# How does the relativistic Doppler effect model change if I want to include general relativistic effects also?

Derivations for the relativistic Doppler effect abound, which extends the description of the Doppler effect to include the effects of relativity in cases where the source or observer is moving at relativistic velocities. This is all pretty well-described in many sources that I have found.

But what if I would like to include the effects of general relativity instead? Assume the following scenario:

• There is a stationary source on the earth emitting light at frequency $f$ (with that frequency measured on the reference frame of the source).
• I launch a rocket that travels to orbit with some velocity profile $\mathbf{v}(t)$.
• The rocket has a detector that continually observes the light source during its flight.
• I would like to know the observed frequency of the light wave from the perspective of the rocket, as a function of time, $f'(t')$ (where $t'$ represents time in the rocket's reference frame).

I see three separate effects that would alter the observed frequency aboard the rocket:

• The classical Doppler effect
• Time dilation on the rocket due to special relativity
• Time dilation on the rocket due to general relativity

All of these will change as a function of time as the rocket accelerates and as it moves within Earth's gravitational field. However, all analyses of the relativistic Doppler effect that I've seen only encompass the first two effects. Is there some model that includes all three?

• Technically you are missing the non-inertial frame and oblateness effects of Earth, but the end result is rather messy. You can find lots of derivations and examples by asking the Almighty Google about "doppler correction gps" for instance. I suggest GPS satellites because their clock drifts (relative to those on Earth) led to some of the first tests of GR. You can also look up "zonal harmonic coefficients" to see how many terms they include when they want to get really precise. Oct 23, 2018 at 20:29

This answer explains how the problem can be formulated in a way that naturally and automatically includes the effects you listed. I'll assume a spherically symmetric, non-rotating earth, and I'll ignore the influence of other bodies like the moon. I'll show how to set up the equations and describe how to use them in principle, but I won't solve them.

Points in space-time will be labelled using a system of coordinates $$t,x,y,z$$. We could describe the motion of the rocket by expressing three of these coordinates as functions of the other one, like this: $$x(t),\,y(t),\,z(t).$$ A more general approach is to specify all four of the coordinates as functions of an auxiliary parameter $$s$$, like this: $$t(s),\,x(s),\,y(s),\,z(s).$$ This defines a worldline, a curve in space-time that represents the rocket's entire history — where it was and when it was there.

The coordinates $$t,x,y,z$$ are just convenient labels for points in space-time. The time actually experienced by a passenger on the rocket is called the rocket's proper time $$\tau$$, which is typically not the same as $$t$$. This whole formulation is based on the following equation, which says how the proper time $$\tau(s)$$ at a given point $$s$$ along the worldline is related to the functions $$t(s),x(s),y(s),z(s)$$ that define the worldline: $$\dot\tau^2=A(r)\dot t^2 - \frac{1}{A(r)}\dot r^2 - (\dot{\mathbf{x}}^2-\dot r^2)$$ with $$A(r) = 1-\frac{\kappa}{r}.$$ An overhead dot means a derivative with respect to $$s$$, as in $$\dot \tau = d\tau/ds$$. The abbreviations $$\mathbf{x}=(x,y,z) \hskip1cm r = \sqrt{x^2+y^2+z^2} \hskip1cm \dot{\mathbf{x}}^2 = \dot x^2+\dot y^2+\dot z^2$$ are also used. The constant $$\kappa$$ is $$\kappa = \frac{2GM}{c^2},$$ where $$G$$ is Newton's constant, $$M$$ is the mass of the earth, and $$c$$ is the speed of light.
This proper-time equation is valid for $$r\geq R$$, where $$r=R$$ represents the surface of the earth. The inequality $$R\gg\kappa$$ ensures that the denominators in the proper time equation are never zero outside the earth. Inside the earth, the equation is different, but we won't need it here.

This proper-time equation implicitly specifies the metric field — the geometry of space-time — external to the earth. This particular metric field is called the Schwarzschild metric. I wrote the proper-time equation here using $$x,y,z$$ coordinates instead of the more traditional "spherical" coordinates (which ironically obscure spherical symmetry). The way I wrote it, the combination $$\dot{\mathbf{x}}^2-\dot r^2$$ corresponds to the usual "angular part" of the metric.

Now, suppose a transmitter is fixed somewhere on the surface of the earth, say at $$\mathbf{x}=(R,0,0)$$. This transmitter is an object with its own worldline, which we can parameterize as $$t(s)=s$$ and $$\mathbf{x}(s)=(R,0,0)$$. This implies $$\dot t = 1$$ and $$\dot{\mathbf{x}} = (0,0,0)$$. Use these in the proper-time equation to get this result for the transmitter's proper time: $$\dot\tau_T^2 = 1-\frac{\kappa}{R},$$ where the subscript $$T$$ means "transmitter." The right-hand side is a constant (independent of $$s$$), so this says that $$\tau_T$$ is proportional to $$s$$, which in turn is proportional to $$t$$.

So far, we have the transmitter's proper time $$\tau_T$$, and we know how to determine the rocket's proper-time $$\tau_R$$ at any point $$s$$ along the rocket's worldline, whatever that worldline may be. (The only constraint on the rocket's world-line is that the right-hand side of the proper-time equation must be positive, so the worldline is timelike.)

The remaining challenge is to relate $$\tau_R$$ to $$\tau_T$$.

We can do this using worldlines that represent the journey of light emitted by the transmitter. If we knew the worldline of every "piece of light" leaving the transmitter, then we could relate $$\tau_R$$ to $$\tau_T$$ like this: for any given point $$t,x,y,z$$ along the rocket's worldline, choose the piece-of-light worldline that passes through that point and also passes through the location of the transmitter $$\mathbf{x}=(R,0,0)$$. Only one piece-of-light worldline can do this, so this determines the specific value of $$\tau_T$$ when this "piece of light" must have left the transmitter in order to arrive at the specified point along the rocket's worldline. In this way, for every value of the rocket's physical time $$\tau_R$$, we can determine the value of the transmitter's physical time $$\tau_T$$ when that "piece of light" was emitted. In other words, we now have $$\tau_T$$ as a function of $$\tau_R$$, written $$\tau_T(\tau_R)$$. If the signal leaving the transmitter is $$\sin(\omega\tau_T)$$ according to the transmitter's clock, then the signal arriving at the rocket is $$\sin\big(\omega\,\tau_T(\tau_R)\big)$$ according to the rocket's clock, where this last expression is regarded as a (probably complicated) function of $$\tau_R$$. This is the total Doppler effect, including the effect of the rocket's velocity, the effect of the rocket's acceleration, and the effect of the earth's gravity. The key message here is that we should not think of these as separate effects, and we don't need to wonder if this list of separate effects is complete, because we naturally derived the full answer in a single, tidy package.

We still need to address one last thing: How do we know which world-lines represent the journey of a "piece of light"? Using the principle described in section 3.19 of General Relativity: An Introduction for Physicists, we can derive the following result starting from the same proper-time equation that was highlighted above. The result says that the worldline any object in free-fall, whether the object is a "piece of light" or a rocket with its engine turned off, satisfies a pair of equations that looks like this: $$\ddot{\mathbf{x}}=-\frac{\kappa}{2}\,\frac{\mathbf{x}}{r^3} \big(b + 3(\dot{\mathbf{x}}^2 - \dot r^2)\big) \hskip1cm \left(1-\frac{\kappa}{r}\right)\dot t = \text{constant},$$ where $$b\geq 0$$ is a constant that depends on how the world-line is parameterized. (These equations are valid for any affine parameterization.) To represent a "piece of light", just set $$b=0$$. I won't show the derivation of these free-fall equations here, because that would double the length of this already-long post.

# Definitions

Let's start with definitions since there are a lot of terms/symbols that are not commonly used by anyone outside the GPS timing community. Most of these notes and comments are taken from Ashby [2003] and Zhang et al. [2006].

• receiver $$\equiv$$ station receiving signal from spacecraft transmitter (subscript r)
• satellite $$\equiv$$ object in orbit that is transmitting signal (subscript or superscript s)
• LOS $$\equiv$$ line-of-sight
• J2 = 1.0826300 x 10-3 $$\equiv$$ Earth's second zonal harmonic.
• $$P_{2}(x) = \tfrac{1}{2} \left( 3 x^{2} - 1 \right)$$ $$\equiv$$ Legendre polynomial of degree two
• ae = 6378.137 km $$\equiv$$ semi-major axis of the World Geodetic System ellipsoid
• aorb $$\equiv$$ semi-major axis of satellite orbit
• B $$\equiv$$ latitude of receiver
• ||r|| $$\equiv$$ distance from Earth center to receiver
• GM = 3.986004418 x 1014 m3 s-2 $$\equiv$$ product of Earth's mass times Newtonian gravitational constant
• $$\Phi_{o}/c^{2}$$ = 6.969290134 x 10-10 $$\equiv$$ potential of the geoid (see IERS reference PDF)
• Ur $$\equiv$$ Newtonian gravitational potential (mass attraction only) at receiver location (see Equation 1 below)
• Us $$\equiv$$ Newtonian gravitational potential (mass attraction only) at satellite location (okay treat Earth like point mass)
• $$\mathbf{v}_{r}$$ $$\equiv$$ velocity of receiver relative to receiver-to-satellite LOS
• $$\mathbf{v}^{s}$$ $$\equiv$$ velocity of satellite relative to receiver-to-satellite LOS
• $$\mathbf{n}_{r}^{s}$$ $$\equiv$$ receiver-to-satellite LOS unit vector
• $$f_{o}$$ $$\equiv$$ nominal signal frequency from satellite
• $$f_{r}$$ $$\equiv$$ signal frequency at receiver from satellite

# Doppler Effects

First we define the gravitational potential at the receiver, given by: $$U_{r} = - \frac{ G M }{ \lVert \mathbf{r} \rVert } \left[ 1 - \left( \frac{ a_{e} }{ \lVert \mathbf{r} \rVert } \right)^{2} \ J_{2} \ P_{2}\left( \sin{B} \right) \right] \tag{1}$$

Now we can define the frequency received from an orbiting satellite as: $$f_{r} = f_{o} \left[ 1 + \frac{ - U_{r} + v_{r}^{2}/2 + \Phi_{o} + 2 G M/a_{orb} + 2 \ U^{s} }{ c^{2} } \right] \ \left[ \frac{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}_{r} \right)/c }{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}^{s} \right)/c } \right] \tag{2}$$

Technically, this is in a non-inertial reference frame and the $$v_{r}^{2}/2$$ must include centrifugal effects such that Equation 2 goes to: $$f_{r} = f_{o} \left[ 1 + \frac{ - U_{r} + \left(\omega^{2} \ \cos^{2}{B} \ \lVert \mathbf{v}_{r} \rVert^{2}\right)/2 + \Phi_{o} + 2 G M \left(a_{orb}^{-1} - \lVert \mathbf{r}^{s} \rVert^{-1} \right) }{ c^{2} } \right] \ \left[ \frac{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}_{r} \right)/c }{ 1 - \left( \mathbf{n}_{r}^{s} \cdot \mathbf{v}^{s} \right)/c } \right] \tag{3}$$ where $$\omega$$ is the angular velocity of Earth's rotation.

# Caveats

There are other errors that come into play like the non-vacuum index of refraction of the magnetosphere, ionosphere, and atmosphere between the satellite and receiver. There are also errors/uncertainties in both the satellite and receiver clocks [e.g., Ashby, 2003; Zhang et al., 2006].

# References

• Ashby, N. "Relativity in the Global Positioning System," Living Rev. Relativity 6, 2003, Online article, https://link.springer.com/article/10.12942/lrr-2003-1. Accessed on 10/26/2018
• Zhang, J., et al., "On the Relativistic Doppler Effect for Precise Velocity Determination using GPS," J. Geodesy 80, pp. 104--110, 2006.