# Relativity - wavelength differences for two observers?

1. May 31, 2010

### jeebs

hi,
my exam is tomorrow and I really need to understand this.

Imagine a region of constant gravitational field, directed "downwards". There is a photon travelling in this downward direction. There are two observers, A and B. At t=0, they do not have any relative velocity. However, B is fixed in position, whilst A is in free-fall. At time t=0, the photon passes the two observers, and they make measurements - both of them agree on the wavelength at this time.
I need to figure out what wavelength B observes at a time t = 0 + T later.
I also should express this in terms of (some or all of): v,c,g,w and T
where v is the relative velocity of the observers, c is obviously the speed of light, g is the constant, uniform gravitational field, T is the time that has elapsed since A was released from rest, and w is the photon wavelength.
Also, the time T is small enough that the velocity v is always far, far smaller than c.

So, what I thought was that, since A is in free-fall, he is completely unaware of any gravitational field. That means that, as far as A is aware, the photon is not going to increase its energy (and therefore shorten its wavelength) due to this gravitational field.

B on the other hand is fixed in place, so that he would notice the photon blue shifting as it travels in the direction of the gravitational field. This means at time T, the wavelength he measures should be shorter than w.

Also, in time T, the relative velocity between the two observers has increased from v=0 to v=gT. The separation of the two observers would therefore be y = vT = gT2.

However, this is where I start to get stuck. All the stuff I have done with doppler shifts has involved light shining from a source, where the observer and the source have some relative motion. In this situation though, the source isn't mentioned, it's just a photon.
This got me thinking about Einstein's postulate of special relativity - that the speed of light for any two inertial observers is the same.

However, in this case, the observers are not inertial. In observer B's frame, he is stationary whilst A accelerates away from him. As a result I am confused about what assumptions I can make, as according to observer B, observer A is a non-inertial observer. Do I still say that they both percieve the same value of c?

I am leaning towards the answer to that being no, because of what I have heard in lectures about the way the speed of light appears to certain observers/coordinate systems to become zero as a photon approaches a black hole's Schwarzschild radius. I must say though I am really not confident about this. There is no Scwarzschild radius in this question, but maybe a similar thing is happening?

What is it about B's velocity relative to A that causes the difference in observed wavelength?

Can anyone help me out here, I really need to understand this and I need to do it soon.
Thankyou.

Last edited: May 31, 2010
2. May 31, 2010

### jamesmo

Have you gotten to the point where you can use the metric? If so start by writing out your metric. Also, if you can make any useful assumptions (such was weak gravity approximation) try that. If you haven't gotten to metrics yet, where are you so that I can offer a starting point?