# Ratio of time dilation relation in different potentials

1. Dec 21, 2015

### sunrah

In the weak field limit, we have

$dt = (1 + 2\phi)^{-\frac{1}{2}}d\tau$

where the usual meaning of the symbols applies. This means that in GR dτ < dt analogous to SR. Let suppose we measure the period dtS of a photon emitted at the surface of the Sun as well as the same photon, i.e. same atomic process, in a lab on Earth where the period is dtE. Therefore we have two relationships w.r.t. to dτ such that

$dt_{S} = (1 + 2\phi_{S})^{-\frac{1}{2}}d\tau$

and

$dt_{E} = (1 + 2\phi_{E})^{-\frac{1}{2}}d\tau$

Taking the ratio we find that

$dt_{E} = \sqrt{\frac{1 + 2\phi_{S}}{1 + 2\phi_{E}}}dt_{S}$

because φ < 0 and |φE| << |φS|

we find that dt_{E} < dt_{S}

This is not what I would expect, surely the Sun's greater gravity means that the coordinate time interval there would be less than the same quantity measured in Earth's lesser gravity? Even though this gives the right relationship for gravitational redshift, I am still very confused.

2. Dec 21, 2015

### Staff: Mentor

Strictly speaking, this is the weak field limit for the case of a stationary spacetime, where the concept of "gravitational potential" $\phi$ is well-defined.

No. There is no analogy to SR here. Gravitational time dilation is not the same as time dilation due to relative motion. Many people try to make this analogy, but it doesn't work and only causes problems and misunderstandings.

Note carefully what these two relationships mean, physically. They are relationships between a coordinate time interval $dt$ and a proper time interval $d\tau$ for the same process. The proper time interval $d\tau$ is what would be measured locally, i.e., by an observer who is at the same spatial location as the emitted photon (on the surface of the Sun or the surface of the Earth). The coordinate time interval $dt$ is what would be measured by an observer at infinity, far away from the Sun and the Earth, out in empty space.

No, you have it backwards. The same process, viewed by the observer at infinity, takes more time (more coordinate time) to complete if it is happening on the surface of the Sun than on the surface of the Earth, because of the Sun's greater gravity--in other words, it is slowed down more (more coordinate time must elapse for the same proper time).

3. Dec 21, 2015

### sunrah

thanks, that's helpful but i'm still asking myself, if "clocks run slower in a G-field" why is the time interval I measure at infinity greater for the process in stronger gravity? If proper time in both cases is the same (I had a problem with this, but it is in my course script so...), then it is the coordinate time interval that must be affected by the field?

4. Dec 21, 2015

### Staff: Mentor

Because "clocks run slower" means less proper time for the same interval of coordinate time, or, conversely, more coordinate time for the same interval of proper time. In this case the interval of proper time is what is being held constant, so the second meaning (more coordinate time for the same interval of proper time) is the relevant one.