Is the Sun invisible at relativistic speeds? Part II

[tionis]

^^This is how the Sun would look from a distance of 4 light-years. If you add all the relativistic effects and stuff, wouldn't the Sun disappear from the visible spectrum? I was told that from a mathematical point of view, it is correct to say that some visible photons would still reach you, but from a physical point of view, and given that the eye needs about 100 photons per second, the Sun does in fact become invisible. Do you guys agree?

In this thread, you may treat the Sun as an ideal black-body if you like, but I would also like a consensus on whether the real Sun becomes invisible (to the human eye) or not.
Edit: In this thought experiment, like in the previous thread, we are considering traveling towards the Sun at close to the speed of light.

 

[Vanadium 50]

What relativistic effects? How fast are you moving?

 

[Dale]

And are you moving towards, away, or transverse to the sun.

Btw, asking for consensus is hopeless. The best you can ask for is that people justify their claims through derivations or references.

 

[tionis]

What relativistic effects? How fast are you moving?

And are you moving towards, away, or transverse to the sun.

Btw, asking for consensus is hopeless. The best you can ask for is that people justify their claims through derivations or references.

Hi vanadium & DaleSpam, yeah, one of the emails I got referred to a ''relativistic beaming'' effect happening at 99.995% the speed of light while moving towards the Sun.

 

[pervect]

What was unclear about the detailed answers in the last thread? In particular the answers given by the paper http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf in figure 4.

Since you're asking the same question, you can expect the same answer. Apparently, you are unhappy with some aspect of that answer, but it's not clear what you're unhappy with, exactly.

Note that to answer your question, you need to know the velocity, which includes both the magnitude and direction, of the observer.

Figure 4 of the paper above gives the answer graphically for a range of velocities directly towards and away from the sun as seen by the naked human eye. I'm assuming you're interested in the general behavior over a range of velocities, which implies that the answer must be a graph. However, if you have some specific velocity in mind, and some specific direction (other than diretly towards or away), the paper has the necessary equations to calculate the answer.

Also if you are using some instrument OTHER than the naked human eye, you need to specify it.

[add]You might also find figure 2 helpful, it's specifically about the sun. It has more details about the perceived spectrum at the expense of restricting consideration to only a pair of velocities.

Read the caption - I'm assuming you CAN get to the paper. Notes in [] are my explanations.

Doppler shifted blackbody spectra and human eye response. The curve labeled 1 represents the sun (5800 K) viewed at rest where it appears yellow. The curve labeled 5 represents the sun as viewed from a spaceship approaching at a speed $\beta = 0.923$ [ed. note - this is a fraction of the speed of light] (so that D=5) [ed note: D is the doppler shift factor, at this speed frequencies are shifted by a factor of 5:1]. The sun appears blue-white and somewhat brighter than at rest. The curve labelled .2 represents the sun as viewed from a spaceship receding at $\beta = 0.923$. The sun appears red, but very dim.

Now if you are interested in velocities other than $\beta = 0.923$ and doppler factors other than 5:1 you'll need to see figure 4. Figure 4 basically says that the brighness, as you approach the sun, initially increases, but as you increase the velocity further, starts to decrease.

 

[tionis]

Pervect, hi. I was hoping you guys could fill-in the details regarding velocities and stuff. If you could walk me through how the Sun would look to the ''naked human eye'' as I approach it from rest all the way to 99.9999...% c, that would be nice. I'm specially interested in what happens as we get closer and closer to c. Does the Sun actually disappears from view? Many experts agree that it does indeed while others do not. Maybe the collective brain power here can get to the bottom of things. I'm not sure if that is clear enough for you. Sorry for not formulating the question in a more rigorous way.

 

[pervect]

Pervect, hi. I was hoping you guys could fill-in the details regarding velocities and stuff. If you could walk me through how the Sun would look to the ''naked human eye'' as I approach it from rest all the way to 99.9999...% c, that would be nice. I'm specially interested in what happens as we get closer and closer to c. Does the Sun actually disappears from view? Many experts agree that it does indeed while others do not. Maybe the collective brain power here can get to the bottom of things. I'm not sure if that is clear enough for you. Sorry for not formulating the question in a more rigorous way.

I added a few remarks to my previous post, which I hope helps. Figure 4 will take you to a velocity all the way up to $\beta = .9998$, a doppler shift factor of 100:1. It's clear what the general trend is from figure 4, and if you look at the equation that is being graphed, you can compute the answer above doppler factors of 100.

 

[tionis]

Pervect, thanks. Does that mean that at speeds above B = .9998 the Sun becomes invisible?

 

[PAllen]

Pervect, thanks. Does that mean that at speeds above B = .9998 the Sun becomes invisible?

NO. It never becomes invisible for black body. It might become invisible if there is low energy cutoff in the sun's emissions, but I am not sure this correct or even known.

 

[Vanadium 50]

Further, as you move towards the sun, it gets brighter, not dimmer.

 

[tionis]

NO. It never becomes invisible for black body. It might become invisible if there is low energy cutoff in the sun's emissions, but I am not sure this correct or even known.

Further, as you move towards the sun, it gets brighter, not dimmer.

PAllen & Vanadium: do you guys agree with this plot?

 

[Vanadium 50]

I have no idea what that plot is saying. Nonetheless, if you can see the sun, and you start moving toward it, you can still see it.

 

[Russell E]

In this thought experiment... we are considering traveling towards the Sun at close to the speed of light... You may treat the Sun as an ideal black-body if you like, but I would also like a consensus on whether the real Sun becomes invisible (to the human eye) or not.

It's a trick question, because obviously at high enough speed the x-rays etc would be so intense that they would fry your eyeballs (not to mention kill you), so the Sun would indeed "become invisible to the human eye". But seriously, a google search of "disappearing sun doppler" turns up

www.mathpages.com/home/kmath693/kmath693.htm

 

[WannabeNewton]

It's a trick question, because obviously at high enough speed the x-rays etc would be so intense that they would fry your eyeballs (not to mention kill you)

Not unless the observer is superman!

www.mathpages.com/home/kmath693/kmath693.htm

Thanks for the link, quite instructive! Would you happen to know if this calculation is from some class? The code "693" makes it seem like it was from a class.

 

[tionis]

It's a trick question, because obviously at high enough speed the x-rays etc would be so intense that they would fry your eyeballs (not to mention kill you), so the Sun would indeed "become invisible to the human eye". But seriously, a google search of "disappearing sun doppler" turns up

www.mathpages.com/home/kmath693/kmath693.htm

Is not a trick question, it's a thought experiment. I'm posting this again 'cause I've spent the past few days really confused by all the contradicting replies I'm getting from scientists all over the world. Also, my friend and I want to make an accurate relativity simulation we can post on youtube, but whatever lol.

 

[PAllen]

Pervect's link seems much more detailed, and claims increase in brightness up to a Doppler factor of 10 for the sun, while (mathpages = Kevin Brown) claims the peak is at half the speed of light. This discrepancy makes me favor the published paper. Further, the paper claims there is only about 3-4 magnitude decrease in the sun's brightness even for the ultra-relativistic speed of Doppler factor of 100. Given the brightness of the sun at rest, if we are talking about an observer at said speed in the solar system, 3-4 magnitude decrease would not matter much. I guess, even based on the paper's results, there might be a point (e.g. Doppler factor of 1 trillion or more) where sun would be too dim to visually see (given magic protection from gamma rays that would instantly convert all matter into subatomic particles).

 

[pervect]

PAllen & Vanadium: do you guys agree with this plot?

First, a quick question that you've never answered (I'm afraid I'm fearing the worst). Can you view the reference link I posted, and did you read it?

You don't define your variables, but I think the idea is that BB is the spectral radiance, as per
http://en.wikipedia.org/wiki/Planck's_law

If you refer to the paper, which I *really hope* you're reading, you'll see that the paper uses the alternate formula from the wiki, where they use $\lambda = \frac{c}{\nu}$ instead of $\nu$, ie in terms of wavelength rather than frequency.

This gives eq 12 of the above paper, which is equivalent to your own but in terms of wavelength rather than frequency.

eq 13 gives the actual received power / unit area, if the stars radius is a and it's distance is R.

Your transformation to the moving frame is incorrect, however.

The spectral radiance in the moving frame is D^3 times the spectral radiance in the stationary frame - see eq 11.

When you integrate this out, you find that the total energy recieved, integrating from lambda from 0 to infinity should scale as D^2. (D being the doppler factor). At least if you can get the integrals to work out - maple doesn't want to do them for me, and I don't even want to attempt to do them by hand.

The appendix to the paper gives a short derivation of why D^2 is correct for the total intensity. You can refer to the "photon arrival" thread for perhaps a clearer discussion, the gist is that the shift in energy per photon gives one factor of D, and that the photon arrival rate also increases by the doppler factor D, giving a total intensity increase of D^2. You can also try the Wiki article on "relativistic beaming" http://en.wikipedia.org/wiki/Relativistic_beaming.

If you evaluate your integral that you give in the moving frame, you should see that the intensity does NOT increase as D^2, but much faster. (If you can get the integral to evalutate, that is - good luck with that!. )

But anyway that is where you appear to have "gone wrong".

[add]
WHile I can't do the integrals myself, I can point out that due to the Steffan-Boltman law
http://en.wikipedia.org/wiki/Stefan–Boltzmann_constant

one expects the total power emitted to be proportional to T^4. Therefore, when you multiply the temperature by D, you get a radiant power increase of D^4. To get the correct transformation law, you need to not only multiply the temperature by D, but divide the intensity by D^2, so that the radiant intensity scales by D^2.

 

[Russell E]

What was unclear about the detailed answers in the last thread? In particular the answers given by the paper http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf in figure 4.

That paper first quotes the black body power density spectrum for the rest frame S, in which the star has radius 'a' and distance R, and then it applies the cubed gamma Doppler factor to transform the power density spectrum to another frame S', but it continues to use 'a' and R as if they are invariant between S and S'. Note that the derivation leading up to their equation 11 doesn't account for the change in the solid angle due to the effects of aberration on 'a' and length contraction on R, and they use equation 11 to transform the power density, going from equation 13 to equation 14. The power spectrum is defined per unit area of emitting surface AND per unit solid angle, and these are not invariant when changing frames. If the densities in S and S' are defined consistently relative to their own frames, another factor is needed to account for the geometric effects, but no such factor appears in the paper. That's why I'd say that although their results are qualitatively correct (with the visible brightness first increasing and then dropping off for greater speeds), they aren't quantitatively correct.

 

[pervect]

That paper first quotes the black body power density spectrum for the rest frame S, in which the star has radius 'a' and distance R, and then it applies the cubed gamma Doppler factor to transform the power density spectrum to another frame S', but it continues to use 'a' and R as if they are invariant between S and S'.

There's a good reason for that - the radiation is only spherically symmetric in the rest frame. What the paper is doing is analyzing the problem in a purely cartesian coordinate system, (one local to the observer) in which there is no solid angle, and the source is essentially pointlike.

They do the boost in said Cartesian coordinate system. Notice that eq 1) uses t,x,y,z - you won't see any solid angles anywhere when they're doing the boost. That's your clue as to why there isn't any consideration of the solid angle.

They do do a conversion, "in passing", to convert from the solid-angle formalism to the simpler to analyze cartesian coordinate system, when they introduce a/R^2. They do this conversion in the star's rest frame, the only place where the radiation is symmetrical.

Note that the derivation leading up to their equation 11 doesn't account for the change in the solid angle due to the effects of aberration on 'a' and length contraction on R, and they use equation 11 to transform the power density, going from equation 13 to equation 14.

They're not concerned with the solid angle, as they are assuming that the star is pointlike.

The power spectrum is defined per unit area of emitting surface AND per unit solid angle, and these are not invariant when changing frames. If the densities in S and S' are defined consistently relative to their own frames, another factor is needed to account for the geometric effects, but no such factor appears in the paper. That's why I'd say that although their results are qualitatively correct (with the visible brightness first increasing and then dropping off for greater speeds), they aren't quantitatively correct.

One is certainly free to analyze the problem in this manner, and it would be instructive. One would expect the results to agree with the (IMO, simpler) approach in which one treats the radiation in a local cartesian frame, as the paper does. It's a matter of choosing the coordinates you like best and which are easier to work with.

If the star is NOT pointlike, there is an additional increase in intensity, due to the fact that the star shrinks its angular size.

However, when you are unable to resolve the disk of the star (which is the case that one would expect), there is no angular size to shrink. The optics of the reciever (in this case, the eye) smear out the star over a greater solid angle than the star actually occupies, due to diffraction and whatnot. The approach taken by the paper treats this case (which is what one expects, one does not expect stars to show a disk). I think it treats it in a simpler manner than introducing the solid angle would -, if you really like spherical coordinates, feel free.

 

[nitsuj]

Is a really dense star not equivalent to this thought experiment? As in a star that's .99 "away" from becoming a black hole is the same as moving at .99 c compared to a "low density" star.

 

[Russell E]

They do do a conversion, "in passing", to convert from the solid-angle formalism to the simpler to analyze cartesian coordinate system, when they introduce a/R^2.

That's the problem. They apply the factor (a/R)^2 to account for the solid angle, but they apply that same factor in both S and S', whereas it represents the solid angle factor only in S. When they write their density function in S' they need to account for the change in the solid angle. (By the way, this has nothing to do with what coordinate system we use.)

...you won't see any solid angles anywhere when they're doing the boost...

Right. That's the problem.

They're not concerned with the solid angle, as they are assuming that the star is pointlike.

It's permissible to assume point-like for purposes of saying all the light is coming from the same direction with the same Doppler shift, but no matter the size or distance of the source we can't neglect the effect of the solid angle on brightness. Two stars with the same surface brightness at the same distance will have different optical brightness viewed from the Earth if one is twice as large as the other, even if they are located many light years away so that we can't resolve either of their disks. If we really wanted to neglect the solid angle, we would have to introduce a delta function for the intensity, so the integral of an infinite intensity over a zero surface area would give a finite result. But they haven't done this.

 

[Dale]

What was unclear about the detailed answers in the last thread? In particular the answers given by the paper http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf in figure 4.

I don't know how I missed that reference last time. Excellent reference, completely answers the question.

 

[tionis]

Can you view the reference link I posted, and did you read it?

Not only did I read it, but I also emailed it to a few professors. They laughed! I guess it's time to update that paper.

Your transformation to the moving frame is incorrect, however.

Yes, I thought so, too. But then I consulted a couple of profs. and they said they couldn't find anything wrong with it. Furthermore, they agreed that when you plot the intensity as a function of photon frequency plus the proper reference frame of the emitter and receiver, along with the coordinate transformations between the two frames and, finally, the transformation from coordinate quantities to a physically-measurable ones, you get an intense UV-Sun.

Excellent reference, completely answers the question.

So, does the Sun become invisible or not?

 

[tionis]

Is a really dense star not equivalent to this thought experiment? As in a star that's .99 "away" from becoming a black hole is the same as moving at .99 c compared to a "low density" star.

Good question, nitsuj. Anyone care to answer??

 

[PAllen]

Good question, nitsuj. Anyone care to answer??

It has some relation to a star moving away near c; none to approaching a star near c. The 'disappearance' moving away from a star is self evident and not under discussion.

 

[PAllen]

Yes, I thought so, too. But then I consulted a couple of profs. and they said they couldn't find anything wrong with it. Furthermore, they agreed that when you plot the intensity as a function of photon frequency plus the proper reference frame of the emitter and receiver, along with the coordinate transformations between the two frames and, finally, the transformation from coordinate quantities to a physically-measurable ones, you get an intense UV-Sun.

Intense UV (or even intense gamma) sun is self evident. This has NO BEARING AT ALL on the question of the intensity of radio emissions blueshifted into visible range. The paper also notes these features (intense UV, ultimately gamma).

 

[Vanadium 50]

The same thing that got the last thread closed is happening again in this one.

Conversations on PF should stay on PF, and not get emailed to third parties for their comments. Apart from being rude, it's difficult to carry on a conversation when half of it is second- or third- hand.

This thread is open, but will be closed again if this continues.

 

[tionis]

Yay! Thanks, Vanadium. It would be awesome if the two threads could be merged into a single one.

 

[Dale]

So, does the Sun become invisible or not?

Yes. See figure 4.

 

[PAllen]

Yes. See figure 4.

Right, but see #16. Using the equations associated with fig. 4, and noting that the visual magnitude of the sun at the distance of Jupiter is -23, and that to become invisible to the naked eye (le'ts not even get into invisible to an telescope), this needs to increase by 29 in magnitude. Using the magnitude formula below figure 4, with 5800 for the sun's temp (noted earlier in paper), and using wolfram alpha to solve, you get that the required Doppler factor is 2 trillion. This corresponds to gamma of 1 trillion. (Note, receding from the sun, a Doppler factor of only .13 would produce the same decrease in brightness - a speed of merely .97 c). If I've counted my 9s right, this would be .9999999999999999999999995 c (24 9s, then 5).

[edit: the above is scenario I thought was under discussion. However, I see that the OP proposed viewing the sun at rapid approach, from 4 light years away in the sun's frame. I can find a figure of .5 for the sun's visual magnitude seen from alpha centauri. Then, to become invisible to the naked eye, an increase of magnitude of 6 would be required. This corresponds to Doppler factor of about 1140, gamma of 570. This corresponds to a speed of .9999985c. Of course, with a pair of binoculars, the sun would still be readily visible. Note, again, for comparison, this decrease in brightness would be achieved traveling away from the sun at just .6c. Underscoring, how, in everyday language, I would say the a star readily disappears traveling away at relativistic speeds. However, for traveling towards it, you need mind boggling ultra-relativistic speeds. ]

 

[pervect]

Not only did I read it, but I also emailed it to a few professors. They laughed! I guess it's time to update that paper.

Interesting, because the results in the paper seem to be in substantial agreement with statements that were earlier represented as having been made by a very famous person.

Now it's being represented that it's being laughed at. This seems like a sudden shift in attitude with little explanation. Unless the laughter is being directed at such a serious analysis of such an abstract problem, perhaps?

I don't have any way of checking up on whether such statements were actually made or not. I'd really like to think that I wouldn't have to "check up" on such things.

I can see from this example why the policy to exclude such third party remarks is a wise one.

On technical grounds, while it's certianly possible that I've made a mistake, I don't see it yet.

I might also add that I still strongly support the approach of analyzing the distant light as a plane wave. In the case of interest, trying to analyze it in terms of the solid angle makes little sense, as the detecting instrument (the eye) can not resolve the solid angle of the source. Therefore , it's wise to consider the amount of energy detected by a detector of cross-sectional area A as the "best" way of determining the intensity of a point source.

ANother way of saying this - geometric optics is the wrong tool for this problem, detection of the pont source is diffraction limited.

Note that given the plane wave approach, one might consider representing the incoming plane wave as being a classical plane wave with a classical E-field and B-field, as a way of cross-checking the analysis.

 

[Dale]

to become invisible to the naked eye, an increase of magnitude of 6 would be required. This corresponds to Doppler factor of about 1140, gamma of 570. This corresponds to a speed of .9999985c. Of course, with a pair of binoculars, the sun would still be readily visible. Note, again, for comparison, this decrease in brightness would be achieved traveling away from the sun at just .6c. Underscoring, how, in everyday language, I would say the a star readily disappears traveling away at relativistic speeds. However, for traveling towards it, you need mind boggling ultra-relativistic speeds. ]

Thanks for running the numbers! That certainly seems reasonable based on the paper.

 

[Russell E]

What was unclear about the detailed answers in the last thread? In particular the answers given by the paper http://cartan.e-moka.net/content/download/248/1479/file/Astronave relativistica.pdf in figure 4.

I looked at that article more closely, and I agree that it gives the correct result. It's mathematically equivalent to the approach I would take; it just looks superficially different, partly because they use the wavelength instead of the frequency form of the power density spectrum, which needlessly complicates their derivation of the leading factor in the transformed density function. Also their verbal description of their derivation is messed up, because (for example) they use the symbols S and S' to denote both reference frames and density spectra.

The reason their neglect of the solid angle ends up not affecting their result is that they neglect it both in their derivation of the transformed spectrum and in their integration of that spectrum. The justification for this comes from viewing the situation entirely from the Sun's rest frame, but their derivation involves changing frames, so you can judge for yourself whether the double neglect was consciously (but tacitly) done or just a fortunate compensation of oversights. Regardless, they arrive at the correct result. It's better to avoid that issue altogether by just evaluating everything in the frame of the Sun, so we don't need to worry about transforming the geometric effects.

 

[Russell E]

...to become invisible to the naked eye, an increase of magnitude of 6 would be required. This corresponds to Doppler factor of about 1140...

Are you using eq 21 from that paper? That equation says the change in magnitude is

2.5 log_10(D) - 26000 K (1/T - 1/DT)

where D is the Doppler factor and T is the temperature of the Sun in its rest frame. Are you taking D=1140 and T=5800 K in this equation? How do you get a magnitude change of 6 from this?

 

[tionis]

Well, it appears a consensus is reachable after all Based on all the number-crunching and stuff, are you guys agreeing, then, that the Sun does not become invisible?

 

[Dale]

Well, it appears a consensus is reachable after all ... are you guys agreeing, then, that the Sun does not become invisible?

No. I don't agree.

 

[tionis]

No. I don't agree.

No, you don't agree that there can be a consensus or that Sun does not becomes invisible?

 

[Russell E]

Well, it appears a consensus is reachable after all Based on all the number-crunching and stuff, are you guys agreeing, then, that the Sun does not become invisible?

In addition to all the other idealizing assumptions and stipulations (such as ideal black body spectrum, ideal eyes with infinite sensitivity, neglecting absorption, neglecting response of actual eyes to total spectrum, etc), you also need to distinguish between these two things:

(1) The limiting value of the visible intensity, as v approaches c, is zero.
(2) For any v less than c, the visible intensity is non-zero.

Both of these are true statements. Statement (1) implies that the intensity in the visible range can be made less than any specified positive value by a speed sufficiently close to c. This would be interpreted by most people as meaning that the "Sun becomes invisible as v approaches c". However, Statement (2) is equally true, and might be interpreted by someone as implying that the Sun never becomes invisible. So, given that both of these statements are true (based on the ideal black body spectrum, etc), would YOU say "the becomes Sun invisible" or not? If you answer this question, it would make it easier for us to know what you have in mind by that phrase.

This is just one more reason why it is a trick question, because the answer can only be given by very carefully stating all the assumptions, idealizations, and interpretations of the various aspects of the question - which you haven't done. In fact (if you don't mind my saying so), you don't even seem to have any interest in understanding the different possible interpretations and contexts your question can have, and how they affect the answer. You've said that you don't want to make any idealizing assumptions, you just want to know "the answer" in the real world - but unfortunately when people then point out that, in the real world the eyes are not infinitely sensitive, and they would destroyed by x-rays, etc, you say "No, it's just a thought experiment, don't worry about those real world effects". Well, you can't have it both ways. You are obviously making many idealizing and unrealistic assumptions (without stating them), but then you object when people point out that the answer depends on what idealizing assumptions you are making.

For example, you haven't even said you want to stipulate an ideal black body spectrum - but you haven't objected to that assumption either - so people can only speculate what you really have in mind. This is important, because the fine distinction noted above applies only to this ideal case where we assume a density spectrum that has non-zero density at all non-zero frequencies. Given this idealized assumption, Statement (2) is self-evident. So the only real question of interest - in this context - is Statement (1), which is what we've been discussing. But it may be that you aren't interested in whether Statement (1) is true, you may only be interested in whether Statement (2) is true, which it obviously is under the stated assumption.

On the other hand, if you say you don't want to assume a spectrum with non-zero density at all non-zero frequencies, and instead you want to assume some cutoff frequency, then again the answer is obvious: In that case we obviously CAN shift all the radiation out of the visible range of frequencies with some v less than c.

On the third hand, if you say you don't know whether the Sun has a cutoff frequency or not, and you want us to tell you... well, that's a completely different question, not specifically related to relativity theory or the Doppler effect or aberration. The production and emission of radiation from the physical processes taking place within and on the surface of a star, and its surrounding atmosphere, is a complicated science, especially at the very extreme ends of the frequency bands that we're discussing here. Is THIS what you are asking about? Or are you asking about the physiology of the human eye, and whether our eyes are infinitely sensitive? I would venture to say they are not, which then (combined with Statement 1 above) implies invisibility for some v less than c.

 

[Dale]

No, you don't agree that there can be a consensus or that Sun does not becomes invisible?

Both. This is the internet, you won't get a consensus, and the sun does become invisible for some v arbitrarily close to c.

 

[Russell E]

...the sun does become invisible for some v arbitrarily close to c.

I think we need to be careful about saying "the sun becomes invisible for some v arbitrarily close to c", because we all agree that the intensity is non-zero for any v less than c, no matter how close. Would you agree that the following two statements are true?

(1) The limiting value of the visible intensity, as v approaches c, is zero.
(2) For any v less than c, the visible intensity is non-zero.

The reason I'm trying to highlight these two different statements is that I suspect the OP will focus on statement (2) and say the Sun does NOT become invisible, whereas most other posters will focus on statement (1) and say the Sun DOES become invisible. So you're talking past each other. To avoid misunderstanding, it's best to explicitly state both facts. (I suppose you might also be assuming some finite sensitivity of the human eye, or a cutoff frequency of real stars, to support your statement, but if so, it would help to say that.)

This is the internet, you won't get a consensus...

Are you saying someone here (or in any external reference) disagrees with statements (1) and (2), assuming an ideal black body spectrum?

 

[Dale]

I think we need to be careful about saying "the sun becomes invisible for some v arbitrarily close to c", because we all agree that the intensity is non-zero for any v less than c, no matter how close. Would you agree that the following two statements are true?

(1) The limiting value of the visible intensity, as v approaches c, is zero.
(2) For any v less than c, the visible intensity is non-zero.

If by "visible intensity" you mean "the amount of energy in the visible spectrum" I would say yes. But at some value v < c that amount of energy, while non-zero, is not visible.

Are you saying someone here (or in any external reference) disagrees with statements (1) and (2), assuming an ideal black body spectrum?

I'm just saying that it is an Internet forum. You could probably get disagreement on whether or not 2+2=4. And a lack of consensus is irrelevant to the facts.

 

[Russell E]

If by "visible intensity" you mean "the amount of energy in the visible spectrum" I would say yes.

Hmmm... the relevant quantity is energy per time, right? So it's the intensity that we're talking about.

But at some value v < c that amount of energy, while non-zero, is not visible.

What are you saying causes that non-zero energy (per time, surely) it to be "not visible"? Are you invoking the limited sensitivity of the human eye? Or are you saying at some point the intensity is negligible compared with background noise? Or something else?

... a lack of consensus is irrelevant to the facts.

Agreed, although in this case I'm not actually seeing any lack of consensus - at least not about statements 1 and 2 (not counting our disagreement over whether it's energy or intensity, and your comment above that I don't yet understand about non-zero intensity in the visible range not being visible).

 

[Dale]

Hmmm... the relevant quantity is energy per time, right?

Yes, sorry about that.

Are you invoking the limited sensitivity of the human eye?

Yes. At some point the non-zero amount of luminous power in the visible spectrum is so low that it is not visible.

 

[PAllen]

Are you using eq 21 from that paper? That equation says the change in magnitude is

2.5 log_10(D) - 26000 K (1/T - 1/DT)

where D is the Doppler factor and T is the temperature of the Sun in its rest frame. Are you taking D=1140 and T=5800 K in this equation? How do you get a magnitude change of 6 from this?

Yes that is what I used. How did I get 6 from that? Via typo, not caught because the result was plausible. The correct answer is more extreme:

To make the sun invisible to the naked eye, heading towards it from 4 ly away, requires Doppler factor of 15600, gamma 7800, speed: 9999999918c.

Meanwhile, making the sun invisible in the same sense, heading away, requires only a speed of .736c.

Of course, heading away, the total EM brightness is reduced. Using the given speed towards the sun, its total brightness is enormously increased, such that any such observer would be fried. However, the brightness in the visible range would have fallen below what is normally considered detectible to the naked eye.

 

[Russell E]

To make the sun invisible to the naked eye, heading towards it from 4 ly away, requires Doppler factor of 15600, gamma 7800, speed: 9999999918c.

That would be the answer if you followed the formulas in that paper, but I don't think it's right. The problem is that, although the paper gives the correct formula for the intensity (equation 16), it involves an integration, and they didn't see how to simplify it, nor did they have the capability (or perhaps the interest) to evaluate that integral to the extremes that we are talking about.

So, beginning with equation 19, their formulas are all based on the least squares curve fit shown in Figure 3. That curve fit extends only up to DT of 10^5, which is nowhere close to the range of values that we are discussing, so there's no reason to think it would be valid in the range we're talking about.

So, to actually answer the question, we have to solve the problem analytically. If you do this, I think you will find that to achieve a drop of six magnitudes you need a Doppler factor of about 7160, which implies a speed of about 0.999999961. (This has one fewer "9" than your answer.)

 

[tionis]

Russell E,

Thanks for your detailed analysis. I'm interested in all possible scenarios as long as they are correct. In your reply you mentioned complications do to the ''extreme ends of the frequency bands.'' Would a sudden optical boom at some v close to c immediately followed by a complete and total optical darkness happen at those speeds?

 

[PAllen]

Russell E,

Thanks for your detailed analysis. I'm interested in all possible scenarios as long as they are correct. In your reply you mentioned complications do to the ''extreme ends of the frequency bands.'' Would a sudden optical boom at some v close to c immediately followed by a complete and total optical darkness happen at those speeds?

Not unless the real sun has an unusual deviation from black body, such that there is sharp peak at some very long wave length before returning to a smooth spectrum. There is no reason to expect such anomaly.

Also, if you read all responses, there is agreement there is no such thing as total optical darkness. There is gradual decline in visibility in the part of the spectrum the eyes can see. Even at the ultra-relativistic speed Russel E and I were discussing, all you need to do is take out binoculars and you will still see the sun. You would need many more 9s to get to a point where even an amateur telescope could not see the sun.

 

[Russell E]

In your reply you mentioned complications do to the ''extreme ends of the frequency bands.'' Would a sudden optical boom at some v close to c immediately followed by a complete and total optical darkness happen at those speeds?

Well, since interstellar space is not a perfect vacuum, the phase velocity of light is actually slightly less than c, and it's possible for an object (e.g., a charged particle) to move faster than the phase speed of light in a medium, resulting in Cherenkov radiation, which has some similarities to the production of a sonic boom when exceeding the speed of sound in a medium. But this has nothing to do with the light emanating from the Sun, so I don't think this can be what you have in mind (unless you're changing the subject).

For the subject at hand, and since you connected it with my comment about what goes on "at the extreme ends of the frequency bands", the only thing I can think of is that you are asking if the Sun's spectrum might contain a blip (relative to the black body spectrum) followed by an abrupt cutoff. There could certainly be a cutoff - as has been mentioned several times - and there could also in theory be other non-ideal features in the spectrum of a star. But that isn't really a relativity question.

It's odd, because in none of the explanations that you've been given has anyone described anything that even remotely suggests "a sudden optical boom". Quite the contrary. It makes me wonder... are you just "free associating"? Or do you have some reason to think there would be an "optical boom" (whatever you think that means)?

 

[Vanadium 50]

I still don't understand why the paper has the output turning over at large beta.

Suppose I am going towards a planet, and on that planet is a man with an Aldis Lamp pointing at me. Every second he flashes the light at my position. Eventually, the pulses reach me, and I start seeing them at one per second.

Now I start moving towards him, and therefore the pulses: I maintain that there is no velocity where I see them coming slower than once per second.

 

[Russell E]

Now I start moving towards him, and therefore the pulses: I maintain that there is no velocity where I see them coming slower than once per second.

Right, the pulses never come slower, they come faster, and in fact, if you are approaching fast enough, the frequency of the pulses (not to mention of the light waves comprising the pulses) will rise above the upper frequency limit of the visible spectrum, into the ultra-violet and then into x-rays, etc, so the intensity in the range of visible frequencies drops.

It's as if you're walking along the railroad track and a train blowing its whistle is approaching you at such a high speed that the whistle is Doppler-shifted above the range of human hearing, so you wouldn't hear it at all... but your dog might save you.