# Gravitational Redshift really verification of GR?

#### thehangedman

If we assume:

$E = mc^{2}$

and for photons:

$E = hv$

Then we can derive an effective mass:

$m = \frac{hv}{c^{2}}$

And using simple classical gravity obtain:

$hv - \frac{GMm}{r} = hv - \frac{GMhv}{c^{2}r} = Constant$

You can derive the constant by evaluating the equation above at the limit as r goes to infinity. This then gives you the gravitational red shift, all without using anything from GR.

So, since this prediction is the same for classical AND GR, how can red shift be used as verification of GR? I'm not questioning GR, just wondering why this is still listed as verification when it's clearly not.

Related Special and General Relativity News on Phys.org

#### kmarinas86

If we assume:

$E = mc^{2}$

and for photons:

$E = hv$

Then we can derive an effective mass:

$m = \frac{hv}{c^{2}}$

And using simple classical gravity obtain:

$hv - \frac{GMm}{r} = hv - \frac{GMhv}{c^{2}r} = Constant$

You can derive the constant by evaluating the equation above at the limit as r goes to infinity. This then gives you the gravitational red shift, all without using anything from GR.

So, since this prediction is the same for classical AND GR, how can red shift be used as verification of GR? I'm not questioning GR, just wondering why this is still listed as verification when it's clearly not.
That's not correct. The gravitational redshift factor corresponds the to change in gravitational time dilation $\frac{d\tau}{dt}-1=\frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}-1$.

#### Passionflower

If you are referring to the Pound and Rebka experiment, it proves gravitational redshift but it does not prove the validity of the Schwarzschild solution (even when we ignore rotation) as the experiment is not accurate enough to do that.