Photon Arrival Rate: Doppler Effect Explained

[Alfred Cann]

Wrong question. Please answer next question.

 

[PeterDonis]

Wrong question.

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

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.

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.

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.

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.

 

[pervect]

Neither the frequency of a light pulse nor the length of a light pulse transforms as \gamma = 1/sqrt(1-\beta^2), \beta = v/c.

Both transform as r = \gamma \left( 1 + \beta \right) (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 \gamma. 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 \beta is positive or negative.

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

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

r^2 = \left( 1 + 2 \beta \right) 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

 

[Alfred Cann]

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?

 

[PeterDonis]

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?

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.

 

[Alfred Cann]

Peter Donis:
Aha! After reading a lot more articles about K correction, I finally realized I had a misconception of what it was. I thought it was simply the factor (1+z)^2; it is not. It is a correction or set of corrections applied to photometric measurements through filters when one uses them to estimate what is called a PHOTOMETRIC redshift. That is a crude method, subject to many errors, which one uses only when one cannot collect enough photons to perform a measurement of SPECTROSCOPIC redshift, the accurate method.
So, the articles did not mean to imply that, in cases of single lines or bolometric measurements, the RECEIVED POWER was not modified by the factor (1+z)^2, as I incorrectly inferred. They merely meant that no K correction needs to be applied to the measurement of REDSHIFT in those cases.

 

[PeterDonis]

the articles did not mean to imply that, in cases of single lines or bolometric measurements, the RECEIVED POWER was not modified by the factor (1+z)^2, as I incorrectly inferred. They merely meant that no K correction needs to be applied to the measurement of REDSHIFT in those cases.

Yes, you've got it.

 

[Naty1]

Very interesting discussion on photon arrival rate...answered my questions from another current discussion...thank you.