Find out velocity of an observer (relativity)

[Istiak]

Homework Statement

An observer O at rest midway between two sources of light at x=0 and x=10m observes the
two sources to flash simultaneously. According to a second observer O’, moving at a constant
speed parallel to the x-axis, one source of light flash 13ns before the other. Which of the
following gives the speed of O’ relative to O?

Relevant Equations

$x\prime = \gamma (x-vt)$

Initial observer is at rest. So $x\prime=0$, and according to question they are 10 meter apart. So lorentz transformation becomes
$vt=x$
$v=\frac{x}{t}$
$=\frac{10 \\ \mathrm m}{13\times10^{-9} \mathrm s}$

But I don't get the expected answer. I believe if I had took $\beta c$ instead of $v$ then I would get the same result. There was a problem into plugging numbers, wasn't it?

 

[PeroK]

What if $O'$ is at rest, but closer to one source of light than the other?

 

[Istiak]

What if $O'$ is at rest, but closer to one source of light than the other?

Then he would see that the closer one's light comes first then another ones

 

[PeroK]

Then he would see that the closer one's light comes first then another ones

What do you think $O$ and $O'$ are measuring? When the flashes of light reach them or when the events that represent the emission of the light take place according to their frame of reference?

 

[Istiak]

What do you think $O$ and $O'$ are measuring? When the flashes of light reach them or when the events that represent the emission of the light take place according to their frame of reference?

I can't think much cause $O$ is at rest and $O\prime$ is moving. So $O\prime$'s position is changing. So I don't know the distance between O' and sources of light. Velocity is unknown also. With two unknown variables I can't find out velocity. According to O' others are moving and he is at rest. But according to O he is at rest including sources of light.

 

[PeroK]

So I don't know the distance between O' and sources of light.

Do you think that's important? Does the time of an event depend on the position of the observer?

 

[Istiak]

Do you think that's important? Does the time of an event depend on the position of the observer?

As far as I can say that is, NO.

Suppose, I sent a signal to you yesterday at 2'O clock. Then I can't say when you will receive it without knowing your position. But if I know that you received the signal at 3'O clock then I can't find out your position without knowing the velocity and vice versa, can I?
If possible then wouldn't you like to show the math?

 

[PeroK]

As far as I can say that is, NO.

That's correct. SR and the Lorentz Transformation relate to reference frames, not to observers. Really $O$ and $O'$ should refer to reference frames and not to isolated observers.

The question emphasises than an observer in $O$ is half-way between the sources and "sees" the light from both sources at the same time. This means that the emission of light from the two sources is simultaneous in a frame of reference where this observer is at rest. That is the important thing. Any other observer at rest in $O$ must measure these events to be simultaneous, whether or not the light from the events reaches the observer at the same time. Or, even if the events emit no light, they still have well-defined time and position coordinates in every reference frame.

You can, therefore, give these two events coordinates of $(0, 0)$ and $(0, 10m)$ in frame $O$.

Now, what happens if you transform these events to a frame $O'$ moving at speed $v$ relative to $O$? Using the Lorentz Transformation.

 

[Istiak]

You can, therefore, give these two events coordinates of $(0, 0)$ and $(0, 10m)$ in frame $O$.

Let $O\prime(0,0)$ So, $x\prime=0=t\prime$ $t=0$ $x=10 m$ (Didn't you write (t,x)?)

Now, what happens if you transform these events to a frame O′ moving at speed v relative to O? Using the Lorentz Transformation.

I answered it already. $O\prime$ will see that $O$ is moving backward (from $O\prime$'s frame of reference) while $O$ will see that $O\prime$ is moving forward (from O's frame of reference)

 

[PeroK]

I answered it already. $O\prime$ will see that $O$ is moving backward (from $O\prime$'s frame of reference) while $O$ will see that $O\prime$ is moving forward (from O's frame of reference)

That's not a coordinate transformation!

 

[Istiak]

That's not a coordinate transformation!

They are observer. Or frame of reference

 

[PeroK]

They are observer. Or frame of reference

Coordinates are things like $(t', x')$. Coordinates are not statements like "O is moving backwards"!

 

[Istiak]

But i haven’t got my answer.

 

[PeroK]

But i haven’t got my answer.

You need to do the Lorentz Transformation of the event $(t = 0, x = 10m)$.