# Special case of relativistic velocity addition law

• B
• IgorIGP
In summary, the conversation discusses the special case of the relativistic velocity addition law and how to calculate v from w, or vice versa, using the equation w = 2v/(1+v^2/c^2). The confusion arises from the fact that there are two roots to the equation, but only one is physically meaningful. The correct formula for v is v = c√[(γw-1)/(γw+1)].
IgorIGP
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?

Last edited:
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
IgorIGP said:
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 approach 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?

IgorIGP said:
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!

## 1. What is the special case of the relativistic velocity addition law?

The special case of the relativistic velocity addition law is when the two velocities being added are equal to the speed of light, c. In this case, the resulting velocity will also be equal to c.

## 2. How is the special case of the relativistic velocity addition law derived?

The special case of the relativistic velocity addition law can be derived from the full relativistic velocity addition formula, which takes into account the effects of time dilation and length contraction at high velocities.

## 3. What are the implications of the special case of the relativistic velocity addition law?

The special case of the relativistic velocity addition law has important implications in the theory of relativity, as it shows that the speed of light is constant and cannot be exceeded by any object. This is a fundamental principle in our understanding of the universe.

## 4. Can the special case of the relativistic velocity addition law be applied to everyday situations?

Yes, the special case of the relativistic velocity addition law can be applied to everyday situations, such as the movement of particles at high speeds in particle accelerators. It is also used in the calculations of spacecraft trajectories and in the study of celestial bodies.

## 5. Are there any exceptions to the special case of the relativistic velocity addition law?

No, the special case of the relativistic velocity addition law holds true in all situations where the two velocities being added are equal to the speed of light. However, it does not apply to velocities that are significantly lower than the speed of light, where classical physics equations can be used instead.

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