One way speed of light and simultaneity

[mananvpanchal]

Hello,

There are two observers. A is stationary and B is on train. Train is moving.
B is at middle of train. If B flashes a beam to front and back, and at the both end there are detectors, then what does B measure one way speed of light in both direction?

Thanks.

 

[Dale]

Assuming that the detectors are synchronized using Einstein's convention then they measure c, by definition.

 

[mananvpanchal]

Ok, We assume left end of train as A, right end is B and middle is M. Observer is at M.
When train is at rest. O flashes two pulse in direction of A and B to synchronize both clocks with O's clock. clock of A and B has some little difference with respect to O's clock because of time taken by light to reach at ends.

Suppose, that O flashes two pulse to both clocks at $t_{o}=0$.
pulse reaches to both clock and set $t_{a}=0$ and $t_{b}=0$, but now O's clock displays $t_{o}=1$. (We can say that the values of t is in time-unit, and train's length is 2 light-time-unit).

Now, train is moving to the right with the speed of 0.5 light-time-unit/time-unit and suppose O flashes two pulse in both direction at $t_{o}=5$. We know that reading of both end clock should be $t_{a}=4$ and $t_{b}=4$ now.

When pulse reach to the clocks, is both clocks reading different or same?
If different that what would be values of $t_{a}$ and $t_{b}$?

 

[Doc Al]

Ok, We assume left end of train as A, right end is B and middle is M. Observer is at M.
When train is at rest. O flashes two pulse in direction of A and B to synchronize both clocks with O's clock. clock of A and B has some little difference with respect to O's clock because of time taken by light to reach at ends.

O is at M?

Suppose, that O flashes two pulse to both clocks at $t_{o}=0$.
pulse reaches to both clock and set $t_{a}=0$ and $t_{b}=0$, but now O's clock displays $t_{o}=1$. (We can say that the values of t is in time-unit, and train's length is 2 light-time-unit).

It would have made more sense to set the clocks equal to 1 when the pulses arrived, that way the three clocks would be synchronized in the train frame.

Now, train is moving to the right with the speed of 0.5 light-time-unit/time-unit and suppose O flashes two pulse in both direction at $t_{o}=5$. We know that reading of both end clock should be $t_{a}=4$ and $t_{b}=4$ now.

"Now" in the train frame.

The question is what would be the reading of the two clocks when pulse reach to those clocks?

I'd say they'd read 5.

Is both clock reading different or same?

The same. Why would you think otherwise?

 

[mananvpanchal]

O is at M?

yes.

It would have made more sense to set the clocks equal to 1 when the pulses arrived, that way the three clocks would be synchronized in the train frame.

yes, you are right. we know the length of train, so can actually.

"Now" in the train frame.

Yes, we are talking about train frame.

The same. Why would you think otherwise?

My problem is different, I am going to elaborate it.

Two event occurs simultaneously, but train observer have no way to prove it. So he has to conclude that the event was not simultaneous because speed of light is always c.

But, if in moving train both light pulse take same time to reach clocks, then why simultaneous events occurred in rest seems one by one in moving frame?

 

[Doc Al]

Two event occurs simultaneously, but train observer have no way to prove it. So he has to conclude that the event was not simultaneous because speed of light is always c.

What two events? Simultaneous according to whom? On what basis does the train observer draw his conclusion?

But, if in moving train both light pulse take same time to reach clocks, then why simultaneous events occurred in rest seems one by one in moving frame?

Simultaneity is frame dependent. In the train frame, the pulses (in your original example) reach the ends of the train at the same time. But to an observer watching the moving train go by, those pulses reach the ends of the train at different times. He also sees the train clocks as being out of synch.

 

[mananvpanchal]

Suppose, train is moving, if lightning happens at the two ends simultaneously with respect to rest frame what is the clocks' reading in train frame? same of different?

 

[Doc Al]

Suppose, train is moving, if lightning happens at the two ends simultaneously with respect to rest frame what is the clocks' reading in train frame? same of different?

Assuming the train clocks are synchronized, the two end clocks will read different times when the lightning strikes them.

 

[mananvpanchal]

Assuming the train clocks are synchronized, the two end clocks will read different times when the lightning strikes them.

Sorry, but I cannot understand this. Please, can you give me some more detail?

 

[Doc Al]

Sorry, but I cannot understand this. Please, can you give me some more detail?

What don't you understand? Are you familiar with the relativity of simultaneity?

According to the ground frame, the train clocks are not synchronized. (The clock in the rear of the train is ahead of the clock in the front of the train.) So if the lightning strikes the ends of the train at the same time according to the ground frame, the train clocks will show different times.

 

[mananvpanchal]

What don't you understand? Are you familiar with the relativity of simultaneity?

I have read it, but I cannot understand it.

(The clock in the rear of the train is ahead of the clock in the front of the train.)

Is this situation changes when train coming to rest observer, going far from rest observer and a moment when it is on the middle of rest observer?

I understand that clocks on the train is not synchronized with rest clock. But the two clocks have same dilation effect with respect to rest clock.
Two train's clock is not synchronized with each other with respect to rest observer is strange to me.
Can you please explain me why does it happen?

 

[Dale]

O flashes two pulse in direction of A and B to synchronize both clocks with O's clock. clock of A and B has some little difference with respect to O's clock because of time taken by light to reach at ends.

Suppose, that O flashes two pulse to both clocks at $t_{o}=0$.
pulse reaches to both clock and set $t_{a}=0$ and $t_{b}=0$, but now O's clock displays $t_{o}=1$.

This is not the Einstein synchronization convention. Using this convention the speed of light outwards would be infinite and the speed of light inwards would be c/2.

Einstein synchronization is described in section 1 of this paper:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Doc Al]

Is this situation changes when train coming to rest observer, going far from rest observer and a moment when it is on the middle of rest observer?

No, the time offset of the two train clocks according to the ground observer does not depend on where the train is in its motion.

I understand that clocks on the train is not synchronized with rest clock. But the two clocks have same dilation effect with respect to rest clock.

That's true: All train clocks display the same dilation according to ground observers.

Two train's clock is not synchronized with each other with respect to rest observer is strange to me.
Can you please explain me why does it happen?

Try reading this explanation: Special Relativity: Synchronizing Clocks

 

[mananvpanchal]

@Doc Al

If train observer synchronize the clocks when train at rest, both observer agree that clocks are synchronized.
If we accelerate the train to some speed and then make speed constant. Now, both frame are inertial.
Train observer doesn't synchronize the clock in the moving frame.
Is both clock still not synchronized with each other with respect to rest observer?

 

[mananvpanchal]

This is not the Einstein synchronization convention. Using this convention the speed of light outwards would be infinite and the speed of light inwards would be c/2.

Einstein synchronization is described in section 1 of this paper:
http://www.fourmilab.ch/etexts/einstein/specrel/www/

Yes, you are right, that is why I followed Doc Al and correct myself by making all three clocks value 0.