- #1
E92M3
- 68
- 0
Homework Statement
Given that the trajectory in frame S is:
[tex]x=\frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)[/tex]
Show that in S' that is moving in the +x direction at speed u:
[tex]x'=\frac{c^2}{a}\sqrt{1+\frac{a^2(t'-A)^2}{c^2}}-B[/tex]
Find the constants A and B.
Homework Equations
Lorentz transformation:
[tex]x'=\gamma (x-ut)[/tex]
[tex]t'=\gamma (t-\frac{ux}{c^2})[/tex]
The Attempt at a Solution
[tex]x'=\gamma (x-ut)[/tex]
plug in x:
[tex]x'=\gamma \left ( \frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)-ut \right )[/tex]
Since we want x' in terms of t'... i used:
[tex]t=\gamma (t'+\frac{ux'}{c^2} )[/tex]
[tex]x'=\gamma \left [ \frac{c^2}{a}(\sqrt{1+\frac{a^2\gamma^2 (t'+\frac{ux'}{c^2})^2}{c^2}}-1)-u\gamma (t'+\frac{ux'}{c^2}) \right ][/tex]
I can group some terms and play around with it:
[tex]x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2\gamma^4 (t'+\frac{ux'}{c^2})^2}{c^2}}-\frac{\gamma c^2}{a}-u\gamma^2 (t'+\frac{ux'}{c^2}) [/tex]
[tex]x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 (t'+\frac{ux'}{c^2}) \right ) [/tex]
[tex]x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t'+\frac{u\gamma^2ux'}{c^2} \right ) [/tex]
[tex]x'-\frac{\gamma^2u^2x'}{c^2}= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right ) [/tex]
[tex]x'\left (1-\frac{\gamma^2u^2}{c^2} \right )= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right ) [/tex]
As you can see, it becomes very messy quickly. Since this is featured in a past paper, I think there should be some trick to do it quickly.
Last edited: