Relativistic Velocity Addition

[kreil]

Homework Statement

Show that the addition of velocities implies the following:

- If $$| \vec V | < c$$ in one inertial frame, then $$| \vec V | < c$$ in any inertial frame

- If $$| \vec V | > c$$ in one inertial frame, then $$| \vec V | > c$$ in any inertial frame

Homework Equations

$$V^{x'}=\frac{V^x - v}{1-\frac{vV^x}{c^2}}$$...(1)

$$V^{y'}=\frac{V^y}{1-\frac{vV^y}{c^2}} \sqrt{1-v^2/c^2}$$...(2)

$$V^{z'}=\frac{V^z}{1-\frac{vV^z}{c^2}} \sqrt{1-v^2/c^2}$$...(3)

The Attempt at a Solution

If |V| < c, then we can write $$V^x = ac$$, for some constant a < 1. Then:

$$V^{x'}=\frac{ac - v}{1-\frac{av}{c}}= c \left ( \frac{ac-v}{c-av} \right )$$

Since no assumptions are made about v (the relative speed between the inertial frames), I'm not sure how I can show this last fraction is less than 1..

Also, is it necessary to show this for each of $V^x, V^y, V^z$ or is $V^x$ sufficient? (If V=c, (1) gives the required answer of c but I don't think (2) or (3) do..)

 

[Quinzio]

Don't know if what I'm saying is correct, but I'd take the composition law for velocities.
Then showing that c is a critical point, that no increment of speed will pass the limit c.
(http://en.wikipedia.org/wiki/Velocity-addition_formula)$$s = {v+u \over 1+(vu/c^2)}$$

Then I'd take the derivative of one of the speeds, let's take u

$$\frac{\partial s}{\partial u} = \frac{1-\frac{v^2}{c^2}}{(1+\frac{uv}{c^2})^2}$$

Then take it to the limit for v --> c

$$\lim_{v \to c} \frac{\partial s}{\partial u} = \lim_{v \to c} \frac{1-\frac{v^2}{c^2}}{(1+\frac{uv} {c^2})^2} = 0$$This shows that:
When an object increases it's speed from zero to c, we can think of it as continuously changing it's reference frame.
You'll arrive to a point at which no matter how you try to increase your speed, the effect of your increment seen from the original frame will be zero.
That is, you cannot pass c, by adding up speed to u.
In the function c is in fact a critical point because it's derivatives goes to zero.
An onbject cannot increase it's speed over c, because any increment of speed will have no effect as seen from a "resting" frame.

Also, is it necessary to show this for each of LaTeX Code: V^x, V^y, V^z or is LaTeX Code: V^x sufficient? (If V=c, (1) gives the required answer of c but I don't think (2) or (3) do..)

I think it's enough $$V_x$$ since any velocity addition can be seen as $$V_x +V_y + V_z$$ taken as 3 separate steps.

 

[kreil]

Don't know if what I'm saying is correct, but I'd take the composition law for velocities.
Then showing that c is a critical point, that no increment of speed will pass the limit c.

I think the problem is simpler than this. For the first part, I just need to show that if an observer in the rest frame observes the velocity of a moving object $V^x$ as less than c, then any observer in an inertial frame moving relative to the rest frame at speed v observes the velocity $V^x'$ as less than c as well.

For example, if $$V^x=c$$, the moving observer sees the object as moving at:

$$V^{x'} = \frac{c-v}{1-v/c}=c \left ( \frac{c-v}{c-v} \right ) = c$$

Which is consistent with c being the maximum allowed speed in any inertial reference frame. Unfortunately when $$V^x$$ is greater or less than c the proof isn't as straightforward.

 

[kreil]

nevermind I got it

 

[paranormal]

To show that the addition of velocities implies that |V| < c in any inertial frame, we can use the Lorentz transformations (1), (2), and (3) to calculate the magnitude of the velocity in the new frame, |V'|:

|V'| = √(V'^x)^2 + (V'^y)^2 + (V'^z)^2

Substituting in the expressions for V'^x, V'^y, and V'^z from the Lorentz transformations, we get:

|V'| = √[(V^x - v)^2 + (V^y)^2 + (V^z)^2] / [1 - (vV^x)/c^2]...(4)

Since we know that |V| < c, we can write V^x = ac, for some constant a < 1. Substituting this into (4), we get:

|V'| = √[(ac - v)^2 + (V^y)^2 + (V^z)^2] / [1 - (av)/c] < c

This shows that |V'| < c in any inertial frame, since the expression inside the square root will always be less than c^2. Therefore, if |V| < c in one inertial frame, it will also be less than c in any other inertial frame.

Similarly, to show that if |V| > c in one inertial frame, then |V| > c in any other inertial frame, we can follow the same steps but use the fact that V^x = ac, for some constant a > 1. This will lead to an expression for |V'| that is greater than c, proving the statement.

To answer your second question, it is sufficient to show this for just V^x, as the components of velocity in the other two directions (V^y and V^z) will also follow the same logic.