Relativity train and clock synchronization

[BHollis]

I'm trying to understand the relativity train thought experiment. One of the key concepts seems to be that the clocks at the front and back of the train are "synchronized" such that the clock at the front of the train is running slightly behind the clock at the back of the train (as seen by an observer outside the train). As I understand it, this is necessary in order for a beam of light shined toward the front of the train, and a beam of light simultaneously shined toward the back of the train to arrive at their respective destinations simultaneously (as seen by an observer on the train).

I've also read that if the two clocks are synchronized together at the middle of the train, the clock that is then taken to the front of the train will lose some time (due to its having run slower as it traveled at the combined speed of the train plus that of moving the clock forward) while the clock at the back of the train will gain some time (due to its having run faster as it traveled at the reduced speed of the train minus that of moving the clock rearward)--which results in the forward clock running slightly behind the after clock.

So far, so good.

But what if a different method is used to synchronize the two clocks, say a mechanical rod extending forward and backwards to both clocks, which is used to start both clocks running at the same time. If the train is stopped, when the two clocks are brought together at the middle of the train, they'll show the same time. But if the train is moving at near the speed of light, wouldn't they show different times? And wouldn't this violate one of the postulates of special relativity?

I'm sure I'm missing something, and I'd appreciate your enlightening me.

 

[bcrowell]

Welcome to Physics Forums!

But what if a different method is used to synchronize the two clocks, say a mechanical rod extending forward and backwards to both clocks, which is used to start both clocks running at the same time.

A rigid rod can't be used to transmit signals instantaneously from one place to another. If you apply a force to one end, the effect propagates along the rod at the speed of sound in the material the rod is made of, which is less than the speed of light.

 

[BHollis]

Welcome to Physics Forums!



A rigid rod can't be used to transmit signals instantaneously from one place to another. If you apply a force to one end, the effect propagates along the rod at the speed of sound in the material the rod is made of, which is less than the speed of light.

Thanks for replying.

But even so, wouldn't the effect be the same at both the front and back of the train--with the net result that the clocks would be tripped "simultaneously," and thereby "synchronized?"

 

[I_am_learning]

I don't think synchronization of the two clocks (which are in same inertial frame) of the train is such a difficult task. I present some methods.

Since the train is moving at constant speed, we (who are inside train) can in every respect assume that everything is stationary. (This isn't necessary but helps to avoid confusion)

1. Place two clocks at respective positions then mark out the mid-point. Flash out a beam of light from there in both direction. Pre-arrange the clock to be started at 0, when a light beam shines off.

2. Press a button, that both starts first clock at 0 and throws a light beam from the fist clock to the second. Pre-arrange the second clock to start at --> distance/c <-- time when beam of light reaches.

Now after the clock is synchronized, To do the experiment: shine off a beam of light from midpoint, just as in 1. But this time prearrange the clock to be paused when the light beam reaches them.
Now you can take your time, bring the clock together and compare there time.
Happy experimenting. :)

 

[Janus]

Thanks for replying.

But even so, wouldn't the effect be the same at both the front and back of the train--with the net result that the clocks would be tripped "simultaneously," and thereby "synchronized?"

In the frame of the train, yes. But not according to our outside observer.

You have to take into account the addition of velocities in Relativity:

$$w =\frac{v+u}{1+\frac{uv}{c^2}}$$

If we say that the velocity of the train is v, and the speed of sound through the rod is u, then the outside observer will see the sound travel with respect to himself at

$$w =\frac{v+u}{1+\frac{uv}{c^2}}$$

in one half of the rod and

$$w =\frac{v-u}{1-\frac{uv}{c^2}}$$

in the other.

If v is .5c and u is .1c, this gives results of

0.571c

And

0.421c


or 0.071c and 0.79c with respect to the train. One impulse will reach its clock and start it before the other.

 

[BHollis]

In the frame of the train, yes. But not according to our outside observer.

You have to take into account the addition of velocities in Relativity:

$$w =\frac{v+u}{1+\frac{uv}{c^2}}$$

If we say that the velocity of the train is v, and the speed of sound through the rod is u, then the outside observer will see the sound travel with respect to himself at

$$w =\frac{v+u}{1+\frac{uv}{c^2}}$$

in one half of the rod and

$$w =\frac{v-u}{1-\frac{uv}{c^2}}$$

in the other.

If v is .5c and u is .1c, this gives results of

0.571c

And

0.421c


or 0.071c and 0.79c with respect to the train. One impulse will reach its clock and start it before the other.

Thanks very much for the response. I'm going to have to think about that.