# Rear Clock Ahead Effect (Special Relativity)

PhDeezNutz
Homework Statement:
I'm reading from "Special Relativity for the Enthusiastic Beginner" by David Morin. Specifically pages 14-16 (first chapter available for free by the publisher) ( https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf )

This is the problem statement as I understand it:

We have a ground frame and a train frame where the train has velocity ##v## relative to the ground and towards the right. There is a person standing somewhere on the train (say ##x## from the left wall and ##y## from the right wall) and the train has a length ##L##. There are two clocks (presumably synchronized in the train frame) positioned at the ends of the train. The person on the train flashes two flashlights in opposite directions and is positioned such that the two flashes of light hit the clocks simultaneously in the ground frame. When the flashes of lights hit the clocks what is the difference in readings according to ground observers?
Relevant Equations:
##d = vt##
##x + y = L##
Time when the left beam hits the left wall in the ground frame

##vt_1-x = c t_1##

##t_1 = \frac{x}{c+v}##

Time when the right beam hits the right wall in the ground frame

##ct_2 = vt_2 + y##

##t_2 = \frac{y}{c-v}##

Setting the times equal to each other with the constraint x+y=L to find x and y

##t_1 = t_2##

##\frac{x}{c+v} = \frac{L-x}{c-v}##

implies

##x = \frac{L(c+v)}{2c}## and ##y = L - x = \frac{L(c-v)}{2c}##

Finding x - y

##x - y =\frac{Lv}{c}##

So basically the person holding the flashlights in the train has to be ##\frac{Lv}{c}## further from the left wall than the right wall in order for the flashes to hit the two end clocks simultaneously in the ground frame

That is to say it takes
##\frac{Lv}{c^2}## more time in the train frame for the left wall to be hit. That means observers in the ground frame when viewing the train clocks will view the rear train clock (the left clock) to be ahead by ##\frac{Lv}{c^2}## when the light beams hit the train clocks in the ground frame.

The part I'm confused about is that Morin says

"Note that the L in the Lv/c 2 result is the length of the train in its own frame, and not the shortened length that you observe in your frame (see Section 1.3.3). Appendix B gives a number of other derivations of Eq. (1.8), although they rely on material we haven’t covered yet." on page 16.

To me this seems inconceivable that we would set things up so that the events are simultaneous in the ground frame by using the train frame length.

Can someone help to clarify why we would use the "proper Length" to make things simultaneous in a different frame?

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To me this seems inconceivable that we would set things up so that the events are simultaneous in the ground frame by using the train frame length.

Can someone help to clarify why we would use the "proper Length" to make things simultaneous in a different frame?
You used the quantity ##L## which is the proper length of the train to do the calculations in the train frame. The answer is correct because the calculations are correct. To substitiute the contracted length ##l## latter into your equation you need to know that ##L = \gamma l##. In which case, the time offset is ##\frac{\gamma lv}{c^2}##

• PhDeezNutz
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PS from an experimental point of view, you could calculate length contraction by:

1) Measuring the speed of the train ##v##.

2) Measuring the offset between clocks at the front and rear (assuming you know these are synchronised in the train frame), ##\Delta t##.

3) Deducing that the proper length is ##L = \frac{(\Delta t) c^2}{v}##.

4) Measuring the contracted length of the train, ##l##.

5) Deducing that the length contraction is ##l/L##.

6) Confirming that this is equal to ##\sqrt{1 - \frac{v^2}{c^2}}##.

• PhDeezNutz
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Time when the left beam hits the left wall in the ground frame

##vt_1-x = c t_1##

##t_1 = \frac{x}{c+v}##
In both of the equations above, ##x## should be ##x/\gamma##, since the the distance ##x## is contracted in the ground frame.

Time when the right beam hits the right wall in the ground frame

##ct_2 = vt_2 + y##

##t_2 = \frac{y}{c-v}##
Likewise, ##y## should be ##y/\gamma## in these two equations.

When you set ##t_1 = t_2##, the ##\gamma##'s cancel.

• PeroK and PhDeezNutz
PhDeezNutz
@PeroK @TSny

I guess this is where my trouble lies

We (the ground observer(s)) are reading moving clocks that are presumably synchronized in the train frame. The events happen simultaneously in the ground frame so why wouldn’t we read the same readings on both moving clocks?

doesn’t the saying go “moving clocks run slow” (with only a dependence on relative speed)? So why wouldn’t the two clocks read the same in the ground frame?

I could totally understand if the clocks were off by ##\frac{Lv}{c^2}## in the train frame but not the ground frame.

Any help is appreciated.

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Gold Member
2022 Award
@PeroK @TSny

I guess this is where my trouble lies

We (the ground observer(s)) are reading moving clocks that are presumably synchronized in the train frame. The events happen simultaneously in the ground frame so why wouldn’t we read the same readings on both moving clocks?

doesn’t the saying go “moving clocks run slow” (with only a dependence on relative speed)? So why wouldn’t the two clocks read the same in the ground frame?

I could totally understand if the clocks were off by ##\frac{Lv}{c^2}## in the train frame but not the ground frame.

Any help is appreciated.
Wherher two clocks are synchronised depends on a prior process of synchronisation.

Either the clocks are synchronised in the train frame (original assumption) and hence can't be syncronised in the ground frame. Or, if they happen to be synchronised in the ground frame, then they won't be synchronised in the train frame.

It all depends on how the experiment was set up.

PhDeezNutz
@PeroK i think I get it now. After reading Morin a little bit more closely I realize if we set the clocks in the train to break when the events happen the rear clock will be ahead by Lv/c^2 according to the train observer. Because the train clocks are broken the ground observer(s) will read the same thing.