How does gravitational redshift and blueshift work?

Question

In a gravitational field, if a source emits a signal from below (at higher potential) every second, the signal will be received above (at lower potential) with a lower apparent frequency because time elapses faster above than below. For example 1 second down = 2 seconds up

But if the time metric is changing, the frequency of the light should increase as it travels up the field as time accelerates and compensate for the difference in measured frequencies so that its frequency at the top would be measured as equal to its frequency at the bottom. (If the source travels up the field it will emit at higher and higher frequencies, so the frequency of the light should also increase and arrive at the top with the same apparent frequency as at the bottom)

So I don't understand the mechanism of gravitational redshift.

Answer 1

The only way to get a really clear answer is to do the calculation carefully yourself. But your question is also about intuition concerning what the calculation is telling us so I'll comment on that.

First of all there is the issue of how to compare a clock at one height with a clock at another. What exactly is being compared with what? I often see statements along the lines of "this clocks registers this and the clock at infinity registers that" and I think such statement are never clear.

Here is an experiment designed to clarify how clocks at different locations in a stationary spacetime can be compared. Take two caesium atoms and use them as the basis of two atomic clocks. Keep one by you as you sit high up, and lower the other on a string to some location low down. Let it stay there a while. Then lift it back up. Meanwhile keep a record of the number of times each caesium-atom-based clock has ticked. You will find that the one that made the journey to down low has a lower count of ticks, and furthermore the difference between the counts gets larger as the sojourn down low is extended for longer periods. This is what it means to say that the clocks up high are going faster.

Now suppose that the clock down low emits a microwave signal. The number $N$ of oscillations of the emitted signal is equal to the number of ticks of the clock. In the time interval during which those $N$ oscillations arrive at the location up high, the clock there ticks more than $N$ times. This is what it means to say there is a red shift.

Now you ask about what happens to the microwaves as they travel upwards. They are following free-fall trajectories so nothing happens to them. By this we mean that as they pass from one side to another of any given local inertial frame, their frequency as observed in such a frame does not change. To be precise, it does not change at either first or second order in distances and times across the local inertial frame.

Finally, what about a source of microwaves located at some intermediate height. It will emit with the characteristic timing of that height. So a source located half way between the down-low and up-high locations will find that the microwaves emitted from below it do not have the same frequency as the ones it may now emit.

At any given height you can of course also consider a source or a detector which is not at rest at that height but in motion (whether upwards or downwards or in some other direction). In that case there is also a Doppler shift to consider. You can always find a Doppler effect that is just the right amount to compensate the shift we call gravitational, and it is quite instructive to look into this.

Answer 2

In a gravitational field, if a source emits a signal from below (at higher

... instead (by convention): lower ...

potential) every second, the signal will be received above (at lower higher potential) with a lower apparent frequency

Plainly, without hedging: if source (sender) and receiver had been held rigidly wrt. each other, with the source lower than the receiver, and the source stated/emitted signals at a particular (constant) emission frequency, then the receiver stated its corresponding reception indications at a particular (constant) reception frequency, which is smaller than the emission frequency of the source.

because time elapses faster above than below.

This unfortunate phrase is so close to nonsense that its use is strongly discouraged; especially use by those, who wouldn't know (yet) how to express the relation between a sender and a receiver correctly (albeit more verbosely).

But [...] I don't understand [yet].

The most rigorous derivation I can offer is to carefully calculate the relations between a sender and a receiver both accelerating uniformly and rigidly wrt. each other in a flat spacetime region (this can be accomplished exactly, see below, by techniques familiar from the study of special relativity); and then to argue with the equivalence principle (as far as it pertains to geometry) that the relations between a sender and a receiver being held rigidly wrt. each other in a curved spacetime region are thus evaluated, too (at least under certain additional conditions, such as the separation between sender and receiver being "small", in some specific sense). Explicitly:

Considering in a flat region

• source $A$ accelerating uniformly, i.e. with constant proper acceleration of non-zero magnitude $g_A$,

• a receiver $B$ constrained to move "in the same direction, straight ahead" of $A$ (such that if $B$ has met and passed certain members of the same inertial frame, then $A$ was going to meet and pass those same members, in the same order), and

• requiring that $A$ finds constant non-zero ping duration $\tau A_{BA}$ wrt. $B$

then it follows that

• $B$ also must have accelerated uniformly, with acceleration magnitude $$g_B = g_A / \text{Exp}[ \, g_A \, \tau A_{BA} / (2 \, c) \, ] \lt g_A,$$

• and $B$ also found constant ping duration $\tau B_{AB}$ wrt. $A$, where

$$ \tau B_{AB} = \tau A_{BA} \, \text{Exp}[ \, g_A \, \tau A_{BA} / (2 \, c) \, ] = \tau A_{BA} \, \text{Exp}[ \, g_B \, \tau B_{AB} / (2 \, c) \, ] \gt \tau A_{BA}.$$

Given $g_A$ of the source, and selecting a receiver $B$ such that $$ 0 \lt g_A \, \tau A_{BA} / (2 \, c) \ll 1, $$

then

$$ 0 \lt \text{Exp}[ \, g_A \, \tau A_{BA} / (2 \, c) \, ] - 1 \ll 1, $$

$$ 1 \lt \frac{\tau B_{AB}}{\tau A_{BA}} \approx 1 + g_A \, \tau A_{BA} / (2 \, c) \approx 1 + g_A \, \tau B_{AB} / (2 \, c) \equiv 1 + g \, \Delta h / c^2 $$

for any suitable "intermediate" acceleration value $g \approx g_A \approx g_B$, and corresponding nominal value of "difference in height" $\Delta h \approx c \, \tau A_{BA} / 2 \approx c \, \tau B_{AB} / 2$ between receiver $B$ and source $A$.

Now suppose that source $A$ has stated $j$ tick indications, at "regular intervals", in the course of having observed $k$ successive ping signal roundtrips to $B$ and back. Therefore $A$'s tick rate has the value $j / (k \, \tau A_{BA})$.

Also, $B$ therefore observed the same number, $j$, of $A$'s tick indications in the course of having observed $k$ successive ping signal roundtrips to $A$ and back.
Consequently $B$'s tick-response rate (reception rate) has value $j / (k \, \tau B_{AB})$, i.e. less than $A$'s tick rate by a factor of $\text{Exp}[ \, g_A \, \tau A_{BA} / (2 \, c) \, ] = \text{Exp}[ \, g_B \, \tau B_{AB} / (2 \, c) \, ] \approx 1 + g \, \Delta h / c^2 $.
-- That's a.k.a. "redshift" of the source $A$'s tick signals, wrt. receiver $B$, which is (therefore) said to have been "held higher" than source $A$.