# Quantifying relativity of simultaneity

1. Jul 24, 2010

### dr.fluis

I have recently attempted quantifying relativity of simultaneity and I was wondering if my attempt has been successful.

Watch the below link before reading, as the calculations are based on the same type of event.

As the video has no parameters, I have made my own.

The velocity of the train: (0.75)^0.5 C or
Length of train: 2 light seconds in the man’s frame of reference.
lighting strike 1: the lightning strike that hits the leading end of the train.
lighting strike 2: the lightning strike that hits the trailing end of the train.
lighting strike occurrence in man’s frame of reference: time on clock 0s

Outline of process:
(a) Determine when light from lighting strike 1 and lighting strike 2 collides with the woman in the man’s frame of reference.
(b) Determine the duration between these collisions and convert from man’s frame to woman's frame using the Lorentz time transformation.

This process should result in the duration between lightning 1 and lighting 2 occurring in the woman’s frame of reference as she is equidistant from the occurrences.

Part A
Collision 1 = 1 light-second / ( speed of light + velocity of train)
= 1 CS/ (1 C + (0.75)^0.5 C)
=0.536 seconds /3dp/
Collision 2 = 1 light-second / ( speed of light - velocity of train)
= 1 CS/ (1 C - (0.75)^0.5 C)
= 7.464 seconds /3dp/

Part B
Collision 2 - Collision 1 = (1 CS/ (1 C - (0.75)^0.5 C)) - 1 CS/ (1 C + (0.75)^0.5 C)
= 6.928 seconds /3dp/
Conversion to woman’s frame of reference = (Collision 2 - Collision 1) x time transformation
= 6.928s x (1 - 0.75)^0.5
= 3.464 seconds /3dp/

In the woman’s frame of reference lighting strike 1 hits the train 3.464 seconds before lighting strike 2

Is this correct?

Last edited by a moderator: Sep 25, 2014
2. Jul 24, 2010

### starthaus

The correct answer is $$\gamma(v)\frac{v \Delta x}{c^2}$$

In your case, this amounts to $$4\sqrt{0.75}=4*0.854=3.416$$

Your symbolic calcuulations are wrong because you are using a strange mix of Galilean and SR kinematics instead of using only relativistic kinematics.. For example, you use (correctly) closing speeds at point A , yet you follow with a strange time contraction at point B.

Last edited by a moderator: Sep 25, 2014
3. Jul 25, 2010

### dr.fluis

So the collisions do not occur in the man's frame of reference as calculated?

4. Jul 26, 2010

### dr.fluis

Is the difference in collisions for the woman's frame, compared to the man's frame a result of time dilation?

5. Jul 26, 2010

### starthaus

No, it is due to what is called "Relativity of simultaneity"

6. Jul 26, 2010

### yuiop

If the velocity of the train in 0.75^(0.5)c in the man's frame then the proper length of the train is 2/sqrt(1-0.75) = 4

A standard formula for the difference in proper times of two clocks at rest with respect to each other and separated by proper distance L (simultaneity of relativity) is Lv/c^2 which is essentially the equation that is quoted by Starthaus where L = $\gamma \Delta x$.

This standard formula gives the difference in proper time between the two events in the woman's frame as Lv/c^2 = 4*0.75^(0.5) = 3.46410162

This implies your method is correct.

In part A you calculate the difference in arrival times of the two flashes at the woman's location as 7.4641-0.5358= 6.9282 as measured in the man's frame.

In part B you use the simple time dilation factor to convert the time interval in the man's frame to the woman's frame using 6.9282*sqrt(1-.75) = 3.4641.

Your method is perfectly valid because the two events (arrival of flash 1 and arrival of flash 2) happen at the same location in the woman's frame.

The reason Starthaus gets a different result is that he made an error in the arithmetic in his execution of the standard formula.

$$4\sqrt{0.75} \ne 4*0.854$$

He then compounds his error by using faulty logic to explain why his answer differs from yours.

7. Jul 26, 2010

### starthaus

This is what happens when you do your physics through "cut and paste", kev. You start writng nonsense like the suff above.
The correct formula is, of course derived very simply from:

$$t'=\gamma(t-\frac{vx}{c^2})$$

so:

$$dt'=\gamma(dt-\frac{vdx}{c^2})$$

For $$dt=0$$

$$dt'=-\gamma\frac{vdx}{c^2}$$

So, the formula I gave is correct. I am quite sure I showed you this derivation before, you should try to remember it next time.

lol,$4\sqrt{0.75}$ is the correct result.

It is good that you used your caculator to find out that $\sqrt{0.75}=.866$, my 2\$ Chinese watch calculator came up with $.854$ , it feels good to see that you are redoing all my calculations :-)

Last edited: Jul 26, 2010
8. Jul 26, 2010

### dr.fluis

Ok so i did get it. Awsome : )

9. Jul 27, 2010

### dr.fluis

If I change the parameter of distance from 2 light seconds to 20 light seconds, is there a problem with the result I get?

The velocity of the train: (0.75)^0.5 C or
Length of train: 20 light seconds in the man’s frame of reference.
lighting strike 1: the lightning strike that hits the leading end of the train.
lighting strike 2: the lightning strike that hits the trailing end of the train.
lighting strike occurrence in man’s frame of reference: time on clock 0s

Outline of process:
(a) Determine when light from lighting strike 1 and lighting strike 2 collides with the woman in the man’s frame of reference.
(b) Determine the duration between these collisions and convert from man’s frame to woman's frame using the Lorentz time transformation.

This process should result in the duration between lightning 1 and lighting 2 occurring in the woman’s frame of reference as she is equidistant from the occurrences.

Part A
Collision 1 = 1 light-second / ( speed of light + velocity of train)
= 10 CS/ (1 C + (0.75)^0.5 C)
=5.36 seconds /2dp/
Collision 2 = 1 light-second / ( speed of light - velocity of train)
= 10 CS/ (1 C - (0.75)^0.5 C)
= 74.64 seconds /2dp/

Part B
Collision 2 - Collision 1 = (10 CS/ (1 C - (0.75)^0.5 C)) - (10 CS/ (1 C + (0.75)^0.5 C))
= 69.28 seconds /2dp/
Conversion to woman’s frame of reference = (Collision 2 - Collision 1) x time transformation
= 69.28s x (1 - 0.75)^0.5
= 34.64 seconds /2dp/

In the woman’s frame of reference lighting strike 1 hits the train 34.64 seconds before lighting strike 2

10. Jul 9, 2012

### Devoney

Sorry to bring up this old topic. But I do not get it (this example).

Could someone please verify or explain otherwise the following to me:

1. Speed of light travels at constant speed (C) relative to observer.
2. If you move at 0.5 C and you flash a light, the speed of light will still move away from you with the speed of 1C.
3. If you move passed a lantern with 0.5C, and it flashes the moment you pass by, the light of that lantern will still move towards you at 1C relative to you, independent of what your own speed is.

To get back at the train example:
4. The flashes occur at the same time, for the women and the man.
5. The women sees both flashes at the same time, because the speed of light travels towards the womens eyes at the same speed. Unlike in the video.

The way I see it the 'difference caused by relativity' is best described as:
6. In the reference frame of the man, the women sees the front flash first. While in the reference frame of the women she sees both flashes happen at the same time.

Can anyone help me out?

11. Jul 9, 2012

### ghwellsjr

I think your problem is that you are actually creating two different scenarios in your mind based on two different reference frames. Whenever you are dealing with relativity scenarios, you need to stick with one frame of reference in which the speed of light is c and describe everything that is happening within the context of that one frame. So if you pick the man's frame and say that the flashes strike the front and rear of the train at the same time (which is not the same thing as saying that he sees them strike at the same time), then the two flashes will travel towards him at the same speed and if he is equidistant from where the two flashes occurred, then later on he will detect them at the same time. Meanwhile, since the train is moving with respect to the man, the woman will detect the flash from the front of the train before she detects the light from the rear. But of course, she will not be at the same place as the man when they both detect the flashes.

On the other hand, if you create a different scenario based on the frame of the woman and say that the two flashes hit the front and rear of the train at the same time (again, nobody sees that happen at this point in time), then the light will travel at c toward her and she will detect them together at a later time. Meanwhile, in the woman's frame, the man on the ground is traveling backwards from her and therefore he will detect the rear flash first and the front flash second.

You could create a third scenario and say that they both detect the two flashes at the same time that they pass each other and then ask the question of when did the flashes occur. Now you have an ambiguous situation because you have not selected a frame of reference in which the speed of light is c. If you choose the woman's frame of reference then you will say that the flashes struck the front and rear of the train simultaneously. But if you choose the man's frame of reference then you will say that they did not strike the front and rear of the train simultaneously.

Does this clarify all your issues?

12. Jul 9, 2012

### Devoney

Hi ghwellsjr,

Yes it totally does. With your explaination I understand that I was thinking correctly about the situation but I neglected to pick 1 reference frame.
It did not know it was a 'convention' to pick just one reference frame.

Thanks for the clarification and the time you took to answer me!

13. Jul 10, 2012

### ghwellsjr

You're welcome and I really appreciate your feedback. Thanks.

14. Jul 12, 2012

### harrylin

It's not just a convention but a mathematical necessity - just like working in meters or inches. You can choose a reference and work with that, and you will get consistent results. However mixing them (without properly transforming from one to the other) leads to erroneous results.