Finding Constants in Relativistic Trajectory Problem

[E92M3]

Homework Statement

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.

Homework Equations

Lorentz transformation:
x'=\gamma (x-ut)
t'=\gamma (t-\frac{ux}{c^2})

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.

 

[kapv89]

Homework Statement

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.

Homework Equations

Lorentz transformation:
x'=\gamma (x-ut)
t'=\gamma (t-\frac{ux}{c^2})

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]

Homework Statement

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.

Homework Equations

Lorentz transformation:
x'=\gamma (x-ut)
t'=\gamma (t-\frac{ux}{c^2})

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.