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Photon arrival rate

  1. Mar 26, 2013 #1
    Let a monochromatic plane wave impinge on a collector of area A, delivering power P. Let the average photon rate of arrival on the collector be B. Now make the collector move toward the source at velocity v.
    By the Doppler effect, the frequency, and hence the energy, of the photons will increase by a factor r = 1+v/c. This increases the collected power by the factor r. So far, so good.
    However, the photon rate of arrival is also increased by r, making the total power increase r^2. That can't be right. I suspect the problem is that the photon interception rate does not really increase, probably because their velocity relative to the collector is still c, but don't understand this.
    If the wave packets get squashed, why doesn't the photon spacing?
     
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  3. Mar 26, 2013 #2

    PeterDonis

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    Actually, it's

    [tex]r = \sqrt{\frac{1 + v/c}{1 - v/c}}[/tex]

    But your question doesn't really depend on the specific form of r; it's just a question about why we don't "double count" r, so to speak.

    No, it isn't.

    Because the wavelength *is* the "photon spacing". More precisely, you're trying to mix together two models of light: the photon model and the wave model. That's not a good idea.

    In the wave model, the wavelength is (roughly speaking) the spacing between wave crests, which gets squashed by the Doppler effect. (The frequency, similarly, is the length of time between wave crests, which gets shortened by the Doppler effect.)

    In the photon model, the "wavelength" is the photon's momentum divided by Planck's constant; it gets changed by the Doppler effect because the photon's momentum gets changed by the Doppler effect. (Similarly for frequency/energy.)

    The two models are not different aspects of light that both get changed; they are two different ways of looking at the same single change.
     
  4. Mar 26, 2013 #3
    I don't quite get it. Let's forget the wave model and stick with photons. Doppler effect increases the energy per photon. Suppose it's very dim light, 30 photons per nS or one photon per cm, on average. Is or is not the rate of photon interception by the collector increased if we make the collector move? If not, why not?
    I'm not trying to fool Mother Nature, just unscrew the inscrutable.
     
  5. Mar 26, 2013 #4

    PeterDonis

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    Energy per photon isn't the same as photons per ns or photons per cm. In the photon model, each photon is a particle with a certain energy (and momentum equal to its energy, if we use natural units in which c = 1). Photons per ns or photons per cm is a measure of particle flux, not energy per particle. (Actually, the correct measure of photon flux is photons per unit time per unit area of the collector, e.g., photons per ns per square cm.)

    No.

    The Doppler effect makes each photon have more energy, but it doesn't change the number of photons in the beam.
     
  6. Mar 27, 2013 #5
    Um. If you emit 300 photons in a 10ns pulse, the pulse will have a length of 3m. To a moving observer (collector) the pulse length will seem shorter, so the photon rate will be higher.

    Edit: Right, you are talking about Doppler effect, not lorentz transform. Silly me.
    Then photon collection rate is still higher though. And yes, power might seem increased more than expected - you'd be collecting more pulses in a given time. But total energy is still not increased by this factor, you just get it in less time.
     
    Last edited: Mar 27, 2013
  7. Mar 27, 2013 #6
    This is a non-relativistic approximation of slowly moving energy collectors:

    A wind turbine that moves slowly against the wind extracts the energy of increasesd number of cubic meters of air in one second. Also every cubic meter of air has more energy, but that energy comes from the thing that is pushing the turbine.

    A solar panel that moves slowly against the light extracts the energy of increased number of photons in one second. Also every photon has more energy, but that energy comes from the thing that is pushing the panel.
     
  8. Mar 27, 2013 #7

    pervect

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    I don't think this is right. Let's look at how the stress-energy tensor of the beam transforms, i.e. the energy density per unit volume.


    The short version is that the energy / unit volume transforms both because the frequency changes, and the number of photons / unit length changes. The total number of photons in the entire beam doesn't change, but the number-flux 4 vector transforms as a 4 vector, so the density of photons isn't the same in the moving frame.

    You can easily compute the number of photons / second from the number of photons per unit length if you prefer that formulation.

    Let's go on to the long version now.

    Use a standard Minkowskii (t,x,y,z) coordinate system and geomeatric units. Assume that the stress-energy tensor is a null dust. Also we know that E = pc. So we can write the stress-energy tensor as

    [tex]T^{ab} = \left[ \begin{array}{cc}
    E & E & 0 & 0\\
    E & E & 0 & 0\\
    0 & 0 & 0 & 0\\
    0 & 0 & 0 & 0\\
    \end{array} \right]
    [/tex]

    (One could do an actual plane wave too, the null dust is a lot simpler).


    A boost towards the beam will give a transformation matrix L of

    [tex]
    L^{a}{}_{b} = \left[ \begin{array}{cc}
    \gamma & \beta \gamma & 0 & 0 \\
    \beta \gamma & \gamma & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    \end{array} \right]
    [/tex]

    Then
    [tex]
    T'^{a b} = L^{a}{}_{c} L^{b}{}_{d} T^{cd}
    [/tex]
    so
    [tex]T'^{00} = L^{0}{}_{0} L^{0}{}_{0} T^{00} + L^{0}{}_{0} L^{0}{}_{1} T^{01} + L^{0}{}_{1} L^{0}{}_{0} T^{10} + L^{0}{}_{1} L^{0}{}_{1} T^{11}=
    [/tex]
    [tex]
    \gamma^2 T^{00} + 2 \beta \gamma^2 T^{01} + \beta^2 \gamma^2 T^{11} = \gamma^2 \left(1 + \beta\right)^2 E
    [/tex]

    You can work out the other components, and confirm that they are all equal , so a null dust boosts into another null dust.

    Note that [itex]\gamma \left(1 + \beta\right) [/itex] is just the relativistic doppler shift, the square of this multiples the energy density.

    [add]
    Another quick-and-easy version of this. if you have n photons in one wavelength of the beam, then when you boost the wavelength shrinks by 1 over the doppler factor, the number of photons in one wavelength stays the same, the volume occupied goes down by the doppler shift factor, and the energy per photon goes up by the doppler shift factor, thus the total energy / volume goes up by the doppler factor squared.
     
    Last edited: Mar 27, 2013
  9. Mar 27, 2013 #8

    Bill_K

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    That's exactly what we should do. The correct transformation of the power for a plane wave is the square of the Doppler factor, r2.

    The "power" is the energy flux, i.e. the Poynting vector E x B. Relativistically, this is the 0i component of the stress-energy tensor. You can get the correct result by applying the Lorentz transformation to a rank-two tensor, or you can get it from the transformation of the E and B fields,
    E → γ(E + v/c x B) and B → γ(B - v/c x E).

    Either way, E x B → γ2(1 + 2v/c + (v/c)2)(E x B) = γ2 (1 + v/c)2(E x B), and γ2(1 + v/c)2 = ((1 + v/c)/(1 - v/c))2 = r2.

    EDIT: Which agrees with pervect's post
     
  10. Mar 27, 2013 #9

    PeterDonis

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    Ah, ok, I missed this. You and Bill_K are right. It's the *density* of photons that matters, not just the number of them.
     
  11. Mar 27, 2013 #10

    PeterDonis

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    Having been corrected on this, I now would like to sow some potential confusion in compensation. :wink:

    Consider a "light sail course change" scenario. I am in my spaceship which has a "light sail", my home base can shine a laser on it to change my course. The base shoots standardized laser pulses of a fixed power density for a fixed time, in the base's rest frame (i.e., the stress-energy tensor of the pulse takes the form pervect wrote down in the base's rest frame, with a fixed energy density E, and each pulse lasts for a fixed time T in the base's rest frame).

    Scenario A: I am sitting at rest with respect to home base. The base fires a single laser pulse at my light sail, which imparts a momentum P to my ship, as seen in the ship's initial rest frame (i.e., a frame at rest with respect to base).

    Scenario B: I am moving towards home base. The base fires a single laser pulse at my light sail, which imparts a momentum P' to my ship, as seen in the ship's initial rest frame (i.e., a frame moving towards home base at my initial velocity, before the pulse).

    What is the relationship between P and P'?

    I'm wondering if this is the question whose answer should be [itex]r[/itex] and not [itex]r^2[/itex]. The power density in the beam, in the ship's initial rest frame, is increased by [itex]r^2[/itex] in scenario B, as compared to A, but the time the beam lasts, in the ship's initial rest frame, is shorter by [itex]r[/itex] in scenario B, correct? So the total energy delivered by the beam to the sail (and hence the total momentum delivered, since they're the same for a light beam) only goes up by [itex]r[/itex]. So [itex]P' = r P[/itex].
     
  12. Mar 27, 2013 #11

    pervect

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    I think this is right. Assuming special relativity, the total energy and momentum in the pulse will transform as a 4-vector.

    And if the pulse is N cycles long (say 100,000) in one frame, it'll be the same number of cycles long in any other frame. But the length associated with this number of cycles will change as you change frames / doppler shift.

    So energy density scales as z^2, but total energy and/or momentum in the pulse scales as z, z being the doppler shift factor
     
  13. Mar 27, 2013 #12
    I do agree. However it could be pedagogical to point out that the difference between the two expressions is due to time-dilation on the part of the moving observer. When the clocks at the observer start ticking slower, the frequency of the incoming light will seem to increase.
     
  14. Mar 27, 2013 #13

    PeterDonis

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    But this can also cause confusion, because in the observer's rest frame, it's the source that's moving, so the source's clocks are ticking slower due to time dilation. That would be expected to cause a *decrease* in frequency, right? But if the source is moving towards the observer, the observed frequency increases.
     
  15. Mar 27, 2013 #14
    Hmm... I would expect that if the source is situated at the north pole of the earth and the receiver moving towards it very fast, your formula would be correct, according to the receiver.

    If, on the other hand, the receiver is fixed at the northpole and the emitter moving fast towards it I would expect the measured frequency to increase by the factor 1/(1-v/c) due to classical dopper shift and decrease by the factor sqrt[1-v^2/c^2] due to time dilation of the source resulting in precisely the same doppler formula you presented.
     
  16. Mar 27, 2013 #15

    PeterDonis

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    I'm not sure I understand. What does the north pole of the Earth have to do with it? If you're just using it to define which frame you're using, you could just say the source's rest frame vs. the receiver's rest frame, couldn't you?
     
  17. Mar 27, 2013 #16
    My apologies for not responding sooner to the many kind people who posted. Well, it seems my simple question about a case of slow (non-relativistic) motion has thrown a cat among the pigeons.
    PETER DONIS. I forgive your bit of pedantry with your relativistic equation for doppler; I think we all agree that in the classical case 1 + v/c is accurate enough.
    Second, the wavelength is not the photon spacing in any sense. Neither is the spacing of wave packets, except in the case of a very dim light. I have realized that I must stop trying to associate each wave packet with a photon. A high wave packet amplitude not only means a high probability of finding a photon there; it can also mean there is probably a whole bunch of photons there. And if the wave packet is, for example, a square pulse, there is a bunch of photons evenly distributed over its duration.
    PERVECT. I cannot forgive your flight of tensors and 4vectors --- totally inappropriate. I don't speak Tensor, and don't believe it's needed for classical physics and not even for a bit of special relativity. However, I am grateful for your surprising opinion that the appropriate factor is indeed r^2 and not r. Apparently, BILL K agrees with you and you have also converted PETER DONIS.
    But I have not changed my mind and still believe, on philosophical grounds, that the correct factor must be r. We are speaking of non-relativistic motion, and all effects must be of order v/c not (v/c)^2. This is reminiscent of the Sagnac effect, also O(r), which was first analyzed relativistically but can also be analyzed classically.

    Here is another thought experiment. Let the source be a laser with a beam expander at least as big as collector area A. Let the distance be 3 m and assume we are working in the near field so we can forget about diffraction. When the collector is stationary, we have a certain rate of arrival R of photons of energy E. The received power is P = RE. Now move the collector at a steady 3 m/s, which is 1E-8 c. The doppler shift is 1E-8 and the factor by which the energy of each photon is increased is 1 + 1E-8 = r. In one second the collector will have moved right up to the source and will have collected not only all the photons it would have collected if stationary, but in addition all the photons that were stacked up in that 3 m space. That number of photons would be R if the space were 3E8 m, but in 3 m is only R* 1E-8. Thus, the total number of photons collected is R(1 + 1E-8) = R*r, and the total power increase is r^2. Gentlemen, I can't buy this; I think it violates conservation of energy.

    If a source radiates energy at a rate P (that's power), a stationary receiver receives energy at the same rate P. If the receiver moves upstream it can collect energy at a higher rate, but only by stealing it from the energy stored in the intervening space, and only for a limited time, until it reaches the source. Note that this is structurally identical with the second part of the argument used in the above example, and yields a power increase factor of only r. I think this is more fundamental than the above example because it considers only energy, and must be right. I will be obliged if one or more if you can pinpoint the fallacy in the examples that imply r^2.

    The right answer must be r for 2 reasons: 1. It's non-relativistic, and 2. Conservation of energy.
     
  18. Mar 27, 2013 #17

    PeterDonis

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    I basically agree with this, and you'll note that none of it has anything to do with the energy or momentum per photon; those correspond to the wave packet frequency/wavelength, not its amplitude--amplitude, as you say, corresponds to photon number (actually photon number is proportional to wave amplitude squared).

    r^2 is not the same as (v/c)^2. If v/c is small, then we have

    [tex]r^2 = \frac{1 + v/c}{1 - v/c} = \left( 1 + v/c \right) \left( 1 + v/c \right) = 1 + 2 v/c[/tex]

    where we have used the small (v/c) approximation and dropped the (v/c)^2 term in the last equality as being too small to matter. So squaring r, in the small (v/c) approximation, just means doubling the v/c term.

    For 1., see above. For 2., see my post #10 and pervect's response to it. The total energy delivered to the collector *does* go like r, not r^2. But the *power* delivered to the collector goes like r^2. When integrating power over time to get total energy delivered, one factor of r is canceled by the fact that the total time the beam is on gets shorter when the source is moving relative to the detector.
     
  19. Mar 28, 2013 #18
    Well I just thought that Hafele-Keating, GPS-satellites etc. tells us that whoever moves faster with respect to a nonspinning earth is the one that will experience most time-dilation. I do not know if this is generally accepted. If the receiver sits still and the source moves with respect to the earths gravitational field it is the emitter who experience time-dilation. If the receiver is moving with respect to the frame of a non-spinning earth and the sender sits still things will be the other way around.

    However, if you assume the velocity of light to be c with respect to the earths gravitational field, which is perhaps a non-standard assumption, the classical dopplershifts of 1+v/c and 1/(1-v/c) will compensate for the assymetry in time-dilation so that you get the same total doppler formula in both cases...
     
  20. Mar 28, 2013 #19

    pervect

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    It turns out that all clocks on the geoid (roughly speaking, at sea level) tick at the same rate.

    The clocks on the equator ARE moving faster, but because of the Earth's shape (called the Earth's figure by geologisits), they are also further away from the center, nd thus at a higher gravitational potential. The higher potential makes them tick faster (this is a GR effect, not included in SR) which compensates for the slowing due to the velocity.

    The two effects exactly cancel. This isn't a coincidence, it's an expected prediction from GR, though I'm not going to try to explain it further in a short post. I could dig up some references in MTW if anyone is interested.
     
  21. Mar 28, 2013 #20
    Yes sure, but that was not the point here. I claim that experience tells us that, given that they are situated at the same gravitational potential, out of two clocks in a near earth environment the clock that moves faster with respect to the earths gravitational field actually does tick slower.

    At the north pole you do not have to take the spinning earth into account, that was the only reason I used the north pole in my example.
     
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