# Speed of rods in lab frame

#### 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?
-----------------------

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.

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.]

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#### 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

Homework Helper
Gold Member
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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member

"Speed of rods in lab frame"

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