I'm having a bit of difficulty puzzling out the correct way to move between frames in the context of a basic Special Relativity problem.

The problem basically states that two bodies ($A$ and $B$) are moving apart at a relative velocity $v$ (and we are assuming only 1-spacial dimension for simplicity), and body $A$ emits a light-pulse at time $t_{\alpha}$ which body $B$ receives at time $t_{\beta}$, and I am asked to simply show that in $A$'s reference frame:

$$t_{\beta} = \frac{t_{\alpha}}{1-\frac{v}{c}}$$

In the initial reference frame of $B$, we observe that $t_{\alpha}^{\prime}=\gamma t_{\alpha}$ and that the light-pulse is emitted at a distance $v\gamma t_{\alpha}$ from $B$, so we receive the light-pulse at:

$$t_{\beta}^{\prime} = \gamma t_{\alpha} + \frac{v\gamma t_{\alpha}}{c}$$

But in the initial reference frame of $A$, we have $t_{\beta} = \frac{1}{\gamma}t_{\beta}^{\prime}$, so:

$$t_{\beta} = t_{\alpha}\left(1+\frac{v}{c}\right)$$

Which is not equal to the expression I was asked to show, so what have I misunderstood?


1 Answer 1


\begin{equation} t_{\beta}^{\prime} = \gamma t_{\alpha} + \frac{v\gamma t_{\alpha}}{c} \tag{01} \end{equation} Your equation (01) is right, but your equation (02) is wrong :

\begin{equation} t_{\beta} \ne \frac{1}{\gamma}t_{\beta}^{\prime} \tag{02} \end{equation}

The right (02) equation is :

\begin{equation} t_{\beta} =\gamma t_{\beta}^{\prime} \tag{02$^{\prime}$} \end{equation} So, multiplying (01) by $\gamma$ yields \begin{equation} t_{\beta} = \gamma^{2}\left( 1 + \frac{v}{c}\right)t_{\alpha}=\dfrac{\left( 1 + \frac{v}{c}\right)}{\left(1 -\frac{v^2}{c^2}\right)}t_{\alpha}=\dfrac{1}{1-\frac{v}{c}}t_{\alpha} \tag{03} \end{equation}

When two events happen at the same point, $\Delta x = 0 $, at $\Delta t$ time apart in the unprimed system (here A) then in the primed system (here B) happen at time apart

\begin{equation} \Delta t^{\prime}=\gamma\left(\Delta t-\frac{v \Delta x}{c^2}\right) \tag{04} \end{equation} that is \begin{equation} \Delta t^{\prime}=\gamma\Delta t \tag{05} \end{equation}

So, suppose that the two systems A and B start at the same space point with $x=0, t=0$ for A and $x^{\prime}=0,t^{\prime}=0$ for B. System B is moving with speed $v$ towards the positive $x$ of A .

We have two events in A :

Event $\sigma $ : Observer A has observer B on $x_{\sigma}=0$ at $t_{\sigma}=0$.

Event $ \alpha $ : Observer A from $x_{\alpha}=0$ at $t=t_{\alpha}$ sends a light-pulse to B.

We have \begin{equation} \Delta x = x_{\alpha}-x_{\sigma}=0 , \qquad \Delta t = t_{\alpha}-t_{\sigma}=t_{\alpha} \tag{06} \end{equation}

So from (05) \begin{equation} \Delta t^{\prime} = t_{\alpha}^{\prime}-t_{\sigma}^{\prime}=\gamma \Delta t =\gamma \left(t_{\alpha}-t_{\sigma}\right) \tag{07} \end{equation}

that is \begin{equation} t_{\alpha}^{\prime}=\gamma t_{\alpha} \tag{08} \end{equation}

That's all right till now.

Now let the following two events in B :

Event $\sigma $ : Observer B has observer A on $x_{\sigma}^{\prime}=0$ at $t_{\sigma}^{\prime}=0$.

Event $ \beta $ : Observer B receives on $x_{\beta}^{\prime}=0$ at $t=t_{\beta}^{\prime}$ the light-pulse from A.

We have \begin{equation} \Delta x ^{\prime}= x_{\beta}^{\prime}-x_{\sigma}^{\prime}=0 , \qquad \Delta t ^{\prime}= t_{\beta}^{\prime}-t_{\sigma}^{\prime}= t_{\beta}^{\prime} \tag{09} \end{equation}

Of course here we have for the inverse Lorentz transformation

\begin{equation} \Delta t=\gamma\left(\Delta t ^{\prime}+\frac{v \Delta x^{\prime}}{c^2}\right) \tag{10} \end{equation}

So, \begin{equation} \Delta t = t_{\beta}-t_{\sigma}=\gamma \Delta t ^{\prime}= \gamma \left( t_{\beta}^{\prime}-t_{\sigma}^{\prime}\right)= \gamma t_{\beta}^{\prime} \tag{11} \end{equation}

and finally

\begin{equation} t_{\beta} =\gamma t_{\beta}^{\prime} \tag{02$^{\prime}$} \end{equation}

When two events happen at the same point of an unprimed system (here A) then in the primed system (here B) \begin{equation} \Delta t^{\prime}=\gamma\Delta t \tag{05} \end{equation}

When two events happen at the same point of an primed system (here B) then in the unprimed system (here A)

\begin{equation} \Delta t=\gamma\Delta t^{\prime} \tag{12} \end{equation}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.