# Confused about relativity and a train

LCSphysicist
Homework Statement:
Two clocks are positioned at the ends of a train
of length L (as measured in its own frame). They are synchronized in the train frame.
The train travels past you at speed v. It turns out that if you observe the clocks at
simultaneous times in your frame, you will see the rear clock showing a higher reading
than the front clock (see Fig. 11.6). By how much?
Relevant Equations:
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We can find the difference of time to light reach both clocks, it should be Lv/c², what i am confused is why the rear clock show a higher reading of the front clock.
Ok, the light take longer time to travel and reach the rear clock, and here is the thing!
What i am interpreting is: When the light reach the clock, it "starts to" run, to read the same hours as a clock localized on the source emissor of light.
So, for example: The light reach the front clock, so it start: 0s, 1s, 2s. And so the light reach the rear, and just now it starts: 0s...
That is, the front clock has a higher reading than the rear, but why is this wrong?

Gold Member
Say LED at the center of the train emit light pulses every second. There are counters at front and rear ends. The both counts increase simultaneously in the train FR. However, in the ground FR, light reaches earlier to the rear end because the rear end intercept the pulses. The front end escapes.

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LCSphysicist
LCSphysicist
Say LED at the center of the train emit light pulse every second. The pulses are counted at front and rear ends. The counts are simultaneous in the train FR. However light reaches earlier to the rear end because rear end encounter the pulses in the ground FR. This is the reason of higher reading.

They need to be simultaneous in our frame, no?. So it need to be
$$df+dr=L$$
$$\frac {df}{c-v}=\frac {dr}{c+v} (*)$$
$$df+df\frac{c+v}{c-v}=L$$
$$2cdf=L(c-v)$$
$$df=\frac{L(c-v)}{2c}$$ (Taking the result from RossMilikan, which was equal to my solution anyway...)

I put (*), it is this equation which implies simultaneous in our frame.
So in the train frame, the light will end taking more time to reach the rear end

This is confused

Homework Helper
Gold Member
2022 Award
They need to be simultaneous in our frame, no?. So it need to be
$$df+dr=L$$
$$\frac {df}{c-v}=\frac {dr}{c+v} (*)$$
$$df+df\frac{c+v}{c-v}=L$$
$$2cdf=L(c-v)$$
$$df=\frac{L(c-v)}{2c}$$ (Taking the result from RossMilikan, which was equal to my solution anyway...)

I put (*), it is this equation which implies simultaneous in our frame.
So in the train frame, the light will end taking more time to reach the rear end

This is confused
It's elementary, surely, that light emitted from the centre of the train will:

a) Reach each end of the train at the same time in the train frame;

b) Reach the rear of the train first in the ground frame.

LCSphysicist
You could also note that the clocks, when observed in the lab frame at time ##t##, will display times ##t' = \gamma(t-\frac{xv}{c^2})## which will differ by ##\Delta t' = -\frac{Lv}{c^2}## where ##L = \gamma \Delta x = \Delta x'##

LCSphysicist
LCSphysicist
I believe i am confusing simultaneously in ground reference and synchronized in train frame. I will take more time to think bout it.

Homework Helper
Gold Member
2022 Award
I believe i am confusing simultaneously in ground reference and synchronized in train frame. I will take more time to think bout it.
Okay, but if the light pulse is used to start the clocks running then:

a) They start at the same time in the train frame;

b) The rear clock starts first in the ground frame.

Staff Emeritus