Two light sources are a distance D apart

[romanski007]

Homework Statement

Two light sources are at rest and at a distance D apart on the x-axis of some inertial frame, O. They emit photons simultaneously in that frame in the positive x-direction. Show that in an inertial frame, O', in which the sources have a velocity v along the x-axis, the photons are separated by a constant distance $$D\sqrt{\frac{c-u}{c+u}}$$

Relevant Equations

$$D\sqrt{\frac{c-u}{c+u}}$$
$$x_2-x_1 =D$$

I let E1 be the event where source 1 emits the photon and E2 for the second source with the respective coordinates in O as $(x_1, t_1$) and $(x_2,t_2)$ such that $t_2=t_1 \because$ simultaneous and $x_2-x_1 =D$.

Using Lorentz transformation I obtained that in O', $$x'_2-x'_1 = \gamma (D-v(t_2-t_1))=\gamma D \ \because t_2 = t_1$$ In the solutions $$t_2-t_1 = \frac{D}{c} $$ however I cannot see where this came from.

 

[PeroK]

Using Lorentz transformation I obtained that in O', $$x'_2-x'_1 = \gamma (D-v(t_2-t_1))=\gamma D \ \because t_2 = t_1$$

Okay, these are the $x'$ coordinates of events E1 and E2. What about the time coordinates $t'$ of these events?

Note that you apparently have length dilation there: $D' = \gamma D$(?)

 

[romanski007]

Okay, these are the $x'$ coordinates of events E1 and E2. What about the time coordinates $t'$ of these events?

Note that you apparently have length dilation there: $D' = \gamma D$(?)

Why do I need $t'$ coordinates if I am only considering the length difference in O' ? Furthermore in O, how does the time difference of $\frac{D}{c}$ emerge if both events are simultaneous in O.

 

[PeroK]

Why do I need $t'$ coordinates if I am only considering the length difference in O' ?

There are many ways to answer that! One answer is why not compute the time coordinates just to see?

Another answer is: what does it mean to say that two moving objects are a certain distance apart? How do you define spatial separation for two moving objects?

Hint: does simultaneity of measurements have anything to do with it?

 

[romanski007]

There are many ways to answer that! One answer is why not compute the time coordinates just to see?

Another answer is: what does it mean to say that two moving objects are a certain distance apart? How do you define spatial separation for two moving objects?

Hint: does simultaneity of measurements have anything to do with it?

Oh so actually it's $t'_1 = t'_2$, so in that case, $x_2-x_1 = D + c(t_2 - t_1 )$ as both sources would emit light simultaneously and hence
$$x_2'-x_1' = \gamma (D + c(t_2 - t_1) - v(t_2 - t_1)$$ and since in O', the event is simultaneous, I got
$$t_2'-t_1' = 0 = \gamma \[ t_2 - t_1 - \frac{v}{c^2} (D + c (t_2 - t_1)) \]$$
which simplifies to $$t_2 - t_1 = \frac{vD}{c(c-v)}$$
Hence substituting for $t_2 - t_1$ I got
$$x_2' - x_1' = \gamma (D + \frac{vD}{c-v} - \frac{v^2 D }{c(c-v})$$
$$x_2' - x_1 ' = \gamma D \frac {c+v}{c}= D \sqrt {\frac{c+v}{c-v}}$$
Did I commit some algebraic mistake or perhaps + and - as an answer were mistaken?

 

[PeroK]

Oh so actually it's $t'_1 = t'_2$, so in that case, $x_2-x_1 = D + c(t_2 - t_1 )$ as both sources would emit light simultaneously and hence
$$x_2'-x_1' = \gamma (D + c(t_2 - t_1) - v(t_2 - t_1)$$ and since in O', the event is simultaneous, I got
$$t_2'-t_1' = 0 = \gamma \[ t_2 - t_1 - \frac{v}{c^2} (D + c (t_2 - t_1)) \]$$
which simplifies to $$t_2 - t_1 = \frac{vD}{c(c-v)}$$
Hence substituting for $t_2 - t_1$ I got
$$x_2' - x_1' = \gamma (D + \frac{vD}{c-v} - \frac{v^2 D }{c(c-v})$$
$$x_2' - x_1 ' = \gamma D \frac {c+v}{c}= D \sqrt {\frac{c+v}{c-v}}$$
Did I commit some algebraic mistake or perhaps + and - as an answer were mistaken?

Have you seen the derivation of length contraction? In any case, you need to be more careful about using $D = x_2 - x_1$. That's only true if $x_1, x_2$ represent the sources at the same time in the unprimed frame. If you take events which are simultaneous in frame $S'$, they do not map to events that are simultaneous in frame $S$, hence do not map to events where $x_2 - x_1 = D$.

You need to take more care.

 

[romanski007]

Have you seen the derivation of length contraction? In any case, you need to be more careful about using $D = x_2 - x_1$. That's only true if $x_1, x_2$ represent the sources at the same time in the unprimed frame. If you take events which are simultaneous in frame $S'$, they do not map to events that are simultaneous in frame $S$, hence do not map to events where $x_2 - x_1 = D$.

You need to take more care.

In relation to the attached diagram, I described the position of $x_1, x_2$ and hence took their difference, and transformed it to the primed reference frame where $t' =$ constant

 

[PeroK]

In relation to the attached diagram, I described the position of $x_1, x_2$ and hence took their difference, and transformed it to the primed reference frame where $t' =$ constant

You have too many constraints there: you demand that $x_2 = x_1 + D$ and that $t'_1 = t'_2$. You cannot have both those equations:
$$t'_1 = t'_2 \ \Rightarrow \ t_1 \ne t_2 \ \Rightarrow \ x_2 \ne x_1 + D$$

 

[PeroK]

PS this is even clearer if you take the first event to be the (common) origin of the frames. Then the second event is: $(0, D)$ in the first frame and $(t'_2, x'_2)$ in the second frame, where $t'_2 \ne 0$.

If you want $t'_2 = 0$, then you need to choose a different event: $(t_1, x_1)$ where necessarily $t_1 \ne 0$ and $x_1 \ne D$.

 

[romanski007]

You have too many constraints there: you demand that x2=x1+D and that t1′=t2′. You cannot have both those equations:
t1′=t2′ ⇒ t1≠t2 ⇒ x2≠x1+D

Yes, the way I worked it out was by letting $x_2 = x_0 + D + ct_2$ and $x_1 = x_0 + ct_1$ so that $x_2 -x_1 = D + c(t_2 - t_1)$

 

[romanski007]

PS this is even clearer if you take the first event to be the (common) origin of the frames. Then the second event is: $(0, D)$ in the first frame and $(t'_2, x'_2)$ in the second frame, where $t'_2 \ne 0$.

If you want $t'_2 = 0$, then you need to choose a different event: $(t_1, x_1)$ where necessarily $t_1 \ne 0$ and $x_1 \ne D$.

I did it in the first case so that
$$x_2'-x_1' = \gamma (x_2 - x_1 - v (t_2 - t_1) = \gamma D \because t_2 - t_1 = 0 $$ but the final answer impliesthat $$t_2-t_1 = \frac{D}{c} $$

 

[PeroK]

Yes, the way I worked it out was by letting $x_2 = x_0 + D + ct_2$ and $x_1 = x_0 + ct_1$ so that $x_2 -x_1 = D + c(t_2 - t_1)$

Okay, yes I see that now. Apologies.

You may have done the wrong transformation. The sources are the "moving frame", so you want the inverse transformation.