# Is the Sun invisible at relativistic speeds? Part II

1. Jun 19, 2013

### 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.

Last edited: Jun 19, 2013
2. Jun 19, 2013

Staff Emeritus
What relativistic effects? How fast are you moving?

3. Jun 19, 2013

### Staff: Mentor

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.

4. Jun 19, 2013

### tionis

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.

5. Jun 19, 2013

### pervect

Staff Emeritus
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.

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.

Last edited: Jun 19, 2013
6. Jun 19, 2013

### 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.

7. Jun 19, 2013

### pervect

Staff Emeritus
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.

8. Jun 19, 2013

### tionis

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

9. Jun 19, 2013

### PAllen

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.

10. Jun 19, 2013

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

11. Jun 19, 2013

### tionis

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

12. Jun 19, 2013

Staff Emeritus
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.

13. Jun 19, 2013

### Russell E

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

14. Jun 19, 2013

### WannabeNewton

Not unless the observer is superman!

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.

15. Jun 19, 2013

### tionis

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 wanna make an accurate relativity simulation we can post on youtube, but whatever lol.

16. Jun 19, 2013

### 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).

Last edited: Jun 19, 2013
17. Jun 19, 2013

### pervect

Staff Emeritus
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 recieved 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".

WHile I can't do the integrals myself, I can point out that due to the Steffan-Boltman law
http://en.wikipedia.org/wiki/Stefanâ€“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.

Last edited: Jun 20, 2013
18. Jun 20, 2013

### Russell E

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.

19. Jun 20, 2013

### pervect

Staff Emeritus
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.

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

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.

20. Jun 20, 2013

### nitsuj

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

21. Jun 20, 2013

### Russell E

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.)

Right. That's the problem.

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.

22. Jun 20, 2013

### Staff: Mentor

23. Jun 20, 2013

### tionis

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.

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.

So, does the Sun become invisible or not?

24. Jun 20, 2013

### tionis

Good question, nitsuj. Anyone care to answer??

25. Jun 20, 2013

### PAllen

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.