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In the May 1923 issue of "The Physical Review", in his paper "A Quantum Theory of the Scattering of X-Rays by Light Elements", Arthur H. Compton included a section titled "Energy Distribution from a Moving, Isotropic Radiator". In that section Compton worked out a mathematical description of how Special Relativity would apply to the emission of blackbody radiation as viewed through the quantum theory. In Compton’s paper, in which he presented his theory of the mutual scattering of photons and electrons, the section offered only a sideshow, but in the field of radiation thermodynamics we must count it an important piece of the puzzle, so I include it here as a tutorial.

"Let us study the energy radiated from a moving, isotropic body," Compton said. Notice that he did not explicitly refer to heat energy, also known to physicists as blackbody radiation. He also noted that, "...the condition of isotropy means that the probability is equal for all directions of emission of each energy quantum." I’ll note in passing that Compton did not use the word "photon" because that denotation for the quantum of electromagnetic radiation was not devised until 1926. In more modern terminology we recast the condition of isotropy as the statement that from the radiating body we get the same flux density of photons in all directions. To recast that statement into operational terms let’s imagine, for simplicity’s sake, that the radiating body has a spherical shape and that we have established a spherical detector shell of much greater radius around it and concentric with it. We then say that in a given elapse of time the same number of photons strikes any given patch of the detector sphere for all patches covering the same amount of area.

Now we introduce our coordinate grid with the center of our radiating body on the origin. We identify the x-axis as defining the direction of relative motion (when we have some), so for a line pointing in any direction we can measure an angle è between that line and the positive x-axis. Thus we can calculate the probability of a photon striking the detector sphere on a ring lying between the angles q and q+dq and we get

(Eq’n 1)

An observer moving in the negative x-direction at some speed v has a different perspective on that scene. In that observer’s inertial frame the radiating body and the detector sphere move in the positive x-direction at the speed v. Although other phenomena come into play and distort the image, we can imagine with Compton that we have an undistorted image of the scene before us. In that scene the detector sphere appears as an oblate spheroid because it has shrunk in the x-direction in accordance with the Lorentz-Fitzgerald contraction, its dimensions shrinking in inverse proportion to the Lorentz factor between the two inertial frames, the Lorentz factor coming from

(Eq’n 2)

If we track a photon that goes from the radiating body to the detector sphere in the stationary frame along a line that makes an angle q0 with the x-axis, then in the moving frame it moves along a line that makes an angle q1 with the x-axis. For both observers the photon must strike the same point on the sphere, so they can calculate their respective angles from the coordinates of that point. The Lorentz Transformation relates the coordinates as measured by the two observers as x0=x1γ and y0=y1, so we have the tangent of q0 as

(Eq’n 3)

in which, for convenience, I have used the usual substitution v=βc. Simple trigonometry tells us that, in consequence of that equation, we also have

(Eq’n 4)

Next we must take the motion of the radiating body into full account in the moving observer’s frame. Assume that the moving observer has built their own spherical detector shell centered on the point that the radiating body occupies at the instant it emits the photon. We note that the photon comes off its source at the angle q1, but also that it shares its source’s motion along the x-axis. It thus moves, for the moving observer, along a line that makes an angle q2 with the x-axis. In accordance with the second postulate of Special Relativity, it must travel along that line at the speed of light. Again we resort, as Compton did, to an act of trigonometry.

Organizing the velocity vectors into a triangle, we exploit the sine law, the statement that dividing the sine of an internal angle by the length of the side opposite that angle yields the same value for all three angles and all three sides of any triangle. We must also exploit the fact the Sin(π-q)=-Sin(q), although we only want the magnitude of the sine, so we ignore the algebraic sign on it. Thus we have

(Eq’n 5)

in which cq represents the component of the photon’s velocity along a line that makes an angle q1 with the x-axis. Next we apply the cosine rule for an obtuse-angled triangle. We get

(Eq’n 6)

Using that result in Equation 5 to solve for Sinq1 and then substituting that into Equation 4 gives us

(Eq’n 7)

That equation allows us to take the angle q2 measured by the observer who sees the radiating body move past their apparatus and convert it into the angle q0 measured on the same photon by the observer who has the radiating body sitting stationary in their frame.

In his next step Compton said that we take the differential of Equation 7 as

(Eq’n 8)

Even a minor acquaintance with the calculus tells us that differentiating Equation 7 does not yield Equation 8, so what did Compton have in mind? Yes, we know that

(Eq’n 9)

gives us an approximation suitable for physics, but only in the minuscule set of angles near the x-axis in this case. And the cosine term changes the differential altogether. But the use to which he put this differential tells us that he wasn’t actually differentiating the sine function of Equation 7; he was calculating the differential width of a band of latitude on his detector sphere and expressing the fact that the Lorentz-Fitzgerald contraction transforms it in the same way in which it transforms the sine.

Compton made use of Equations 7 and 8 in calculating the probability that an emitted photon would strike a given latitude on the detector sphere. We imagine that someone has painted a stripe of minuscule width along one line of latitude on the sphere (a line that sits a constant distance from the x-axis, which defines the direction of relative motion). Both observers must assign the same probability to the event of the emitted photon striking that stripe, so our observers calculate

(Eq’n 10)

That equation applied to each and every photon that the radiating body emits.

But the body does not emit merely one photon: it emits photons in great streams. An observer moving with the body measures that it emits photons at a rate of N0 photons per second. Our moving observer knows that the stationary observer’s clock ticks off dilated time, so they expect to measure the rate of emission in their frame as

(Eq’n 11)

Of that flux the amount radiated into the band lying between the angles q2 and q2+dq2 we calculate as dN2=N2p(q2)dq2. However the radiating body approaches the stationary observer with a component of velocity equal to βCosq2, so they must modify dN2 by multiplying it by the reciprocal of 1-βCosq2. That correction comes from the fact that the radiating body’s motion effectively compresses the space occupied by a give burst of photons.

Each photon carries an energy of hνq, in accordance with Planck’s hypothesis. If the observer moving with the radiating body measures the photon’s frequency as ν0, then the stationary observer would apply the relativistic Doppler shift to calculate

(Eq’n 12)

Compton commented on the observation that the frequency of each emitted photon would differ for the two observers, noting that such a difference seems inconsistent with the principle of conservation of energy. He responded by noting in a footnote that "When one considers, however, the work done by the moving body against the back-pressure of the radiation, it is found that the energy principle is satisfied. The conclusion reached by the present method of calculation is in exact accord with that which would be obtained according to Lorenz’s (sic) equations, by considering the radiation to consist of electromagnetic waves."

Finally we calculate the intensity of the radiation that our moving observer receives from the radiating body. The intensity Iq received at the angle q2 a distance R from the moving body conforms to

(Eq’n 13)

But an observer moving with the radiating body would measure the intensity of the radiation a distance R from the body as

(Eq’n 14)

Note that Compton has not envisioned his observers measuring the intensity of the radiation along the same line (which would make the distance R subject to the distortions of the Lorentz Transformation), but rather saw them measuring the intensity at the same given distance from the radiating body, presumably along different lines. Thus the two observers compare the results of their measuring the intensity of the radiation coming off the body through

(Eq’n 15)

Finally, imagine that the body under study emits pure blackbody radiation. As the moving observer passes it, they snap a quick full-spectrum photograph of the radiation emitted in one direction, q2. In accordance with the Principle of Relativity, no one looking at that picture can tell whether the source of the radiation was moving or not: they see only a blackbody spectrum. But our two observers can compare the spectra that they have obtained and they find that the peak frequencies on the spectra differ from each other in accordance with Equation 12. But Wien’s law tells us that the peak frequencies in the spectra of two bodies (or in the same body at different times) stand proportional to the absolute temperatures of the bodies in accordance with

(Eq’n 16)

If we now compare Equations 12 and 15 in light of that statement, we find that

(Eq’n 17)

which conforms to the Stefan-Boltzmann law. Compton used that result to verify his calculation of the relativistic emission of radiation.

habg

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