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I have question regarding the following velocity transform formula:

$$v_b = \frac{v_a - u}{1-uv_a/c^2}$$

$u$ here refers to the speed that one frame is moving relative to another. How do I decide whether to use a positive or negative value of $v$.

On top of that, when do we use the normal time dilation formula $t = \gamma t'$ compared to $t_B = \gamma(t_A −ux_A/c^2)$ and $l = l'/\gamma$ compared to $x_B =γ(x_A−ut_A)$.

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  • $\begingroup$ For the velocity equation: As a general rule of thumb, equations like this should make sense in the non-relativistic limit as well. Taking this limit you would obtain $v_b = v_a - u$. Now you should be answer the question yourself... $\endgroup$
    – DJBunk
    Commented May 3, 2014 at 20:19

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It's not clear what $v_a$ and $v_b$ represent. Let's back up and change the notation. We will work in 1 spatial dimension.

There are two reference frames, $\mathcal O$ and $\mathcal O'$.

The velocity of the frame $\mathcal O'$ in the $\mathcal O$ frame is $v$.

The velocity of a particle relative to the primed frame $\mathcal O'$ is $u'$. Using this notation, it's clear which frame the velocity is relative to.

The velocity $u$ of the particle relative to the unprimed frame $\mathcal O$ is related to $u'$ by the relativistic velocity addition formula:

$$u = \frac{u' + v}{1 + \frac{u'v}{c^2}} $$

Remember, these are velocities so the signs will take care of themselves.

On top of that, when do we use the normal time dilation formula

The formula

$$\Delta t = \gamma \Delta t' $$

comes from Lorentz time coordinate transformation

$$\Delta t = \gamma(\Delta t' + \frac{v\Delta x'}{c^2})$$

when $\Delta x' = 0$. Thus, if $\Delta t'$ is the elapsed time of a clock at rest in the primed frame, $\Delta x' = 0$ and $\Delta t = \gamma \Delta t'$.

In other words, $\Delta t = \gamma \Delta t'$ only holds for the time between two co-located events in the primed frame.

Similarly, $l = \frac{l'}{\gamma}$ only holds for the distance between to simultaneous events in the primed frame.

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  • $\begingroup$ for the velocity transform, how do i know which is considered prime? i can switch u and u' in the formula and say that the other is prime right? $\endgroup$
    – user10024395
    Commented May 3, 2014 at 23:58
  • $\begingroup$ @user136266, as I wrote, the primed frame has velocity $v$ in the unprimed frame and we know the particle velocity $u'$ in the primed frame. If, instead, we know the particle velocity $u$ in the unprimed frame and we wish to know $u'$, a little algebra yields $u' = \frac{u - v}{1 - \frac{uv}{c^2}}$ $\endgroup$
    – Alfred Centauri
    Commented May 4, 2014 at 0:09
  • $\begingroup$ yup and the problem i am facing now is that the problem will not specifically say this is the velocity in the prime or unprime frame, i have to decide for myself and so i am just wondering what makes it a prime or unprime frame $\endgroup$
    – user10024395
    Commented May 4, 2014 at 0:39
  • $\begingroup$ @user136266, as I also wrote, "It's not clear what $v_a$ and $v_b$ represent". There simply isn't enough context to give you a definitive answer. In other words, one must specify the reference frame the velocity is relative to. $\endgroup$
    – Alfred Centauri
    Commented May 4, 2014 at 0:45

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