Relativistic Beaming (shooting a laser at 90 degrees from your line of motion)

[khfrekek92]

Homework Statement

A ship is traveling at a speed of .8c. When the ship is 90 degrees from the target, it shoots a laser downwards. What angle must it shoot the laser to hit the target?

Homework Equations

$\mu_{x}=0$
$\mu_{y}=-c$
$\mu_{x}'=$$\frac{mu_{x}-v}{1-\frac{v*mu_{x}}{c^{2}}}=-v$
$\mu_{y}'=$$\frac{mu_{y}}{1-\frac{v*mu_{x}}{c^{2}}}=-.6c$

The Attempt at a Solution

I have tried everything I can think of, but nothing seems to work..
I have gotten to this point and I'm fairly sure this far is at least right. Does anyone know where to proceed from here?

Thanks in advance,
khfrekek

 

[anathema]

Hello khfrekek,

Thank you for your post. I am a scientist and I would be happy to help you with this problem.

To solve this problem, we can use the equations of special relativity. We know that the ship is traveling at a speed of 0.8c, which means that its velocity is 0.8 times the speed of light. This means that the ship is moving at a very high speed and we need to take into account the effects of time dilation and length contraction.

First, let's define a coordinate system where the target is located at the origin and the ship is moving along the x-axis. This means that the ship's velocity vector is in the positive x-direction.

Next, we can use the Lorentz transformations to find the velocity of the laser beam as seen by an observer on the target. The Lorentz transformations are given by:

\mu_{x}'=\frac{\mu_{x}-v}{1-\frac{v*\mu_{x}}{c^{2}}}
\mu_{y}'=\frac{\mu_{y}}{1-\frac{v*\mu_{x}}{c^{2}}}

where \mu_{x} and \mu_{y} are the x and y components of the velocity of the laser beam as seen by the ship, and \mu_{x}' and \mu_{y}' are the x and y components of the velocity of the laser beam as seen by the target.

We know that the ship is traveling at a speed of 0.8c, so we can plug this value into the equations:

\mu_{x}'=\frac{\mu_{x}-0.8c}{1-\frac{0.8c*\mu_{x}}{c^{2}}}=\frac{\mu_{x}-0.8c}{1-0.8\mu_{x}}=-0.8c
\mu_{y}'=\frac{\mu_{y}}{1-\frac{0.8c*\mu_{x}}{c^{2}}}=\frac{\mu_{y}}{1-0.8\mu_{x}}=-0.6c

Now, we can use the Pythagorean theorem to find the magnitude of the velocity of the laser beam as seen by the target:

\sqrt{\mu_{x}'^{2}+\mu_{y}'^{2}}=\sqrt{(-0.8c