Relativistic Trajectory

E92M3

1. The problem statement, all variables and given/known data
Given that the trajectory in frame S is:
$$x=\frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)$$

Show that in S' that is moving in the +x direction at speed u:
$$x'=\frac{c^2}{a}\sqrt{1+\frac{a^2(t'-A)^2}{c^2}}-B$$

Find the constants A and B.
2. Relevant equations
Lorentz transformation:
$$x'=\gamma (x-ut)$$
$$t'=\gamma (t-\frac{ux}{c^2})$$

3. The attempt at a solution
$$x'=\gamma (x-ut)$$

plug in x:
$$x'=\gamma \left ( \frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)-ut \right )$$

Since we want x' in terms of t'... i used:
$$t=\gamma (t'+\frac{ux'}{c^2} )$$

$$x'=\gamma \left [ \frac{c^2}{a}(\sqrt{1+\frac{a^2\gamma^2 (t'+\frac{ux'}{c^2})^2}{c^2}}-1)-u\gamma (t'+\frac{ux'}{c^2}) \right ]$$

I can group some terms and play around with it:

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2\gamma^4 (t'+\frac{ux'}{c^2})^2}{c^2}}-\frac{\gamma c^2}{a}-u\gamma^2 (t'+\frac{ux'}{c^2})$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 (t'+\frac{ux'}{c^2}) \right )$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t'+\frac{u\gamma^2ux'}{c^2} \right )$$

$$x'-\frac{\gamma^2u^2x'}{c^2}= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

$$x'\left (1-\frac{\gamma^2u^2}{c^2} \right )= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

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:
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kapv89

1. The problem statement, all variables and given/known data
Given that the trajectory in frame S is:
$$x=\frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)$$

Show that in S' that is moving in the +x direction at speed u:
$$x'=\frac{c^2}{a}\sqrt{1+\frac{a^2(t'-A)^2}{c^2}}-B$$

Find the constants A and B.
2. Relevant equations
Lorentz transformation:
$$x'=\gamma (x-ut)$$
$$t'=\gamma (t-\frac{ux}{c^2})$$

3. The attempt at a solution
$$x'=\gamma (x-ut)$$

plug in x:
$$x'=\gamma \left ( \frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)-ut \right )$$

Since we want x' in terms of t'... i used:
$$t=\gamma (t'+\frac{ux'}{c^2} )$$

$$x'=\gamma \left [ \frac{c^2}{a}(\sqrt{1+\frac{a^2\gamma^2 (t'+\frac{ux'}{c^2})^2}{c^2}}-1)-u\gamma (t'+\frac{ux'}{c^2}) \right ]$$

I can group some terms and play around with it:

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2\gamma^4 (t'+\frac{ux'}{c^2})^2}{c^2}}-\frac{\gamma c^2}{a}-u\gamma^2 (t'+\frac{ux'}{c^2})$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 (t'+\frac{ux'}{c^2}) \right )$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t'+\frac{u\gamma^2ux'}{c^2} \right )$$

$$x'-\frac{\gamma^2u^2x'}{c^2}= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

$$x'\left (1-\frac{\gamma^2u^2}{c^2} \right )= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

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

kapv89

1. The problem statement, all variables and given/known data
Given that the trajectory in frame S is:
$$x=\frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)$$

Show that in S' that is moving in the +x direction at speed u:
$$x'=\frac{c^2}{a}\sqrt{1+\frac{a^2(t'-A)^2}{c^2}}-B$$

Find the constants A and B.
2. Relevant equations
Lorentz transformation:
$$x'=\gamma (x-ut)$$
$$t'=\gamma (t-\frac{ux}{c^2})$$

3. The attempt at a solution
$$x'=\gamma (x-ut)$$

plug in x:
$$x'=\gamma \left ( \frac{c^2}{a}(\sqrt{1+\frac{a^2t^2}{c^2}}-1)-ut \right )$$

Since we want x' in terms of t'... i used:
$$t=\gamma (t'+\frac{ux'}{c^2} )$$

$$x'=\gamma \left [ \frac{c^2}{a}(\sqrt{1+\frac{a^2\gamma^2 (t'+\frac{ux'}{c^2})^2}{c^2}}-1)-u\gamma (t'+\frac{ux'}{c^2}) \right ]$$

I can group some terms and play around with it:

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2\gamma^4 (t'+\frac{ux'}{c^2})^2}{c^2}}-\frac{\gamma c^2}{a}-u\gamma^2 (t'+\frac{ux'}{c^2})$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 (t'+\frac{ux'}{c^2}) \right )$$

$$x'= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t'+\frac{u\gamma^2ux'}{c^2} \right )$$

$$x'-\frac{\gamma^2u^2x'}{c^2}= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

$$x'\left (1-\frac{\gamma^2u^2}{c^2} \right )= \frac{c^2}{a}\sqrt{\gamma^2+\frac{a^2 (\gamma^2t'+\frac{\gamma^2ux'}{c^2})^2}{c^2}}-\left ( \frac{\gamma c^2}{a}+u\gamma^2 t' \right )$$

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

E92M3

"a" is just a constant.

"Relativistic Trajectory"

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