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IgorIGP

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- TL;DR 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 which 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?

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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