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EM waves

  1. Jul 18, 2004 #1
    I'm reading an astronomy textbook, and I'm not sure about some things. The textbook says the EM waves are caused by accelerating charges. I don't understand how this is. I'm not sure, but my memory and intuition tells me that only charges moving in a sinuosidal (can never spell it) manner.

    Also regarding synchrotron emission, since the magnetic field does no work on the charge and the KE of the particle does not decrease, where does the energy in the radiation come from?

    And one more question. From what exactly does the radiation from a blackbody originate?

  2. jcsd
  3. Jul 19, 2004 #2


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    Any acceleration can be decomposed into sinusoids, if that helps you swallow it. However, fundamentally, sinusoidal behavior is not a requirement for EMR.

    One of the features that characterizes synchrotron radiation is that the motion of the charge is relativistic. In such a consideration, the electric and magnetic fields do not conveniently separate. You could speak of certain components of the electromagnetic field tensor as electric components and magnetic components, however, there are really just different components of the same field, and they can be rotated. What happens in the rest frame of the charge? Does the magnetic field accelerated the charge from rest, thereby doing work? No. In the rest frame, the components that were characteristically magnetic actually rotate into components that are characteristically electric. AFAIK, there is a decrease in KE, as well as gravitational PE.

    Firstly, realize that black-bodies are not believed to literally exist (except possibly for some exotic forms of matter). The model of the black body is derived from a resonant cavity. Modes of raditation bounce back and forth within the cavity. The temperature of the wall of the cavity determines the probablity that it will be able to absorbe or reflect a photon of a certain frequency mode back into the cavity, thus eliminating or preserving that frequency in the resononace. These two basic mechanisms together produce the familiar black-body spectral curve.
  4. Jul 19, 2004 #3
    Thanks for the help turin.

    But as far as black body radiation goes, from what I understand, a lot of things, although not perfect blackbodies, are pretty close in behaviour. Given that these objects are not cavities, how does one explain the radiation distribution for these?
  5. Jul 20, 2004 #4


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    The spectra of many of the very hot objects seen out in space are very close to what we would expect to see in a black body. It's not even limited to astronomical bodies; the spectrum emitted by a red-hot stove-top will also closely approximate the spectrum of a black-body.

    The way I understand it is that the high temperatures excite modes of small enough wavelength that the inter-particular spacings (and even the particles themselves) act like little cavities. The wavelength is so small that the consituents themselves actually absorb the radiation and hold onto it long enough to obscure its identity. Density also plays a roll in this mechanism of black-body impersonation. If the particles that compose the body are packed closely enough together, then they can release their acquired energy to adjacent particles in different amounts than the strict energy levels of there electrons (whereas in a sparse instersellar gas, for instance, you see sharp lines because the atoms in the gas are operating independently, and thus the emission is due to electron orbital transitions.)

    When the body releases the energy, it does so according to the statistics. That is, the frequency at which it releases the energy has a temperature dependence. At a particular temperature, there is a frequency at which a photon is most likely to be emitted because it corresponds roughly to the thermal quantum kBT. Of course, the temperature is statistical, so the radiated frequency is statistical.

    This may not be entirely accurate, but it is how I sleep at night.
    Last edited: Jul 20, 2004
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