Train clocks in special relativity

Let's say there is a train passing in front of me with a velocity $v=3c/5$ and proper length $L_0$ and let's say there are two clocks on the train: one in the front, and one in the back. When the back clock passes me, our clocks both indicate 0 and I want to find out then what will the clock in the front indicate in my frame of reference.

I used the Lorentz transformation $$t'=\gamma \left(t-\frac{vx}{c^2} \right)$$ where I have calculated $x=4L_0/5$ and $\gamma=5/4$ and where $t=0$ and $v=3c/5$. According to these values, I get $$t'=-\frac{3L_0}{5c}.$$ The problem I have with this is how can the clock in the front of the train be early on the clock in the back. Isn't it supposed to be the opposite?

Here's a spacetime diagram on rotated graph paper
which may help visualize the result you obtained and help develop a strategy for getting the result from time-dilation and length contraction.

The train moves at (3/5)c, so that $\gamma=5/4$.
The rear of the train has the GREEN worldline.
The front of the train has the BLUE worldline.
The proper length $L_0$ of the train is given by OY=10, where OY is simultaneous in the train-frame.

Your t' is $-\frac{3}{5}L_0=-6$, which corresponds to the event X that is 6-ticks before event Y on the front worldline.

To use length contraction,
you will use triangle OYX (or something similar to it), where (OX)=(OY)/$\gamma$, which is 8=10/(5/4).

To use time-dilation.
you will use triangle XYZ (or something similar to it), where (ZY)=$\gamma$(XY), which is 7.5=(5/4)6.

Geometrically, the gamma-factor $\gamma$ is the ratio between the adjacent leg and the hypotenuse [which is $\cosh\theta$ in terms of the rapidity $\theta$].

(Following Michael Fowler's story in the URL you posted, (http://galileoandeinstein.physics.virginia.edu/lectures/synchronizing.html )...
The pink worldline is the center of the train.
If it emits light signals,
the rear-worldline receives the signal 2.5 ticks later in your frame, and
ZY=7.5 is equal to the difference of those two times.) Hint: What we want to know is $t$, not $t'$. Because we measure the time of the clock in front of train in our frame. In addition, the clocks in the train should be synchronized in the frame of the train. Okay.

EDIT : Your calculation is correct. I made mistakes. My approach is just about the back of the train.

So, you confused two situations.

First, what you solved is about this. When $t=0$, every people in $x$-axis check the train's clock and they shared their result. Then, the clock in the front side is what you calcualted.

Second, What the link said and you confused about is the time(measured in our frame) in front of the train when $t'=0$

Two situations are obviously different unless the train has no length. In the second situation, you could get the result that the clock of the back is earlier than the clock of the front side. (In here, time is measured in our frame.)

So, the hint is for the second situation.

It is because, when you see the syncronization of the time in the train, in our frame, we saw the signal light to the back side arrive faster than to the front side.

• Isn't t' the time on the train as viewed in our frame of reference? – Racconn Mar 12 '18 at 5:01
• $t'$ is the time when the event occured in the frame of the train. Inverse transformed, corresponsing $t$ will be what we measure(in our frame) – ChoMedit Mar 12 '18 at 5:03
• Ok so this $t'$ I calculated is the time on the train when the back passed in front of me right? – Racconn Mar 12 '18 at 5:05
• So the clock in front of the train is going to be ahead of the one in the back by $\frac{vx}{c^2}$ – Racconn Mar 12 '18 at 5:42
• Oh. I made some mistakes. I'll be modified my answer. I think your calculation is correct. – ChoMedit Mar 12 '18 at 6:03