Relativistic Trajectory

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1. The problem statement, all variables and given/known data
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.
2. Relevant equations
Lorentz transformation:
[tex]x'=\gamma (x-ut)[/tex]
[tex]t'=\gamma (t-\frac{ux}{c^2})[/tex]



3. 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:
15
0
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
Lorentz transformation:
[tex]x'=\gamma (x-ut)[/tex]
[tex]t'=\gamma (t-\frac{ux}{c^2})[/tex]



3. 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.
what is 'a' (i.e. the small a) , is it the accn of the body
 
15
0
1. The problem statement, all variables and given/known data
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.
2. Relevant equations
Lorentz transformation:
[tex]x'=\gamma (x-ut)[/tex]
[tex]t'=\gamma (t-\frac{ux}{c^2})[/tex]



3. 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.
what is 'a' (i.e. the small a) , is it the accn of the body

takin it as a const its already pretty complicated
 
69
0
"a" is just a constant.
 

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