Built on Facts

You’re taking your morning shower and a thought occurs to you. “In classical electrodynamics, an accelerating charge radiates. In general relativity, acceleration is equivalent to a gravitational field. Therefore a stationary charge should radiate simply by virtue of being in a gravitational field. What’s up with that?”

You wonder about what the radiated power would be for a given gravitational field. You figure maybe you could use the Larmor formula with the Stefan-Boltzmann law to estimate the equivalent thermal radiation but you don’t remember either one of those equations exactly and you’re pretty sure you’d have to finagle some spatial factors anyway (the Stefan-Boltzmann law has a factor of surface area).

One alternative is to try to construct a quantum field theory in curved spacetime, but this is ludicrously tough even if you’re not in the shower without pen and paper. But we might be able to just juggle some constants around and get an estimate. We’ll start off with the assumption that the effective temperature of the radiation by the charge is proportional to the local acceleration due to gravity. In other words, that the temperature is just the acceleration multiplied by some constants that we don’t know:

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Well, what is that stuff? In terms of units, we’ve got:

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We need to pick the “stuff” so that this works out. Remember that the kelvin is the unit of temperature, 0 K being absolute zero with room temperature (roughly) around 300 K. Well, we need to pick a fundamental constant involving temperature. Boltzmann’s constant k fits the bill. Its units are J/K (ie, energy per temperature), so our expression needs to have k in the denominator so that the kelvin comes out on top:

That meters/seconds needs to be killed off. There’s pretty much only one universal constant involving speed, and that’s the speed of light c. Again, it needs to be in the denominator to cancel out the meters and seconds:

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With units:

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There’s still a “seconds” that we need to get rid of, and there’s also that “joules” that we introduced when we decided to drag Boltzmann’s constant into things. Are there any constants involving joules and seconds that might help us out? There is indeed a doozy of a constant: Planck’s constant h. We end up with units that cancel properly:

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With the units written explicitly:

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That might reassure us that the radiation emitted by a charge accelerating at 9.8 m/s^2 is pretty small, since h is so tiny and c is so large. k is pretty tiny as well, but the tininess of h is much greater. Or less, depending on how you look at it. In any case in our estimate T is around 10^-19 K for an acceleration equivalent to the earth’s surface gravity.

So you get out of the shower and Google the Unruh effect, you’ll see that the actual equation derived with much pain and suffering by Bill Unruh is:

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We’re off by a factor of 4 pi^2, or about 40, but in terms of order of magnitude that’s pretty good. We didn’t even have to get out of the shower.

Comments

  1. #1 Raskolnikov
    October 20, 2010

    Nice one.

    I have a question though, you started talking about classical electromagnetism. How is quantum mechanics suddenly entering the picture through h? I mean, I understood your mathematical reasoning, but since you also expect the presence of radiation classically, isn’t it surprising that you need to implicate quantum mechanics?

    Now, I know that nature doesn’t care about the distinctions we make between our various theories, but still, I think there’s something to be understood here.

  2. #2 Raskolnikov
    October 20, 2010

    Could it be related to the problem of self-energy and the “size” of elementary particles?

  3. #3 Eric Lund
    October 20, 2010

    If the Wikipedia page is to be believed (a dubious assumption), then Planck’s constant enters for the same reason it turns up in calculating the blackbody spectrum. Once you assume that the radiation can be described by an effective temperature, you need to make sure there is no ultraviolet catastrophe in your theory, even if the “ultraviolet catastrophe” would actually be deep in the ULF range.

  4. #4 Neuenschwander
    October 20, 2010

    When I get writers block I always take a shower. I often refocus and come up with some new ideas. Ritual therapy I guess, plus I smell better.

  5. #5 Uncle Al
    October 20, 2010

    1.74 solar-mass 2.1499 ms pulsar PSR J1903+0327 enjoys 1.8×10^11 surface gees – not nearly enough acceleration to be charged particle radiationally interesting. Its equator spins at 11% of lightspeed. Still not enough acceleration. Bremsstrahlung is not optimistic. Looks like we need a very fat gamma ray photon to kick an electron’s butt, or an ultrarelativistic e-e head-on collision.

  6. #6 Raskolnikov
    October 20, 2010

    Hi Eric,

    that does make sense, but I kinda remembered that another problem with radiation from an accelerated charge is that it comes from interaction of the charge with its own retarded field. The problem is that when you try to write down an equation for it, you somehow have to introduce a particle size, if I remember correctly. That’s why I thought Planck’s constant might enter that way as well. I guess that is the factor that cuts off the possible emission of arbitrarily high frequency waves.

  7. #7 Josh
    October 20, 2010

    The classical emission of radiation from a moving charge is a completely different effect than the Unruh temperature. A classical charge only emits if there is a jerk, that is, a time-dependent acceleration. You can see this from the Abraham-Lorentz force law in which the reaction force from emitting classical radiation goes as the time-derivative of the acceleration. If uniformly accelerated charges radiated, then everything at rest on the surface of the earth would be radiating like crazy.

    The Unruh effect is purely qunatum in nature and occurs for a detector that follows a uniformly accelerated trajectory in the vacuum of a quantum field theory. It doesn’t really matter what the quantum field theory is, as long as it respects special relativity. There doesn’t even have to be a notion of electric or any other kind of charge. In any case, this detector will observe a background gas of the quanta (whatever they are) at the Unruh temperature.

    At any rate, nice dimensional analysis to get the approximate form for the Unruh temp.

  8. #8 Paul Murray
    October 20, 2010

    Thanks for this – I think I asked it as a question here a while ago.

    @7 As for classical emission being depended on change in acceleration, this means that a charge in constant circular motion doesn’t rad … no, of course not. We are talking about a permanent magnet in that case, and permanent magnets don’t radiate power.

  9. #9 Jesse
    October 21, 2010

    Very cool post, I love this kind of surprise elegance that pops up now and then.

    I want to comment on your specific example; it made me smile. I was lucky enough to be a student of Bill Unruh a few years ago at the University of British Columbia. It wasn’t until the very end of the semester that any of the students clued into the fact that he has an ‘effect’ named after him. He was definitely one of the best professors I ever had – great, intuitive teaching style, extremely friendly, generous with office hours, and funny. On day one he introduced himself, “I’m Bill Unruh… Ruh as in ‘rue the day’ and Un as in… not.”

    Before the final exam he held a series of extended office hours which were attended by half the class – close to 50 students crammed into a tiny room for hours and hours going over relativity and QM problems on the blackboard until late at night, and he personally bought pizza for every one of us. Not once during the entire course did he bring up his own work or renown.

    Great prof, great person. Thanks for the nostalgia!

  10. #10 agm
    October 21, 2010

    Bravo. And well-written too, glad to have you back good sir.

  11. #11 Raskolnikov
    October 21, 2010

    @ Josh : Yeah, you’re right, forgot about the Abraham-Lorentz law. Maybe I should check that chapter in my electromagnetism book before opening my mouth again. ;)

    @ Jesse : I met professor Unruh at a conference in Dresden. I remember asking him about his name, I suspected he was from german descent, since “Unruh” in german means “unrest”. He confirmed as much. He also was talking pationately about the movie “the Third Man” and how he visited Vienna to find all the film locations. Funny guy.

  12. #12 Rach
    October 21, 2010

    I’m puzzled; you say that a stationary charge in a gravitational field will radiate energy. Where does the energy come from? Can it continue indefinitely?

  13. #13 Bill K
    October 21, 2010

    Sorry Matt, better go back and take another shower. God, this is wrong in so many ways its hard to know where to start. The effect you’re considering has nothing at all to do with Unruh radiation, they’re completely different.

    Unruh radiation is a quantum effect, saying that to an accelerated observer the vacuum state appears to contain a bath of thermal radiation. That is, he believes himself surrounded by black body radiation at a certain temperature T, very similar to Hawking radiation. Note the Wikipedia page mentions that anyway this does NOT apply to an observer sitting on the surface of the Earth, or going around in a circular path.

    What you’re visualizing is a charged particle sitting at rest on the Earth’s surface, and wondering if it emits electromagnetic radiation. The answer is no. One way to see this is to realize that the radiation must escape to infinity, and in the rest frame at infinity the particle is not accelerating.

    Yes it is locally accelerating, but from that point of view one must also consider how it is being supported. To remain still it must be supported by a vertical electric field, e.g. between the plates of a parallel plate capacitor. But then the two plates are also charged. And also accelerating, hence they also radiate. Carrying opposite charges, their radiation tends to cancel. The plate nearest the center of the Earth has a slightly greater acceleration, hence emits a slightly greater radiation, just enough greater to also cancel the radiation coming from the charge. Net result: zero.

  14. #14 Taylor
    October 26, 2010

    This may be a silly complaint, but try \left( and \right) for properly sized parentheses.

  15. #15 TBRP
    October 28, 2010

    Dimensional analysis can be interesting and informative, but remember not to abuse it.

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