Red-Shifting vs. Luminosity Question

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Main Question or Discussion Point

Is red-shifted light brighter than what its black-body emission would be at the same temperature?

As a figurative example, a star, neutron star, or other astronomical body has a temperature of 6,000 K, and is so close to its Schwarzschild radius that its light is red-shifted by a factor of 3, meaning that it appears to an observer outside the gravity well to have a temperature of 2,000 K, and a measure of the equivalent black-body temperature of its light confirms this. Atomic spectral emission lines are red-shifted by a factor of 3. (Is red-shifting stated this way?) Doppler red-shifting, such as from the expansion of the universe, would be the same.

Photon energy and flux are decreased by the amount of red-shift. However, luminosity increases by the fourth power of temperature. Photon energy is directly proportional, flux varies by the third power, so the end result varies by the fourth power.

In red-shifting from 6,000 K to 2,000 K, the photon wavelength increases by 3, so the energy decreases by 3, and the flux decreases by 3 due to time dilation inside the gravity well. Cumulatively, these are decreases by the inverse square, or 9.

There still remains the square of photon flux that is not red-shifted. Therefore, outside of the gravity well, the star's light would be observed to be at the color temperature of 2,000 K, but shining 9 times brighter than what its black-body emission would be at that temperature.

The equivalent black-body temperature for the same energy of luminosity is 2,000 K(√3), or 3,464 K. Assuming it is far enough out to not be red-shifted itself, a planet with a semimajor axis where it would receive 1.0 Earth's received energy flux from the Sun, but at 3,464 K, will receive that same energy from the star's light at 2,000 K, but 9 times brighter.

Momentarily ignoring the impossibility of a star to be so dense, do I have the correct assessment of this situation? If not, why not? If yes, it seems to me highly unlikely that I'd be the first to see it, so I assume red-shifting is already being considered with the "un-red-shifted square?"

I don't think I'm putting forth any new theory, just pointing out something fairly easily seen, but seldom discussed. I just don't see any way that, if red-shifting and either gravitational time dilation or the slowing of arriving Doppler-shifted photons are both inversely proportional, and together they result in the energy of the light spectrum varying by the inverse square, and the black-body emitted light energy varies by the fourth power of temperature, there's any way to avoid having an "un-red-shifted square."

I'll refrain from discussion of the implications of this, until I have firmer ground to stand on. I don't want to be premature.
 
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  • #2
Bandersnatch
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Let's consider cosmological redshift, mainly because I'm familiar with it.
(btw, the 'red-shift factor' you've been using is normally expressed as z+1, at least in cosmology, with z being the measured redshift; the meaning is the same)
The surface brightness of an emitter in an expanding universe is reduced by a factor of ##(1+z)^4##. One (1+z) comes from stretching of wavelengths, one comes from time delay between peaks, and two come from 'aberration' (the change in observed size) across two dimensions of observation (for details see: Tolman surface brightness test).

By the Stefan-Boltzman law, we have ##j^*=\sigma T^4##.
Comparing the un-redshifted (0) black body with the redshifted one (rs):
$$\frac{j^*_0}{j^*_{rs}}=\frac{\sigma T^4_0}{\sigma T^4_{rs}}$$
and given that $$j^*_{rs}=\frac{j^*_0}{(z+1)^4}$$
we get
$$T_{rs}=\frac{T_0}{z+1}$$
So, for example, some 3000 K plasma emits radiation that is later observed at redshift 1089 as the cosmic microwave background radiation. Its observed temperature will be 2.7 K.
And, it's the same as what we'd get from Wien's law.
So there's no problem there.

Now, I don't know if there's a similar 'aberration' effect with gravitational redshift, but let's assume for the sake of the argument that there isn't, and only the first two factors are in play. We end up with
$$T_{rs}=\frac{T_0}{(z+1)^{1/2}}$$
only now the z+1 is under a square root, which means the observed temperature of the redshifted body will be lower by a smaller factor.
But, if we try using Wien's law for the same redshift, we get a different temperature. This discrepancy is what I think is your argument hinges on.

Only I don't think this means you can get lower flux and lower energy result in a brighter body - that'd be like saying you made the body dimmer and brighter at the same time.
What I think might be going here is either of: a) gravitational redshifts have a similar, or even identical, extra ##(z+1)^2## factor that is being omitted; b) naively using either/or Wien's law or S-B law is not kosher when redshifting is introduced, in which case there being no problem with the cosmological case would be a coincidence.
In either case this looks like a snafu-type situation.
 
  • #3
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I was saying that the observed red-shifted photon flux would be much higher than it should be, given the color temperature of the observed light, and what the black-body photon flux emission should be at the temperature. A more extreme case would be even more red-shifting, say from 20,000 K to 1,000 K, resulting in very red light, but too bright to look at.

It looks like a good chance that gravitational and Doppler red-shifting operate differently. Does cosmological red-shifting operate the same as a straight Doppler effect, of two objects moving apart at the same relative velocity as the cosmological case, but in non-expanding space? If not, then cosmological, Doppler, and gravitational red-shifting would be three different cases.

I believe that relativity states that velocity acceleration and being acted upon by gravity are indistinguishable. However, this assumes non-expanding space, which is of course not true for cosmological red-shifting.

I was assuming that cosmological and Doppler red-shifting are the same, an assumption which someone only casually familiar with the former would make. Would mixing Wien's and S-B laws cause the same problems with straight Doppler red-shifting, as with cosmological red-shifting?

btw, what's the power level of the CMB? Perhaps, a few milliwatts over the whole sky?
 
  • #4
Ibix
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I believe that relativity states that velocity acceleration and being acted upon by gravity are indistinguishable.
It doesn't say that. In fact it says almost the opposite. The equivalence principle says that free falling in the absence of gravity is indistinguishable from free falling in a gravitational field using local measurements. That means that if we are in a closed box and experience acceleration we also cannot tell if we are accelerating in free space or are at rest on the surface of a planet (or star).

The angular diameter of the star is affected by its gravitational field (gravitational lensing, essentially), so I suspect that there are probably some subtleties around angular aberration, similar to the point Bandersnatch is making. By neglecting them, I think that you are implicitly integrating flux over a sphere enclosing the star. Since this is most definitely not a "closed box" local measurement, it wouldn't particularly surprise me if we could see effects of the gravitational field.

(Obviously we could have a black body in our closed box and move around it, but sometimes we'd be below it and sometimes above - not so with the star which would always be below us.)
 

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