Is the Sun invisible at relativistic speeds?

[tom.stoer]

@samshorn: no, this is is not what the formula tells us. There is not simply a shift like ω → ω' = ω+Δω, but the star looks hotter with T'(v) > T(v=0) when approaching the star, and therefore it looks brighter for every single frequency, b/c the number of observed photons for every single frequency increases!

Please have a look at the references.

Note that this was the mistake I made, too: I was thinking that when approaching the sun a photon emitted in the IR is observed in the visible spectrum, and that for the far IR there are much less photons emitted than in the visible spectrum; in the end we observe less photons. This reasoning is wrong. The formulas for n(ω,T) and u(ω,T) tell a different story. The Doppler shift of every single frequency ω can be "absorbed" in a new, velocity-dependent temperature T'(v). So it looks as if the star gets hotter. But for an hotter black body the number of photons is increased for every single frequency ω, therefore the star looks brighter for every single frequency (not only in total).

 

[pervect]

I think this is not what the formula says. There is not simply a shift like ω → ω' = ω+Δω, but the star looks hotter with T'(v) > T(v=0) when approaching the star, and therefore it looks brighter for every single frequency.

Well, it would be good to have the expression to compare with the literature, but otherwise it's not really needed - you can work the problem out without even introducing the $\Omega$ coordinate.

Unfortunately the problem in the literature isn't quite the one that the OP is stating - I'm not sure if it makes a difference yet. I'd rather like to see "Distribution of Blackbody Cavity Radiation in a Moving Frame of Reference" , which appears to have a more detailed calculation, but I don't have access.

I do have some concerns, but not enough time to track them all down.

 

[tom.stoer]

Unfortunately the problem in the literature isn't quite the one that the OP is stating - I'm not sure if it makes a difference yet.

Where's the difference? the point-like source? the sun instead of an idealized black body?

I'd rather like to see "Distribution of Blackbody Cavity Radiation in a Moving Frame of Reference" , which appears to have a more detailed calculation, but I don't have access.

For referenmces please have a look at post #4; the standard derivation is for CMB, but b/c dΩ is there and the trf. is known, this applies to other sources as well.

Another paper I found is http://arxiv.org/abs/1007.4539v1 In section LORENTZ TRANSFORMATION OF THE TEMPERATURE FIELD the transformation is discussed in detail. I gues we can all agree on the formulas, so it's only the interpretation which could be subject to discussion.

Please let me know where you see problems

 

[pervect]

The problem of the sun's appearance is a different (and much simpler) problem than the CMB background. The stress-energy tensor is different, too. I'm not sure if you ever looked at the thread I referenced previously, https://www.physicsforums.com/showthread.php?t=681172

Using standard t,x,y,z coordiates (don't need to use spherical coordinates, it just makes life difficult) we get the stress energy tensor for the case of interest , a "null dust" of a beam of light headed for us.

T^{ab} = \left[ \begin{array}{cc}<br /> E &amp; E &amp; 0 &amp; 0\\<br /> E &amp; E &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; 0\\<br /> \end{array} \right]<br />

For the CMB, you'll get something, different - the stress energy tensor of a perectg fluid, with rho = 3P. Letting rho = E, we would write:

T^{ab} = \left[ \begin{array}{cc}<br /> E &amp; 0 &amp; 0 &amp; 0\\<br /> 0 &amp; E/3 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; E/3 &amp; 0\\<br /> 0 &amp; 0 &amp; 0 &amp; E/3\\<br /> \end{array} \right]<br />

Relativisitic visualization papers, such as http://arxiv.org/pdf/physics/0701200v1.pdf, describe what happens to intensity - but there is a referenced paper that's fairly easy to find, and even better!In search of the ’’starbow’’: The appearance of the starfield from a relativistic spaceship
John M. McKinley and Paul Doherty

one link is:
http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf

Equations 10 and 11, and the associated discussion, gives what I derived in terms of the stress-energy tensor and get the same result for I.

Equation 21, in particular, gives the difference in magnitude of a star due to the motion, weighted for the approximate optical sensitivity of the human eye.

log_{10} W = 10400 K (1/T - 1/DT) - log_{10} D

D being the doppler factor $\gamma(1+\beta \cos \theta)$

There are also some plots of particular stars of certain temperatures, given in fig 4.

An inspection of eq 21 and inspection of fig 4, and taking the limit as D goes to infinity seems to suggest that Kip was right (not too surprising), though I don't quite see yet where I went wrong.

You can clearly see in fig 4 the intensity dropping as D increases, though the drop is hardly dramatic.

 

[tom.stoer]

The problem of the sun's appearance is a different (and much simpler) problem than the CMB background.

I do not talk about CMB, only abolut black body radiation. It should be irrelevant where the bb radiation comes from. If a different approach yields different results then something must be fundamentally wrong (the isotropic bb radiation should be at least reasonable for a very large and nearby star, omitting geometry effects due to point like emitter, small discs etc.). Using bb radiation with increasing intensity per frequency results trivially in increasing integrated intensity. The only difference could be due to dΩ for pointlike emitters.

Anyway - I would propose to discuss the bb isues first, b/c they are widely accepted and used in the astronomy / astrophysics community.

 

[Samshorn]

@samshorn: no, this is is not what the formula tells us... The Doppler shift of every single frequency ω can be "absorbed" in a new, velocity-dependent temperature T'(v). So it looks as if the star gets hotter. But for an hotter black body the number of photons is increased for every single frequency ω, therefore the star looks brighter for every single frequency (not only in total).

Remember that the intensity function is usually expressed per unit area of the emitting surface (so it doesn't apply to a "point source"). If we approach the source at extreme relativistic speed, the area of the emitting surface shrinks due to aberration. It's as if you are looking at the aperture of a cavity radiator of a certain temperature, but the size of the aperture is shrinking, and hence the energy received is reduced for a given temperature. So I don't think the intensity per unit area can be used as a measure of the brightness of the star, because the emitting surface area is shrinking.

 

[tom.stoer]

Starting with a point-like emitter is strange.

If there is a nearby, huge star, the whole forward-dΩ (-dΩ') stays bright (becomes brighter) under Lorentz-trf.; so I don't see what you want to use instead of the bb radiation formula

$\frac{d\Omega}{e^{\hbar\omega/kT} - 1}$

which takes this dΩ into account.

If you look at a certain Ω'(v) which is the image of Ω(0) via a Lorentz trf. with (velocity v) where Ω(0) is fully covered by the star (!) then within these Ω and Ω' the standard trf. for bb radiation should apply. You problems seem to result from taking the black sky into account.

 

[mfb]

I think we have two different scenarios here.
pervect is treating the star as point-source and looking at the total luminosity (in the visible range), while tom.stoer and me are considering the surface brightness.

I am a bit surprised that this leads to a different behavior in the ultra-relativistic range.
Edit2: No, I am not longer surprised. It explains everything.

Edit: Oh, took me so long to check the references, calculations and posts that I missed two new posts.

 

[pervect]

I do not talk about CMB, only abolut black body radiation.

The reference in #4 , whose formula you ere using, WAS talking about the CMB.

It should be irrelevant where the bb radiation comes from.

It matters if the radiation is in a beam, or of it's randomly going in all directions (like the CMB). The transverse components don't transform like the ones pointed along the line of travel.

If a different approach yields different results then something must be fundamentally wrong (the isotropic bb radiation should be at least reasonable for a very large and nearby star, omitting geometry effects due to point like emitter, small discs etc.). Using bb radiation with increasing intensity per frequency results trivially in increasing integrated intensity. The only difference could be due to dΩ for pointlike emitters.

Did you read any of the references I posted?

 

[mfb]

So the sun would be bright blue. It's angular size
would become small as you approached c. So bright blue and small is how
it would look.

I think that is in agreement with all our analyses.

 

[tom.stoer]

The reference in #4 , whose formula you ere using, WAS talking about the CMB.

Yes, but it doesn't matter if you are close enough and if you restrict the analysis to Ω and Ω' such that no black sky is contained. First I want to understand the bb radiation issue, then we may include the geometric effects.

It matters if the radiation is in a beam, or of it's randomly going in all directions (like the CMB). The transverse components don't transform like the ones pointed along the line of travel.

see above

Did you read any of the references I posted?

I checked http://arxiv.org/pdf/physics/0701200v1.pdf but there is not enough math to see what they are really doing. http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf is damaged and does not open on my computer.

 

[tom.stoer]

If you are traveling fast toward the sun, the thermal radiation from the
sun gets blue shifted. That makes it correspond to a hotter and hotter
temperature. ... So the sun would be bright blue. It's angular size
would become small as you approached c. So bright blue and small is how
it would look.

That's what I expect after all the discussions.

 

[tionis]

I'm waiting for Prof.Thorne reply to what Prof. Gott said. I'll keep you posted.

 

[Samshorn]

Professor Richard Gott replied:
... It's angular size would become small as you approached c. So bright blue and small is how
it would look.

Right, but the question is, which of those effects wins? The intensity per unit area of the emitter goes up, but the area goes down... so in the limit as we approach c, do we receive more energy per second from the star (in the visible range), or less? In other words, does the star fade from sight, or become more and more visible?

 

[tionis]

Right, but the question is, which of those effects wins? The intensity per unit area of the emitter goes up, but that area goes down... so in the limit as we approach c, do we receive more energy per second from the star (in the visible range), or less? In other words, does the star fade from sight, or become more and more visible?

I know, Samshorn. I'm nonplussed, too. Prof. Thorne said the Sun would ultimately disappear from view, which is what I'm trying to get @. But both men are leading experts on relativity, so...

 

[mfb]

Right, but the question is, which of those effects wins?

In terms of the total light you get, the smaller size wins.

We cannot get a black disk, but we can get a disk that is so small that you don't see it any more, even with the increased brightness per area.


Some more mathematics: If I consider the ultra-relativistic limit here,
$$n(\omega,\Omega)d\omega d\Omega \approx \frac{\omega T_{eff}}{2\pi^2}d\omega d\Omega$$ and
$$T_{eff} \approx 2T_* \gamma$$

Therefore, the brightness increases with the relativistic gamma-factor, while the area scales with $\gamma^{-2}$ (?). This would give a total luminosity (in the visible range) which scales with $\gamma^{-1}$ for very high speeds.

 

[Samshorn]

In terms of the total light you get, the smaller size wins. We cannot get a black disk, but we can get a disk that is so small that you don't see it any more, even with the increased brightness per area.

I agree, and this same conclusion can be reached simply by applying a Lorentz transformation to the light energy impinging on the observer, as explained in post #61, given that the star's spectrum in its own rest frame drops off more rapidly than can be compensated by the Doppler intensification.

 

[tom.stoer]

I agree, and this same conclusion can be reached simply by applying a Lorentz transformation to the light energy impinging on the observer, as explained in post #61, given that the star's spectrum in its own rest frame drops off more rapidly than can be compensated by the Doppler intensification.

Sorry to repeat myself, but your reasoning

As our speed toward the Sun increases, the visible light comes from the lower frequency range of the Sun's spectrum (Doppler shifted up to the visible frequency range). Remember that the energy of a pulse of light under a Lorentz transformation increases in exactly the same ratio as the frequency. So, for example, when our speed toward the Sun doubles the frequencies (i.e., when we are seeing light emitted by the Sun at half the visible frequencies), it will also be doubling the energies. However, at half the frequency, the spectral energy is extremely low, so doubling it doesn't make it very big. The blackbody spectrum eventually drops exponentially, and this drop prevails over the Doppler energy increase. Remember that as the temperature of a blackbody increases (as we approach the Sun at higher and higher speed), the peak frequency of the spectrum increases, and it will eventually pass out of the visible range. So instead of seeing the light, we'll just be getting fried with x-rays, etc..

cannot be correct.

You are talking about spectrum and Doppler shift only w/o taking into account geometric effects. But the shrinking of the emitter IS a geometric effect. By your reasoning from #61 even isotropic bb radiation would appear darker to the approaching observer, but we know that it appears brighter. So even if the geometric reasoning is correct, it is not contained in #61. Or could you please tell me how to change the reasoning from #61 to get the behavior for CMB? Where does the difference hide?

 

[Samshorn]

Sorry to repeat myself, but your reasoning cannot be correct. You are talking about spectrum and Doppler shift only w/o taking into account geometric effects.

You only need to worry about the size of the source if you have taken the approach of first working out the transformed intensity of the source per unit area, in which case you then need to determine the transformed area of the source. You might be tempted to take that approach, thinking that the theorem about black bodies transforming to black bodies provides a shortcut to the answer, but it actually is the long way around. It's better to just look at the energy impinging on the observer from the direction of the star, and apply the Lorentz transformation, and note that the energy content drops off faster than can be compensated by the Doppler intensification.

By your reasoning from #61 even isotropic bb radiation would appear darker to the approaching observer, but we know that it appears brighter. So even if the geometric reasoning is correct, it is not contained in #61. Or could you please tell me how to change the reasoning from #61 to get the behavior for CMB? Where does the difference hide?

The CMB is a different scenario. With a star we have energy impinging on the observer from a specific direction so that it all shares essentially the same Doppler shift and can be analyzed (for its energy content) on that basis. But for the CMB we have a surrounding impingement from all directions, some from behind, all being swept forward by aberration, but with a whole range of Doppler shifts. So the simple reasoning based on a single Doppler shift for the impinging energy that works for a star would not be applicable to that case.

 

[Samshorn]

OK. Here is Prof. Thorne's reply: I thought you were traveling away from the sun...

That's puzzling, because his original reply said "The sun emits infrared radiation, which will get shifted into the visible part of the spectrum...", and so on. This sure makes it sound like he was talking about someone traveling toward (not away from) the sun. Why would infrared get shifted to visible if you were traveling away from the sun?

It's also a bit strange that he says he agrees with Gott, because it isn't obvious what Gott's answer really is. Gott just said there are two factors (increased intensity per unit area, and decreased area), but didn't say which one prevailed. So, when Thorne says he agrees with Gott, it isn't clear (to me) what he is agreeing to, i.e, is he saying the star fades from sight (as the FAQ says and as some of us have concluded here in this discussion), or is he saying it becomes more visible (meaning the energy received in the visible range increases)?

 

[tionis]

I have forwarded your question :

Right, but the question is, which of those effects wins? The intensity per unit area of the emitter goes up, but the area goes down... so in the limit as we approach c, do we receive more energy per second from the star (in the visible range), or less? In other words, does the star fade from sight, or become more and more visible?

to Dr. Gott, Samshorn.

 

[Evo]

A reminder to members - if you post e-mails, please remove all personal contact information first.

Thanks.

 

[tom.stoer]

@Samshorn, pervect:

I'm not sure if you ever looked at the thread I referenced previously, https://www.physicsforums.com/showthread.php?t=681172 ...

In the meantime I did, and I worked out the null-dust and the plane wave example for the Poynting vector i.e. the T⁰ⁱ components. After a Lorentz trf. I find - as expected - the factor

$r = \frac{1-v}{1+v}$

But this does not help for the case of a star b/c
1) in the time averaged Poynting vector the Doppler shift is not visible
2) the homogeneous plane wave or null-dust do not show any effect due to geometry

I understand the concerns regarding isotropic bb radiation not being an appropriate model for a (nearly) pointlike star. But I sill do have concerns not taking the Planck spectrum into account. Even for a pointlike star the spectrum is thermal for one single direction of wave propagation.

 

[pervect]

I checked http://arxiv.org/pdf/physics/0701200v1.pdf but there is not enough math to see what they are really doing. http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf is damaged and does not open on my computer.

That's really the best paper. :-( It computes just what you're looking for - or very close.

The digital object identifier (doi) for this paper is: http://dx.doi.org/10.1119/1.11834, but it may be behind a paywall.

I don't think I can do the paper justice in a post, but I can provide a quick summary:

The total photon arrival from the source, integrated over the entire view, scales as r. So the photon arrival rate increases if you're moving towards the object.

You seem to have missed a square root in your presentation of r, I'll assume it's just a typo unless otherwise argued about.See http://en.wikipedia.org/wiki/Relativistic_Doppler_effect for example, for the derivation of r Just In Case.The total energy delivered scales as r^2, because $E = h \nu$ and $\nu$ gets doppler shifted. The usual defintion of intensity is via delivered energy, so the intensity scales as r^2.

Relativistic aberation makes the angle subtended by the object smaller - this causes the object's apparent area to shrink by a factor of r^2. This would make the surface brightness scale as r^4 (the same energy is delivered in a smaller area). But if you are far enough away so that the object is smaller than the optical resolution of the telescope that you look at it through, this effect won't matter. YOu'll be limited by your optics, and you'll only see a r^2 increase.

Usually, stars don't show a disk , the telescope can't resolve the surface, and we talk about the "stellar magnitude" based on the total energy received, the entire visual field maps to what's effectively a point. So that's my starting assumption. With this assumptoin we get a r^2 brightness enhancent - and a doppler shift.

The authors actually work out the received brightness in the human visual range, by using a crude model of the eye's black and white frequency response.

I'll upload a few screenshot, which I think constitutes "fair use" for educational purposes, so you can at least get some information.

One post graphs their results.

Doppler factors > 1 represent motion towards the star. You can see that eventually the brightness starts to decline.

The other screen snapshot represents the equation for the unweighted spectrum S(r). You multiply this by your "response function", to filter out invisible frequencies, and that integral gives your brightness. The authors used an gaussian weighting function to represent the sensitivity of the human eye, rather than a crude square step function (which has a value of 1 if the frequency is visual and 0 if it isn't).

But I don't feel up to presenting it at the level of detail that includes their approximate visual weighting function specifically, you'll need to track down the original paper for that.

You might be wondering why the factor in front of the integral is r^-2, if the intensity scales as r^2. If you perform the integral though (the paper does it), you'll see that the end result DOES scale as r^2.

WHile you can see that the authors integrated in terms of wavelength rather than frequency, it appears that my earlier mistake was a failure to scale $d\nu$ properly under the transform. I scaled what multiplied it properly, but not $d\nu$ itself. The paper doesn't use the same approach, they use $\lambda$ instead.

 

[tom.stoer]

Thanks for the summary.

I will check the references the again. The PDF that seemed to be broken on my notebook is displayed correctly on the iPad ;-)

I agree to all the r and r^2 factors; and I guess we agree on the main ideas.

My key idea is that if you use plane waves or something like that the object always gets darker b/c the visible part of the spectrum is shifted to the UV and there is nothing from the IR to replace that. But in reality there is a Planck-type spectrum and you get an enhancement in the visible part due to replacement from the IR which is shifted and enhanced due to the hotter effective temperature T'(v). I think we agree in that idea.

I found some other references discussing images of nearby stars. The images are not distorted due to length contraction (spheres a mapped to spheres) but they take different Doppler shifts due to the extension of the disc into account. That's quite interesting.

http://www.vis.uni-stuttgart.de/~weiskopf/publications/acmtog99.pdf
http://www.tempolimit-lichtgeschwindigkeit.de/sphere/sphere.pdf

 

[pervect]

The IR does shift up to the visible, but when you do the math correctly, if you assume a black body spectrum it just isn't enough to replace what you've lost. So the intensity slowly goes down.

I don't know for sure how good an approximation the black body spectrum is, but I would think it'd be a good approximation at low frequencies.

 

[tom.stoer]

The IR does shift up to the visible, but when you do the math correctly, if you assume a black body spectrum it just isn't enough to replace what you've lost. So the intensity slowly goes down.

Can you please give me a hint where exactly I can find this formula

 

[Samshorn]

To evaluate the asymptotic behavior at ultra-relativistic speeds, we can focus on just the very low frequency range of the star’s spectrum, so the energy received (from a star assumed to radiate a black body spectrum of a certain temperature) by an eye at a certain location with frequencies near n is proportional to n^2. Hence the ratios of the energies impinging on the eye at a low frequency n1 and an even lower frequency n2 is (n2/n1)^2. If we give the eye some (additional) speed v directly toward the star, such that the n2 frequency is Doppler shifted up to n1, we will scale up the energy by n1/n2 (since energy scales like frequency), so the energy now being received at frequency n1 is n2/n1 times the energy that was being received at that frequency without that increased speed. Thus the energy drops off asymptotically in proportion to the frequency, so the factor in terms of speed v is just the Doppler factor sqrt[(1-v)/(1+v)].

 

[tionis]

seems to suggest that Kip was right

Yup!

 

[Samshorn]

Yup!

Well, again we have the confusion between considering the actual spectrum of the Sun, versus an ideal black body spectrum. Carroll says "There is a lower limit to the frequency of light emitted by the Sun, although it's down in the radio regime." So he is not considering the question for an ideal black body that emits out to infinitely long wavelengths. I think everyone agrees that, if there is a lower limit on the Sun's frequencies, then obviously we could Doppler shift its emissions up above the visible frequency range. That is self-evident. The more challenging question that we've been discussing is what happens in the theoretical case of an ideal black body.

By the way, I think that posing vaguely and ambiguously worded questions to random "experts", without clearly explaining the background and intent of the question, is not a very efficient way of seeking enlightenment. I would guess that all of the "experts" that have been cited would quickly agree on the answer, if only the question was posed to them in a clear way - for example, distinguishing between the actual Sun versus an ideal black body. Again, if we're talking about the actual Sun, which has a lower limit to emitted frequencies, the answer is self-evident.