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1. Aug 1, 2018

### jordankonisky

I understand that a radio telescope can be tuned to receive radio waves generated by neutral hydrogen atoms present in galactic gas, for example, within the spirals of the milky way. I think I understand that the incoming radio waves will be a mixture of red- and blue-shifted photons depending on the whether the photon sources are moving towards or away from the dish. But what about the phases of the collection of photons? Since the sources may differ in distance and time of photon release, is the signal a collection of photons in different phases? And, if so, how does the telescope take this into account in processing the data? Or perhaps the phase does not matter at all. Am I thinking about this in the right way?

2. Aug 1, 2018

This is something that I don't know that the people who are experts in the Quantum Theory will be in agreement on, but I believe, even in a laser, the phases of the individual photons, (assuming you can assign a phase to an individual photon), would be completely random. If the individual photons were in phase, this would necessarily violate energy conservation, because the energy (irradiance (watts/m^2)) of the electromagnetic wave is proportional to the square of the electric field amplitude, while the energy must also be proportional to the number of photons $N$. If the photons are in phase, we have a real dilemma, because $E_{total}=NE_o$ making the energy proportional to $N^2$. Alternatively If the phases are random, the phasor diagram becomes a 2-D random walk, with $E_{total} \approx \sqrt{N} E_o$, making the energy proportional to $N$ as it needs to be. $\\$ Hopefully this post doesn't get "pinged" for being a personal theory, but it is the best answer I have for what otherwise is a very big dilemma.

Last edited: Aug 1, 2018
3. Aug 1, 2018

### Tom.G

Huh?

4. Aug 1, 2018

I thought the OP is asking a reasonable question. The radio telescope is basically a large antenna that looks in one particular direction, and radio waves will get picked up as a sinusoidal voltage on an oscilloscope. You can expect fluctuations in amplitude and phase if you have an electronic filter/amplifier system that looks at a particular frequency. $\\$ The amplitude and phase of the radio waves is a result of r-f (radio frequency) photons that are getting picked up by the receiver at any given time. There will be fluctuations in photon counts.

Last edited: Aug 1, 2018
5. Aug 1, 2018

### kimbyd

The phase of the photons is only of interest when combining the signals from multiple radio telescopes. With multiple detectors, the relative phase between the different detectors can be used for interferometry.

The absolute phase is irrelevant in all contexts that I'm aware of. But the relative phase can be of tremendous interest.

6. Aug 2, 2018

### vela

Staff Emeritus
The telescope doesn't detect individual photons. It detects the combined effect of the massive number of photons incident on the telescope, which we call a radio wave. The fact that the photon phases are random just means the light coming from the sources is incoherent.

7. Aug 2, 2018

### f95toli

Yes, or at least this would be true for the hydrogen hyperfine (1420 MHz)
However, the next generation of radio telescopes will have detectors that ARE capable of detecting single photons, albeit at a higher frequency (say above 100 GHz or so, which is still radio astronomy). The detectors for these are currently under development (kinetic inductance detectors or improved transition edge sensors) and at least in the lab they are able to detect individual photons as "event"; similar to what has been possible with optical photons for a long time.

Hence, the question from the OP sort-of makes sense.

That said, the "phase" of a single photon is not a well-defined concept meaning in this context it is probably easier to just use classical theory; the full QM theory would work but would just agree with the classical result for e.g. coherent radiation.

8. Aug 2, 2018

Very interesting. For the OP @jordankonisky : When the classical r-f electromagnetic wave is viewed on an oscilloscope as a sinusoidal signal, it may be the result of thousands or millions of photons or more contributing to the sinusoidal r-f signal. When a single photon is detected, (at least this is how it is done in the U-V (ultraviolet)), it causes a disturbance on the detector and hundreds of thousands of electrons may be collected as a result. There is no observation of any sinusoidal electromagnetic field for this latter case.

Last edited: Aug 2, 2018
9. Aug 2, 2018

### jordankonisky

Thanks all. I think I am starting to get it.

Based on responses, allow me to restate my question. Consider the receiving parabolic dish of a radio telescope and the collection radio waves of same wavelength and phase that arrive perpendicular to the vortex line. No matter where each of these waves lands on the dish, each will then travel the same distance from its landing position on the dish to the telescope focus point. Thus, the collective beam will arrive at the focus point both in phase and in focus. On the other hand, members of a collection of radio waves of same wavelength and phase that arrive at an angle to the vortex line will not arrive on the dish at the same time, and, therefore, not all will arrive at the focus point simultaneously.

Suppose, a radio telescope is tuned to detect radio waves derived from neural hydrogen atoms at some region of the sky. My question is “do all the radio waves from a specific area of the sky arrive at the surface of the dish in phase?” It would seem to me that the flood of radio waves comprises of a collection of radio waves of same wavelength, but in all possible phases such that the dish sees radio waves of same wavelength, but in all phases, all of which reach the focus point.

10. Aug 2, 2018

To the OP @jordankonisky : For this particular problem of all the photons(scratch that=all of the wavefront) arriving in phase at the collection point, my fascination as well as the part that puzzled me was the energy conservation. It seemed to violate energy conservation because the various portions of the wave would constructively interfere as they were brought to focus=all having equal path distance. It is perhaps a somewhat routine optics calculation in diffraction theory of imaging, that solves the dilemma=the focused spot size, (and it does have a finite diffraction-limited spot size), decreases in area as the area of the parabolic dish is increased. (It actually took me a quite a while to figure this one out, after I had an advanced Optics course with diffraction theory). (And if the wavefront does not have constant phase, energy would still be conserved in the focusing=the spot size would simply be somewhat larger than the diffraction-limited spot size, and the focused intensity (watts/m^2) somewhat lower). $\\$ See also the Insights article that I authored about a year ago: https://www.physicsforums.com/insights/diffraction-limited-spot-size-perfect-focusing/ $\\$ And, yes, I think the wavefront can be treated classically as a plane wave with constant phase across the whole plane as it arrives at the telescope dish. Alternatively, as @kimbyd has pointed out, it is possible to do interferometric measurements, with multiple telescopes, where the intensity and/or phase across the whole plane is found to not be constant. (I need to review the Hanbury Brown -Twiss experiment, which I think was the first experiment in stellar interferometry).$\\$ Hopefully this helps answer your question, and doesn't get too far off topic.

Last edited: Aug 2, 2018
11. Aug 2, 2018

### Staff: Mentor

12. Aug 2, 2018

### marcusl

You have gotten some good responses from vela, f95toli and kimbyd. I take issue with this, however:
Stimulated emission preserves phase, so the EM field builds up coherently. This is why lasers are such a "powerful" source (in numerous senses).

13. Aug 2, 2018

The laser has basically a single phase, but it cannot be the result of $N$ photons all having the same phase. It may be completely incorrect to assign a sinusoidal electric field and phase to a single photon, but if such a model is used, the in-phase superposition of $N$ photons results in a violation of the conservation of energy. The energy density in the electromagnetic wave is proportional to $E^2$, and that would make the energy of the $N$ photons proportional to $N^2$.

14. Aug 2, 2018

### marcusl

You are mixing classical and quantum descriptions. Stick to one. In the quantum world, stimulatied emission preserves quantum numbers of incoming and outgoing photons. Classically, the phase and frequency of the emitted radiation matches that of the stimulating wave. I think you are misapplying energy conservation as well. The atoms in the excited state supply the additional energy.

15. Aug 2, 2018

I welcome other people's inputs on this. I have put considerable time and effort into the previously mentioned energy calculations. I actually prefer the classical picture, but it is important, especially when working with photodiodes (in the IR and visible) to have consistent photon models as well.

16. Aug 2, 2018

### jartsa

If we pretend that every hydrogen atom sends a tiny classical radio-wave at random times, then if 10 of those identical waves hit a radio-telescope at the same time and in-phase, the absorbed energy is only ten times one waves energy, right?

17. Aug 2, 2018

Yes. $\\$ Edit: Classically, one can work with sources that have larger amplitudes such as 10 r-f antennas, each (by themselves) radiating equal powers in a spherically symmetric pattern. By proper array design, there may be locations (solid angles) where the signals (sinusoidal electric fields) are all in phase with each other, so that the energy density\power per unit solid angle is 100x higher than that of a single source. In any case the total power radiated over the $4 \pi$ steradians remains 10x that of a single radiator, and that means there are necessarily solid angles where some destructive interference/cancellation occurs in the signals. The energy and number of photons is conserved, but by adjusting the geometry/relative position of the sources, the available energy/number of photons can be steered into specific directions rather than getting a spherically symmetric pattern.

Last edited: Aug 2, 2018
18. Aug 3, 2018

### sophiecentaur

furthermore, you cannot talk in terms of different photons hitting different parts of the receiving dish. The energy in each photon is spread all over the dish (all of space, actually) and the wave interferes with itself to produce a probability density of where it will be detected) After enough photons have arrived, there will be a peak in received signal power at the feed from a source on the boresight .
I have skimmed through this thread an I haven't yet found the term "bandwidth". However you choose to model the source, it will have a bandwidth and the spectrum of the resulting wave that arrives at your telescope will have a peak and a gaussian distribution about that peak. At one-photon-at-a time rate, the same spectral shape would build up. If you want to discuss photons that are created by stimulated emission, the situation is different because the coherence is greater.
There is a problem with trying to think of this as a wave and a particle phenomenon at the same time. For instance, what would be the pulse shape of this burst of waves that you want each photon to consist of? If they are all identical in nature then that would imply that the little pulsed oscillators that generate them must have identical 'circuit characteristics'. That's nonsensical and it's been discussed endlessly in threads about ' how big is a photon?', how long does a photon last for'? or 'what is the length of a photon?".
A classical model with a set of random pulsed oscillators is OK and the result will give you a centre frequency that's noise modulated. But it's too far to go to involve photons.

19. Aug 3, 2018

### jartsa

Well, can't we tell that same story, with some small modifications, about ten photons emitted by ten hydrogen atoms?

By the way, I meant to show that the same 'problem' exists with classical radio-waves too. (It's an interesting problem)

Last edited: Aug 3, 2018
20. Aug 3, 2018

### sophiecentaur

Each photon will have its own probability function if it's spontaneous emission (omnidirectional pattern). If there's stimulated emission then the radiation pattern / probability function will have a maximum in the direction of the progressive phase of the atoms.