# Gravitational redshift equation

1. Apr 28, 2010

### Zman

Gravitational redshift is given by the following approximate equation;

$$\frac{\lambda}{\lambda_o} = 1 - \frac{GM}{r c^2}$$

From http://scienceworld.wolfram.com/physics/GravitationalRedshift.html

Where $$\lambda$$ is the shifted wavelength and $$\lambda_o$$ is the rest wavelength.
r is the distance from the gravitating body with mass M
The photon is being emitted from the surface of M directly away from the centre of M.

As r is increased and M constant, the redshift is increased as I expected. The photon has to climb further which reduces its energy which is expressed as a larger wavelength or lower frequency.

But with r held constant and M increased, I expected the energy loss of the photon to be increased at r. The photon now travels through a stronger gravitational field and should lose more energy than when travelling through a weak gravitational field.

But the equation above tells me that if r is held constant and M increased, then the gravitational redshift is reduced.

Where am I going wrong?

2. Apr 28, 2010

### yuiop

r is the radius at which the photon is emitted and had rest wavelength $\lambda_o$.
r is not the location of the observer or the photon in its travels. The location of the observer in this equation is always at infinity.

It seems that the wolfram website you quote got it wrong. The Wiki website http://en.wikipedia.org/wiki/Gravitational_redshift gives the redshift (z) as:

$$z= \frac{1}{\sqrt{(1 - \frac{2GM}{r c^2})}} -1 = \frac{\lambda - \lambda_o}{\lambda_o}$$

So:

$$\frac{\lambda}{\lambda_o}=\frac{1}{\sqrt{(1 - \frac{2GM}{r c^2})}}$$

Expressed like this the redshift ratio is unity when there is no redshift (eg when M=0) and tends towards infinite as r tends towards 2GM/c^2.

More generally:

$$\frac{\lambda}{\lambda_o}=\frac{\sqrt{(1 - \frac{2GM}{R c^2})}}{\sqrt{(1 - \frac{2GM}{r c^2})} }$$

where r is the location where the photon is emitted and has wavelength $\lambda_o$ and R is the location where the photon is received and the observer at R measures the the wavelength to be $\lambda$

If the observer is below the source (R < r) the observer measures the wavelength to be smaller and the frequency and energy of the photon to be higher. This situation is known as blue shift.

Last edited: Apr 28, 2010
3. Apr 28, 2010

### Zman

Thank you very much for that.

It also clears up an issue I had with the binomial expansion of;

$$\frac{\lambda}{\lambda_o} = \frac{1}{\sqrt{(1 - \frac{2GM}{r c^2})}}}$$

According to my understanding it should give;

$$\frac{\lambda}{\lambda_o} = 1 + \frac{GM}{r c^2}$$

And not

$$\frac{\lambda}{\lambda_o} = 1 - \frac{GM}{r c^2}$$

$$\frac{\lambda}{\lambda_o} = 1 + \frac{GM}{r c^2}$$ gives results that make sense (after your superb explanation) whereas the ‘1 – factor’ does not.

My original reason for doing this was to get a rough idea of how much gravitational redshift we should expect to see when viewing Neutron star.

A typical neutron star has a mass of very roughly 2 solar masses, with a corresponding radius of about 12 km

After plugging in these figures I came up with a gravitational redshift of 1.26

Not as big as I would have have thought.

4. Apr 28, 2010

### George Jones

Staff Emeritus
Careful: 1) the shift is a blueshift, not a redshift; 2) As r is increased with M constant, the blue shift decreases.
The photon is falling from infinity to a position with radial coordinate r.
Can you answer this question now?

Yes, it is.
I don't think so.

5. Apr 28, 2010

### George Jones

Staff Emeritus
Using the physical interpretation that I gave in my previous post,

$$\frac{\lambda}{\lambda_0} =\sqrt{1 - \frac{2GM}{r c^2}}},$$

and the first two terms of a binomial expansion give the expression on scienceworld's website.

6. Apr 28, 2010

### Zman

Thank you George for that.

I now have a clearer understanding of this issue.

But I would say that the equation that you used refers to gravitational blue shift (just as you said)
and the inverse of the equation that you used refers to gravitational red shift.

Without your contribution I would have remained confused about the equation found on scienceworld's website because I also found it in other respectable sources. It was too common to be a mistake.