How does an arbitrary choice of the origin affect real observations?

[larpal]

TL;DR Summary

how can the choice of origin have a real-world affect, if the choice is arbitrary?

Problem I can't seem to get my head around, if anyone can help - consider:
- a stationary frame S containing an infinite line of clocks in the x direction
- an (initially) stationary frame S', also containing an infinite line of clocks in the x direction
- the clocks in S and S' are synchronized with each other
- S' becomes a moving reference frame (moving in the positive X direction)
- the Lorentz transformation says that for observers in S, clocks in the +x direction of S': t' < t, and for clocks in the -x direction: t' > t
- so at some point (the origin), t' = t
- but the origin can be chosen arbitrarily
- so where is the physical point where t' = t?

If two observers A and B in S do a simultaneous (in the S frame) observation of the moving clocks currently adjacent to them (call them A' and B'), then both A and B should agree that the time shown by A' is greater than the time shown by B'. If A picks themself as the origin, at t=0, then the time of A' should be 0, and the time of B' should be negative. But if B picks themself as the origin, then the time of B is 0, and the time shown by A' should be positive. But shouldn't all observers in S agree on the times shown by the clocks at A' and B'?

 

[Orodruin]

It cannot.

What you are describing is just an arbitrary assignment of what is zero time. Not a measurement of times.

 

[FactChecker]

Once the reference frames are inertial, pick a location, O, and denote the clock times at the initial instant there in the two reference frames as $t_O$ and $t_O'$, respectively. For the remainder of the problem, times $t-t_O$ and $t'-t_O'$ will be reset clocks in their respective reference frames. They will be synchronized in their respective reference frames and the initial reset times at the location, O, will be 0.

 

[Mister T]

TL;DR Summary: how can the choice of origin have a real-world affect, if the choice is arbitrary?

- so where is the physical point where t' = t?

Clock synchronization is not physical. It's a convention so it has no physical significance.

 

[Orodruin]

TL;DR Summary: how can the choice of origin have a real-world affect, if the choice is arbitrary?

- S' becomes a moving reference frame (moving in the positive X direction)

This is also not correct. If S’ is an inertial frame and is to be treated as such, it cannot suddenly become moving. If it does, then it is not inertial. It is a self contradiction.

An observer can start moving and describe events using their new reat frame (or the initial one), but an inertial frame moves at constant velocity relative to all other inertial frames.

 

[Ibix]

TL;DR Summary: how can the choice of origin have a real-world affect, if the choice is arbitrary?

- S' becomes a moving reference frame (moving in the positive X direction)

I was about to make the point Orodruin just made. Additionally, if what you actually mean is "there are two lines of clocks at rest and synchronised, then one line starts moving", that would be possible, but the clocks would not in general be synchronised in their new rest frame.

 

[Dale]

The actual law of physics that governs the reading on a clock is obtained from the metric as* $$\Delta \tau = \frac{1}{c}\int_P \sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$ This is a completely general expression that works in any spacetime for any valid clock, regardless of gravity or acceleration, etc. The time on the clock is $\tau$ and the spacetime path that the clock travels on is $P$, $g$ is the metric and $x$ is the 4D spacetime coordinates.

The important point is not the details of the expression, but the fact that the time on the clock is $\tau$ but the physical law gives you $\Delta \tau$.

So for each clock $\tau=\tau_{initial} + \Delta \tau$. If you choose to define $\tau_{initial}=\tau(t=0)$ then when you change your origin you will potentially change the $\tau_{initial}$ for each clock. The end result will be that the physical location will not change, but the coordinate representation of that location will.

*edited, see below

 

[Orodruin]

$$\Delta \tau = \frac{1}{c}\int_P \sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$

Fixed. Assuming you intended a -+++ metric.

 

[Dale]

Oops, yes

 

[larpal]

Sorry for the imprecision of my language. Maybe just one more try, from a more intuitive angle:

- there are two parallel lines of clocks, say infinitely long, at rest. All the clocks read 10:00. Rest frame S.
- one line of clocks starts moving in the positive X direction and becomes the new rest frame S' (steady brief acceleration resulting in inertial frame S')
- at some point later, say 10:15 in S, the S' clocks in the +x direction will show earlier times than the clocks in the -x direction, for any observer in S
- which of the clocks in the S frame sees the S' clock adjacent to it as also saying 10:15: one of them (the origin, but origin is arbitrary), all of them, or none of them? T coordinates may be invisible and arbitrary, but the clock shows something definite.

If the answer is all observers in S see the clock adjacent to them in S' as reading the same time as theirs, then observer A sees the clock adjacent to A (in the S' frame) reading 10:15, but sees the clock adjacent to B reading say, 10:13

But since B sees the clock adjacent to B as reading 10:15, then A and B disagree as to the readings of the clocks in S'. Is that the correct interpretation? A and B disagree on what any given S' clock is currently showing (after correcting for the propagation time of light when viewing the remote clock)?

 

[PeroK]

Sorry for the imprecision of my language. Maybe just one more try, from a more intuitive angle:

- there are two parallel lines of clocks, say infinitely long, at rest. All the clocks read 10:00. Rest frame S.
- one line of clocks starts moving in the positive X direction and becomes the new rest frame S' (steady brief acceleration resulting in inertial frame S')

The problem with this idea is that if the clocks all accelerate simultaneously in frame S, then they do not accelerate simultaneously in a frame that is instantaneously moving with the clocks. If the clocks are to remain synchronized in S', then their acceleration is asynchronous in frame S.

To see this, consider two clocks accelerating from rest.

 

[PeterDonis]

- there are two parallel lines of clocks, say infinitely long, at rest. All the clocks read 10:00. Rest frame S.
- one line of clocks starts moving in the positive X direction and becomes the new rest frame S' (steady brief acceleration resulting in inertial frame S')

This is actually not a good way to frame the scenario, for two reasons:

(1) It brings in acceleration, which, however brief, means you have objects moving non-inertially, which complicates the analysis;

(2) It is impossible to have an infinite line of clocks (your S' clocks) that all start out in sync in the original rest frame, all briefly accelerate, and all end up in sync in the new rest frame. That is forbidden by the kinematics of SR. (For some insight into why, look up "Rindler horizon" and the Bell spaceship paradox.)

A better framing, which also makes the answer to your question obvious, is to have two infinite lines of clocks, S and S', each of which are all at rest relative to the other clocks in the same line, but the two lines are in relative motion in the $x$ direction, with S' moving in the positive $x$ direction relative to S. Then you have to specify, at instant $t = 0$ in S, which of the S' clocks shows the same time as the S clock it is next to at that instant; the usual convention is to have the S' clock at $x' = 0$ (the spatial origin of S') be next to the S clock at $x = 0$ (the spatial origin of S) at $t = 0$ in S, and showing the same time; that means that the S' clock at $x' = 0$ is showing $t' = 0$ at that event, where those two clocks meet.

Once you have made that specification, it should be obvious that none of the other S' clocks will show the same time at either $t = 0$ or $t' = 0$ as the S clocks they are next to at those events (which are not the same because of relativity of simultaneity).

An even better way of framing all this is to draw a spacetime diagram showing the S clock worldlines and the S' clock worldlines, which will form a grid of parallelograms, and looking at how the crossing points match up.

 

[Ibix]

- one line of clocks starts moving in the positive X direction and becomes the new rest frame S' (steady brief acceleration resulting in inertial frame S')

...and they will not be synchronised in the new frame.

Possibly the point you are missing is that a Lorentz transformation is not the same thing as acceleration. An everyday example of the difference is to wait at traffic lights and watch cars coming towards you at 30mph. You can do a Lorentz transform to a frame with $v=30\mathrm{mph}$ and describe the resulting frame where you have velocity $-30\mathrm{mph}$, but you are still standing at the lights. You have not accelerated, you have just changed the description of the world.

So when you talk about accelerating one of your lines of clocks, that's a very different thing to switching between one set of clocks and another that were already in inertial motion. The clocks that one could use to implement a relativistic frame have always been in motion at the same speed (or at least, for the whole region of spacetime you're studying).

 

[Dale]

There is one bad piece of the specification:

the new rest frame S' (steady brief acceleration resulting in inertial frame S')

The accelerated clocks are not inertial (by virtue of having accelerated). So their rest frame cannot be inertial. So if S' is an inertial frame then it is necessarily not the rest frame of the accelerated clocks. Specifically, if all of the accelerated clocks accelerate when they read 10:00, then each accelerated clock is not at rest in S' prior to when it reads 10:00. Each accelerated clock begins not at rest in S', but becomes at rest in S' after it reads 10:00.

To be clear I will not refer to S clocks or S' clocks, but to "inertial" clocks and "accelerated" clocks. Each accelerated clocks is at rest in S before it reads 10:00 and at rest in S' after it reads 10:00. And before they read 10:00 all clocks, inertial and accelerated, are synchronized in the S frame where they are all at rest.

- at some point later, say 10:15 in S, the S' clocks in the +x direction will show earlier times than the clocks in the -x direction, for any observer in S

Actually, for the specific scenario that you have proposed, all of the accelerated clocks display the same reading at any given time in S.

- which of the clocks in the S frame sees the S' clock adjacent to it as also saying 10:15: one of them (the origin, but origin is arbitrary), all of them, or none of them? T coordinates may be invisible and arbitrary, but the clock shows something definite.

None of them. As you have specified the scenario at 10:15 in S the accelerated clocks will all read $10\colon 00 +0\colon 15 \ \sqrt{1-v^2/c^2}$. This is independent of the choice of origin for S or S'.

Indeed, "the reading of the accelerated clock next to inertial clock A when inertial clock A reads $10\colon 15$" is the outcome of a measurement. All coordinate systems agree on the outcome of any measurement, even if they do not interpret the measurement as having the same meaning.

 

[pervect]

Sorry for the imprecision of my language. Maybe just one more try, from a more intuitive angle:

- there are two parallel lines of clocks, say infinitely long, at rest. All the clocks read 10:00. Rest frame S.
- one line of clocks starts moving in the positive X direction and becomes the new rest frame S' (steady brief acceleration resulting in inertial frame S')

You need to provide more details about the process of how you accelerate the clocks to compute any answer. If you start accelerate all the clocks at the same instant as defined by the notion of simultaneity in frame S, you can get an answer, but you need to specify the details.

Note that starting the acceleration at some instant of time in frame S is a different problem from starting the acceleration at some instant of time in S', due to the relativity of simultaneity.

The fact that you omit the important detail of exactly how you accelerate the clocks suggests to me that you suffer from the common blind spot of the issue of the relativity of simultaneity. Unfortunately this is both very common and tends to take up a lot of time as people resist getting the point :(.

This issue is also frequently discussed in the context of "Bell's spaceship paradox". See https://en.wikipedia.org/wiki/Bell's_spaceship_paradox or other sources. If you have two spaceships connected by a string , and they both accelerate identically in frame S, starting at the same time in frame S, the string will break. It's possible to find an acceleration profile that won't break the string, the sort of acceleration that preserves lengths is known as "Born rigid motion". Note that the breaking the string is a physical effect, independent of frame.

 

[cianfa72]

The actual law of physics that governs the reading on a clock is obtained from the metric as* $$\Delta \tau = \frac{1}{c}\int_P \sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$$ This is a completely general expression that works in any spacetime for any valid clock, regardless of gravity or acceleration, etc.

This is basically the clock hypothesis. When we claim that the rate of a clock doesn’t depend on its acceleration, we are actually talking of its proper acceleration (that's invariant across frames).