What is the Transformation of Four-Velocity between Frames?

[Kaguro]

Rod R1 has a rest length `1m and rod R2 has rest length 2m. R1 is moving towards right with velocity v and R2 is moving towards left with velocity v with respect to lab frame. If R2 has length 1m in rest frame of R1, then v/c is what?
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I let vel of R1 be u, of R2 be w w.r.t lab and velocity of R2 w.r.t. R1 be w'.
Then u = v, and w = -v.

Then by relativistic velocity addition formula:
w' = (w - u)/(1- (u*w)/c^2)
=> w' = -2v/(1+v^2/c^2) -----------(1)

Now, according to Lorentz contraction:
L/L0 = √(1 - (w'/c)^2 )
But L/L0 = 1/2

Solving, I find w'/c = (√3)/2 ------------(2)

From simplifying (1) and (2):
v^2 + (4/√3) vc + c^2 = 0
So, v = -c/√3 or v = -√3 *c.

But... why is answer negative?
Even if I reject the 2nd one and accept the magnitude of first, I get v = 0.577c...

But the exact answer given is v = 0.6c.

Where did I go wrong?Also, the solution given says:

Where did this one come from??

Any help will be appreciated.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[Ibix]

The negative comes in because you neglected a solution for $w'$. It could be $w'=\pm\sqrt 3/2$, and your equation 1 implies that $w'$ is negative if $v$ is positive.

Otherwise I have to say I agree with you.

 

[Kaguro]

The negative comes in because you neglected a solution for $w'$. It could be $w'=\pm\sqrt 3/2$, and your equation 1 implies that $w'$ is negative if $v$ is positive.

Oh yeah.. You're right!

 

[Ibix]

The expression for length contraction in the photo is weird. You get different behaviour as $v\rightarrow c$ and $v\rightarrow -c$. That's not right, unless it's a specialisation of something with a hidden assumption that $v>0$. What book did it come from?

 

[Kaguro]

What book did it come from?

It is a book for IIT-JAM solved papers.

 

[Ibix]

So just a book of practice problems? In that case (unless someone else comes up with something we've both missed) I'd just bear in mind that some of the answers might not be perfect.

 

[robphy]

The first equation is not the Lorentz Contraction equation since the right-hand-side is the Doppler factor.

 

[Kaguro]

So just a book of practice problems? In that case (unless someone else comes up with something we've both missed) I'd just bear in mind that some of the answers might not be perfect.

Yeah i thought so too...

thanks everyone.

 

[PeroK]

Yeah i thought so too...

thanks everyone.

You can also do this with rapidities. In this case, in the frame of R1, we have a rapidity of $2\theta$, where $\theta$ is the rapidity associated with speed $v$. And:

$2 = \gamma_1 = \cosh 2\theta = 2\cosh^2 \theta - 1$

Hence $\gamma^2 = \cosh^2 \theta = 3/2$

Then, we have:

$\frac{v}{c} = \tanh \theta = \sqrt{1 - sech^2 \theta} = \frac{1}{\sqrt{3}}$

Or, more generally, we can relate the gamma factor for $v$ to the gamma factor, $\gamma_1$, in the reference frame of rod 1:

$\gamma^2 = \cosh^2 \theta = \frac{\cosh 2\theta +1}{2} = \frac{\gamma_1 +1}{2}$

And:

$(\frac{v}{c})^2 = 1 - \frac{1}{\gamma^2} = \frac{\gamma_1 - 1}{\gamma_1 + 1}$

In this case, with $\gamma_1 = 2$, we get:

$(\frac{v}{c})^2 = \frac13$

 

[robphy]

When one is comfortable with all of the intermediate steps,
one can streamline @PeroK 's calculation by v=\tanh(\rm arccosh(2)/2)
https://www.wolframalpha.com/input/?i=tanh(arccosh(2)/2)=\displaystyle\frac{ 1}{\sqrt{3}}

 

[PeroK]

Here's another approach, using transformation of the four-velocity between frames.

In the lab frame, R1 has four-velocity $(\gamma c, \gamma v)$. The time component when transformed to the frame of R2 is:

$\gamma_2 c = \gamma (\gamma c + \frac{v}{c} \gamma v) = \gamma^2 (1 + v^2/c^2) c$

We want $\gamma_2 = 2$, hence:

$\frac{1 + v^2/c^2}{1- v^2/c^2} = 2$ etc.