Reciprocal time dilation

[fuentes1979]

TL;DR Summary

Reciprocal time dilation

Hello everyone,
I would be thankful ir someone explained where I am mistaken in this reasoning on Einstein’s mental experiment on Special Relativity.

Taking the railway as K reference frame, a light beam is thrown in the x positive direction the moment the train passes in the same positive direction at v= 2/3*c velocity. After a time t=1 sec taken from K frame, an observer situated in x=300000 km will receive the light beam.

Changing frames to K’ frame (the train) through Lorentz transformations we have that the said observer receives the light at x’ =(x-vt)/sqr(1-v2/c2) of 128972 km, bigger than the 100000 predicted by classical mechanics, so the lenghts are contracted in K’ as predicted by special relativity.
As relates time, t’=(t-vx/c2)/sqr(1-v2/c2) equals 0.4472 sec, so time is dilated in K’ as predicted by theory. As 128972/0.4472=c, everything is Ok.

Now, if we see the same situation considering the train as the K reference frame “at rest”, what we have is the light beam thrown the moment a moving observer at the railway passes with v=-2/3*c velocity. Now (from K) x= 300000 km, t=1 sec. From K’: x’= 670817 km, bigger than the 500000 km predicted by classical mechanics, so again the lenghts are contracted in the now moving reference frame ok K’, at the railway.

As for the time t’ though, it is now equal to 2.236 seconds according to: t’=t-(vx/c2)/sqr(1-v2/c2). Again 670817/2.236 equals c, but the problem is that now the time is not dilated, but contracted, in the moving reference frame of K’ (now the railway) as seen from K (now the train). That does not make sense as according to special relativity predictions, moving frames always show dilated times as mensures from frames at rest.

I cannot see which is my mistake. Could you please help me with this?
Thank you very much for your patience.

 

[Orodruin]

That’s easy: what you are computing is not time dilation. It is the coordinate transformation of events, which requires the full Lorentz transformation (which you applied). The concept of time dilation refers to particularly the time elapsed on the clock of a moving observer relative to the coordinate time of a frame where the observer is moving at non-zero speed.

 

[Ibix]

Orodruin beat me to the physics answer, but I'd started typing the rest.

A suggestion: use $v=\frac 35c$ or $\frac 45c$, since these give rational $\gamma$ factors. Also consider measuring distances in light seconds and time in seconds - this is a unit system where $c=1$ and most of your numbers will become much shorter and easier to type and read.

Also, please use LaTeX for maths. There's a guide linked below the reply if you aren't familiar with it. If you ever find that LaTeX appears not to be working on a page, there's a known bug where it doesn't work if you are the first person using it on a page. Enter some LaTeX in the reply box, preview it (the LaTeX will not render), refresh the page, and LaTeX will work.

 

[fuentes1979]

Thanks Orodruin and Ibix for your responses.
As for Orodruin reply, I cannot see the difference between, using your words “the time elapsed on the clock of a moving observer relative to the coordinate time of a frame…” and the situations where (1) time of the event “the light beam arrives to observer” is 1 second in the clock of the railway, at rest, and 0.4472 sec from the train’s clock, moving; and (2) time of the same event is 1 second in the train’s clock,now at rest, and 2.236 seconds in the railway clock, now moving.
As I look at it, situation (1) shows time elapsed on the clock of a moving observer, as is predicted by theory, but situation (2) seems to show the opposite result, time “contracting” int the moving clock, which must be erroneous, according to special relativity, unless I am mistaken (I definitely am, but do not know where).
Thank you very much again.

 

[Ibix]

Time dilation is the ratio of tick rates between two clocks. You are calculating coordinates of events. They're two different things.

I also think you are calculating coordinates of different events, which is probably confusing you even further.

Changing frames to K’ frame (the train) through Lorentz transformations we have that the said observer receives the light at x’ =(x-vt)/sqr(1-v2/c2) of 128972 km, bigger than the 100000 predicted by classical mechanics, so the lenghts are contracted in K’ as predicted by special relativity.
As relates time, t’=(t-vx/c2)/sqr(1-v2/c2) equals 0.4472 sec, so time is dilated in K’ as predicted by theory. As 128972/0.4472=c, everything is Ok.

Ok, you've taken an event at $x=1\mathrm{ls}$ and $t=1\mathrm{s}$ and Lorentz transformed the coordinates into a frame moving with $v=\frac 23c$ and got $x'=\frac 1{\sqrt{5}}\mathrm{ls}$ and $t'=\frac 1{\sqrt{5}}\mathrm{s}$. Fine.

Now, if we see the same situation considering the train as the K reference frame “at rest”, what we have is the light beam thrown the moment a moving observer at the railway passes with v=-2/3*c velocity. Now (from K) x= 300000 km, t=1 sec. From K’: x’= 670817 km, bigger than the 500000 km predicted by classical mechanics, so again the lenghts are contracted in the now moving reference frame ok K’, at the railway.

Here you seem to have switched labels on your frames, defeating the point of labelling them. In your train's rest frame you seem to have taken $x'=1\mathrm{ls}$ and $t'=1\mathrm{s}$ and then transformed into the platform frame, getting $x={\sqrt{5}}\mathrm{ls}$ and $t'={\sqrt{5}}\mathrm{s}$.

Note that I've used primed labels for coordinates in the train's rest frame and unprimed for those in the platform frame throughout. So it's immediately obvious that you are considering two different events because the $x'$ coordinates are not equal in the two cases and neither are the $t'$, $x$, and $t$ coordinates.

Finally, the point is that the Lorentz transforms relate the coordinates of events recorded using one frame to the coordinates of those events recorded using another frame. This is not the same as time dilation, which is to do with clock rates. If you want to derive time dilation from the Lorentz transforms, you need to consider two events that are at the same place in one or other frame (i.e., have the same $x$ or $x'$ coordinate). The usual way to do this is to have your light pulse be emitted at $(x,t)=(0,0)$, be reflected off a mirror at $(x,t)=(L,L/c)$ (where $L$ is a constant you may choose) and return to the spatial origin at $(x,t)=(0,2L/c)$. You will find that the transformed $t'$ coordinate is lower by a factor if $\gamma$ in either direction because the $\pm vx/c^2$ term is zero.

 

[fuentes1979]

Now I understand. I was just mixing things. Thank you very much for your very clear explanation. The main point I was missing was relativity of simultaneity.

 

[PeroK]

The main point I was missing was relativity of simultaneity.

I've waited 11 years on here to hear someone say that!

 

[fuentes1979]

What do you mean PeroK? I guess it is a quite common error isn’t it?

 

[PeroK]

What do you mean PeroK? I guess it is a quite common error isn’t it?

The error is common. The recognition of the error is rare!

 

[Ibix]

What do you mean PeroK? I guess it is a quite common error isn’t it?

Take a look at Orodruin's signature. If you can't see it, rotate your phone into landscape mode or look on a laptop.

People seem to accept time dilation and length contraction, but are often highly resistant to accepting that "now" can mean different things in different frames, even though they can often see that their understanding of relativity is contradictory without it. That you got there by post #6 is a welcome change.

 

[weirdoguy]

There should be some special badge for people who achieve that xD

 

[Orodruin]

Take a look at Orodruin's signature. If you can't see it, rotate your phone into landscape mode or look on a laptop.

And if that still doesn't do it, here is a screenshot:

There should be some special badge for people who achieve that xD

How about the Orodruin award ...

 

[Ibix]

How about the Orodruin award ...

A volcano, sinister, on a field, argent, blazoned as a Minkowski diagram?

 

[Mister T]

What do you mean PeroK? I guess it is a quite common error isn’t it?

It's not only a common error, it's a stubborn misconception. That is, when the error is pointed out or discovered, the misconception usually persists and the error keeps getting repeated. It's well known that the relativity of simultaneity is one of the hardest concepts to learn. We have this mistaken notion that there's the equivalent of a giant clock in the sky that keeps track of a universal time that is the same for everyone. When learning the relativity of simultaneity students often retain this mistaken notion, thinking that observers who experience a difference in simultaneity are experiencing a variation from the real, or true, time.

 

[pervect]

A volcano, sinister, on a field, argent, blazoned as a Minkowski diagram?

A pair of crossed telescopes in the center of the shield (google calls it the fess or heart point), Two erupting volcanoes in lower dexter and sinister positions (google finds dexter base and sinister base), and a flying spaceship at the top (google finds middle chief). Some wavy motion lines to indicate the spaceship is flying to the right. https://www.theclanbuchanan.com/glossary-of-heraldic-terms