# Train question.

1. Apr 30, 2007

### Santural

Simultaneity

Let's imagine there is a train moving at the speed of light. When in the middle, two lightning flashes appear on it's left and right. Why is it that it sees the right flash before the left?
And if it is moving left to right, why does it see the left before the right?

It is to do with the train's motion? It's position relative to the flash? If so, how?

P.S: This is not a homework question, just a general question.

Last edited: May 1, 2007
2. Apr 30, 2007

### bernhard.rothenstein

train at the speed of light?

We should have a hard imagination in order to accept that a train could move at the speed of light

3. May 1, 2007

### pervect

Staff Emeritus
As has been mentioned once or twice or even more times before, physical objects can't move at the speed of light.

Attempting to sneak the idea that they can in through an assumption in a question leads to confusion and long pointless threads. Technically, it's an example of a fallacy, known as the "loaded question", see for instance

http://en.wikipedia.org/wiki/Fallacy_of_many_questions

What may be confusing to the Original Poster is why objects can't move at the speed of light, or even faster. This is a much more productive issue to discsus, my favorite explanation here involves the way velocities add. But I won't go into much more detail unless someone indicats an interest, and can't find it in one of the other threads where it has been talked about.

Last edited: May 1, 2007
4. May 1, 2007

### Santural

Well, in that case, let's imagine the train is moving at 0.8886c.

5. May 1, 2007

### neutrino

What do you mean by "when in the middle"? Who is the observer here and how is that person moving relative to the train?

6. May 1, 2007

### Santural

Slightly changed it for better understanding.
A rocket ship traveling left to right and near the speed of light relative to the earth observes two lights flashing on earth (simultaneously). If the rocket observer is directly between them when they flash, the observer sees the one on the right flash before the one on the left.
If the rocket ship in the above question is moving right to left the one on the left flashes before the one on the right.

Why?

7. May 1, 2007

### matheinste

If the observer sees the flashes as simultaneous how can he see one before the other.

Matheinste.

8. May 1, 2007

### JesseM

If the flashes were simultaneous in the Earth's frame, they're not simultaneous in the rocket's frame--for example, if the rocket is moving to the right in the Earth's frame, in the rocket's frame the light on the right side flashed at an earlier time than the light on the left side. If the rocket has clocks at the front and back which are synchronized in its own frame using the Einstein synchronization convention (for example, the rocket-observer could have synchronized them by setting off a flash at the center of the rocket, and making sure both clocks read the same time when the light from the flash reaches each one, since light is assumed to travel at c in both directions in the rocket's frame and should therefore take the same time to reach both clocks), then in the Earth-frame these two clocks will be measured to be out-of-sync (in the Earth's frame, if a flash is set off at the center of the rocket and the rocket is moving right, naturally the light will catch up to the left clock before it catches up to the right clock, under the assumption that light travels at c in both directions in the Earth's frame).

9. May 1, 2007

### JesseM

If I'm understanding the scenario, "simultaneously" referred to when the flashes happened in the Earth's frame, "sees" referred to when the light from each flash actually reached the rocket-observer's eyes.

10. May 1, 2007

### pervect

Staff Emeritus
A much better question! I think it has been answered fairly well by other posters.

11. May 1, 2007

### pervect

Staff Emeritus
Right, that's what I would have assumed.

Simultaneous events are events that are assigned the same time coordinate, in some particular coordinate system.

Seen is ambiguous, but usually means the time assigned to the arrival of light.

The important point, which can't be stressed enough, is that the set of events regarded as simultaneous is one coordinate system is not simultaneous in another.

This can be drawn on a space-time diagram, for instance the dotted and dashed lines in the diagram below represent the set of events regarded as simultaneous by two different observers. Visual inspection shows that they are not the same set of points.

#### Attached Files:

• ###### Relativity_of_simultaneity.png
File size:
527 bytes
Views:
96
12. May 1, 2007

### Santural

Thanks! I understand, but let me clarify in case of any errors:
-Simultaneity is true only to a certain rest frame.
-In any other rest frame, observers would not experience similar simultaneity as the other frame.

Thanks! And my mistake, the Speed of Light shall be for light alone.

P.S. JesseM, you got the essence of the question, I should have posted it better.

13. May 1, 2007

### JesseM

Yup, I think you've got it. To give you a formula, if there are two clocks which are at rest with respect to each other and a distance of x apart in their own frame, and the clocks are synchronized in their own frame, then if in my frame the two clocks are moving at speed v along the axis between them, I will measure the back clock's time to be vx/c^2 ahead of the front clock's time.

14. May 1, 2007

### Santural

Gotcha! Thanks for the formula!

15. May 2, 2007

### paw

JesseM, I'm not sure I follow how the position of your co-moving clocks affects the time measured between them in your frame.

Imagine two spaceships (a and b) travelling on a parallel course away from Earth at 0.5c measured in the Earth centered frame. The distance between ships is negligable and perpendicular to the direction of travel. Synchronized clocks are located on the Earth, ship a and ship b.

In the frame where ship a is at rest, Earth is seen to be moving away at 0.5c and a clock on earth would be seen to be running 25% slow. Ship b is at rest in ship a's frame so it too should observe Earth clock running 25% slow.

Additionally, since ship b is at rest in a's frame it's clock should be observed to be syncronous with ship a's clock. Do we agree so far?

In the Earth centered frame ship a's clock is observed to be running 25% slow. So is ship b's clock. The same argument holds true even if the ships are moving along the axis between then doesn't it? I mean, I don't see any 'x' in the Lorentz transform.

So where does the position (x) come in? I can't see how position can matter as long as ship a and ship b are at rest wrt each other. Am I missing something?

16. May 2, 2007

### neutrino

No, it does not. If the clocks are separated by a distance x (in the clocks' rest frame) along their direction of motion, then the leading clock lags behind trailing clock by an amount xv/c2, where v is their relative speed with respect to the observer.

Last edited: May 2, 2007
17. May 2, 2007

### Johnny R

Because the speed of light is c and not instantaneous. The same reason that if a person in front of you is running from left to right between two baseball players and they both throw a baseball at the same speed towards the person, the ball thrown from the right will nail the person in the head first.

18. May 2, 2007

### JesseM

I think you're confusing "synchronized" with "ticking at the same rate". The Earth does indeed observe each clock to be running at the same rate, but that's not what I'm talking about. Two clocks can be running at the same rate but out-of-sync--for example, one clock can be consistently 1 hour behind the other, so that when the first clock reads 2 the second reads 1, when the first clock reads 3 the second reads 2, and so forth. The clocks of the two ships a and b will be out-of-sync in this sense, assuming they synchronized their clocks in their own frame using the Einstein synchronization procedure. The reason is that the synchronization procedure assumes the speed of light is the same in all directions in the ship's rest frame, so that they could synchronize their clocks by setting off a flash at the midpoint of the line between them, and making sure both clocks read the same time when the light from the flash reaches them. Since the Earth assumes that light moves the same speed in both directions in its frame, this means that in the Earth's frame the light must reach one clock before the other, since in the Earth's frame one ship is moving towards the point the flash happened and one is moving away from that point. This means that if the ships set their clocks to read the same time at the moment the light from the flash hits them, in the Earth's frame one clock's time will naturally be ahead of the other.
Are you sure you're not thinking of the time dilation equation? The Lorentz transform definitely has an x coordinate in it:

$$x' = \gamma (x - vt)$$
$$y' = y$$
$$z' = z$$
$$t' = \gamma (t - vx/c^2)$$
where $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$

Last edited: May 2, 2007
19. May 2, 2007

### JesseM

That's not a good analogy, because in the frame of the runner, the two balls have different speeds. And if you forget about the relativity of simultaneity, and falsely imagine that the two flashes also happened at the same time in the train's frame, then since both light signals should have a velocity of c in the train's frame, you should predict both light signals reach the center at the same moment.

20. May 2, 2007

### paw

Ok, I think I see what you're saying. In the Earth (observer) frame, both clocks are moving, so light takes a little less time travelling from a to b than it does from b to a (assuming a is the lead clock). So....

In the Earth (observer) frame, if they are perpendicular to the direction of travel the ships clocks would agree?

At any angle in between they'd disagree by some factor between 0 and xv/c2 given by the angle?

This effect must disappear in the limit as x approaches zero, correct?

Now I thought time dilation was completely reciprocal. Since ship a and ship b are at rest wrt each other there is no time dilation effect between them. So how is this reconciled with the fact that an Earth observer does measure a's clock as lagging behind b's?