 #1
PhDeezNutz
 508
 271
 Homework Statement:

I'm reading from "Special Relativity for the Enthusiastic Beginner" by David Morin. Specifically pages 1416 (first chapter available for free by the publisher) ( https://scholar.harvard.edu/files/davidmorin/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_1x = 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}{cv}##
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{Lx}{cv}##
implies
##x = \frac{L(c+v)}{2c}## and ##y = L  x = \frac{L(cv)}{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?
##vt_1x = 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}{cv}##
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{Lx}{cv}##
implies
##x = \frac{L(c+v)}{2c}## and ##y = L  x = \frac{L(cv)}{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?