# Classical Relativity and the Speed of Light

Tags:
1. Aug 11, 2014

### knowLittle

1. The problem statement, all variables and given/known data
Let's assume that the classical ideas of space and time are correct, so that there could only be one frame, "ether", in which light traveled with same speed in all directions.

Assume that the earth's speed relative to the ether frame is our orbital speed around the sun.

a.) What would be the observed speed (on earth) of a light wave traveling parallel to v?
b.) " " ... traveling anti-parallel to v?
c.) What if it were travelling perpendicular to v (as measured on earth)?
2. Relevant equations
$c = 2.9979 \times 10^8$ m/s
$v= 3\times10^4$ m/s
3. The attempt at a solution
Question C is my problem
Apparently, this is just...

^ c
|
| ----> v

However, I have seen that the hypotenuse is calculated as $\sqrt{c^2-v^2}$ it confuses me a bit.
I know that we are using vectors and v's direction would be ideally facing c, but I am still lost.

Or, are they assuming that :
^ c
|
| <----- v

In order to write the hypotenuse normally and then change the sign in v?
$\sqrt{c^2 + (-v^2)}= \sqrt{c^2-v^2}$ , but if this is the case, why not do something like this:
Code (Text):

^ c
|
|
------> v

And, then write the hypotenuse without any change of sign?
Help?

Last edited: Aug 11, 2014
2. Aug 11, 2014

### sonicharmony

If you were traveling in a car going speed v, what would be the observed speed of a second card moving in the same direction in the lane next to you? What about if the other car was moving in the opposite direction in the lane next to you? For question C, think of your classic vector drawing. The vector AB has a component in the a direction, and a component in the b direction. AB is the hypotenuse of that triangle:

So the Observed Speed V$_{observed}$ = $\sqrt{c^{2}+v^{2}}$