co-ordinates in which the equations of Newtonian mechanics hold good" and "If at the point A of space there is a clock ... If there is at the point B of space another clock ..." so clearly the points A and B are points in space which are by definition stationary in the Newtonian coordinate system.

 

[Jonathan Stanley]

we can simply see what he actually said.

Correct; we're using the translations from his original work. Earlier you wrote:

A and B are different in the two examples.

It isn't clear to me if you were saying there was an error with the equations I wrote, or refactor, etc. This is why I am using the word "agree." You have no disagreement with what I wrote?

 

[Dale]

It isn't clear to me if you were saying there was an error with the equations I wrote, or refactor, etc. This is why I am using the word "agree." You have no disagreement with what I wrote?

Insofar as you are referring to Einstein's work then what you wrote is clearly wrong. Any disagreement between you and me is irrelevant because we can simply look at Einstein's paper. A and B from Einstein's work explicitly refers to points in space (v=0), not ends of a rod as you wrote in the OP.

 

[PeroK]

Correct; we're using the translations from his original work. Earlier you wrote:

It isn't clear to me if you were saying there was an error with the equations I wrote, or refactor, etc. This is why I am using the word "agree." You have no disagreement with what I wrote?

Although it is a marvellous document, it will be difficult to learn SR from it. The lack of a diagram, for example, may have led to your confusion. It's great to read the 1905 paper, but you'd be better off with a modern approach if you want to learn SR.

And, it's a bit late to be looking for errors in the 1905 paper - about 113 years too late!

 

[Jonathan Stanley]

A and B from Einstein's work explicitly refers to points in space (v=0), not ends of a rod as you wrote in the OP.

See section 2 of the original docs:
"Let us suppose that the two clocks synchronous with the clocks in the system at rest are brought to the ends A, and B of a rod, i.e., the time of the clocks correspond to the time of the stationary system at the points where they happen to arrive ; these clocks are therefore synchronous in the stationary system."

 

[Dale]

Oops, I stand corrected. You are right.

Note that in section 2 the clocks A and B on the ends of the rod are not synchronized by using the light pulses from one end of the rod to another as in your rod clocks. They are synchronized by stationary clocks which were in turn synchronized by the method in section 1.

 

[Jonathan Stanley]

Oops, I stand corrected. You are right.

Note that in section 2 the clocks A and B on the ends of the rod are not synchronized by using the light pulses from one end of the rod to another as in your rod clocks. They are synchronized by stationary clocks which were in turn synchronized by the method in section 1.

I'm not sure what distinction this imposes on my original post. Maybe if I restate more verbosely using Einstein's terms:

Einstein wrote in section 1:

For example, a ray of light proceeds from A at A-time $t_A$ towards B, arrives and is reflected from B at B-time $t_B$, and returns to A at A-time $t_A\prime$. According to the definition, both clocks are synchronous, if $t_B - t_A = t_A\prime - t_B$​

and in section 2:

$t_B - t_A = \frac{r _{AB}}{c - v}$
and
$t_A\prime - t_B = \frac{r _{AB}}{c + v}$​

Since we have the equations for $t_B - t_A$ and $t_A\prime - t_B$, we can rewrite Einstein's definition of synchronism ($t_B - t_A = t_A\prime - t_B$) as

$\frac{r _{AB}}{c - v} = \frac{r _{AB}}{c + v}$

 

[Ibix]

Perhaps you could just say where you are going with this. So far all you seem to be doing is quoting bits and pieces of Einstein's 1905 paper and asking if we agree with them. The answer is yes, why wouldn't we? There is sometimes some confusion due to losing track of what part is being quoted (edit: speaking for @Dale - I think that's what happened in #8 - hope he doesn't mind).

 

[Dale]

I'm not sure what distinction this imposes on my original post.

The distinction is that the A and B times in section 2 are not used to determine simultaneity (as was done in section 1). Simultaneity is determined in the same way as in section 1 and then the A and B clocks are set to read the same time as stationary clocks that they are next to at each point. They don’t keep time themselves like normal clocks.

 

[Jonathan Stanley]

Einstein establishes definitions in section 1 which form the basis for section 2 equations. I was trying to be more succinct, but maybe I can try rephrasing again. Apologies for the verbosity; I appreciate your interest.

In section 1, we have two fixed points (A and B) according to Newtonian stationary system. So, given light speed constancy, the time for light to go between the points in either direction is equal.

In section 2, we have two moving points (A and B) according to Newtonian stationary system. So, given light speed constancy, the time for light to go between the points is different in each direction.

Einstein does not introduce the theory of coordinate and time transformation until section 3. Since we're not at section 3 yet; we're remaining in the Newtonian stationary system at this point. When light is emitted from the rod's end A, affix a stationary point $P_0$ at the axes where the light is emitted. As the light moves along its path together with the rod and reaches the rod end B, record the reflection point $P_1$. Once the reflected light returns to A, record the meeting point $P_2$. Let $d _{0\rightarrow1}$ be the distance between $P_0$ and $P_1$ and let $d _{2\leftarrow1}$ be the distance between $P_1$ and $P_2$. Notice that light’s distance of travel from A to B is not necessarily equal to $r _{AB}$. If we consider the rod’s movement away from $P_0$ until the moment we establish $P_1$, we see that $d _{0 \rightarrow 1}$ is greater than $r _{AB}$. Similarly, if we consider the rod’s movement passing $P_1$ until the moment we establish $P_2$, we see that $d _{2\leftarrow1}$ is less than $r _{AB}$. $d _{0 \rightarrow 1}$ and $d _{2\leftarrow1}$ and $r _{AB}$ are all equal only when given a rod with zero velocity.

Einstein concludes section 2 with:

Taking into consideration the principle of constancy of the velocity of light, we have $t_{B}-t_{A}=\frac {r_{AB}}{c-v}$ and $t'_{A}-t_{B}=\frac {r_{AB}}{c+v}$, where $r_{AB}$ is the length of the moving rod, measured in the stationary system. Therefore the observers stationed with the watches will not find the clocks synchronous, though the observer in the stationary system must declare the clocks to be synchronous. We therefore see that we can attach no absolute significance to the concept of synchronism​

If Einstein's definition of synchronism requires fixed points, then we cannot use $\frac {r_{AB}}{c-v} = \frac {r_{AB}}{c+v}$ to determine the presence or absence of synchronism. We have to use the fixed points $P_0$, $P_1$, and $P_2$. Of course, if we use those "P" points, then observers stationed at those points with their clocks, would actually declare the clocks synchronous. Also, this would mean we would have to continue to attach absolute significance to the concept of synchronism.

If Einstein's definition of synchronism allows moving points, then it must accept moving points in its equations. In other words, we would have synchronism as $\frac {r_{AB}}{c-v} = \frac {r_{AB}}{c+v} = t_B - t_A = t_A\prime - t_B$. As we saw, this definition is structurally invalid. For example, this definition could be refactored to $c=c−\frac{v^2}{c}$

Again, we're not to section 3 yet, so it's all according to the Newtonian stationary system. Per the question from @Ibix, it seems like the bigger challenge is understanding why the Lorentz factor works despite this issue in sections 1 and 2.

 

[Dale]

I would take that conclusion of section 2 as a motivated preview of the results to be formally developed in section 3.

 

[Ibix]

Again, we're not to section 3 yet, so it's all according to the Newtonian stationary system. Per the question from @Ibix, it seems like the bigger challenge is understanding why the Lorentz factor works despite this issue in sections 1 and 2.

I don't see your problem. In section 2, Einstein says that we could put clocks on the ends of the rods that simply repeated times they read off nearby "stationary" clocks. Then points out that such clocks can only be synchronised for an observer moving with the rod if v=0. It's not as elegant as the train/platform thought experiment, but it's the same thing.

The problem I see with what you've written is that there's no such thing as a fixed point. There's only a fixed point in a chosen reference frame. Einstein is simply pointing out that the same synchronisation procedure applied to different notions of what constitutes a "fixed point" yields different notions of "synchronised" if the speed of light is frame invariant.

 

[Jonathan Stanley]

I don't see your problem. In section 2, Einstein says that we could put clocks on the ends of the rods that simply repeated times they read off nearby "stationary" clocks.

That is not what was written. Section 2 is regarding a "stationary system" but the rod and clocks are not stationary.

"We further imagine that there are two observers at the two watches, and moving with them"​

The problem I see with what you've written is that there's no such thing as a fixed point. There's only a fixed point in a chosen reference frame.

Again, we have not reached section 3 yet. Einstein explicitly stated that he established the definitions according to classical mechanics stationary system. But you can see See the beginning of section 1.

"Let us have a co-ordinate system, in which the Newtonian equations hold. For distinguishing this system from another which will be introduced hereafter, we shall always call it "the stationary system. If a material point be at rest in this system, then its position in this system can be found out by a measuring rod, and can be expressed by the methods of Euclidean Geometry, or in Cartesian co-ordinates."​

This was a good call on his part, otherwise it would likely just be considered circular fallacy. I have it laid out the issue in very simple steps on my last post, but anyone is welcome to read it directly from Einstein's original work. I think it helps to quote the specific if there is any dispute about that, but it is pretty straightforward. I welcome other thoughts as well.

 

[Dale]

I have it laid out the issue in very simple steps on my last post,

Frankly, it doesn’t seem like a big issue to me. He proceeds to explicitly derive the transformation, and the relativity of simultaneity is clearly in the transformation. So the comment appears to be merely a small preview of the upcoming section. Section 3 as a whole seems designed to motivate the subsequent derivation more than anything.

 

[Ibix]

That is not what was written.

It's exactly what's written.

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.

Again, we have not reached section 3 yet.

So what? The idea that a "fixed point" is meaningless goes back to Galileo. Einstein is writing for the leading physicists of his day, who would be well aware of the notion. And Einstein is explicitly working with observers moving with the rod without explicitly introducing a coordinate system for them. Again, so what? You don't need coordinates to do physics. He simply shows that observers who regard the rod as "at rest" cannot agree that clocks synchronised to the stationary system are synchronised for them. He develops the idea into a complete coordinate transformation in the next section.

 

[Jonathan Stanley]

The next sentence from that translation may help.

"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. We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks."​

Note that the clocks are placed at the ends of the rod (ie traveling with the moving rod). That's why the end of the sentence says "at the places where they happen to be". Einstein also describes earlier that the clocks always have to be in the immediate proximity of the observer (otherwise the observer cannot observe the clocks). We can also look at the other prominent translation which more clear in the paragraph you quoted:

"Let us suppose that the two clocks synchronous with the clocks in the system at rest are brought to the ends A, and B of a rod, i.e., the time of the clocks correspond to the time of the stationary system at the points where they happen to arrive ; these clocks are therefore synchronous in the stationary system. We further imagine that there are two observers at the two watches, and moving with them, and that these observers apply the criterion for synchronism to the two clocks."​

It is clear that in section 2, Einstein is asking the reader to imagine a Newtonian stationary system with two clocks moving with the ends of the moving rod. It is not stationary clocks within a stationary system.

So what? The idea that a "fixed point" is meaningless goes back to Galileo. Einstein is writing for the leading physicists of his day, who would be well aware of the notion. And Einstein is explicitly working with observers moving with the rod without explicitly introducing a coordinate system for them. Again, so what? You don't need coordinates to do physics. He simply shows that observers who regard the rod as "at rest" cannot agree that clocks synchronised to the stationary system are synchronised for them. He develops the idea into a complete coordinate transformation in the next section.

Einstein asks the reader to imagine a Newtonian stationary system, so that's what we must do to move forward with the theory. We certainly cannot invalidate Einstein in order to validate Einstein. Since Einstein does explicitly introduce the Newtonian stationary coordinate system in section 1 and reference it continually in section 2, so that's what we're to use.

"Let us have a co-ordinate system, in which the Newtonian equations hold. For distinguishing this system from another which will be introduced hereafter, we shall always call it "the stationary system.""​

 

[Jonathan Stanley]

Frankly, it doesn’t seem like a big issue to me. He proceeds to explicitly derive the transformation, and the relativity of simultaneity is clearly in the transformation. So the comment appears to be merely a small preview of the upcoming section. Section 3 as a whole seems designed to motivate the subsequent derivation more than anything.

I think you have an interesting point here. If I'm understanding you, the fact that Einstein provides the Lorentz transformation in his work is what makes it so compelling. Since we've tested the Lorentz transformation, and know it works, then whatever details that lead up to seem less consequential. For example, we can prove math with a system of logic, but it is really a long exercise for something we already know works. It's much more reasonable to just rely on the overwhelming evidence that math works than to negotiate someone's formal proof of math and potential intricacies of that proof.

 

[PeroK]

What's the point at issue in this thread?

 

[Dale]

I think you have an interesting point here. If I'm understanding you, the fact that Einstein provides the Lorentz transformation in his work is what makes it so compelling. Since we've tested the Lorentz transformation, and know it works, then whatever details that lead up to seem less consequential. For example, we can prove math with a system of logic, but it is really a long exercise for something we already know works. It's much more reasonable to just rely on the overwhelming evidence that math works than to negotiate someone's formal proof of math and potential intricacies of that proof.

Largely I agree with that, but I would add that the derivation in the following section does not rely on the comment at the end of section 2. He mentions it but does not put it into the upcoming derivation. So the mention does not impact the legitimacy of the following derivation