Special relativity - measure of a rod and simultaneity

[Gom]

Homework Statement

The problem requests to find Lorentz contraction by using two different methods, the first one being Lorentz transformations, and the second, by following some steps described below.

Relevant Equations

Lorentz transformations, Lorentz contraction

Hi, I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, $O$ and $O'$, the last one moving with relative velocity $v$ in the $x$ direction. There's a rod with length $L'$ fixed to frame $O'$, such that front end $A$ is at $x'=0$, and back end $B$ is at $x'=L'$. Clocks are synchronized at time $t=t'=0$, when position is $x=x'=0$. An observer from $O$ measures the rod, and the result is $L$. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as $x'=L'$, we have $L'=\gamma L$, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$

Now I need to find the result $L'=\gamma L$, but with another method. First, the problem requests to find the $\Delta t'$ (from the point of view of $O'$) since front end $A$ of the rod is measured by $O$, until back end $B$ is measured by $O$. Of course, both events are simultaneous for $O$, then $\Delta t=t_B-t_A=0$, so $t_B=t_A=t$. This is what I did (I'm not sure if it's correct)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$
$$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that back end $B$ is measured before front end $A$.

After that, the problem requests to find the position of coordinate origin at $O$, at the moment when back end $B$ is measured by $O$, as seen by $O'$; and also the position of coordinate origin at $O$, at the moment when front end $A$ is measured by $O$, as seen by $O'$. Finally, with this and $\Delta t'=-\gamma vL/c^2$, I should find again the difference of length between $L$ and $L'$.

Please let me know if I wasn't clear. Thanks.

 

[PeroK]

Homework Statement:: The problem requests to find Lorentz contraction by using two different methods, the first one being Lorentz transformations, and the second, by following some steps described below.
Relevant Equations:: Lorentz transformations, Lorentz contraction

Hi, I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, $O$ and $O'$, the last one moving with relative velocity $v$ in the $x$ direction. There's a rod with length $L'$ fixed to frame $O'$, such that front end $A$ is at $x'=0$, and back end $B$ is at $x'=L'$. Clocks are synchronized at time $t=t'=0$, when position is $x=x'=0$. An observer from $O$ measures the rod, and the result is $L$. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as $x'=L'$, we have $L'=\gamma L$, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$

Okay, so you transformed the event $(0, L)$, which is the end of the rod at time $t = 0$ to get the x-coordinate of the rod in $O'$. And, as the rod is stationary in $O'$ that end remains at $\gamma L$ independent of time.

Now I need to find the result $L'=\gamma L$, but with another method. First, the problem requests to find the $\Delta t'$ (from the point of view of $O'$) since front end $A$ of the rod is measured by $O$, until back end $B$ is measured by $O$. Of course, both events are simultaneous for $O$, then $\Delta t=t_B-t_A=0$, so $t_B=t_A=t$. This is what I did (I'm not sure if it's correct)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$
$$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that back end $B$ is measured before front end $A$.

First of all, what the problem calls the "back end" seems to be the front end to me. It's ahead in the direction of motion. The rod could be moving backwards, of course.

In any case, what you've found is that the event $(0, L)$ in $O$ transforms to the event $(-\frac{\gamma vL}{c^2}, x')$ in $O'$.

In other words, a clock at the end of the rod in $O'$ would read $t' = -\frac{\gamma vL}{c^2}$ as that end of the rod passes some marker at the point $x = L$ in $O$ and a local clock in $O$ reads $t = 0$.

That's one example of the relativity of simultaneity.

After that, the problem requests to find the position of coordinate origin at $O$, at the moment when back end $B$ is measured by $O$, as seen by $O'$; and also the position of coordinate origin at $O$, at the moment when front end $A$ is measured by $O$, as seen by $O'$. Finally, with this and $\Delta t'=-\gamma vL/c^2$, I should find again the difference of length between $L$ and $L'$.

Please let me know if I wasn't clear. Thanks.

I'm not sure why the problem asks you to do this. It seems a little muddled to me.

An alternative method to derive length contraction is as follows:

1) Calculate the trajectory of each end of the rod in frame $O$ using the data from frame $O'$,

2) Calculate the length of the rod in $O$ using distance between where the front and back of the rod are at the same time in frame $O$.

 

[Gom]

Thanks PeroK. The exercise doesn't allow to use another method, so I will keep looking for it. However, I would like to know about the method you suggested. Is there any bibliography where I can find it explained?