Doppler Effect for Arbitrary Motion

Consider an emitter and receiver moving (in vacuum) along arbitrary world lines, and suppose the emitter sends an electromagnetic wave at time t_e, and this reaches the receiver at time t_r, as depicted in the figure below (with two space dimensions).

The coordinates t, x, y (and z) comprise an inertial coordinate system, and the spatial coordinates of the emitter are given as functions of time by x_e(t), y_e(t), z_e(t), and of the receiver by x_r(t), y_r(t), z_r(t). The differential proper times along these world lines are

From these relations we get

Therefore, the ratio of the rate of the proper time of the receiver at the time of reception to the corresponding proper time of the emitter at the time of emission is

To determine the ratio dt_r/dt_e we make use of the fact that light follows null intervals, which means that

Hereafter we omit the arguments for the space coordinates (and velocities), with the understanding that the receiver coordinates are at the reception time t_r, and the emitter coordinates are at the emission time t_e. Differentiating both sides with respect to t_e, we get

Now we divide through by 2(t_f – t_e), and note that (x_r – x_e)/(t_r – t_e) is the x component of the velocity vector c for the pulse of light emanating from emitter at t_e to receiver at t_r, and similarly for the y and z components. Let c_x, c_y, c_z denote these components (so we have c_x² + c_y² + c_z² = 1). Also, notice that dx_f/dt_f is the x component of the velocity vector v of the receiver at the time of reception, and dx_e/dt_e is the x component of the velocity vector u of the emitter at the time of emission, and likewise for the y and z components. Therefore, the above equation can be written in the form

Solving this for dt_r/dt_e gives

Combining this with the reciprocal of equation (1), and noting that the frequency ratio is the reciprocal of the proper time ratio, we get the formula for the Doppler shift

where u is the velocity vector of the emitter at the moment of emission, v is the velocity vector of the receiver at the moment of reception, and c is the velocity vector of the light pulse from emission to reception, all evaluated in terms of the reference system of inertial coordinates. Using the trigonometric formula for the dot products, and recalling that the magnitude of c is 1, equation (2) can also be written as

where θ_c,v and θ_c,u are the angles between the subscripted vectors (again, evaluated in the reference inertial frame). These equations confirm that the Doppler shift for a given pulse of light from one event to another depends only on the velocities of the emitter (at the time of emission) and receiver (at the time of reception), not on the accelerations.

To illustrate the use of these equations, consider the question of whether there is a Doppler effect for light propagating from an emitter to a receiver, both attached to the rim of a circular disk that is rotating at uniform speed about its center. (This situation arises in a Sagnac apparatus.) Now, it is obvious from first principles that there can be no Doppler shift in this case, because, in terms of the rest frame of the central axis, the path length from emitter to receiver is not changing, and the emitter and receiver both have the same speed so there is no difference between their time dilation factors. Nevertheless, people are sometimes confused by the Doppler formulas above, thinking that those formulas predict some Doppler effect in this case. They typically agree that, in terms of the frame of the axis, we have u = v, so the factor under the square root in (3) is unity, but they think that the leading factor must not be unity, because they fail to recognize that, in this case, θ_c,v = –θ_c,u , so the cosines are equal, and hence the leading factor is unity as well. To understand why those angles are equal and opposite, consider the figure below.

The triangle on the chord is isosceles, so the angles that u and v make with c are equal and opposite. Therefore, the Doppler formula predicts no Doppler shift for this situation (as was obvious from first principles).

To describe the analysis in more explicit detail, in terms of the inertial coordinate system in which the central axis is at rest, let R denote the radius of the disk, V the speed of the rim, (t_e,x_e,y_e) the coordinates of the emission event, and (t_r,x_r,y_r) the coordinates of the reception event.  The velocity vectors u and v of the emitter (at emission) and receiver (at reception) are tangent to the circle, perpendicular to the position vectors, so we have

The velocity vector for the pulse of light from emission to absorption is the chord between them

where ρ_e = (x_e,y_e) and ρ_r = (x_r,y_r). Consequently we have

Inserting these values into equation (2) confirms that the frequency ratio is 1 in this situation.

In pure vector notation, which makes explicit that we can allow the disk to have any orientation (not specializing to the disk in the xy plane), let ω denote the angular rotation vector, and let ρ_e = (x_e,y_e,z_r) and ρ_a = (x_r,y_r,z_r). Then we have u = ω x r_e, v = ω x ρ_r, and c = (ρ_r – ρ_e)/|ρ_r – ρ_e|. In these terms, we have

Noting the scalar triple product identities A∙(B x A) = 0 and A∙(B x C) = –C∙(B x A), this confirms (again) that c∙u = c∙v, and hence the frequency ratio is 1.

One might wonder if the frequency ratio given by the Doppler formula (2) really is invariant under general Lorentz transformations. This is easily verified, by applying the relativistic velocity composition formulas for a frame moving with speed W in the negative x direction. Letting U, V, and C denote the velocity vectors of the emitter, receiver, and light pulse (respectively) in the transformed frame, we have

Substituting these vector components into (2), we get

This confirms that we get the same frequency ratio by applying equation (2) to the transformed coordinates as we do with the original coordinates. This enables us to specialize the equation if we wish. For example, if we transform to the inertial coordinates in terms of which the emitter is at rest at the moment of emission, then in terms of the transformed coordinates we have u = 0 and equation (2) reduces to

On the other hand, if we transform to the inertial coordinate system in terms of which the receiver is at rest at the moment of reception, then in terms of these transformed coordinates we have v = 0 and equation (2) reduces to

These specialized forms are most convenient when either the emitter or receiver is at rest in some inertial frame, and the velocity of the other is known explicitly in terms of that frame. It’s common for the kinematics of a situation to be specified in terms of some inertial frame, rather than in terms of an accelerating frame. For situations in which both the emitter and the receiver are accelerating (i.e., neither is at rest in an inertial frame), it’s likely that their velocities will be specified in terms of some inertial frame, and it’s most convenient to use equation (2). In such a case, it is still permissible in principle to use (4) or (5), but only for single pulses, and we must continually change our frame of reference for successive pulses, so this isn’t a particularly sensible approach. Needless to say, if we are given the velocity vectors u, v, and c in terms of some arbitrary frame, and we wish to convert to the frame in which (say) u = 0, the transformed vector v is not simply the vector difference v – u of the original vectors; we must use relativistic velocity composition to determine the appropriate vector v, and we must also apply the same Lorentz transformation to give the transformed vector c, accounting for aberration of the light ray.

It’s worth noting that, in different frames, the two factors contributing to the Doppler shift will have different values, even though the product of those factors is invariant. This is why the partition of the Doppler effect into a “path length change” factor and a “time dilation” factor is not absolute. For example, with the emitter and receiver on the rim of a rotating disk discussed above, if we work in terms of the inertial coordinates of the axis, there is no time dilation effect and no path length change effect. Both factors are 1. But if we consider the same situation in terms of the inertial coordinates in which the emitter is instantaneously at rest at the moment of emission, we find that there is a time dilation effect, but there is also a path length change effect, and these two effects cancel out, so the product of the two factors is still unity.

We showed formally that the frequency ratio given by (2) is invariant under Lorentz transformations by actually making the substitutions, but we can make this invariance more obvious by expressing the frequency ratio in terms of 4-vectors. Let us define two 4-vectors consisting of the spacetime coordinates of the emitter and receiver at the times of emission and reception respectively

Since these two events are light-like separated, we have

where the dot product here signifies the inner product, which is to say, A∙B = A^TηB where

Now, the usual product rule for differentiation applies to dot products, so we can take the differential of (6) to give

which implies

The 4-velocity of the emitter is V_e = dX_e/dτ_e and of the receiver is V_r = dX_r/dτ_r where τ_e and τ_r denote the corresponding proper times of the emitter and receiver. Making these substitutions into the above equation, and recalling that the frequency ratio is the reciprocal of the proper time ratio, we get

This is equivalent to equation (2), but since it is written in terms of dot products of 4-vectors, the invariance under Lorentz transformations is “self-evident”, at least if we have already convinced ourselves that dot products are invariant under Lorentz transformations. Recall from the note on Preserving the Intervals that a Lorentz transformation L is defined by the property that η = L^TηL. Therefore, given any dot product A∙B = A^TηB of 4-vectors, we can apply an arbitrary Lorentz transformation L to each of the vectors A and B, and then take the dot product, which gives

To see the equivalence of (7) and (2), we can write out the dot products in (7) as follows:

Factoring the first terms out of the numerator and denominator, and making use of the ordinary 3-vectors c, u, and v defined previously, this can be written as

which is identical to equation (2).

Incidentally, equation (7) corresponds to the equation for the Doppler shift given in section 2.8 of Misner, Thorne and Wheeler’s “Gravitation”. However, Misner et al use the 4-momentum P of a photon instead of X_r – X_e, and don’t actually demonstrate that the ratio of energies is equal to the ratio of frequencies. They simply invoke the relation E = hν from quantum theory (rather than proving that energy and frequency transform identically, as Einstein did in his 1905 paper on special relativity), and also tacitly make use (without proof) of the fact that dot products of 4-vectors are invariant under Lorentz transformations. Therefore they don’t actually present a comprehensive derivation, which has misled many of their readers. They then use their formula to determine that there is no Doppler shift between an emitter and receiver attached to the rim of a rotating disk (as we showed above). After “deriving” the ratio of dot products, they say “this ratio is most readily calculated in the inertial laboratory frame”, and they revert from 4-vectors down to the scalars and 3-vectors in terms of the rest frame of the axis (which they call a Lorentz frame), and note that, in this particular frame, the speeds of emitter and receiver are equal (so the time dilation factor under the square root is unity), and the angles that u and v make with c are equal, so the path length is not changing (in terms of these coordinates), and therefore the frequency ratio is unity. This matches our previous reasoning exactly, but…

At this point Misner, et al, weirdly say “Notice that this solution made no reference whatsoever to Lorentz transformations – they have not even been discussed yet in this book! The power of the geometric, coordinate-free viewpoint is evident!” This is highly misleading, first because they haven’t actually presented a genuine solution (they show only an energy ratio, and smuggle in the frequency ratio via quantum theory), second because their reasoning tacitly relies on the fact that dot products are invariant under Lorentz transformations (which they haven’t even discussed yet), and third because what they have presented is not coordinate-free. Granted they present the general expression for the Doppler shift in terms of 4-vectors, applicable to any system of inertial coordinates, but this does not give the solution to the problem. In order to actually determine that there is no Doppler shift in this situation they specialize to the Lorentz frame of the axis, and explicitly write down the velocity 3-vectors in terms of that coordinate system, noting that they have equal magnitudes and make equal angles with the light vector, statements that are true only for the coordinate-dependent variables in this particular system of coordinates. So, the “coordinate-free solution” of Misner, et al, of the “photon centrifuge” in that section is not really a solution (i.e., it does not represent a complete derivation of the Doppler effect), and it is not “coordinate-free”.

Of course, from the symmetry and time-invariance of the situation, one can see immediately that there is no Doppler shift, even without doing a calculation. It’s possible that Wheeler, et al, conflated the exploitation of symmetry and invariance with the concept of coordinate-freeness. Indeed, Wheeler famously said “Never make a calculation until you know the answer… try a simple physical argument (symmetry! invariance! conservation!) before every derivation…” One of his students, Feynman, recalled that he had once taken some calculations regarding radiation reaction to Wheeler, who immediately pointed out several reasons why Feynman’s idea wasn’t right. Feynman later wrote “What bothered me was, I thought he must have done the calculation. I only realized later that a man like Wheeler could immediately see all that stuff when you give him the problem. I had to calculate, but he could see.”

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