Can Relativistic Velocity Addition Ensure Speeds Stay Within Limits?

[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