Red shift, blue shift, or stupidity?

Quantum1332

This might be a stupid question, but it came to me the other day while reading. Since the speed of light is the same in all frames what would happen if you were chasing a light beam at .999% c, so you were heading right for it (but could never catch it) and it is ahead of you traveling at c. If you were to include the doppler effect what color would the light beam appear?

jtbell

Since you'd never catch up with the light, you'd never see it at all!

Doc Al

I don't understand the scenario. If the "light beam" is moving away from you, you won't see it at all.

[jtbell beat me too it!]

Quantum1332

if it were possible to see it?

JesseM

The doppler shift equation is based partly on the time dilation of the source in your frame--in other words, if a source is emitting wave peaks at a frequency of 5*10^14 per second (visible light) in its own rest frame, then because of time dilation, when it's moving with respect to you it will emit less peaks per second in your frame. In the limit as the velocity of a clock in your frame approaches c, the time between ticks of the clock approaches infinity, no matter how long between ticks in the clock's own frame...so analogously, in the limit as a source's velocity relative to you approaches c, the time between peaks of any signal it emits in your direction approaches infinity too. So I think the answer you're looking for is that the light from a source moving at c would be infinitely redshifted, even though that only makes sense as a statement about limits since a massive light-emitting source cannot actually move at c.

clj4

Black.

Black is black,
But I want my baby back.
For grey is is grey,
Since you went away.

pervect

If you were heading away from the source emmiting light of a known frequency at .999c, so that the light from the source could catch up with you, then you would just use the relativistic doppler shift formula to convert the velocity into a redshift:

http://hyperphysics.phy-astr.gsu.edu...v/reldop2.html

You would compute

[tex]
\sqrt{\frac{1+\beta}{1-\beta}}
[/tex]

to get sqrt(.001/1.999) and a redshift frequency factor of .022 (or a wavelength incrase of 44.7).

As various posters have pointed out, it doesn't make any sense to ask what the frequency is if you can't observe it.