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A black body in empty space in the CMB rest frame should (discounting starlight) come to an equilibrium temperature of around 2.7 K.

However, if the body is moving relative to the CMB frame, one side of the sky will appear hotter, and the other side colder, and it is not immediately obvious that these effects will balance out; in fact, it seems like the hot side should win, as absorbing higher-frequency photons from the front side should decelerate the body unless its speed is maintained by some other force, gradually converting its kinetic energy in the CMB rest frame into heat.

To find the equilibrium temperature it seems that I would need to integrate over the surface of a sphere, applying the formula for the relativistic Doppler effect to find out what the apparent temperature of the sky is in every direction at a given velocity. However, I am unsure how to properly account for the relativistic aberration, which causes the view to contract towards the front and expand in the back, which would alter the distribution of radiative power over the surface of the black body even apart from frequency-shifting effects.

What's the right way to approach this?

EDIT: Here's the best I've got so far:

http://arxiv.org/pdf/1503.05845v3.pdf gives a formula for the thermal distribution of the sky in the forward hemisphere (as measured in the rest frame). A model that covers the entire sky would be better, but I figure this will at least serve as a good first-order estimator. (Turns out to be pretty much the same formula you can get from Wikipedia, but getting it from an academic paper on interstellar travel makes me feel more confident that it's actually the correct formula to use.)

So, half of the incident radiation looks like this:

$T = T_0 \frac{\sqrt(1 - b^2)}{(1-b~cos(t))}$, where $b = v/c$.

To find the equilibrium temperature, we need to integrate the power received over the whole sky; or, in this case, half the sky, 'cause the model is only valid for backwards-moving photons. To do that, we raise the temperature to the 4th power; unit conversion constants don't matter 'cause we'll just be converting this back into a temperature later with a quartic root. The total power from half the sky should then be given by

$P = T_0^4 (1-b^2)^2 \int_{0}^{\pi/2} (sin(\theta)/(1-b~cos(\theta)))^4 d\theta$

This is a really nasty integral, and but with the help of Wolfram Alpha I get:

$P = T_0^4 \frac{(1-b^2)^2}{6~b^4} (\frac{6 (3 b^2-2) arctanh(\frac{b+1}{\sqrt(b^2-1)})}{(b^2-1)^{3/2}}-\frac{b (-2 b^4+5 b^2-6)}{1 -b^2}+3\pi)$

But this looks very wrong. It, too, peaks before b = 1. Maybe it should do that, but that seems very counterintuitive to me. This makes me think that something, somewhere along the way, has gone very wrong. It also starts out predicting less-than-baseline power when b gets close to zero, but that's to be expected- it's ignoring the whole aft hemisphere, so the approximation should only be good for large values of b.

Soldiering on for now, we can get back to a temperature by taking the fourth root:

$T = T_0 \frac{\sqrt(1-b^2)}{1.5651~b} (\frac{6 (3 b^2-2) arctanh(\frac{b+1}{\sqrt(b^2-1)})}{(b^2-1)^{3/2}}-\frac{b (-2 b^4+5 b^2-6)}{1 -b^2}+3\pi)^{1/4}$

And that predictably also looks very wrong. See this graph.

Now, my first thought was that maybe we can't actually ignore the aft hemisphere and still get good results; photons coming in mostly from the side but with a slight forward component will end up coming in from the front, and those transverse photons might end up providing a significant fraction of the total power flux at high beta. But even without that, the power should never go down; even if they are concentrated into a smaller viewing cone, you don't start hitting fewer photons by going faster, and the ones you do hit will always be higher energy.

So, where am I going wrong?

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  • $\begingroup$ Maybe I'm missing something but if we visualize a giant blackbody cavity at an equilibrium temperature of, say, 2.7K, then a stationary body inside it will also equilibrate to the same temperature. If that body then moves at some high or even relativistic speed within the cavity, doesn't it have to stay at the same equilibrium temperature? Otherwise, it seems that it will get hotter without limit (if net flux to it is +) or else get colder until it reaches 0K (if net flux is -), and either option seems inconsistent with having a body in a blackbody cavity at some equilibrium temperature. $\endgroup$ – Samuel Weir Apr 14 '16 at 23:59
  • $\begingroup$ @SamuelWeir It doesn't have to get hotter without limit. Radiation carries momentum as well as energy, so whenever the body absorbs energy from a photon, it will also accelerate. Unless I'm missing something, the limit is that it ends up at rest wrt the radiation rest frame, having converted all of it's kinetic energy in that frame into heat. So, the net flux should be positive, but not unbounded, unless an external force keeps the body at constant velocity. $\endgroup$ – Logan R. Kearsley Apr 15 '16 at 0:06
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    $\begingroup$ @CuriousOne The radiation is not isotropic, but there is still a well-defined total power that the body will absorb, corresponding to an equilibrium temperature at which it will radiate that much power back out. It should, in principle, be no different than calculating the equilibrium temperature of a black body orbiting the sun. Could you elaborate on the treatment with ray optics? I'm not sure where you're going with that. $\endgroup$ – Logan R. Kearsley Apr 15 '16 at 4:08
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    $\begingroup$ @CuriousOne Yes. And? I don't see the relevance. It still has a well-defined temperature at radiative equilibrium. I made no claim about thermal or thermodynamic equilibrium- I just want to know how to find the black body temperature at radiative equilibrium, which is one necessary condition for thermodynamic equilibrium, but can occur in the absence of thermodynamic equilibrium. Objects in radiative but not thermodynamic equilibrium still have a well-defined temperature. $\endgroup$ – Logan R. Kearsley Apr 15 '16 at 6:44
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    $\begingroup$ @CuriousOne That's why I specified "black body"- absorptivity and emissivity of 1. But fine, if thinking in terms of an equivalent temperature bothers you, go ahead and ignore the temperature and just think of it in terms of radiant intensity- how much power do you get from the whole sky? I can do a fourth root to convert to an effective average temperature myself. $\endgroup$ – Logan R. Kearsley Apr 15 '16 at 7:00
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An arbitrary speed is tricky, but the ultra relativistic case is easier.

If you look at this What's the max speed a man-made satellite can travel in space before its circuitry stopped working? SE answer, I noted that the *specific intensity * (in W m$^{-2}$ sr$^{-1}$ Hz$^{-1}$ of the CMB radiation is boosted by a factor $z^3$, where $z$ is the usual redshift factor, and if $z \gg 1$, then $z^3 \simeq 8\gamma^3$. The radiation arrives in beamed fashion from a small spot on the sky in the forward direction, with an angular diameter of $\sim 1/z$ and "looks" like a blackbody spectrum with temperature of $z T_0$ (if the CMB has a rest-frame temperature of $T_0=2.7$K). The solid angle subtended is reduced by a factor $z^2$.

The overall effect of these transformations is that if in the rest frame the object received a flux $$f(\nu) = 4\pi B_{\nu}(T_0),$$ where $$B_{\nu}(T_0) = \frac{2h \nu^2}{c^2} \frac{1}{\exp(h\nu/kT_0)-1},$$ then in the moving frame it receives $$f^{\prime}(\nu) = \frac{4\pi}{z^2} B_\nu(zT_0)$$

Integrating over all frequencies we get the usual Stefan's law and find that the total received flux at all frequencies is boosted by a factor of $z^2$. To balance this, the radiated power must also go up by a factor of $z^2$ and so the temperature of the object must increase by a factor of $z^{1/2}$.

I see from your edit that you may be interested in a more general case where you cannot assume that $z \gg 1$. In which case you're on your own!

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  • $\begingroup$ Thanks! Not entirely on my own- now I can check that any formula I come up with matches this in the ultrarelativistic limit. That'll definitely help! $\endgroup$ – Logan R. Kearsley Apr 15 '16 at 9:02

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