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If you emit a single photon, wait an interval, emit a second photon, wait the same interval, and keep repeating, is the resultant stream of photons describable by the frequency at which you emitted each one? Or does each photon have its own "frequency", independent of the rate of emission? I thought the former description explained the Doppler effect.

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How is colour perceived when shot by a single action “photon pistol”?

The same way it's perceived normally.

First: Suppose you had a single action photon pistol. You fire photons at an observer with super-human sensitivity to light. Her vision is so sensitive she can perceive the difference between 2 photons.

No problem. But note that the photon has energy E=hf, where f is frequency. Each photon has a frequency. So your superhuman observer would see one photon as (say) a green flash. Then she'd see the next photon as another green flash.

If the interval between each fired photon equals 1, let that coincide with the perception of green light.

No, don't. That's assuming the photon itself doesn't have a frequency, and that photon frequency is instead the frequency at which photons arrive.

Now, if you decreased or increased the interval, photon by photon, and by just the right degree, would she see blue and red respectively?

No. She'd see a slower or faster succession of green flashes. You'd have to increase the energy of each photon for her to see blue flashes. Or decrease the energy of each photon for her to see red flashes.

Second: This time you fire the photons from a massive body where the gravity you feel exceeds the gravity felt by a now distant observer. The interval between firings in your view remains "1" as in the first question. But if the observer's gravity was less than yours by just the right amount, would the frequency of each photon perceived increase so that she sees blue light?

This is flawed on a number of counts:

1) Gravitational redshift depends on the difference in gravitational potential between two locations rather than the difference in gravitational force.

2) When you fire photons upwards, the other observer at a higher location sees them as being redshifted as opposed to blueshifted.

3) Note what Einstein said: "An atom absorbs or emits light at a frequency which is dependent on the potential of the gravitational field in which it is situated". The photon is emitted at a lower frequency at the lower location. You don't notice this because you and your clocks are "running slow" at the lower location, because of gravitational time dilation. The higher observer measures the photon to have a lower frequency than you because his clocks are running faster than yours. But the photon frequency didn't actually decrease as the photon ascended. E=hf and conservation of energy applies.

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  • $\begingroup$ I posed "Now, if you decreased or increased the interval, photon by photon, and by just the right degree, would she see blue and red respectively?" and you replied "No. She'd see a slower or faster succession of green flashes". Do you think she would still see a slower or faster succession of green flashes if the emitter was receding or approaching her? If not then you need to explain the meaning of frequency in order to answer my question. Otherwise how do you account for the Doppler shift $\endgroup$ – argonaut Jan 9 '17 at 10:38
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    $\begingroup$ @argonaut: if she approached the emitter she'd see each individual photon blueshifted, and she'd see the interval between flashes diminish. $\endgroup$ – John Duffield Jan 9 '17 at 17:10
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nobel prize winner and superstar Richard Feynman says it is tricky to fully understand quantum mechanics, Beware wasting too much time trying !! The best simple way to think of photon is to imagine a wavepulse or train of a quantum "psi" field (or Electromagnetic field). This pulse may have a finite length L and inside you get many many short periods of oscillation corresponding to the light wavelength lambda or frequency. This pulse can travel through a medium or lenses or whatever and it can travel through like a particle following a quite localised spatial trajectory. The photon frequency doesnt correspond to the interval between pulses (as imagined in question) but rather to the oscillation inside one individual pulse. The photon is thought of as somewhere unknown within the pulse, until a detection localises it. The photon picture has rather peculiar aspects if one thinks deeply. for example 2 laser beams from different sources can interfere in usual way and we need to interpret this as a single photon can simultaneously pop out of BOTH guns much like a photon can pass through 2 slits and interfere with itself on passing through making familiar diffraction pattern.

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If you emit a single photon, wait an interval, emit a second photon, wait the same interval, and keep repeating, is the resultant stream of photons describable by the frequency at which you emitted each one?

No. The frequency of a photon is uniquely tied up with the increment/quantum of energy that generated it , by E=h*nu, where h is the Planck constant and E the energy and nu the frequency.

A photon is a quantum mechanical entity , an elementary particle in the standard model of particle physics. It can be created by an atomic transition of electrons between two orbitals,, atomic or molecular or lattice probability locuses. Also by accelerating or decelerating charged particles a quantum of energy is lost, and appears as the photon.

The intervals of single photon emission have nothing to do with the frequency of the wave that a large ensemble of photons will build up, nu of the photon is the nu of the emergent classical wave. See this link single photon at a time double slit experiment, where the interference pattern in the accumulation of photons has the frequency of the E=hnu of the photon.

The reason the same frequency appears in the E=hnu and in the built up classical EM wave wave is due to the fact that both are solutions of maxwell's equations. The wave functions describing the photon

enter image description here

are solutions of a quantized maxwell equations, and the E nd B fields affect the wavefunctions whose complex conjugate square give the probability density for the photons. Example, the interference pattern in the above link:

singlephot

which is the probability density for the experiment "single photon scattering on two slits".

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Each photon has its own frequency. Say you emitted a 630 nanometer photon (1.97 eVolts) once every 2.1 femtoseconds (the time it takes a 630nm photon to complete one cycle). You would see a stream of photons that would have the same frequency as the individual photons. But you could also consider releasing multiple photons in that 2.1 femtosecond period, or you could release a set of different frequency photons in a very small timeperiod and watch them "beat" of form cycles related to their frequency relative to the frequency of 630nm photons. Also note that all photons travel at the same speed independent of their frequency.

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  • $\begingroup$ When you say "the time it takes a 630nm photon to complete one cycle" you must mean the frequency that the EM field oscillates per second. That's an "internal" frequency. My original question mentions the Doppler effect. Suppose you are moving away from the light emitter as you receive photons. Here "frequency" describes the lengthening periods of reception; you perceive light is red-shifted because of it. Sure, light has its own "internal" frequency, but it turns out this "external" frequency (for want of an expression) is consistent with wavelength in a moving frame, hence Doppler shift. $\endgroup$ – argonaut Feb 4 '17 at 23:28
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I think that when we consider light as a corpuscular phenomenon, the frequency of light depends on photon intervals and by changing the interval, the frequency (light's color) may alter. I have a reason for this claim:

If you consider Einstein's light clock consisting of a laser gun, a mirror and a detector from the viewpoint of the lab frame (M) where the light clock is set in motion, and if you (as observer N that is stationary with respect to the clock) assume the detector and the laser gun emit different photons (reds) towards the lab observer in a perpendicular direction to the path of light clock's main photons (light-blues) exactly at the time that each main photon is emitted towards and received from the mirror by the gun and detector respectively; you (this time, as the lab observer) will see that the interval of photons exactly complies with the relativistic Doppler effect.

enter image description here

For the second question, remember that frequencies are greater on massive bodies. If it takes, e.g., ten seconds for the observer on the body with less gravity to perceive the two successive photons, it takes one second for instance due to time dilation for the observer on the massive body, thus the recent observer calculates a greater frequency rather than the observer in low gravity.

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    $\begingroup$ Sammy, I understand your concern about the physiology potentially involved in this question. However, these questions are not really about the mystery of colour perception, but rather about the nature of photons in a stream that we can manipulate....continued $\endgroup$ – argonaut Jan 7 '17 at 10:44
  • $\begingroup$ So we ask: if we create a stream, photon by photon, are very short intervals consonant with the idea of high frequency (and therefore short wavelengths) and are very long intervals consonant with low frequency (and therefore long wavelengths)? And could a stream of gamma rays produced in similar fashion-photon by photon-from a lower gravitational field than our earth be perceived by us as radio waves? It seems the answers should be yes by definition, but I’m not so sure. That’s why I asked. $\endgroup$ – argonaut Jan 7 '17 at 10:45
  • $\begingroup$ Ugh...error... its radio waves being sent to us that appear as gamma rays $\endgroup$ – argonaut Jan 7 '17 at 10:57
  • $\begingroup$ If radio waves being sent from distant massive stars are received as gamma rays on Earth, we would have been dead long ago!!... $\endgroup$ – Mohammad Javanshiry Jan 7 '17 at 11:48
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    $\begingroup$ Note - the magnetic field only protects from charged particles - gamma rays are not charged, and don't care about magnetic fields. $\endgroup$ – Floris Jan 8 '17 at 4:17

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