Light move away from the emitter at 1.00 c ?

1. Jul 8, 2009

keepitmoving

if an object moving at .99 c emits a light, does that light move away from the emitter at 1.00 c ?

2. Jul 8, 2009

Bob_for_short

Re: .99c

Yes, because the light does not know that the source is in motion.

3. Jul 8, 2009

keepitmoving

Re: .99c

if the emitter suddenly stops, does the light know it? If the light doesn`t know it, the light will be moving at c plus the previous speed of the emitter.

4. Jul 8, 2009

Hootenanny

Staff Emeritus
Re: .99c

No it won't. A stationary observer would still measure the speed of light as c. The invariance of the speed of light is a fundamental concept in special relativity.

5. Jul 8, 2009

ZikZak

Re: .99c

The light will not ever be moving at c plus the speed of the emitter. Light rays travel at c. Every light ray in vacuum will be measured to be traveling at c by every inertial observer. The emitter measures the light ray to recede from him at c. The receiver measures the light ray to approach him at c. Any and all passers-by to this experiment measure the speed of the light ray as c. It makes no difference whatsoever what the various observers' velocities are relative to each other; they all, each and every one of them, measure the speed of that light ray as c. Not c + the velocity of the emitter, not c + the velocity of the receiver, not c + anything. C.

That is what it means for the speed of light to be constant.

6. Jul 9, 2009

Fredrik

Staff Emeritus
Re: .99c

In a way it will, but with a different definition of "plus". In SR you have to add up velocities like this:

$$u\oplus v=\frac{u+v}{1+\frac{uv}{c^2}}$$

If u=0.99c and v=c, the result is

$$u\oplus v=\frac{0.99c+c}{1+\frac{0.99c^2}{c^2}}=\frac{1.99c}{1.99}=c$$

What if u isn't 0.99c? Let's try it again, with u arbitrary and v=c:

$$u\oplus v=\frac{u+c}{1+\frac{uc}{c^2}}=\frac{c\left(\frac u c+1\right)}{1+\frac{u}{c}}=c$$

7. Jul 9, 2009

pesto

Re: .99c

Fredrik, what equation is that? Does it have a name I can google? :)

8. Jul 9, 2009

9. Jul 9, 2009

Staff: Mentor

Re: .99c

That's the relativistic addition of velocity formula.

10. Jul 9, 2009

pesto

Re: .99c

Thank you. That's handy.

11. Jul 13, 2009

pesto

Re: .99c

Does the $$u \oplus v$$ means the velocity of the emitter relative to the observer plus the velocity of the photon relative to the emitter? If I could try your patience for one more stupid question, what's a plus sign with a circle around it? Just a sign for adding vectors or what?

Thanks again.

12. Jul 13, 2009

malawi_glenn

Re: .99c

v is the velocity relative the observer for objectA and u is the velocity of the objectB relative objectA, always. And $u \oplus v$ means the velocity of object B as measured by the observer. $\oplus$ means "composition law for velocities under boosts"

In the wiki article this $u \oplus v$ is just called "s".

Last edited: Jul 13, 2009
13. Jul 13, 2009

pesto

Re: .99c

Woah! Back to the old drawing board. Thank you for the explanation. I though $$u \oplus v$$ or "s" was the velocity of objectB relative to the observer!

Last edited: Jul 13, 2009
14. Jul 13, 2009

malawi_glenn

Re: .99c

here is really good explanation what is included http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

forget about my last post, I was not focused on this thread, Sorry for that.

I have changed it now so it should be correct now, I had many irons in the oven :-) (and maybe I was confused by the non-standard notation \oplus ) ;-)

The point of the calculation given by Fredrik is that light always move with speed c for all observers.

Last edited: Jul 13, 2009
15. Jul 13, 2009

DrGreg

Re: .99c

Sorry, that's confused.

If A, B and C are 3 objects all moving along the same straight line
• u is the velocity of B measured by A
• v is the velocity of C measured by B
• $u \oplus v$ is the velocity of C measured by A
The symbol $\oplus$ is not standard notation. Some may use it but others don't. It's just a way of denoting a different way of "adding" velocities. I'd prefer to call it "composition" rather than "addition".

16. Jul 13, 2009

malawi_glenn

Re: .99c

yup I know, it was an error, I fixed it.

regarding "composition" vs. "addition" I actually mentioned that $\oplus$ stands for "composition" right? why adding that?

Cheers

17. Jul 13, 2009

DrGreg

Re: .99c

Sorry, my intentions weren't clear. That comment was really aimed at Pesto in answer to post #11.

In case anyone reading this thread is confused, I was correcting an error in post #12, but then malawi_glenn corrected his own error at the same time, making my comment out of date by the time I posted it. But it's too late to undo that now without causing even further confusion.....

18. Jul 13, 2009

malawi_glenn

Re: .99c

it's ok, it do happens sometimes that a person is editing his posts meanwhile someone is pointing out the misstake they did - no big deal

19. Jul 13, 2009

pesto

Re: .99c

Thanks again. Fixed my post too :).