Twin Paradox: Calculating Stan's Time Elapsed According to Moe

[topsquark]

TL;DR Summary

I am trying to calculate the time elapsed for the stationary twin solely based on the traveling twins perspective. I've never seen it done this way and I'm screwing something up.

I know, I know, yet another Twin Paradox thread.

(My apologies if this is already on the Forum somewhere. I found a similar discussion but with no answer to my question. I'd appreciate a link if someone has already given an answer to it.)

I'm trying to construct the time elapsed for Stan ("stationary frame") as observed by Moe ("moving frame.") Assume the usual conditions for the paradox. Let Moe leave Earth (A) with a speed of 0.8 (c = 1) and travel out 4 light years (P). Let's also assume that Moe is going to jump on another rocket 4 light years out that is traveling back to Earth at 0.8 (and assume this wouldn't kill him) and returns to Earth (D).

At event B, Stan sends a light beam out to event P to meet Moe. When Moe gets to P and switches direction, he sends a light beam back to Earth at event C.

The time elapsed for the whole trip for Stan is $T = 2 \cdot \dfrac{4}{0.8} = 10$ light years. According to Moe, it takes $T' + T'' = \dfrac{T}{ \gamma } = 6$ light years.

That's the usual thing. What I am trying to do is figure out how to calculate this according to Moe. Between the outward and inward trips he sees Stan age 6 light years. Moe sees Stan age 4 light years (almost) instantly when he switches coordinate systems 4 light years out.

The shaded triangle somehow gives the missing 4 years. But is there a way to get that from the triangle? I'm thinking it has something to do with the area of the triangle, and I could theoretically do all sorts of numerology on it to get the answer. But I haven't a clue how to derive it. (And, frankly, just how to find the area of the triangle in Minkowski space would be another question! My Differential Geometry is fairly primitive.)

Thanks!

-Dan

 

[PeterDonis]

The shaded triangle somehow gives the missing 4 years.

No, it doesn't, because the two sides going to P are lightlike, but to account for the "missing 4 years" they would need to be spacelike lines of simultaneity from P in the two applicable inertial frames (outbound and inbound).

Read this page of the Usenet Physics FAQ article on the twin paradox:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gap.html

Actually, read the whole article, all pages. It should be required reading for anyone before asking any questions whatever about the twin paradox.

What I am trying to do is figure out how to calculate this according to Moe.

And, as you have already been told in previous threads, before you can even do that you need to define what "according to Moe" means. That means you need to define a single reference frame for Moe to use, in which he is at rest (since that's what you mean by "according to Moe"), and that covers all of the applicable region of spacetime, i.e., it covers all of both Moe's and Stan's worldlines between the events where they meet.

What you are doing, instead, is using two frames, the "outbound" inertial frame and the "inbound" inertial frame, neither of which covers all of the applicable region (since the "outbound" frame stops working at P, and the "inbound" frame only starts working at P), but ignoring that fact and not trying to "stitch" them together into a single frame that covers the entire scenario. Of course that's not going to work.

 

[PeterDonis]

Moe sees Stan age 4 light years (almost) instantly when he switches coordinate systems 4 light years out.

No, that is not what Moe "sees", i.e., that is not what the light signals coming from Moe to Stan actually show him. Read the "Doppler shift" page in the Usenet Physics FAQ article I linked to for a correct description of what Moe and Stan actually see (the article uses the names Stella and Terence).

If you were to construct a valid single reference frame for Moe as I described previously, this frame would indeed assign a large change in Stan's age to a small interval of coordinate time in that frame when Moe turns around. But this is a coordinate-dependent quantity and has no physical meaning. Moe's turning around doesn't do anything to Stan.

 

[Dale]

TL;DR Summary: I am trying to calculate the time elapsed for the stationary twin solely based on the traveling twins perspective. I've never seen it done this way and I'm screwing something up.

I'm trying to construct the time elapsed for Stan ("stationary frame") as observed by Moe ("moving frame.")

If you want to do that then you are far better off doing a constant proper acceleration Moe rather than the standard instantaneous turnaround.

 

[topsquark]

If you want to do that then you are far better off doing a constant proper acceleration Moe rather than the standard instantaneous turnaround.

Good thought. I'll work on that for a bit. Thanks!

-Dan

 

[PeterDonis]

I'll work on that for a bit.

We had a recent thread on it:

https://www.physicsforums.com/threads/twin-paradox-with-accelerated-motion.1047089/

 

[topsquark]

Just a quick update. I'm working on several things at once and I want to spend some careful time with the Rindler coordinates and do the job right. I'll be back soon either to check my solution or with more questions. :)

-Dan

 

[member 728827]

If you want to do that then you are far better off doing a constant proper acceleration Moe rather than the standard instantaneous turnaround.

@Dale, seems pretty much @topsquark wants to use the Moe's metric to measure Stan elapsed time, isn't it?

 

[topsquark]

I have completed my initial review of the Rindler metric. I was going to post my derivation and proof of the interval formula and ask someone to check the work, but once I got the hang of it, it went pretty easily.

My only remaining question about it is if there is a way to derive the interval directly? The source I used focussed on how to find the Rindler transformation equations, but said nothing about how to derive the interval. I suppose I could go to the GR field equation but I suspect that there may be a simpler method for this specific case. I just can't think of it.

Either way, I think I now have a good enough answer to my OP to continue for now.

Thank you, all!

-Dan

 

[PeterDonis]

My only remaining question about it is if there is a way to derive the interval directly?

I don't understand. If you have the metric in your new coordinates, you have the interval in those coordinates.

 

[topsquark]

I don't understand. If you have the metric in your new coordinates, you have the interval in those coordinates.

Sorry, I did word that badly, didn't I? I should have given more details. The analogy I'm thinking of is using Minkowski 4-vector methods to derive the Lorentz transformations, rather than, say, Einstein's derivation. The 4-vector method does this far more efficiently. The derivation I have for the Rindler coordinates is based on defining the instantaneous speed of the accelerating frame in terms of a stationary one, and then deriving the transformations step by step, much like Einstein's work. I was wondering if there might be a Linear Algebraic method to derive the Rindler metric similar to the 4-vector approach to derive the Minkowski metric. (That is, without going full out GR on it.)

-Dan

 

[Dale]

My only remaining question about it is if there is a way to derive the interval directly? The source I used focussed on how to find the Rindler transformation equations, but said nothing about how to derive the interval.

Yes, I like this source for that:

https://arxiv.org/abs/physics/0412024

It describes how you can calculate the metric in radar coordinates based on the accelerometer reading of the observer whose worldline is generating the radar coordinates.

 

[PeterDonis]

using Minkowski 4-vector methods to derive the Lorentz transformations

the 4-vector approach to derive the Minkowski metric

This sounds circular. You can't use 4-vectors based on the Minkowski metric to derive the LT, but also use 4-vectors to derive the Minkowski metric.

A specific reference to whatever source you got these things from would help.

 

[PeterDonis]

That is, without going full out GR on it.

I don't know what this means either. We are talking about flat spacetime, where GR is not needed.

 

[PAllen]

Yes, I like this source for that:

https://arxiv.org/abs/physics/0412024

It describes how you can calculate the metric in radar coordinates based on the accelerometer reading of the observer whose worldline is generating the radar coordinates.

A nuance is that the metric at any point is determined by proper acceleration in the future as well as the past. The further away an event is from the defining observer, the further in the future/past is necessary to determine the metric at that event. These are the integration bounds in eq. 7, the main result of the paper. This contrasts with Fermi Normal (which cover a much smaller region, in general), but are not affected by distant past and future (there is no similar integral over past/future in the Fermi metric, given in eqn. 22 of the paper).

 

[Dale]

A nuance is that the metric at any point is determined by proper acceleration in the future as well as the past.

Yes, that is correct. The same is true about the radar coordinates themselves. You have to be able to send a radar pulse out and then receive the radar echo back to even assign an event its radar coordinates. The coordinates themselves depend on your motion during that entire period, as does the metric.

 

[topsquark]

This sounds circular. You can't use 4-vectors based on the Minkowski metric to derive the LT, but also use 4-vectors to derive the Minkowski metric.

A specific reference to whatever source you got these things from would help.

I admit that my Minkowski 4-vector (well, in 1+1 space-time for simplicity) derivation is currently, in fact, a bit circular. I'm not finished with it. If you are telling me that it can't be done I'll accept that, but I have a suspicion that it might be possible under certain assumptions. However, I haven't looked at it that closely and this is only a feeling.

For consumption, here's my derivation:
I am using the metric
$\eta = \left ( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right )$
Consider the usual frames O and O' where O' has a speed of v in the $+x^1$ direction with respect to O. Further consider an object in O' that is at rest.

I begin by assuming a differential transformation of the form ${dx^{ \prime }}^{ \mu } = \Lambda ^{ \mu }_{ \phantom{ \mu } \nu } x^{ \nu }$
$\left ( \begin{matrix} {dx^{ \prime }}^0 \\ {dx^{ \prime }}^1 \end{matrix} \right ) = \left ( \begin{matrix} \Lambda ^0_{ \phantom{0} 0} & \Lambda ^0_{ \phantom{0} 1} \\ \Lambda ^1_{ \phantom{1} 0} & \Lambda ^1_{ \phantom{1} 1} \end{matrix} \right ) \left ( \begin{matrix} dx^0 \\ dx^1 \end{matrix} \right )$

Divide both sides by $dx^0$:
$\left ( \begin{matrix} \dfrac{{dx^{ \prime }}^0}{dx^0} \\ \dfrac{{dx^{ \prime }}^1}{dx^0} \end{matrix} \right ) = \left ( \begin{matrix} \Lambda ^0_{ \phantom{0} 0} & \Lambda ^0_{ \phantom{0} 1} \\ \Lambda ^1_{ \phantom{1} 0} & \Lambda ^1_{ \phantom{1} 1} \end{matrix} \right ) \left ( \begin{matrix} \dfrac{dx^0}{dx^0} \\ \dfrac{dx^1}{dx^0} \end{matrix} \right )$

Since the object in O' is stationary $\dfrac{{dx^{ \prime }}^1}{dx^0}= 0$ and we know that $\dfrac{dx^1}{dx^0} = v$, so
$\left ( \begin{matrix} \dfrac{{dx^{ \prime }}^0}{dx^0} \\ 0 \end{matrix} \right ) = \left ( \begin{matrix} \Lambda ^0_{ \phantom{0} 0} & \Lambda ^0_{ \phantom{0} 1} \\ \Lambda ^1_{ \phantom{1} 0} & \Lambda ^1_{ \phantom{1} 1} \end{matrix} \right ) \left ( \begin{matrix} 1\\ v \end{matrix} \right )$

Now, hereabouts I have to make my two assumptions. The first feels rather natural, but I have yet to sit down and prove it:
1) $\Lambda ^1_{ \phantom{1} 0} = \Lambda ^0_{ \phantom{} 1}$. My sketchy argument is that the metric is diagonal. I need to do more work on this.

2) Here's the possible circularity. I have to presume the time dilation formula: $\dfrac{{dx^{ \prime }}^0}{dx^0} = \sqrt{1 - v^2}$. For the purposes of my first draft this is fine, but I haven't yet gone back to check to see "how bad" this assumption is. In the end I may just have to accept the circularity and use the Minkowski approach merely to show that it produces a consistent result.

The rest is straightforward:
$\left ( \begin{matrix} \sqrt{1-v^2} \\ 0 \end{matrix} \right ) = \left ( \begin{matrix} \Lambda ^0_{ \phantom{0} 0} & \Lambda ^0_{ \phantom{0} 1} \\ \Lambda ^0_{ \phantom{0} 1} & \Lambda ^1_{ \phantom{1} 1} \end{matrix} \right ) \left ( \begin{matrix} 1\\ v \end{matrix} \right )$

We also know that $\eta _{ \alpha \beta } = \eta _{ \mu \nu } \Lambda ^{ \mu }_{ \phantom{ \mu } \alpha} \Lambda ^{ \nu }_{ \phantom{ \nu } \beta }$ and chosing $\alpha = \beta = 0$ gives
$-1 = - \left ( \Lambda ^0_{ \phantom{0} 0} \right )^2 + \left ( \Lambda ^0_{ \phantom{0} 1} \right )^2$

Solving the two matrix equations plus the above equation simultaneously gives the solution
$\Lambda ^0_{ \phantom{0} 0} = \Lambda ^1_{ \phantom{1} 1} = \gamma, ~ \Lambda ^0_{ \phantom{0} 1} = -v \gamma$

where $\gamma = \dfrac{1}{\sqrt{1 - v^2}}$.

(The bulk of the derivation should be familiar. I merged a couple of sources and modified them slightly to use my own notations.)

(I have to run. Any further typos will have to go uncorrected.)

-Dan

 

[topsquark]

Sorry about that. I had to get to work.

Fortunately I have plenty of spare time here. :)
I was able replace the circular part in the derivation in my last post. The problem,of course, is the $\dfrac{{dx^{ \prime }}^0}{dx^0}$. I was able to fix this by simply deriving a condition for the time dilation equation:

Consider the frames O and $O^{ \prime }$ from above again. Shine a light in $O^{ \prime }$ up the x' axis. Then we have, by assumption:

$d{x^{ \prime }}^0 = \Lambda ^{0}_{ \phantom{0} 0} dx^0 + \Lambda ^{0}_{ \phantom{0} 1} dx^1$

$d{x^{ \prime }}^1 = \Lambda ^{1}_{ \phantom{1} 0} dx^0 + \Lambda ^{1}_{ \phantom{1} 1} x^1$

Since the speed of light (c = 1) is the same in both frames we have that $dx^0 = dx^1$ and ${dx^{ \prime }}^0 = {dx^{ \prime }}^1$. Inserting these into the two equations gives
$d{x^{ \prime }}^0 = \Lambda ^{0}_{ \phantom{0} 0} dx^0 + \Lambda ^{0}_{ \phantom{0} 1} dx^0 = \left ( \Lambda ^{0}_{ \phantom{0} 0} + \Lambda ^{0}_{ \phantom{0} 1} \right ) dx^0$

$d{x^{ \prime }}^0 = \Lambda ^{1}_{ \phantom{1} 0} dx^0 + \Lambda ^{1}_{ \phantom{1} 1} x^0 = \left ( \Lambda ^{1}_{ \phantom{1} 0} + \Lambda ^{1}_{ \phantom{1} 1} \right ) dx^0$

Setting these equal gives the condition
$\Lambda ^{0}_{ \phantom{0} 0} + \Lambda ^{0}_{ \phantom{0} 1} = \Lambda ^{1}_{ \phantom{1} 0} + \Lambda ^{1}_{ \phantom{1} 1}$

Using this equation to replace the $\sqrt{1 - v^2}$ equation in the system in the prior post gives the solution sets
$\Lambda ^{0}_{ \phantom{0} 0} = \Lambda ^{1}_{ \phantom{1} 1} = -\gamma, ~ \Lambda ^{0}_{ \phantom{0} 1} + \Lambda ^{1}_{ \phantom{1} 0} = v \gamma$

and
$\Lambda ^{0}_{ \phantom{0} 0} = \Lambda ^{1}_{ \phantom{1} 1} = \gamma, ~ \Lambda ^{0}_{ \phantom{0} 1} = \Lambda ^{1}_{ \phantom{1} 0} = -v \gamma$

I am choosing the second solution as it belongs to the proper orthochronous subgroup of the Poincare group.

So apparently I don't have to make an assumption about the equality of the diagonal elements of the Lorentz transformation.

My question (and hopefully I've finally made it clear enough, now!) is how to change this derivation to reflect the constant acceleration system? (If that's even possible.) Clearly I need to change my assumption as to how the differential transformation equations work but what change would that be?

-Dan

 

[PeterDonis]

I am using the metric

If you assume that that metric looks the same in the primed frame as in the unprimed frame (which in itself requires that both frames are inertial frames), that alone is sufficient to pin down the Lorentz transformations as the only possible ones. This has been known, at the very least, since Minkowski published his "spacetime" paper in, IIRC, 1907.

Conversely, if you assume that the Lorentz transformations are the proper transformations between any two inertial frames, that alone is sufficient to pin down the Minkowski metric as the only possible one. This has also been known since the Minkowski "spacetime" paper.

None of the assumptions you make in any of your derivations are required to prove the above implications in either direction.

 

[PeterDonis]

how to change this derivation to reflect the constant acceleration system?

A non-inertial frame is not the same as an inertial frame, and there is no general rule for how to derive transformations to or from non-inertial frames. It depends on the specific non-inertial frame. Even for a frame in which an observer with constant proper acceleration is at rest, there are multiple possible ways to define such a frame.

 

[member 728827]

I don't know if this can help you in you endeavor, but as I read about it recently, I write down what I understood. Sorry for the non very technical vocabulary.

The following equations describe the hyperbolic worldline of an uniformly accelerated observer (c=1):

$T = \frac{1}{g} \cdot \sinh{(\,g \cdot t\,)} \quad (1)$
$X = \frac{1}{g} \cdot \cosh{(\,g \cdot t\,)} \quad (2)$

From an accelerated observer point of view, first such observer has to layout an appropriate frame suited to his needs, and after he can transform from/to an inertial frame or whatever.

Would seem logical for such observer to setup his frame as a conceptual grid of coordinate points that accelerate all at his own proper acceleration, but this is a bit of a problem, because of the "Bell string paradox", blah, blah... such a grid would break apart as the distances between the coordinate points change and points move relative one to the others, including the observer... what a mess.

I think that some clever people would say that such a grid is a congruence with a positive expansion factor, and that's not good for a coordinate system. Let's say would be an "elastic" frame, so to speak.

Then for example, to define the Rindler coordinates $(t,x)$, we need to layout a set of observers all of which agree with the simultaneity of events. Our non-inertial frame will be a collection of remote observers spread all over, which report back about what is happening locally for them. From equations (1) and (2), it follows easily that the line of simultaneity for any event in the worldline, passes through the origin. This is true for any worldline defined by an arbitrary g acceleration

We can fill in the whole space with observers labelled $O_x$, each of which has a constant proper acceleration $g_x$, and a turnaround point located at $x=\frac{1}{g_x}$. So in each observer’s MCRF, the worldlines of all the other observers maintain a constant position, and always measure each other to be at rest with respect to them, forming a rigid lattice in a Born rigid motion fashion. That's good.

Each "local" observer $O_x$ in the frame will have its own proper time, but as they all agree in the simultaneity of events, is then possible to establish an unique frame coordinate time. So to speak, each observer will adjust its clock to give the agreed common coordinate time.

We can see easily that the following applies:

$g_x \cdot \tau_x=g_o \cdot \tau_o \quad (3)$

and we can use $g_o \cdot \tau_o$ in all (1) and (2) equations as a common coordinate time:

$T = \frac{1}{g_x} \cdot \sinh{(\,g_x \cdot \tau_x \,)} = \frac{1}{g_x} \cdot \sinh{(\,g_o \cdot \tau_o \,)} = x \cdot \sinh{(\,g \cdot t \,)} \quad (4)$

$X = \frac{1}{g_x} \cdot \cosh{(\,g_x \cdot \tau_x \,)} = \frac{1}{g_x} \cdot \cosh{(\,g_o \cdot \tau_o \,)} = x \cdot \cosh{(\,g \cdot t \,)} \quad (5)$