# Special Relativity Questions

1. Nov 2, 2009

### cfrogue

1) When Einstein said,

Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train.
http://www.bartleby.com/173/9.html

Did he specifically mean M and M' were at the same point on the x-axis of the embankment observer when the lightning flashed at both A and B?

2) Does any have the calculations for the time when the lightning will strike the front and back of the train of the moving observer calculated from the frame of embankment observer.

2. Nov 2, 2009

### yuiop

Yes. Note that "when lightning flashes at both A and B" is defined from the point of view of the embankment observer.

See the formula I gave in post#4 of this thread : https://www.physicsforums.com/showthread.php?t=351140

3. Nov 2, 2009

### cfrogue

OK, I would like to be like very confident.

M and M' will be at the same point on the x-axis of the embankment observer?

4. Nov 2, 2009

### cfrogue

Sorry to bother you but I am not able to see that equation.

How do I look at it?

5. Nov 2, 2009

### cfrogue

I see it now.

6. Nov 2, 2009

### cfrogue

This has got t and t' in it.

I meant to ask what is the calculation for the arrival time of the light to the front and back of the moving observer using only information from the stationary observer and calculated from the stationary observer.

For example, at the end of chapter two, Einstein wrote

tB - tA = rAB/(c-v) and t'A - tB = rAB/(c+v) where rAB would be the contracted train length.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Are the equations like the above which includes (c-v) and (c+v)?

He said in Chapter 9,
Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.
http://www.fourmilab.ch/etexts/einstein/specrel/www/

so I am wondering from the reference to the railway embankment, would (c-v) and (c+v) describe this hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A.

Now, of course the train actually sees a constant c and so does the embankment observer in their respective frames.

7. Nov 2, 2009

### yuiop

Yes, the (c-v) applies to "riding on ahead of the beam of light coming from A" and (c+v) applies to "hastening towards the beam of light coming from B".

So let us say the rest length of the train is 200m and the train is moving at a velocity v=0.8c. The length contracted length of the train from the embankment observer's point of view (POV) is 120m. If the train observer is at the centre of the train then we are concerned with half the length of the train (60m). From the embankment POV, the flash from event A at the back of the train takes 60/(1-0.8) = 60/0.2 = 300 seconds to arrive at the centre of the train and the flash from event B at the front of the train takes 60/(1+0.8) = 60/1.8 = 33.333 seconds to arrive at the centre of the train, as timed by stationary clocks on the embankment.

8. Nov 2, 2009

### cfrogue

Perfect.

Thanks,

Now, I would like to ask another question.

Would you consider it crackpottery to deny using (c+v) and (c-v) for this specific situation, of couse, when determining the time the lightning strikes for the front and rear of the moving frame from the view of the "rest" frame?

I am having trouble with someone that claims to know special relativity that asserts Yes, the (c-v) applies to "riding on ahead of the beam of light coming from A" and (c+v) applies to "hastening towards the beam of light coming from B" is false.

And this same individual claims M and M' are not coincident for the train experiment as Einstein asserted.

Thank you Kev.

9. Nov 2, 2009

### cfrogue

I have another question also.

With the Lorentzian Transformation construction, does there is exist dependencies on the fundamental component of clock synchronization of a frame?

Further, does Einstein does depend on points of a frame meeting points of another frame with this LT construction?

In particular Einstein said just at the begining of his LT derivation chapter 3,

To any time of the stationary system K there then will correspond a definite position of the axes of the moving system

http://www.fourmilab.ch/etexts/einstein/specrel/www/

10. Nov 4, 2009

### yuiop

Have a look at this link: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2

I will use the notation used in the hyperphysics link, as Einstein seems to use notation that is no longer generally used. For example he uses $\beta$ to refer to $1/\sqrt{(1-v^2/c^2)}$ whereas $\beta$ is more commonly used to refer to v/c these days and $\gamma$ more commonly refers to $1/\sqrt{(1-v^2/c^2)}$. In keeping with the commonly used notation, I will call the stationary frame k and the moving frame k'.

Now let us say that all clocks in k have been synchronised with each other in k and likewise all clocks in k' have been synchronised with each other. If we consider ourselves to be frame k the first thing we would notice is that the clocks in k' do not appear to synchronised from our point of view and some are ahead of our clocks and some are behind. Now hyperphysics states that the reference frames coincide at t=t'=0. The first problem is that because all the clocks in k' appear to be showing different times to us, we would be wise to pick one that shows the same time as the clocks in our frame and call that time zero in both frames. The two clocks that are alongside each other momentarily in the two frames and showing the same time give a convenient location for the origins of the two frames so that x.y.z.t = x',y',z',t' = 0. Now if we uses the transformation equation given in the link , e.g:

$$t = \gamma \left(t' + \frac{vx'}{c^2}\right)$$

it can be seen that there is a dependency or assumption of zeroing the two clocks when the origins of the two frames coincide. However, if we are only concerned with differences between two events using the equations shown here: http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html#c2 e.g:

$$(t_2 - t_1) = \gamma\left(t_2 ' + \frac{vx_2 '}{c^2}- t_1 ' - \frac{vx_1 '}{c^2}}\right)$$

then there is no requirement to have the two clocks at the origins of the two frames showing the same time when the origins coincide.

As for the statement "To any time of the stationary system K there then will correspond a definite position of the axes of the moving system" it is self evident that if the moving frame is moving at a constant velocity there will be a correlation between the relative position of the origins of the two frames and the time in the stationary system. Whether or not there is an offset in the times of the origin clocks when the origins of the spatial axes of the two frames coincide, there will still a "definite position" for the origin of the moving frame for any given time in the stationary system. At any given time t in the stationary frame the origin of the moving frame will be at v*t+t_offset. Obviously choosing an t_offset of zero is more convenient, but not an absolute prerequisite.

11. Nov 4, 2009

### cfrogue

You are very clear with your answer and this makes perfect sense.

12. Nov 4, 2009

### cfrogue

OK, I have another question.

Please consider a different type of experiment like the one for LT.

Let L be the length of a train when at rest with observer M and let M' be at the center of this train.

Now, let M' be in relative motion v along the positive x-axis.

It just so happens when M and M' are coincident along the x-axis, a light flashes from M toward the back of the train of M'.

Now, it seems the length of the moving train is L/λ for length contraction.

I would like to figure out when, in the proper time of M, the light will stike the back of the train of M'.

Since the train is L/λ from the frame of M, then the back of the train will be a distance L/(2λ) from M' as far as M is concerned.

So, for light to meet the back of the train, while light travels ct, it travels the distance L/(2λ) minus the relative motion of the train since the back of the train, from the perspective of M, is hastening toward the light as Einstein put it given a constant speed of light.

So,

ct = L/(2λ) - vt.

ct + vt = L/(2λ)

t(c + v) = L/(2λ)

t(c + v) = L/(2λ)

t = L/(2λ(c + v))

Please correct any errors from the above. It seems to be what Einstein said in section 2 where he used (c+v) and (c-v) from the perspective of the at rest frame M.

13. Nov 4, 2009

### yuiop

Looks good to me :)

Likewise a light signal going forwards from the centre of the train to the front of the train will appear to take t = L/(2λ(c - v)) from the perspective of the rest frame of M and of course from the point of view of riders on the front and back of the train, the forward and backward light signals from M' will appear to arrive at the ends of the train at the same time.

14. Nov 4, 2009

### cfrogue

Thank you very much, you are excellent!