Special Relativity - clocks on two trains

[Dale]

In your calculation, what are the primed and the unprimed variables?

In the first section after the picture you are calculating two spacetime intervals and setting them equal to each other. Which spacetime intervals are being calculated and why are they equal?

In the second section, it is unclear where the initial formula arises.

 

[Mike_bb]

Top right corner says $V=2|\vec{v_1}|$.

Thanks for your note. It's my mistake, indeed.

And you still didn't explain what $V$ is supposed to represent physically.

$V$ is relative velocity of moving trains.

 

[Mike_bb]

In your calculation, what are the primed and the unprimed variables?

$\Delta t$ is primed variable, $\Delta t'$ is unprimed variable.

In the first section after the picture you are calculating two spacetime intervals and setting them equal to each other. Which spacetime intervals are being calculated and why are they equal?

Left side of equality: view from left train (rest frame); right side view from right train. They are equal because of invariance of interval.

In the second section, it is unclear where the initial formula arises.

Yes, it may be incorrect formula for calculating "effect of simulteneity".

 

[Dale]

$\Delta t$ is primed variable, $\Delta t'$ is unprimed variable.

Yes, but what do they mean? Typically primed variables represent one frame and unprimed variables represent another. But which is which? Especially since there are three frames of potential interest here.

Left side of equality: view from left train (rest frame); right side view from right train. They are equal because of invariance of interval.

Views of what? The spacetime interval is calculated between two events. Which two events? To invoke the invariance of the spacetime interval it must be the same two events on both sides.

 

[robphy]

$\Delta t$ is primed variable, $\Delta t'$ is unprimed variable.

That sounds reversed to me.

 

[A.T.]

That sounds reversed to me.

Avoiding answering the obviously intended question and still getting it wrong.

 

[Mike_bb]

Yes, but what do they mean? Typically primed variables represent one frame and unprimed variables represent another. But which is which? Especially since there are three frames of potential interest here.

$\Delta t$ is time on the left clock (in the left train which at rest). $\Delta t'$ is time on the right moving clock.

Views of what? The spacetime interval is calculated between two events. Which two events? To invoke the invariance of the spacetime interval it must be the same two events on both sides.

Event1: reading the left clock $\Delta t$ when distance between left and right clock is $\Delta x$
Event2: reading the right clock $\Delta t'$ when distance between left and right clock is $\Delta x$ ($\Delta x' =0$)

 

[Dale]

Ok, I think now there is enough information to analyze things. I will go back and look.

In the future, please place all of this type of information in the first post. Not scattered over 37 posts. Clarity is important.

 

[Mike_bb]

Ok, I think now there is enough information to analyze things.

Ok. Could you write formula of "effect of simulteneity"? I have found many different formulas but I suspect these are wrong.

 

[Mike_bb]

Ok, I think now there is enough information to analyze things. I will go back and look.

In the future, please place all of this type of information in the first post. Not scattered over 37 posts. Clarity is important.

$\Delta T=\frac{vL}{c^2*Y}=\frac{0.8c*8c}{c^2*1.6}=4sec$. It's correct answer.

 

[Mike_bb]

I used these formulas for calculating "effect of simulteneity" but I obtained wrong result again. I can't understand where the mistake is.

 

[Ibix]

I used these formulas for calculating "effect of simulteneity" but I obtained wrong result again. I can't understand where the mistake is.

View attachment 374921

I think because of simultaneity. The event on the right clock that is simultaneous with "left clock reads 2s" according to the left clock's rest frame is not "right clock reads 2s". I think you are assuming that the separation is $\Delta x/\gamma$, but this is not correct.

I do not think there is a standard formula for "the effect of simultaneity". It depends too much on the state of motion of the other clock, which affects both the distance and clock rates. If you have two clocks that are at rest with respect to each other and synchronised in their rest frame then you could write down the difference between their readings taken simultaneously according to some other frame (I think that's what @Point Conception wrote down earlier in the thread), but it's of limited utility.

 

[Mike_bb]

I do not think there is a standard formula for "the effect of simultaneity".

Perhaps. I found different formulas and explanations how to find "value of desynchronization" but they didn't work. The simplest working formula was $t_A-t_B$

 

[Dale]

My calculations are here. The rest frame - left train. I obtained wrong result (5.51sec instead of 4sec) .
Could anyone explain where did I go wrong? Thanks!


View attachment 374902

At the end of that first section you have Δx=L=8c, which Δx should be a length but 8c is a speed.

$8c$=8*300000. I denoted "c" as dimensionless unit in this case (for convenience).

In the second section the formula ΔT uses a capital T instead of lower case t. Is there any significance to that?

This formula is used for calculating "effect of simultaneity". I don't know how to say this in English.

What is your V supposed to represent?

$V = \frac{2v_1}{1+v_1^2/c^2}$

$V$ is relative velocity of moving trains.

Yes, but what do they mean? Typically primed variables represent one frame and unprimed variables represent another. But which is which? Especially since there are three frames of potential interest here.

Δt is time on the left clock (in the left train which at rest). Δt′ is time on the right moving clock

Views of what? The spacetime interval is calculated between two events. Which two events? To invoke the invariance of the spacetime interval it must be the same two events on both sides.

Event1: reading the left clock Δt when distance between left and right clock is Δx
Event2: reading the right clock Δt′ when distance between left and right clock is Δx (Δx′=0)

OK, so the events are described in terms of $\Delta x$ which is in the unprimed (left) train's frame. In the left frame the worldline of the left clock is $(c t, x) = (c t, 0)$ and the worldline of the right clock is $(c t, V t)$ where $V=2 v/(1+v^2/c^2)$. I think you are using $v=0.4 c$ so $V=0.69c$. Anything using $V=0.8c$ will be wrong.

Now, at $t=10$ we can simply substitute into the worldlines to get the events. Event 1 is $E_1=(c 10, 0)$ and Event 2 is $E_2=(c 10, V 10)=(c 10, c 6.9)$. So $c\Delta t = 0$ and $\Delta x = 6.9 c$. This gives a spacetime interval of $$c^2 \Delta t^2 - \Delta x^2 = 0 - 100 V^2 = -47.6 c^2$$

This is great but, since we do not already know $\Delta t'$ nor $\Delta x'$ it doesn't really get us much information. It is definitely not correct that $\Delta x'=0$.

We can transform into the primed frame to find that information. Taking the Lorentz transform of $E_1$ and $E_2$ we get $E_1'=(13.8 c, - 9.5 c)$ and $E_2'=(7.2 c, 0)$. So $c\Delta t'=6.6 c$ and $\Delta x'= -9.5 c$ which gives a spacetime interval of $$c^2 \Delta t'^2 - \Delta x'^2 = -47.6 c^2$$

I am not getting anything that looks like your $\Delta T$, but it is hard to tell with some things using numbers and some things using variables. But the big mistakes that I see are assuming that $\Delta x'=0$ and that $V=0.8c$, which lead to some other follow on errors.

 

[Mike_bb]

Anything using $V=0.8c$ will be wrong.

Why?

 

[A.T.]

I think you are using v=0.4c so V=0.69c. Anything using V=0.8c will be wrong.

Why?

You are using the separation speed V as the relative speed. See:
https://en.wikipedia.org/wiki/Velocity-addition_formula#Special_relativity

 

[Dale]

Why?

Because $V=0.69c\ne 0.8c$

 

[PeterDonis]

Could you write formula of "effect of simulteneity"?

First we would need to know what "effect of simultaneity" even means. That's not a standard term that appears anywhere in relativity.

 

[Mike_bb]

Because $V=0.69c\ne 0.8c$

If $V=0.8c$ $v$ can be equal: $v=0.5c$

First we would need to know what "effect of simultaneity" even means. That's not a standard term that appears anywhere in relativity.

It follows from Lorentz transformation:

 

[Dale]

If $V=0.8c$ $v$ can be equal: $v=0.5c$

Certainly. You can change numbers around. If you are changing numbers, let me suggest also changing to $c=1$. This is using natural units, like years for time and light-years for length. Or meters for length and light-meters for time.

 

[Mike_bb]

Certainly. You can change numbers around. If you are changing numbers, let me suggest also changing to $c=1$. This is using natural units, like years for time and light-years for length. Or meters for length and light-meters for time.

I don't understand what you mean. I changed nothing. I just said that $v$ can be equal $0.5c$ and then $V=0.8c$ (but you wrote that it's wrong(?)).

 

[Dale]

I don't understand what you mean. I changed nothing. I just said that $v$ can be equal $0.5c$ and then $V=0.8c$ (but you wrote that it's wrong(?)).

$V=0.8 c$ is wrong with $v=0.4c$.

I changed nothing.

You changed $v= 0.4c \rightarrow v=0.5c$. If you believe this was not a change then it was yet another unclear point. You really must aim for substantially more completeness and clarity in your writing.

 

[PeroK]

Here's the thing. Suppose we have a clock, A, at the origin of an inertial reference frame. Another clock, B, moving with constrant velocity, of magnitude $v$, passes the origin
and both clocks are synchronized - to 0, say.

When the clock B is a distance $d$ from the origin, it will be measured to read $\frac{d}{\gamma v}$ in the original inertial frame. While the clock A will read $\frac d v$.

We can then look at the difference in the two clocks readings, which is $\dfrac{d}{\gamma v}(\gamma -1)$.

The OP wants to call this "the effect of simultaneity".

 

[A.T.]

So in your scenario:
- What are the two frames F and F'?
- What are the two spatially separated events, that are simultaneous in F, but not simultaneous in F' ?

 

[Mike_bb]

So in your scenario:
- What are the two frames F and F'?
- What are the two spatially separated events, that are simultaneous in F, but not simultaneous in F' ?

I solved my problem (big thanks to Dale). But if you want to know:
1. Frame F - left train (at rest). Frame F' - right train(moving).
2.
Event1: reading the left clock Δt when distance between left and right clock is Δx
Event2: reading the right clock Δt′ when distance between left and right clock is Δx

 

[PeterDonis]

It follows from Lorentz transformation

Is what you put in the quote box here a quote from somewhere? Where?

 

[Mike_bb]

Is what you put in the quote box here a quote from somewhere? Where?

These are my words. What's wrong with this?

 

[PeterDonis]

These are my words. What's wrong with this?

Normally if you put something in a quote box like that, it means you're quoting a reference. If it's just your words, you can just post them directly, like the rest of your words in your posts.

 

[PeterDonis]

It follows from Lorentz transformation

Ok, so you already have the formula you were asking for, for what you are calling "effect of simultaneity" (which doesn't seem like a good name for it, but you've given the math so we know what you mean). So what's the problem?

 

[Mike_bb]

Normally if you put something in a quote box like that, it means you're quoting a reference. If it's just your words, you can just post them directly, like the rest of your words in your posts.

There is no quote box: