# Is the Sun invisible at relativistic speeds?

1. Jun 8, 2013

### tionis

At which speed does the Sun disappear from the visible part of the spectrum if I were to travel towards it at high speed? Let's assume there are no other stars in the visible universe and I'm a few light-years away.

2. Jun 9, 2013

### tom.stoer

You can use the Planck formula for the spectral density u(ω).

Just use the Doppler shift formula ω → ω'(v) for the velocity v. Then apply it to the formula u → u' = u(ω'). You get a v-dependent frequency spectrum u' where the maximum is shifted towards shorter / longer wave length.

With an idealized black body spectrum u' will never be zero, but the brightness in the visible part of the spectrum for ω' will of course depend on v. I think this can be visualized quite easily using Excel, Mathematica, etc.

3. Jun 9, 2013

### Staff: Mentor

To make the sun invisible, I would try to move away from it.
I don't think it is possible if you travel towards it, as relativistic effects will increase the power, probably cancelling the effect of the weaker emissions in the corresponding frequency range.

4. Jun 9, 2013

### tom.stoer

mfb is right, one has to take the Lorentz transformation property of the frequency spectrum u(ω) into account. In my post I assumed implicitly that u transforms as a Lorentz scalar, i.e. that the only effect is the frequency shift ω → ω'; this could be wrong, which means that one will find additional γ-factors etc. multiplying u'.

http://arxiv.org/abs/0910.0164

An observer moving within the black body radiation (like for CMB) would have to integrate dΩ over the whole sphere, which means she has to apply (17). An observer receiving the radiation from the sun i.e. from a very small dΩ has to apply (2) directly.

Another paper which contains useful results could be

http://www.ita.uni-heidelberg.de/research/bartelmann/Lectures/theoAstro/theoAstro.pdf [Broken]

Last edited by a moderator: May 6, 2017
5. Jun 9, 2013

Staff Emeritus
A blackbody in motion will still appear to be a blackbody; this corresponds to only a temperature shift. (By a factor sqrt((c-v)/(c+v)). ) So if one wishes to move fast enough, the sun will appear redder and darker, and given enough velocity, can be made arbitrarily red and dark. I would not call this "invisible" so much as "opaque", though.

6. Jun 9, 2013

### tionis

So... is that a yes or a no lol? Forget the CMB and all that. Does the Sun become invisible the faster I approach c? In other words, when I look out the window, am I going to be able to to say ''holy sh*t, the Sun just disappeared?!! ''

Also, what happens to neutrinos coming from the Sun? Do they gain mass as I move closer to c?

Does the mass of the Sun increases from my frame of reference, too?

Can the Sun become a black hole from my frame?

7. Jun 9, 2013

### tom.stoer

Why should the sun become a bh??

There is a frequency shift, and I think there are enough hints how to calculate it.

8. Jun 9, 2013

### tionis

Doesn't the Sun's mass increase from my frame of reference?

Tom, I wanna be able to say with full confidence that the Sun does indeed disappears from view as my speed goes to c. A simple yes or no will do. I'm not looking for complicated maths or anything lol.
Yes or no?

9. Jun 9, 2013

### tom.stoer

The energy increases, but that does not create a bh

Have a look at Vanadium's post

10. Jun 9, 2013

### tionis

OK. We got that one out of the way.

Already did. He agrees that it becomes invisibly ''opaque.'' What say you?

11. Jun 9, 2013

### tionis

12. Jun 9, 2013

### Staff: Mentor

You can use the formulas given here:
Just plug in numbers, and compare it to the static case (v=0) to see if the sun gets dimmer or brighter (hint, hint).

No, masses do not depend on the reference frame. Energies do.

13. Jun 9, 2013

### tionis

mfb, this isn't homework. I'm not a physicist or a physics student. I don't know how to ''plug in numbers'' or anything... I just want a simple answer to a simple question: does the Sun become invisible if you move at relativistic speed? I don't even care to know at which specific percentage of c this would happen (though I woudn't mind lol).

Got it. Thank you.

14. Jun 10, 2013

### tom.stoer

ok, simple answer: the power spectrum of a star has a maximum

https://en.wikipedia.org/wiki/File:Solar_Spectrum.png

for the sun the maximum of the radiation power is at a wavelength λ ≈ 500 nm; in addition the power spectrum tends to zero for short wavelength λ → 0 and long wavelength λ → ∞

depending on the velocity v of the observer relative to the sun, the observed spectrum is blue- / red-shifted, so what the observer will see is not range of the maximum but a shifted range of wavelengths which comes from the range of shorter / longer wavelengths (for blue / red shift); but in these ranges the emitted power becomes smaller than in the Originale visible range for v=0

blue shift: a fast-moving observer approaching the sun will see wavelengths which have been emitted by the sun at very large wavelength (infrared and beyond, far right of the diagram); the power spectrum tends to zero, so the sun becomes darker w.r.t. the observer

red shift: a fast-moving observer moving away from the sun will see wavelengths emitted by the sun at very small wavelength (UV and beyond, close to λ = 0 in the diagram); again the power spectrum tends to zero, so the sun becomes darker w.r.t. the observer

Last edited: Jun 10, 2013
15. Jun 10, 2013

### tionis

Darker as in ''Whoa! The Sun is invisible. My whole universe just went black!'' kind of dark? Or darkened, but not quite invisible?

16. Jun 10, 2013

### tom.stoer

the faster you move the more the wavelengths are red-/blue-shifted; the higher the shift is the less photons you have in the visible (shifted) range of the power spectrum (and the darker the star gets)

17. Jun 10, 2013

### pervect

Staff Emeritus
Note you'll see a relativistic intensification effect if you move towards the sun - that will fight the other effect.

18. Jun 10, 2013

### tom.stoer

Can you elaborate on that? Is it due to dΩ → dΩ'? Does it apply ta point-like source well? Note that the total number of photons while travelling through a certain region of space does not change, however Lorentz contraction / time dilation will affect the result.

However the effect we a discussing so far is due to u(ω), whereas you are talking about additional kinematical corrections, aren't you?

19. Jun 10, 2013

### pervect

Staff Emeritus
You can actually see the effect if you watch the visualizations in http://www.anu.edu.au/physics/Searle/ [Broken]

Last edited by a moderator: May 6, 2017
20. Jun 10, 2013

### tom.stoer

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;
Due to the factor of ω2 in the Planck formula for u(ω,T) * Ω the two terms β2 and β-2 just cancel.

So the remaining effect is the Doppler shift in the exponent in u(ω,T), which means that the Doppler shift ω → ω' can be re-interpreted as a different temperature T → T'.

To summarize, the spectrum of a moving blackbody with temperature T appears identical to a stationary blackbody with different temperature T'.

Last edited: Jun 10, 2013
21. Jun 10, 2013

Staff Emeritus
Getting back to the original question, it all depends on what you mean by invisible. The sun will never become transparent no matter how fast you can go. The best you can do is make it a dark sphere. You tell me how dark you want it, and I'll tell you how fast you need to go.

22. Jun 10, 2013

### Staff: Mentor

I usually assume that everyone here can use a calculator (or WolframAlpha, or whatever).

No, it will not. Relativistic effects dominate, as you can see in the paper you linked.

Our sun has a surface temperature of (roughly) 6000K, and it is clearly visible, even from a distance of several light years.

From this paper, the number density of photons can be expressed as
$$n(\omega,\Omega)d\omega d\Omega=\frac{\omega^2}{2\pi^2\left(\exp(\omega/T_{eff}(\theta))-1\right)}d\omega d\Omega$$
with the effective temperature of
$$T_{eff}(\theta)=\frac{T_* \sqrt{1-v^2}}{1-v\cos(\theta)}$$
θ is the angle between our direction of motion and the star, if we move towards it, θ=0 and cos(θ)=1.
T* is the temperature of our light source, 6000K.
If the effective temperature increases, the object gets brighter (in visible light, and all other frequency ranges) - this should be clear, but it can be calculated as well.

So let's look at some numbers:
v=0: Teff = 6000K
v=0.1: Teff = 6600K - the sun gets brighter if we approach it at 10% the speed of light
v=0.5: Teff = 10400K - it looks blue now
v=0.9: Teff = 26200K
v=0.99: Teff = 84600K
The temperature is always increasing with the speed, so the sun gets brighter and brighter.

What happens if we move away?
v=-0.1: Teff = 5400K - the sun gets dimmer
v=-0.5: Teff = 3500K - that begins to look red
v=-0.8: Teff = 2000K - the sun looks red
v=-0.99: Teff = 425K (152 °C, 305 °F) - the sun looks like a black disk

23. Jun 10, 2013

### tom.stoer

@mfb, @all: sorry for the stupid mistake I made; everything was ready ...

24. Jun 10, 2013

### willem2

If you look at planck's law;

$$I(\nu,T) = \frac {2 h \nu^3} {c^2} \frac {1}{e^\frac{h \nu} {k T} -1}$$

Where I is the intensity, h is plancks constant, c the speed of light, $\nu$ the frequency, k boltzman's constant, and T the temperature

it's clear that the intensity increases if the temperature increases at all frequencies.
so the intensity also increases at all frequencies if you move towards a radiating black body.

25. Jun 10, 2013

### tionis

Well, a dark sphere in my hypothetical universe (devoid of other stars and the CMB, etc) is the same as invisible, is it not? I mean, it would still be there, but not visible to my eyes, right?

Why the descriptive gap? Isn't the Sun supposed to switch to the invisible part of the spectrum at some point after 10400K thus becoming invisible?

Awesome! Good to know.