estimating/counting photons in two frames


by mathfeel
Tags: frames, photons
mathfeel
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#1
Feb27-10, 06:36 PM
P: 181
Suppose there is some EM wave in the vacuum with frequency and field strength [tex]\omega,E[/tex]

In the frame of someone moving along with the light, the frequency and field become:
[tex]\omega^{\prime} = \alpha \omega\,, E^{\prime} = \alpha E\,,
\alpha = \sqrt{\frac{1-\beta}{1+\beta}}[/tex]

Suppose both observers want to estimate the photon number. They do:
[tex]n \propto E^2/\hbar \omega[/tex]
in their respective frame and will come up with a number that differ by a factor [tex]\alpha[/tex]

So photon number is not a Lorentz scalar or function of one? That's not a big deal. But usually there'd be other related quantity (like time and space is related) that is transforms with n. What it is?
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bcrowell
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Feb27-10, 07:41 PM
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I believe that observers in inertial motion relative to one another always agree on the number of quanta. To get a disagreement, you have to have an acceleration, which leads to the Unruh effect ( http://en.wikipedia.org/wiki/Unruh_effect ), and the accelerations involved are so large that there has never been any way to experimentally confirm it. So if you're convincing yourself that different inertial observers in flat space disagree on n, then I think you've made a mistake in your calculation.

I don't think the field strength transforms the way you're saying. It transforms like three of the components of the electromagnetic field strength tensor.

Also, when you take the energy to be proportional to field strength squared, you're implicitly assuming that the volume over which you're integrating is fixed. It's not, because of length contraction.
mathfeel
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#3
Feb28-10, 03:18 AM
P: 181
Quote Quote by bcrowell View Post
I believe that observers in inertial motion relative to one another always agree on the number of quanta. To get a disagreement, you have to have an acceleration, which leads to the Unruh effect ( http://en.wikipedia.org/wiki/Unruh_effect ), and the accelerations involved are so large that there has never been any way to experimentally confirm it. So if you're convincing yourself that different inertial observers in flat space disagree on n, then I think you've made a mistake in your calculation.

I don't think the field strength transforms the way you're saying. It transforms like three of the components of the electromagnetic field strength tensor.

Also, when you take the energy to be proportional to field strength squared, you're implicitly assuming that the volume over which you're integrating is fixed. It's not, because of length contraction.

You are right. I forgot about volume. It's actually more interesting now because over a volume, I have to think about simultaneity.

This field transformation is correct. I first derived it using four-potential, but I just checked that it agrees with Griffiths:
[tex]E^{\prime}_x = \gamma (E_x - \beta B_{y}) = \gamma (1-\beta) E_x = \sqrt{\frac{1-\beta}{1+\beta}} E_x[/tex]

since E_x = B_y (in Gaussian unit) for wave in vacuum (traveling in the z direction, linearly polarized in the x direction).


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