Special Relativity: an ideal experiment

[Hak]

Let me preface this by saying that I do not claim to dismantle relativity.
That said, in reading the Feynman I came across an ideal experiment that is not entirely clear to me. The context is the usual two spacecraft moving at relatively constant speed. With the reasoning I give below Feynman tries to explain why the lengths along the axes perpendicular to the direction of motion are the same in the two frames of reference:

How do we know that the perpendicular lengths do not change? Men can agree to make marks on each of the rods along the $y$-axis [the relative velocity is along the $x$-axis], when they pass each other. By symmetry the two marks must be found at the same $y$ and $y'$ coordinates, because otherwise, when they come together to compare the results, one mark will be above or below the other, and so we might know who was actually in motion [contradicting the principle of relativity]

What is not clear to me is the dynamics of the "reunion" between the two: to reunite at least one of the two must have changed frame of reference, so how can I know that in the new frame of reference its lengths along the $y$-axis will be the same as before it changed frame of reference?
The problem is that I cannot do any kind of math since Maxwell's transformations have as an assumption that $y=y'$.
I hope I have been clear. Let me know, thank you very much for any clarification.

 

[Ibix]

What is not clear to me is the dynamics of the "reunion" between the two: to reunite at least one of the two must have changed frame of reference

No.

Build a washer and a bullet that fits exactly through the hole. Shoot the bullet through the washer. If there's lateral length contraction, one frame says the bullet doesn't fit and one says there's room to spare. No frame change needed.

 

[Hak]

No.

Build a washer and a bullet that fits exactly through the hole. Shoot the bullet through the washer. If there's lateral length contraction, one frame says the bullet doesn't fit and one says there's room to spare. No frame change needed.

Ok, thank you so much.
I had thought we cannot justify the thesis using Lorentz transformations. The fact that $y=y'$ is an assumption of the Lorentz transformations, if we take the Lorentz transformations for granted we do not have to justify it.
Instead, I wanted to justify the fact that the heights are equal in the two frames of reference from the principle of relativity alone.

Suppose we have two frames of reference that we call $S$ and $S'$, in motion with respect to each other at relative velocity $\vec v$. In both systems there is a rod of rest length $L_0$, oriented perpendicular to $\vec v$. In the $S$ system there is rod 1, and in the $S'$ system there is rod 2. Suppose that, at the time when the two rods are side by side, the $y$-coordinates of their lower extremes coincide. Suppose further that on the upper extremes two chalks are fixed.
By "length at rest" I mean the length measured in the system in which the rod is stationary.

Let us put ourselves in the $S$ system. Rod 1 is stationary and has length $L_0$, rod 2 is moving at speed $v$ and has length $L$ which, without assuming Lorentz transformations, is unknown. We expect $L$ to depend on $L_0$ and the velocity, in a way we need to find out.
Suppose absurdly that $L(v)<L_0$: then, at the moment when the rods meet (suppose they pass side by side), the chalk fixed at the top of rod 2 makes a mark on rod 1. The chalk fixed at the top of rod 1, on the other hand, being higher up, makes no mark.
After the encounter we therefore have: rod 1 with a chalk mark, rod 2 with no mark.

Let us now review the same phenomenon from the point of view of an observer in the $S'$ system. Here rod 2 is stationary and has length $L_0$, rod 1, on the other hand, is moving at speed $v$. What length does rod 1 have? The transformation law of lengths must be the same as before: if a rod is moving at speed $v$ with respect to the observer, its length must still be the same $L(v)$ as before; if not, we would have an experiment to distinguish the $S'$ system from the $S$ system: the transformation law of perpendicular lengths would be different in the two systems. This is prevented from us by the principle of relativity.
So let us return to the system S': rod 2 has length $L_0$, rod 1 has length $L(v)<L_0$. So the chalk at the top of rod 2 makes no mark, the chalk at the top of rod 1 makes a mark on rod 2.
After the match we have: rod 1 with no mark, rod 2 with a chalk mark.

So we have a paradox: for an observer in the $S$ system rod 1 has a chalk mark, for an observer in the $S'$ system the same rod 1 is clean. But this is a paradox! The property "being chalk-marked" should not depend on the frame of reference. If you don't like the chalk version, put in something more radical, like a knife cutting the rod in two; the substance does not change.

Thus we have shown that $L(v)$ is not less than $L_0$.

By similar reasoning that I will not rewrite, we show that $L(v)$ is not greater than $L_0$.
We then conclude that $L(v)=L_0$, i.e., that the length of the rod measured by an observer in motion at speed $\vec v$ perpendicular to the rod itself is always the same.

Actually, it seems to me to be spinning as reasoning. But how do I know that perpendicular lines in $S$ are still perpendicular in $S'$? Specifically, in the frame of reference of $S$, are two rods placed along $x'$ and $y'$ perpendicular? I have absolutely no idea how to answer this.

Thanks.

 

[PeroK]

Let me preface this by saying that I do not claim to dismantle relativity.
That said, in reading the Feynman I came across an ideal experiment that is not entirely clear to me. The context is the usual two spacecraft moving at relatively constant speed. With the reasoning I give below Feynman tries to explain why the lengths along the axes perpendicular to the direction of motion are the same in the two frames of reference:

What is not clear to me is the dynamics of the "reunion" between the two: to reunite at least one of the two must have changed frame of reference, so how can I know that in the new frame of reference its lengths along the $y$-axis will be the same as before it changed frame of reference?
The problem is that I cannot do any kind of math since Maxwell's transformations have as an assumption that $y=y'$.
I hope I have been clear. Let me know, thank you very much for any clarification.

They both have, for example, a ruler exactly a metre long, oriented in the y direction

 

[Hak]

They both have, for example, a ruler exactly a metre long, oriented in the y direction

See post #3. What can you say about that?

 

[Ibix]

Instead, I wanted to justify the fact that the heights are equal in the two frames of reference from the principle of relativity alone.

That's exactly what I did...

 

[Hak]

That's exactly what I did...

That's what I meant, I may have expressed myself wrongly in post #1. But what about the final question in post #3? It seems sensible as a question to me, even though it is probably trite and dumb, but I don't find a solid proof.

 

[A.T.]

Specifically, in the frame of reference of $S$, are two rods placed along $x'$ and $y'$ perpendicular.

Look at the Lorentz Transformation.

 

[PeroK]

See post #3. What can you say about that?

Post #3 is too long.

 

[Ibix]

That's what I meant, I may have expressed myself wrongly in post #1. But what about the final question in post #3? It seems sensible as a question to me, even though it is probably trite and dumb, but I don't find a solid proof.

Shoot a bullet with a + shaped cross section through a matching hole.

 

[Hak]

Shoot a bullet with a + shaped cross section through a matching hole.

OK, thanks, so?

 

[PeroK]

Shoot a bullet with a + shaped cross section through a matching hole.

Or, observe the whole experiment using a frame of reference in which both have equal and opposite velocities.

 

[Orodruin]

The fact that y=y′ is an assumption of the Lorentz transformations, if we take the Lorentz transformations for granted we do not have to justify it.

No, it is not an assumption. You can derive it from basic principles starting from the Lorentz transformations preserving the form of the Minkowski metric.

 

[Nugatory]

What is not clear to me is the dynamics of the "reunion" between the two: to reunite at least one of the two must have changed frame of reference

All that is necessary is for both observers to exchange messages of the form "A chalk mark at height H appeared at time zero" to carry out Feynman's analysis - and you will notice that Feynman does not mention any sort of reunion.

Your quoted comment suggests that you do not yet clearly understand the distinction between using a particular reference frame and being at rest in that frame.
We do not have to be at rest in a given frame to use that frame. Changing our state of motion does not "change our frame of reference", it just means that if we were at rest in some frame we no longer are.

Go back to @PeroK's post #12 in this thread, consider how when you choose to analyze the problem using a frame in which neither is at rest the symmetry becomes clear, and also how that analysis is equivalent to Feynman's.

 

[Hak]

But how do I know that perpendicular lines in $S$ are still perpendicular in $S'$? Specifically, in the frame of reference of $S$, are two rods placed along $x'$ and $y'$ perpendicular? I have absolutely no idea how to answer this.

Or, observe the whole experiment using a frame of reference in which both have equal and opposite velocities.

I was not able to prove what I wanted by following your advice, only to justify it, perhaps not even that. However, I still want to report the procedure.
The rod placed along $x'$ is parallel to the velocity of $S$ as seen from $S'$. In fact, more, let us say that the line on which it lies is coincident with the direction of $\vec v$. When I look at it from $S$ it will remain on the line of the direction of $\vec v$ (at least you must grant me that coincident lines go in coincident lines).

The rod placed along $y'$, seen from $S'$ is perpendicular to the line of $\vec v$. Viewed from system $S$ will it still be perpendicular to $\vec v$? If not, it should be tilted by an angle $\alpha$, fixed by theory; however, theory could not tell me which way it would be tilted, whether so \ or so /. In fact, an inclination \ turns into an inclination / with a rotation of 180° around the x-axis. Translated: if two observers, both in the system $S$, are one standing and one upside down, they see one the tilt \ and the other the tilt /. Therefore, it is not possible for the theory to predict the "sign" of the tilt to me, since this would be different for two observers in the same system. If the rods, viewed from the $S$ system, were not perpendicular, they could then form either an obtuse or acute angle, without the theory being able to predict to me which one. The only alternative in which there are no such "unpredictable sign" problems (obtuse and acute angle) is the case in which the angle continues to be right. It is therefore at least reasonable that it continues to be right-angled.

Alternatively, if the angle varied, the plane perpendicular to $v$ would become a cone and the straight lines in the plane from before would become broken, but this is absurd because straight lines must go into straight lines (a straight line is the trajectory of a "ray of light," and if I saw that this is not a straight line I would understand that I am moving in contradiction to the principle of relativity). In fact, I am not convinced that the line is necessarily the trajectory of a ray of light (it is, but it is not obvious to me to state this from the postulates of relativity alone). Let me explain further: let us take our vertical rod in $S'$. Let us assume that a beam of light travels through it. When we look from $S$, the instantaneous position of the rod will no longer be the trajectory of the same light ray (it would be only if the speed of light were infinite) since the rod is moving to the right at speed $v$.

How to fix this correctly?

 

[Sagittarius A-Star]

The rod placed along $y'$, seen from $S'$ is perpendicular to the line of $\vec v$. Viewed from system $S$ will it still be perpendicular to $\vec v$?

Yes. All atoms of the rod at $x'=0$ will have the same $x=vt$. This follows from the LT.

The fact that $y=y'$ is an assumption of the Lorentz transformations, if we take the Lorentz transformations for granted we do not have to justify it.

As others mentioned, $y=y'$ is not an assumption of the Lorentz transformations.

The derivation of $y'=y$ and $z'=z$ is part of the derivation of the LT from the 2 postulates. See for example Rindler "Introduction to Special Relativity", 2nd edition.

 

[Dale]

No, it is not an assumption. You can derive it from basic principles starting from the Lorentz transformations preserving the form of the Minkowski metric.

@Hak this is actually my preferred approach from a conceptual standpoint. Everything in SR can be derived from the Minkowski metric: $$ds^2 = - c^2 dt^2 + dx^2 + dy^2 + dz^2$$

The Lorentz transforms then are the class of transformations that preserves the form of this metric. They include $y=y'$, not as an assumption, but as a consequence of the definition of the transform. This is not the way such things are taught, but whenever you think there is a circular chain then just ignore it. There exists a conceptually clean derivation from this one equation, even if the math is tedious and not physically insightful. Anything that suggests either a self-contradiction in SR or a circularity is immediately dismissed with this approach.

 

[Orodruin]

This is not the way such things are taught

Wait, what? I am teaching relativity wrong?

From this follows:
$$- c^2 t^2 + x^2 + y^2 + z^2 = - c^2 {t'}^2 + {x'}^2 + {y'}^2 + {z'}^2$$
If you define the coordinate systems $S$ and $S'$ according to the standard configuration, then $y=0$ must entail $y'=0$ and with the above equation follows
$$y^2 = {y'}^2$$
The solution $y'=-y$ can be excluded, because $v \rightarrow 0$ must continuously lead to $y'=y$. Thefore, $y'=y$ for all $v<c$.

Not really though. The $y = y’$ is a choice among many possible. However, it is the choice that corresponds to an irrotational Lorentz transform, which is the standard configuration. For example, $y’ = -y$ would work perfectly fine with $z’ = -z$. It is just not a pure boost.

 

[Sagittarius A-Star]

Not really though. The $y = y’$ is a choice among many possible. However, it is the choice that corresponds to an irrotational Lorentz transform, which is the standard configuration. For example, $y’ = -y$ would work perfectly fine with $z’ = -z$. It is just not a pure boost.

You are correct.

 

[Dale]

Wait, what? I am teaching relativity wrong?

Hahaha, no, of course not. I should have said "usually". I do prefer your approach, but it certainly is not the usual approach

 

[Hak]

I was not able to prove what I wanted by following your advice, only to justify it, perhaps not even that. However, I still want to report the procedure.
The rod placed along $x'$ is parallel to the velocity of $S$ as seen from $S'$. In fact, more, let us say that the line on which it lies is coincident with the direction of $\vec v$. When I look at it from $S$ it will remain on the line of the direction of $\vec v$ (at least you must grant me that coincident lines go in coincident lines).

The rod placed along $y'$, seen from $S'$ is perpendicular to the line of $\vec v$. Viewed from system $S$ will it still be perpendicular to $\vec v$? If not, it should be tilted by an angle $\alpha$, fixed by theory; however, theory could not tell me which way it would be tilted, whether so \ or so /. In fact, an inclination \ turns into an inclination / with a rotation of 180° around the x-axis. Translated: if two observers, both in the system $S$, are one standing and one upside down, they see one the tilt \ and the other the tilt /. Therefore, it is not possible for the theory to predict the "sign" of the tilt to me, since this would be different for two observers in the same system. If the rods, viewed from the $S$ system, were not perpendicular, they could then form either an obtuse or acute angle, without the theory being able to predict to me which one. The only alternative in which there are no such "unpredictable sign" problems (obtuse and acute angle) is the case in which the angle continues to be right. It is therefore at least reasonable that it continues to be right-angled.

Alternatively, if the angle varied, the plane perpendicular to $v$ would become a cone and the straight lines in the plane from before would become broken, but this is absurd because straight lines must go into straight lines (a straight line is the trajectory of a "ray of light," and if I saw that this is not a straight line I would understand that I am moving in contradiction to the principle of relativity). In fact, I am not convinced that the line is necessarily the trajectory of a ray of light (it is, but it is not obvious to me to state this from the postulates of relativity alone). Let me explain further: let us take our vertical rod in $S'$. Let us assume that a beam of light travels through it. When we look from $S$, the instantaneous position of the rod will no longer be the trajectory of the same light ray (it would be only if the speed of light were infinite) since the rod is moving to the right at speed $v$.

How to fix this correctly?

Thanks to all, but... what about this?

 

[Dale]

The only alternative in which there are no such "unpredictable sign" problems (obtuse and acute angle) is the case in which the angle continues to be right. It is therefore at least reasonable that it continues to be right-angled

That is not just reasonable, but that is a proof by symmetry.