Special case of relativistic velocity addition law

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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?

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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
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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