# Special case of relativistic velocity addition law

• B

## Summary:

How can I express v(w) from $$w =\frac{2\cdot v}{1+\frac{v^{2}}{c^{2}}}$$
- special case of relativistic velocity addition law (where v1=v2=v)?

## Main Question or Discussion Point

Hi gentlemen. My model is simple. Imagine, I watch a body which approaches me with velocity of w wich value I can measure . I know that this a body has velocity v in a some reference system which aproaches me with same velocity of v. This velocity needs to be calculated (I can not mesure it).
So when I know v the value of w can be expressed as
$$w =\frac{2\cdot v}{1+\frac{v^{2}}{c^{2}}}$$
and when I do not know v but know w I should resolve the reverse task and... As I can see there is not a complicated quadratic equation, but I am confused by the ambiguity of the solution-2 roots, and the relative bulkiness of the expressions for these roots. That looks like
$$v_{1,2} =\frac{c^{2}\pm\sqrt{c^{4}-v^{2}\cdot c^{2}}}{v}$$
May be I made something (or everything) wrong? When it is a right formula what means the fact of two roots?

Last edited:

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Ibix
I don't get the 2 in the denominator, but otherwise it looks right. I suspect that if you check (edit: feed your solution for ##v## into your expression for ##w## - do you get ##w##?) and correct that, you'll find that one of the solutions exceeds light speed for all ##w##. That's a valid solution to the maths, but not physically meaningful - usually called a "spurious solution".

A trivial example: I have a square of area 4 - what's it's side length? The answer is ##\sqrt 4=\pm 2##. We discard the negative solution as physically meaningless. The same is happening in your question.

• Dale and IgorIGP
PeroK
Homework Helper
Gold Member
Summary: How can I express v(w) from $$w =\frac{2\cdot v}{1+\frac{v^{2}}{c^{2}}}$$
- special case of relativistic velocity addition law (where v1=v2=v)?

Hi gentlemen. My model is simple. Imagine, I watch a body which approaches me with velocity of w wich value I can measure . I know that this a body has velocity v in a some reference system which aproach me with same velocity of v. This velocity needs to be calculated (I can not mesure it).
So when I know v the value of w can be expressed as
$$w =\frac{2\cdot v}{1+\frac{v^{2}}{c^{2}}}$$
and when I do not know v but know w I should resolve the reverse task and... As I can see there is not a complicated quadratic equation, but I am confused by the ambiguity of the solution-2 roots, and the relative bulkiness of the expressions for these roots. That looks like
$$v_{1,2} =\frac{c^{2}\pm\sqrt{c^{4}-v^{2}\cdot c^{2}}}{2\cdot v}$$
May be I made something (or everything) wrong? When it is a right formula what means the fact of two roots?
The ##2v## on the denominator is wrong. Apart from that the formula looks correct. Only one solution is valid, though.

PS You seem to have swapped the meaning of ##v## and ##w## at some point. Assuming ##w## is the initial velocity and ##v## is the relativistic half of that, then you should get:

##v = \frac{c^2}{w}(1- \frac{1}{\gamma_w})##

Or, with a bit more algebra:

##v= c \sqrt{\frac{\gamma_w -1}{\gamma_w + 1}}##

• IgorIGP
Thank you very much gentlemen! I am sorry of my error. After reducing the fraction I've typed numerator in a reduced edition and the denominater remains in an old one. Thanks I've corrected the mistake with pleasure.
So you think the one root has physical meaning only?

PeroK
Homework Helper
Gold Member
Thank you very much gentlemen! I am sorry of my error. After reducing the fraction I've typed numerator in a reduced edition and the denominater remains in an old one. Thanks I've corrected the mistake with pleasure.
So you think the one root has physical meaning only?
Yes. You set the problem up assuming ##0 < v < w##.

• IgorIGP
That were great answers, thak you very much!