# Modern Physics/Gravitational Red Shift

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## The red shift

Spacetime diagram for explaining the gravitational red shift.

Light emitted at a lower level in a gravitational field has its frequency reduced as it travels to a higher level. This phenomenon is called the gravitational red shift.

We can see why this happens by using the principle of equivalence. Being in a gravitational field is equivalent to being in an accelerated frame, so knowing how the doppler shift works in such a frame will tell us how it works in a gravitational field.

We view the process of light emission and absorption from the the unaccelerated or inertial frame, as shown in the figure above. In this reference frame the observer of the light is accelerating to the right, as indicated by the curved red world line, which is equivalent to a gravitational force to the left.

The light is emitted at point A with frequency $\omega$ by a source which is stationary at this instant. At this instant the observer is also stationary in this frame. However, by the time the light gets to the observer, they have a velocity to the right which means that the observer measures a Doppler shifted freqency $\omega'$ for the light. Since the observer is moving away from the source, $\omega'<\omega$, as indicated above.

The relativistic Doppler shift is given by

$\frac{\omega'}{\omega} = \sqrt{ \frac{1 - U/c}{1 + U/c}} ,$

so we need to compute U/c. The line of simultaneity for the observer at point B goes through the origin, and is thus given by line segment OB. The slope of this line is U/c, where U is the velocity of the observer at point B. From the figure we see that this slope is also given by the ratio X' /X.

Equating these, eliminating X in favor of L = √(X2 - X2), which is the actual invariant distance of the observer from the origin, and substituting into the previous equation results in our gravitational red shift formula:

$\begin{matrix} \frac{\omega'}{\omega} & = & \sqrt{ \frac{X - X'}{X + X'} } & = & \sqrt{\frac{(L^2 + X'^2 )^{1/2} - X'}{(L^2 + X'^2 )^{1/2} + X'} }\\ & =& \frac{(L^2 + X'^2 )^{1/2} - X'}{L} & =& \frac{X-X'}{L} \end{matrix}$

If X′ = 0, then there is no redshift, because the source is collocated with the observer. On the other hand, if the source is located at the origin, so X′=X, the Doppler shifted frequency is zero. In addition, the light never gets to the observer, since the world line is asymptotic to the light world line passing through the origin. If the source is at a higher level in the gravitational field than the observer, so that X′ < 0, then the frequency is shifted to a higher value, i. e., it becomes a blue shift.

To see how this doppler shift relates to the strength of gravity, g, and the distance h between the source and the observer, first note that

$L=\frac{c^2}{g} \qquad X-X'=L-h$

Making these substitutions gives

$\frac{\omega'}{\omega} = 1-\frac{gh}{c^2} \qquad (1)$

So the redshift is proportional to gravity.

Since this doppler shift doesn't depend on the type of wave we can conclude that it is actually caused by time dilation, just like the doppler shift due to relative motion.

That is, gravity slows down time.

## Energy and frequency

In equation 1 gh is the change in gravitational potential energy so the change in frequency is proportional to change in potential energy, which suggests there might be a connection with energy conservation.

However, since we haven't yet established any connection between frequency and energy, we can't simply apply a energy conservation argument. Instead, we can argue in reverse, finding out what the energy-frequency relationship must be is energy is to be conserved.

Suppose we have two identical systems, both at rest in a uniform gravitational field g with initial energy E separated by a vertical distance h.

The system has mass E/c2, giving it potential energy, so the total energy of the two systems is initially

$E_i=2E - \frac{gh}{c^2}E$

where the second term is due to the lesser potential energy of the lower system.

The lower system emits a burst of waves, frequency ω, energy E(ω). When this reaches the upper system the waves have been red-shifted to frequency ω', energy E(ω'). This energy is absorbed by the upper system.

The total energy is now

$E_f=E-E(\omega)+E+ E(\omega') - \left(E-E(\omega)\right)\frac{gh}{c^2}$

Since we want to preserve energy conservation, these two equations must give the same result. Equating them, we get

$E(\omega') = E(\omega)(1-\frac{gh}{c^2})$

Comparing this with the doppler shift we see that

$\frac{E(\omega')}{E(\omega)} = \frac{\omega'}{\omega}$

which can only be true if E(ω) is proportional to ω

I.e, energy conservation implies energy is proportional to frequency, which is one of the axioms of quantum theory.

We could equally well have started with the quantum result and proved the gravitational red shift must exist. Either theory requires the other for consistency.

Since energy and frequency are each the temporal components of a four-vector, their being proportional implies the four-vectors themselves, and their spatial components, are also proportional. So, for waves, momentum is proportional to k.

Remember too, we saw earlier, when we looked at Hamilton's equations, that classical mechanics would be equivalent to a theory of anisotropic waves, in the geometrical optics limit, if energy were proportional to frequency and momentum to wave number. This proportionality isn't just required for energy conservation, it would make possible a theory uniting waves and particles.

None of this actually proves the proportionality, doing that requires experiment, but it does make it a natural assumptiom, which is indeed confirmed by experiment.

Because of all this, from now on we'll assume that energy and frequency are related in this way, with the constant of proportionality being $\hbar$

$(c\mathbf{p}, E) = \hbar (c\mathbf{k}, \omega)$

## Gravity and curvature

The gravitational redshift also implies that space is curved. We can see this by considering a rectangle in space-time.

Without gravity, if we start at some point A, wait for time t, then move at light-speed to the right for a distance h, we get to the same place, B, as if we move at light-speed for a distance h then wait for time t at rest with respect to A.

With gravity, if we follow the first path, we rise a distance of ct then wait for time t. On the second path, we begin by waiting for time t, but this is dilated by gravity. To an observer at B we appear to be waiting for a time t(1+gh/c2) before we start, so we end up at B later than on the first path.

Thus, with gravity, it matters which order we add distance vectors in. This can't happen if space is flat, so space must be curved.

To describe how it's curved we'd need the techniques of General Relativity.