# Event times in different reference frames in the context of Special Relativity

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?

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

$$t_{\beta} \ne \frac{1}{\gamma}t_{\beta}^{\prime} \tag{02}$$

The right (02) equation is :

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

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

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

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 $$\Delta x = x_{\alpha}-x_{\sigma}=0 , \qquad \Delta t = t_{\alpha}-t_{\sigma}=t_{\alpha} \tag{06}$$

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

that is $$t_{\alpha}^{\prime}=\gamma t_{\alpha} \tag{08}$$

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 $$\Delta x ^{\prime}= x_{\beta}^{\prime}-x_{\sigma}^{\prime}=0 , \qquad \Delta t ^{\prime}= t_{\beta}^{\prime}-t_{\sigma}^{\prime}= t_{\beta}^{\prime} \tag{09}$$

Of course here we have for the inverse Lorentz transformation

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

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

and finally

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

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

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

$$\Delta t=\gamma\Delta t^{\prime} \tag{12}$$