Is the Sun invisible at relativistic speeds?

In summary: Can the Sun become a black hole from my frame?No. The sun is a white dwarf.Why should the sun become a bh??I'm not sure. It's just an interesting question.
  • #36
Too hasty, I was still editing my post; now it's completed
 
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  • #37
Thanks, Tom. I will let him know lol.
 
  • #38
tionis said:
Thanks, Tom. I will let him know lol.
seriously, we all agree - after some discussions - that the sun looks brighter; "lol" is not appropriate; if you do believe in Kip Thorne only, then there is no reason to start this thread and let us discuss about it
 
  • #39
tom.stoer said:
seriously, we all agree - after some discussions - that the sun looks brighter; "lol" is not appropriate; if you do believe in Kip Thorne only, then there is no reason to start this thread and let us discuss about it

lol What are you on about?
 
  • #40
tionis said:
lol What are you on about?
about taking this discussion and the time we invest seriously
 
  • #41
tom.stoer said:
about taking this discussion and the time we invest seriously

lol But I do. Seriously.
 
  • #42
I think there may be some confusion here, because some people are answering the question for a hypothetical black body source, and other people are answering it for the actual Sun, which does not emit a perfect black body spectrum. In fact, the Sun has several absorption bands, i.e., wavelengths at which it emits virtually zero radiation. So it deviates a lot from a smooth black body spectrum. If we set our speed of approach such that one of these absorption bands is aligned with the visible spectrum, the radiation in the visible spectrum would go down. But then it would go back up again as we increase our speed still further - at least until reaching the next absorption band. At some point, beyond radio waves, the Sun's spectrum may drop to virtually zero - I don't know, but I wouldn't assume it adhered to the theoretical black body spectrum out to infinitely long wavelengths. A perfect black body is an idealization that doesn't actually exist.
 
  • #43
Samshorn said:
I think there may be some confusion here, because ... other people are answering it for the actual Sun, which does not emit a perfect black body spectrum.
Nobody did that; we were all talking about a perfect black body
 
  • #44
Wait a minute, I'm talking about the real Sun here.
 
  • #45
But you never mentioned red-shift fine-tuning shifting absorption bands exactly to the visible spectrum
 
  • #46
tom.stoer said:
But you never mentioned red-shift fine-tuning shifting absorption bands exactly to the visible spectrum

How am I suppose to know all that? You guys are the physicists here. I merely posted a question based on some online article I read...
 
  • #47
So let's talk about the perfect black body spectrum
 
  • #48
tom.stoer said:
So let's talk about the perfect black body spectrum

I don't know what that is.
 
  • #49
tom.stoer said:
Nobody did that; we were all talking about a perfect black body

tionis said:
Wait a minute, I'm talking about the real Sun here.

This is the confusion I was talking about. Also, I suspect Kip Thorne hadn't clearly thought about the power spectrum - he didn't mention a black body spectrum - so it's unclear whether he was thinking of some actual cutoff limit for the real Sun's wavelengths, or if he assumed a blackbody spectrum (tacitly) and just overlooked the intensity amplification effect of approaching speed (as Tom did originally), which he also didn't mention.

tom.stoer said:
But you never mentioned red-shift fine-tuning shifting absorption bands exactly to the visible spectrum

It isn't just absorption bands. The real Sun may (for all you know) have a cutoff wavelength, beyond which it emits essentially no radiation, and no longer conforms at all to the theoretical blackbody spectrum.

tionis said:
How am I suppose to know all that? You guys are the physicists here. I merely posted a question based on some online article I read...

What article? That might help clear up the confusion.
 
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  • #50
Samshorn said:
This is the confusion I was talking about. Also, I suspect Kip Thorne hadn't clearly thought about the power spectrum - he didn't mention a black body spectrum - so it's unclear whether he was thinking of some actual cutoff limit for the real Sun's wavelengths, or if he assumed a blackbody spectrum (tacitly) and just overlooked the intensity amplification effect of approaching speed (as Tom did originally), which he also didn't mention.

Samshorn, help me out please. Who's right??
 
  • #52
tom.stoer said:
I think we should try to be more exact.

There is a Doppler shift ω → ω' = βω;
There is a trf. of Ω → Ω' = β-2Ω;

I'm not sure what you mean. What' trf. an abbreviation for, and what does the variable Ω represent?

Since we're trying to be exact, the relativistic doppler shift is

[tex]z =\sqrt{\frac{1+\beta}{1-\beta}} = \frac{1+\beta}{\sqrt{1-\beta^2}} = (1+\beta)\gamma[/tex]

where ##\beta = v/c##

The frequency and wavelength of a pulse of light is shifted by a factor of z. As a consequence, the length of a light pulse is shifted by a factor of 1/z.

The energy density (usually denoted ##T^{00}## in GR) scales as z^2, a factor of z comes from the shift in frequency (giving each individual photon more energy) and another factor of z comes from an increased photon arrival rate. I worked out the details of the transformation of the stress-energy tensor T in the previous thread.

Wikki gives, for a black body

[tex]
I = \left( \frac{2\,h\,\nu^3}{c^2}\right) \left(\frac{1}{e^{\frac{h\,\nu}{k\,T}}-1} \right)
[/tex]

where I is the energy/unit time, or the power. Given the energy /unit volume ##\rho##, I would be equal to ##\rho## * (volume) / (time) == ##\rho## * (area) * (length)/(time) = ##\rho## * (area) * c. So I and ##\rho## are proportional , I = ##\rho A c##

Neither A or c should be affected by the lorentz transform, so I should transform in the same manner as ##\rho##.

Thus we can say that for a black body (I'm not sure how good an approxiation the sun is of one)

[tex]
I = (1+\beta)^2 \gamma^2 \left( \frac{2\,h\,\nu_0^3}{c^2}\right) \left(\frac{1}{e^{\frac{h\,\nu_0}{k\,T}}-1} \right)
[/tex]

where ##\nu_0## is the frequency of emission.

Substituting ##\nu_0 = \frac{\nu}{\gamma \, \left(1+\beta\right)}## where ##\nu## is the frequency of reception should give the received intensity law as a function of received frequency, if I haven't made any errors.

This gives a black-body like spectrum, I think - but multiplied by a diferent scale factor.
 
  • #53
Hmm, either a bunch of replies snuck in, or I didn't update before I responded - a lot of this has been covered already.

It looks to me, though, like the relativistic emissions should turn out to be

(1/z) * black_body, with Teff = zT

The increasing Teff makes it brighter, the factor of 1/z makes it dimmer.
 
  • #54
Pervect, you mean Kip is right after all?
 
  • #55
Hmmm- well, doing a series expansion, I'm currently disagreeing with Kip :-(. A series expansion indicates that at the limit as ##\nu## goes to zero, the blackbody radiation goes up quadratically, and the factor that fights it, 1/z, is only linear.
 
  • #56
I'd suggest getting back to professor Thorne and seeing if he agrees with our analysis, or if we've made some silly blunder, if that's feasible.

(add) It'd be helpful to provide more details than to just say "they say you're wrong". SOme of the math, or a link to the thread, or something.
 
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  • #57
I think you miss the trf. of dΩ
 
  • #58
I"m guessing that trf is "Lorentz transform" and dΩ is solid angle? I didn't use that approach at all, I just used the energy density per unit volume, assuming a plane wave (which should be a good approximation far away from the sun).

I'll have to read the cited paper to see how to handle the non-plane wave case - we may be talking about different things.

(add)
I'm basically using the flux, http://en.wikipedia.org/wiki/Jansky, to determine the brightness of a point source. This is just energy / meter^2 (where we constrain the energy to be in the visual band).

As I mentioned, the area, A, of the detector isn't affected by the Lorentz transform if you're moving directly towards the source.

This approach gives the "apparent mangnitude" of a star when you take an apporpriate logaraithm of it.
 
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  • #59
I agree, of a point source you don't expect a dΩ contribution; however the classical Statement is that the spectral density for a black body is u(ω,T) is transformed into another black body u(ω,T') with a velocity-dependent temperature T'(v); this requires to use the solid angle in combination with the spectral density to cancel the per-factors between Ω and u.

Refer to #4, #20 and #22
 
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  • #60
pervect said:
Hmmm- well, doing a series expansion, I'm currently disagreeing with Kip :-(. A series expansion indicates that at the limit as ##\nu## goes to zero, the blackbody radiation goes up quadratically, and the factor that fights it, 1/z, is only linear.

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.

By the way, this is more or less consistent with what is described in the FAQ article cited by the OP. That article talks about moving at high speed toward the Orion belt, and how the appearance of two of the stars would change: "The red Betelgeuse gets its large infrared spectral energy distribution shifted into the visible region, while the blue Rigel gets its spectral energy distribution shifted out to UV and x-ray wavelengths until it fades away." Of course, if we keep increasing the speed, Betelgeuse will fade away too, i.e., it will be shifted up into the x-ray region, not visible to the human eye. I think that article uses some confusing words, though, when it earlier says regarding the Doppler effect: "This results in higher flux intensity, and causes stars in front to get brighter, and from rear to get fainter." The words brighter and fainter can be misleading, because they tend to connote visibility, but really what they mean there is that the overall energy flux increases. This increase causes it to pass out of the visible range, so it gets fainter (in the visible sense) as it gets brighter (in the total energy sense).
 
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  • #61
@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).
 
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  • #62
tom.stoer said:
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.
 
  • #63
pervect said:
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?

pervect said:
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
 
  • #64
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.

[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]

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:

[tex]T^{ab} = \left[ \begin{array}{cc}
E & 0 & 0 & 0\\
0 & E/3 & 0 & 0\\
0 & 0 & E/3 & 0\\
0 & 0 & 0 & E/3\\
\end{array} \right]
[/tex]

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.

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

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.
 
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  • #65
pervect said:
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.
 
  • #66
tom.stoer said:
@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.
 
  • #67
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.
 
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  • #68
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.
 
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  • #69
tom.stoer said:
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?
 
  • #70
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.
 
<h2>1. Is the Sun actually invisible at relativistic speeds?</h2><p>No, the Sun is not actually invisible at relativistic speeds. It may appear to be invisible due to the effects of time dilation and length contraction, but it is still emitting light and energy. </p><h2>2. How fast would I have to travel for the Sun to become invisible?</h2><p>The speed at which the Sun would become invisible depends on the observer's frame of reference. However, for an observer on Earth, the Sun would appear to be invisible at speeds close to the speed of light, which is approximately 299,792,458 meters per second. </p><h2>3. Why does the Sun appear invisible at relativistic speeds?</h2><p>The Sun appears invisible at relativistic speeds due to the effects of time dilation and length contraction. Time dilation causes time to slow down for objects moving at high speeds, which means that the light emitted by the Sun would appear to be moving slower and thus, the Sun would appear dimmer. Length contraction also causes objects to appear shorter in the direction of motion, which could make the Sun appear smaller and less visible. </p><h2>4. Would the Sun still provide heat and energy at relativistic speeds?</h2><p>Yes, the Sun would still provide heat and energy at relativistic speeds. The Sun's energy is not dependent on its appearance, but rather on its nuclear fusion processes. Even if it appears to be invisible, the Sun would still be emitting energy and providing heat to its surrounding environment. </p><h2>5. Can we ever travel at relativistic speeds to see the Sun become invisible?</h2><p>It is theoretically possible for humans to travel at relativistic speeds, but it would require a tremendous amount of energy and advanced technology. Additionally, the effects of time dilation and length contraction would make it difficult to accurately perceive the Sun's appearance at such high speeds. Therefore, it is unlikely that we will ever be able to witness the Sun becoming invisible due to our own travel. </p>

1. Is the Sun actually invisible at relativistic speeds?

No, the Sun is not actually invisible at relativistic speeds. It may appear to be invisible due to the effects of time dilation and length contraction, but it is still emitting light and energy.

2. How fast would I have to travel for the Sun to become invisible?

The speed at which the Sun would become invisible depends on the observer's frame of reference. However, for an observer on Earth, the Sun would appear to be invisible at speeds close to the speed of light, which is approximately 299,792,458 meters per second.

3. Why does the Sun appear invisible at relativistic speeds?

The Sun appears invisible at relativistic speeds due to the effects of time dilation and length contraction. Time dilation causes time to slow down for objects moving at high speeds, which means that the light emitted by the Sun would appear to be moving slower and thus, the Sun would appear dimmer. Length contraction also causes objects to appear shorter in the direction of motion, which could make the Sun appear smaller and less visible.

4. Would the Sun still provide heat and energy at relativistic speeds?

Yes, the Sun would still provide heat and energy at relativistic speeds. The Sun's energy is not dependent on its appearance, but rather on its nuclear fusion processes. Even if it appears to be invisible, the Sun would still be emitting energy and providing heat to its surrounding environment.

5. Can we ever travel at relativistic speeds to see the Sun become invisible?

It is theoretically possible for humans to travel at relativistic speeds, but it would require a tremendous amount of energy and advanced technology. Additionally, the effects of time dilation and length contraction would make it difficult to accurately perceive the Sun's appearance at such high speeds. Therefore, it is unlikely that we will ever be able to witness the Sun becoming invisible due to our own travel.

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