Photon Arrival Rate: Doppler Effect Explained

In summary, the power delivered to the collector is increased by a factor of r due to the Doppler effect, but the total energy delivered is not increased by the same amount.
  • #1
Alfred Cann
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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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  • #2
Alfred Cann said:
By the Doppler effect, the frequency, and hence the energy, of the photons will increase by a factor r = 1+v/c.

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.

Alfred Cann said:
However, the photon rate of arrival is also increased by r

No, it isn't.

Alfred Cann said:
If the wave packets get squashed, why doesn't the photon spacing?

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.
 
  • #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.
 
  • #4
Alfred Cann said:
Doppler effect increases the energy per photon. Suppose it's very dim light, 30 photons per nS or one photon per cm, on average.

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.)

Alfred Cann said:
Is or is not the rate of photon interception by the collector increased if we make the collector move?

No.

Alfred Cann said:
If not, why not?

The Doppler effect makes each photon have more energy, but it doesn't change the number of photons in the beam.
 
  • #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.
 
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  • #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.
 
  • #7
PeterDonis said:
The Doppler effect makes each photon have more energy, but it doesn't change the number of photons in the beam.

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.
 
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  • #8
Actually, it's r=√(1+v/c)/(1−v/c) 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.
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
 
  • #9
pervect said:
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.

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.
 
  • #10
PeterDonis said:
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.

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].
 
  • #11
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
 
  • #12
PeterDonis said:
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.

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.
 
  • #13
Agerhell said:
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.

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.
 
  • #14
PeterDonis said:
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.

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.
 
  • #15
Agerhell said:
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.

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?
 
  • #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.
 
  • #17
Alfred Cann said:
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.

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).

Alfred Cann said:
We are speaking of non-relativistic motion, and all effects must be of order v/c not (v/c)^2.

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.

Alfred Cann said:
The right answer must be r for 2 reasons: 1. It's non-relativistic, and 2. Conservation of energy.

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.
 
  • #18
PeterDonis said:
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?

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 Earth's 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 Earth's 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...
 
  • #19
Agerhell said:
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.

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.
 
  • #20
pervect said:
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.

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 Earth's 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.
 
  • #21
Agerhell said:
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.

The H-K experiment is a different experiment than what we have been talking about in this thread. First of all, in the H-K experiment, the altitude of the moving clocks changes, and that affects the results. But you could run a similar experiment, in principle, with clocks that stay on the ground but move in opposite directions, to eliminate the altitude effect.

You are also proposing putting the Earthbound clock at the north pole, which makes the scenario significantly different from the standard H-K experiment, where the Earthbound clock is at a fairly low latitude and two moving clocks go westbound and eastbound around the Earth; in the standard H-K experiment, the westbound clock ends up with *more* elapsed time than the earthbound clock (and this would still be true if the altitude effect were eliminated).

More importantly, in the H-K experiment, the "moving" clocks go all the way around the Earth and then meet up again with a clock that remains at rest relative to the Earth. That makes it more like a "twin paradox" type scenario; there is more going on in this experiment than just Doppler shift. (The Doppler shift that the Earthbound observer would see, if each moving clock were emitting light towards him, would *change* during the experiment; he would see a Doppler redshift during the first part, when both clocks are moving away from him, and he would see a Doppler blueshift during the last part, when both clocks are moving towards him, having gone around the world in opposite directions.)

GPS satellites are different again because they are at a high enough altitude that they run faster than ground clocks even when their motion relative to the Earth is taken into account.
 
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  • #22
To Peter Donis: All right; you've convinced me. The clincher was your pointing out that r^2 is O(v/c).
I can analyze it two ways. Consider I am flying toward a pulsed radar (and the cw radar case is is included by considering it a train of contiguous pulses). Both the pulse repetition frequency and the carrier frequency inside the pulses is increased by r, and the pulse width is reduced by r. If the energy per pulse were preserved, the pulse peak power would increase by r and the total received power would increase by r. But from quantum mechanics, the increase in carrier frequency increases the energy of all the photons and thus the energy in the whole pulse by r. Alternatively, from SR, the mass of the pulse and therefore its energy, is increased by r.
This leads to the surprising conclusion that, when the pulse is compressed, its voltage is increased by r, not its power.
I am now looking for a physical reasoning that would give this result. Perhaps this effort is doomed because either QM or SR must be invoked. I can think of an analogy, which may be irrelevant. Let a burst of tone be recorded on a magnetic tape, then played back at 2 speeds with ratio r. At the higher speed, the burst length is reduced by r, and the voltage increased by r. I can already see a flaw; if the read head is magneto-resistive instead of magneto-inductive, no voltage increase will occur. I will think about pulse-compression radar; maybe I can find a hint there.
Thanks to Peter Donis and Pervect.
 
  • #23
Alfred Cann said:
I am now looking for a physical reasoning that would give this result. Perhaps this effort is doomed because either QM or SR must be invoked.

I'm not sure why that would "doom" the effort. Wouldn't an argument based on QM or SR count as "physical reasoning"? If not, why not?
 
  • #24
Sorry. Should have said 'classical reasoning'. I'm thinking of Maxwell's equations. Also studying Bill K's post about ExB. I seem to recall there were some suspicions about funny behavior with motion before 1900 and these were later considered possible precursors of the Lorenz-Fitzgerald contractions and of SR. Need to look this up.
 
  • #25
Agerhell said:
Yes sure, but that was not the point here.

I'm not quite sure what your point was then, sorry. I re-read your post and I'm still not quite sure what your point was. If you think it's important, perhaps you could repeat it?

I don't see any huge difference in our viewpoints, except for the word "actually" when you say "actually runs faster".

Time dilation, when it is measured, has to be measured relative to some particular coordinate system.

My remarks are relevant to TAI time, international atomic time, http://en.wikipedia.org/w/index.php?title=International_Atomic_Time&oldid=541760037

Perhaps you are implicitly thinking of Geocentric Coordinate Time (TCG) http://en.wikipedia.org/w/index.php?title=Geocentric_Coordinate_Time&oldid=543611924 to measure time dilation.

But if you think that time dilation exists independently of a choice of coordinate system, I'm afraid I have to disagree.
 
  • #26
There was a mistake in my post #22. The reference to a mass increase from SR is spurious. So, the only mechanism used so far is QM. Still looking for a classical derivation.
 
  • #27
Alfred Cann said:
The reference to a mass increase from SR is spurious.

Not really, because you equated it with an energy increase, which means you were talking about relativistic mass, not rest mass. Light has relativistic mass even though it has no rest mass, since relativistic mass is just total energy. SR does say the energy of the pulse increases if the source and detector are moving towards each other.
 
  • #28
I thought we were finished, but, not so. On further reflection I realized that the relativistic mass increase to which you (Peter Donis) referred is q = √(1-v^2/c^2), orders of magnitude smaller than r, and certainly not an alternative explanation for it. Even worse, if we're really going to do SR, our motion makes the source slow down by the same factor q, which wipes out the mass increase.
So, I was right when I said I was wrong to invoke SR.
 
  • #29
In the Wikipedia article on Redshift, under 'Observations in Astronomy', I found the heartening comment that 'Both photon count rate and photon energy are redshifted'. But my joy was short-lived; apparently this dimming only applies to light measured in a filter.
The article on K correction emphatically states that it does not apply to a single line, nor to bolometrically measured total light. I don't understand this at all. My reasoning (with which you agreed around 3/28) was specifically for a single line.
 
  • #30
Alfred Cann said:
the relativistic mass increase to which you (Peter Donis) referred is q = √(1-v^2/c^2)

No, it isn't. I wasn't talking about the source; I was talking about the light pulse itself. It has relativistic mass, i.e., energy, even though it has no rest mass. The energy of the light pulse increases by r, as I said before. The relativistic mass, i.e., energy increase of the light source is a different thing, and doesn't come into play here.

Alfred Cann said:
if we're really going to do SR, our motion makes the source slow down by the same factor q, which wipes out the mass increase.

Huh? What "our motion" are you talking about? The relative velocity v we've been talking about is the source's velocity relative to us, which is the same thing as our velocity relative to the source; it has nothing to do with whether we're moving relative to something else. And you only count v once; you don't count it once because the source is moving and a second time because we are moving.
 
  • #31
Wrong question. Please answer next question.
 
  • #32
Alfred Cann said:
Wrong question.

Well, you did post the other question as well, so I answered it.

Alfred Cann said:
Please answer next question.

Looking at the Wikipedia articles, I think they are talking about something different than what we're discussing here. See below.

Alfred Cann said:
I found the heartening comment that 'Both photon count rate and photon energy are redshifted'. But my joy was short-lived; apparently this dimming only applies to light measured in a filter.

I don't think that's what the article is saying. I think it is saying if measurements through a filter are all you have, you have to make some extra assumptions and corrections to translate the filter measurements to an estimate of redshift. I don't think the article is trying to claim that light from the same source measured some other way besides a filter will not be redshifted.

Alfred Cann said:
The article on K correction emphatically states that it does not apply to a single line, nor to bolometrically measured total light. I don't understand this at all.

It's because these are special cases in which you don't need to apply the corrections that you need to apply when you only have a portion of the total spectrum. see below.

Alfred Cann said:
My reasoning (with which you agreed around 3/28) was specifically for a single line.

When you say a "single line", I think what you really mean is that the light beam is coherent, i.e., it is emitted from the source with a single known frequency, so that the beam's total energy is just its frequency times the photon number (times Planck's constant if we are using ordinary units). But your reasoning also assumed that we have a perfect detector that detects the entire beam energy, as transformed to the detector's frame (i.e., with the appropriate redshift/blueshift due to the relative velocity of the source and the detector).

In other words, your reasoning, which which I did (and do) agree, applies to an idealized case where (1) we know the emitted spectrum prior to the measurement (because we know something about the source independently of the measurement that allows us to model its emitted spectrum), and (2) we detect the entire redshifted/blueshifted spectrum, so we can do a direct comparison of the detected spectrum with the known emitted spectrum to derive an estimate of the redshift.

In real life, the two cases in which these idealized conditions come close to being realized are the ones mentioned in the Wikipedia article as not requiring a K correction: measurement of an emission line, where we know the emitted spectrum and it's easy to measure the entire detected spectrum because it's narrow band; and bolometric measurement, which covers all wavelengths/frequencies (but in this case you would still need some independent assumption about the emitted spectrum, for example that it is a black body at a certain temperature--but you would still have to know what temperature, so I think the Wikipedia article is leaving some things out here).

In general, though, you don't detect the entire spectrum, you only detect a portion of it, as in a filter; and you don't really have independent knowledge of the emitted spectrum, you have to make assumptions about it. That's why corrections have to be made for such cases. That certainly doesn't mean that redshift is only detected in such cases.
 
  • #33
Neither the frequency of a light pulse nor the length of a light pulse transforms as [itex]\gamma[/itex] = [itex]1/sqrt(1-\beta^2)[/itex], [itex]\beta = v/c[/itex].

Both transform as [itex]r = \gamma \left( 1 + \beta \right)[/itex] (Or to be more precise, the length tranforms as 1/r).

Because the speed of iight is constant, one expects the frequency and the length of a light pulse to always vary inversely. If a light pulse contains 1000 cycles in one frame, it contains 1000 cycles in any frame - an observer can count them. The wavelength of 1 cycle is c / frequency, so the length of 1000 cycles is 1000 * (c / frequency).

Thus the behavior and length transformation of a beam of light is simply different from that of a rigid rod. The later does transform as the reciprocal of [itex]\gamma[/itex]. The former does not.

This may or may not be intuitive, but it can be confirmed directly from the Lorentz transform. The Wiki articles on relativistic doppler shift may also be helpful. There may be some minor sign differences in my treatment as opposed to Wikki's, which can be resolved by whether or not [itex]\beta[/itex] is positive or negative.

http://en.wikipedia.org/w/index.php?title=Relativistic_Doppler_effect&oldid=549045215

[itex]\beta[/itex] is order 1 in v/c, [itex]\gamma[/itex] is order 2. r , as their product, is of order 1.

[itex]r^2 = \left( 1 + 2 \beta \right)[/itex] to order 1.

Relativistic time dilation provides a second order correction to the first order doppler shift.

Trying to apply concepts of "relativistic mass" naively does give the wrong answer. The solution is not to use relativistic mass naively. I would mostly recommend not using relativistic mass at all, personally.

Applying the formalism of the stress energy tensor does give the right answer as well (though it may be a bit advanced for some readers).

http://en.wikipedia.org/w/index.php?title=Stress–energy_tensor&oldid=541691083
 
  • #34
Peter Donis: In response to your post #32.
I don't care if it's a single line as from a laser or a cw radar, or just a fairly narrow line as in the emission spectrum of a star, I assume a photodiode or similar detector that has a flat response over the frequency range of interest, or even a bolometer (partly your assumption 2). I don't need to know the emitted spectrum (your assumption 1).
I believe that, when motion is introduced, not only does the frequency change (which my detector does not indicate), but both the photon energy and count rate are changed by the factor 1+z (to use the conventional notation), and therefore the received power (which this kind of detector indicates) is changed by (1+z)^2. Why is that not true?
 
  • #35
Alfred Cann said:
I don't need to know the emitted spectrum (your assumption 1).

Why not? In the idealized cases we've been talking about, you know it because you specified it in the scenario; but you still need to know it.

For example, I see hydrogen Lyman alpha lines redshifted by some factor. How do I know what the factor is? Because I know the emitted frequency of Lyman alpha lines, which means I know the emitted spectrum. If I didn't know the emitted line frequencies (i.e., the frequencies seen by a detector at rest right next to the emitter), how could I possibly know what the redshift factor was when I detect them from a moving source?

Alfred Cann said:
the received power (which this kind of detector indicates) is changed by (1+z)^2. Why is that not true?

I thought we agreed that it was true; the received *power* (i.e., energy received per unit time, or the energy density of the beam as pervect put it) changes by r^2. But the total received *energy* (integrated over time) changes by r. See post #17.
 
<h2>1. What is the Photon Arrival Rate?</h2><p>The Photon Arrival Rate is the number of photons per unit time that arrive at a specific location. It is a measure of the intensity of light at that location.</p><h2>2. What is the Doppler Effect?</h2><p>The Doppler Effect is the change in frequency of a wave (such as light) due to the relative motion between the source of the wave and the observer. This effect can cause a shift in the wavelength of light, resulting in a change in color.</p><h2>3. How does the Doppler Effect affect the Photon Arrival Rate?</h2><p>The Doppler Effect can cause a change in the frequency of light, which in turn affects the Photon Arrival Rate. If the source of light is moving towards the observer, the frequency increases and the Photon Arrival Rate also increases. If the source is moving away, the frequency decreases and the Photon Arrival Rate decreases.</p><h2>4. What factors can affect the Photon Arrival Rate?</h2><p>The Photon Arrival Rate can be affected by the speed and direction of the source of light, the distance between the source and the observer, and the properties of the medium through which the light is traveling (such as air or a vacuum).</p><h2>5. How is the Photon Arrival Rate measured?</h2><p>The Photon Arrival Rate is typically measured using a photodetector, which counts the number of photons that arrive at a specific location over a set period of time. This measurement can be used to determine the intensity of light at that location and any changes due to the Doppler Effect.</p>

1. What is the Photon Arrival Rate?

The Photon Arrival Rate is the number of photons per unit time that arrive at a specific location. It is a measure of the intensity of light at that location.

2. What is the Doppler Effect?

The Doppler Effect is the change in frequency of a wave (such as light) due to the relative motion between the source of the wave and the observer. This effect can cause a shift in the wavelength of light, resulting in a change in color.

3. How does the Doppler Effect affect the Photon Arrival Rate?

The Doppler Effect can cause a change in the frequency of light, which in turn affects the Photon Arrival Rate. If the source of light is moving towards the observer, the frequency increases and the Photon Arrival Rate also increases. If the source is moving away, the frequency decreases and the Photon Arrival Rate decreases.

4. What factors can affect the Photon Arrival Rate?

The Photon Arrival Rate can be affected by the speed and direction of the source of light, the distance between the source and the observer, and the properties of the medium through which the light is traveling (such as air or a vacuum).

5. How is the Photon Arrival Rate measured?

The Photon Arrival Rate is typically measured using a photodetector, which counts the number of photons that arrive at a specific location over a set period of time. This measurement can be used to determine the intensity of light at that location and any changes due to the Doppler Effect.

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