Phase of radio waves received by a radio telescope dish

  • #51
sophiecentaur
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When considering photons, 1 photon+1 photon=2 photons, but their electric fields don't add in a similar manner.
Yes. Energy will be conserved but where and when that energy turns up will depend on the situation.
That equation looks pretty much like a standard diffraction / interference formula with some extra jiggery pokery. I still have a problem with actually assigning phase and position to the photon and that equation seems to be doing that. (from a fuzzy inspection)
 
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Charles Link
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Yes. Energy will be conserved but where and when that energy turns up will depend on the situation.
That equation looks pretty much like a standard diffraction / interference formula with some extra jiggery pokery. I still have a problem with actually assigning phase and position to the photon and that equation seems to be doing that. (from a fuzzy inspection)
See the previous post=post 50=a part I added, from Sakurai's Advanced Quantum Mechanics book. This results from some algebra he did just before that with some photon creation and destruction operators. The uncertainty relation may be telling us that we won't be able to reach any firm conclusions on the phases of the individual photons.
 
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  • #54
Charles Link
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I would like to make a couple additional comments to this discussion which I think mathematically are quite reasonable. There seems to be two different types of "laser" type (or coherent r-f signal) patterns being considered here=one is a plane wave and the other is a narrow pattern that has a finite beam spread/divergence. For both cases, if a given signal amplitude is considered, there needs to be a correlation of the phases of the signal across/throughout the beam (i.e. the 3-D beam pattern)=whether it is considered to be of a photon nature or simply as a completely classical picture. ## \\ ## For both cases, (of a plane wave or a beam pattern with a finite divergence), to double the intensity of the signal, the amplitude at each location only needs to increase by a factor of ## \sqrt{2} ## if the amplitudes are in phase everywhere. Somehow, a factor needs to get introduced where a factor of ## N ## in energy is a factor of ## \sqrt{N} ## in the resultant electric field amplitude everywhere. ## \\ ## One way to achieve this is by overlaying the layers and giving each layer a random phase. Another way, is simply to keep all the superimposed layers in phase, (or some arbitrary phase), with each other, and have a mathematics where ## E_{total}=\sqrt{N} E_o ##, where ## E ## is electric field amplitude, and the usual mathematics with phasor diagrams where ## E ## fields add is no longer applicable. The Q.M. description, (see also post 50), seems to use this latter approach. (edit: see the next paragraph. In this present paragraph, I was treating the photon as a fundamental sinusoidal disturbance with a given amplitude and having some phase. These disturbances would then add linearly, taking into account the phases as they are added. For random phases, the amplitude would then be proportional to ## \sqrt{N} ## . This picture of a photon may or may not be realistic.)## \\ ## After pondering this question further, one additional item occurred to me that perhaps resolves some or most of this: The energy density being proportional to the square of the amplitude is how waves behave in general in linear materials, so that one should not expect to lay a layer of (wave) energy in phase on a linear material and have the electric field amplitude double. This ## \sqrt{N} ## factor is normal behavior for a wave in a linear material.
 
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sophiecentaur
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Somehow, a factor needs to get introduced where a factor of NN N in energy is a factor of √NN \sqrt{N} in the resultant electric field amplitude everywhere.
I think you are worrying overmuch about this. Conventional Electromagnetic theory and RF Engineering took care of this, way before QM reared its head. We mostly accept that the two follow parallel paths and that, when we care to check, we can verify this. So why not have confidence to 'let go' and follow classical antenna theory and allow it to give the answers that you want?
I am not aware of any difference in the performance of a radio dish when it is part of a comms link (coherent signals) or when its used for looking at the power coming from astronomical objects (incoherent - mostly).
 
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  • #56
Charles Link
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@sophiecentaur The r-f approach is essentially the Kirchhoff diffraction theory which is also widely used and what I am most familiar with. I have found it very useful in working with photodiodes to have a photon model that is in reasonable agreement with the diffraction theory. I am satisfied, after a close inspection and review of the Q.M. ,(see post 50), with how the quantum theory actually is in close agreement to what the r-f engineers came up with long before the Q.M. was ever invented. And thank you for your inputs. :smile:
 
  • #57
davenn
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I think you are worrying overmuch about this. Conventional Electromagnetic theory and RF Engineering took care of this, way before QM reared its head. We mostly accept that the two follow parallel paths and that, when we care to check, we can verify this. So why not have confidence to 'let go' and follow classical antenna theory and allow it to give the answers that you want?

Couldn't agree more ... and in addition to over worrying/overthinking ... just making it plain overly complicated !

Classical physics was enough to answer the OP's question and unfortunately it seriously deviated into QM and the OP was lost way back in mid page 1
Cant say I blame him/her, even I got lost with the QM side of this thread :smile:

@Charles Link .... It's probably the best to remember the KISS method when answering threads :wink:

Tho it may have been a good discussion that eventuated, it didn't really answer the OP's questions in a straightforward way
as not everyone is well versed in QM

Regards
Dave
 
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