- #1

Gom

- 2

- 3

- 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.

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

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.