# Proving c in Special Relativity

1. Mar 10, 2006

### kahless2005

I have a starship moving at a speed v, which is larger than .5c, away from Earth in the x direction. A blast of its phaser banks (assume highly focused light) travels at an angle (phi)'.

From this I need to prove that an astronomer who observes the phaser blast moves at a speed of c, the speed of light.

I have it worked out that the angular velocity of the phaser blast from the starship in the x direction is v'sub(x) = c * cos ((phi)') and in the y direction is v'sub(y) = c * sin((phi)').

Any help would be appreciated!

2. Mar 10, 2006

### kahless2005

Okay, I have some more work to the problem. I found that the velocity function in the x direction is vsub(x) = (v'sub(x) + u) / (1 + (u* v'sub(x)/c)) and that works out to be:
(c * cos ((phi)') + u) / ( 1 + u * cos((phi)'))

Now what?

3. Mar 10, 2006

### Andrew Mason

If this is intended as an exercise in using the Lorentz transformations, it is somewhat circular. The Lorentz transformations are based on the principle that the speed of light is c in all inertial frames of reference.

Here you have a light pulse fired from a moving observer, so we know on principle that the speed of light relative to all observers will be c.

What is more interesting is to work out the angle that the phasor pulse will have relative to the earth observer. It will be less than $\phi$.

AM

4. Mar 10, 2006

### kahless2005

Proving c in Special Relativity or Corbomite Manuevar

Thanks Andrew

Then next step has the phaser banks replaced by a light source.

Now I am assuming that the light source radiates in all directions in the starship's frame of reference. And this I assume denotes grounds for the headlight effect for the astronomer's frame of reference.

The problem now asks me to find $\phi$naught on Earth.

I already have found the equation:
cos($\phi$) = (cos($\phi$')+ u/c) / (1 + (u* cos($\phi$')/c))

Do I simply solve for $\phi$, or is there some equation for headlight effects? My book and my notes only have a mention of headlight effect, that I can find.

5. Mar 13, 2006

### Andrew Mason

Think of the perceived angle as proportional to the ratio of speed in the y direction to speed in the x direction (accurate only for small angles):

$$\phi ' \approx \frac{\frac{dy'}{dt'}}{\frac{dx'}{dt'}}$$

$$\phi \approx \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Since dy'/dt' = dy/dt, the ratio of angles is:

$$\phi ' / \phi \approx \frac{\frac{dx}{dt}}{\frac{dx'}{dt'}} = \frac{1}{\gamma}$$

AM