Is the Sun invisible at relativistic speeds?

[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.

 

[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.

 

[mfb]

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.

 

[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

 

[Vanadium 50]

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.

 

[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?

 

[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.

 

[tionis]

Why should the sun become a bh??

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

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

Tom, I want to 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?

 

[tom.stoer]

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

The energy increases, but that does not create a bh

Tom, I want to 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?

Have a look at Vanadium's post

 

[tionis]

The energy increases, but that does not create a bh

OK. We got that one out of the way.

Have a look at Vanadium's post

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

 

[tionis]

I agree on the frequency shift, but I have to think about some factors.

Please do and get back to me. Thx!

Anyone else? I found this site http://math.ucr.edu/home/baez/physics/Relativity/SR/Spaceship/spaceship.html, but I'm not sure if it answers my question 'cause they take into consideration the CMB and other galaxies.

 

[mfb]

You can use the formulas given here:

http://arxiv.org/abs/0910.0164

Just plug in numbers, and compare it to the static case (v=0) to see if the sun gets dimmer or brighter (hint, hint).

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?

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

 

[tionis]

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).

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).

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

Got it. Thank you.

 

[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

 

[tionis]

the sun becomes darker w.r.t. the observer

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

 

[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)

 

[pervect]

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

 

[tom.stoer]

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

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 traveling 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?

 

[pervect]

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

There was a thread about this on the thread [Photon Arrival Rate]

 

[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'.

 

[Vanadium 50]

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.

 

[mfb]

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

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

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: T_eff = 6000K
v=0.1: T_eff = 6600K - the sun gets brighter if we approach it at 10% the speed of light
v=0.5: T_eff = 10400K - it looks blue now
v=0.9: T_eff = 26200K
v=0.99: T_eff = 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: T_eff = 5400K - the sun gets dimmer
v=-0.5: T_eff = 3500K - that begins to look red
v=-0.8: T_eff = 2000K - the sun looks red
v=-0.99: T_eff = 425K (152 °C, 305 °F) - the sun looks like a black disk

 

[tom.stoer]

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

 

[willem2]

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

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.

 

[tionis]

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.

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?

So let's look at some numbers:
v=0.5: Teff = 10400K - it looks blue now <--------------
v=0.9: T_eff = 26200K <---------------------------------
v=0.99: T_eff = 84600K <--------------------------------

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?

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

Awesome! Good to know.

 

[mfb]

The description does not change after 10000K - the light stays blue and gets more and more intense. If you look directly at the sun (close enough to see it as a disk), it will appear white as the human eye cannot see colors correctly if the light is too bright.

Well, a dark sphere in my hypothetical universe (devoid of other stars and the CMB, etc) is the same as invisible, is it not?

Without background sources: right. If you move away from the star so fast that the light shifts to infrared, you can still measure this infrared radiation.

 

[tionis]

mfb,

Vanadium said:

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.

You are saying:

The description does not change after 10000K - the light stays blue and gets more and more intense.

Do you see the conflict here? Who am I suppose to believe?

 

[mfb]

If you travel away from the sun, it can become invisible (or very dark, if you like) to the human eye.

 

[tionis]

If you travel away from the sun, it can become invisible (or very dark, if you like) to the human eye.

I'm sure it does, but that is not what we are talking about here. Do you disagree with Vanadium's reply? Are you saying the Sun actually gets brighter and brighter as we approach it @ close to c?

 

[mfb]

Are you saying the Sun actually gets brighter and brighter as we approach it @ close to c?

Yes.

I don't know if that is in disagreement with Vanadium's post.

 

[mfb]

That was my first thought as well, but the calculations done in http://arxiv.org/abs/0910.0164 and my own estimate disagree with that first thought, at least for a perfect blackbody.

 

[tom.stoer]

That was my first thought as well, but the calculations done in http://arxiv.org/abs/0910.0164 and my own estimate disagree with that first thought, at least for a perfect blackbody.

I had the reference available, but I made the same mistake as Kip Thorne; checking the relevant resources there is agreement that relative speed v will turn the black body radiation u(ω,T) of an emitter (e.g. the sun) at rest into black body radiation u(ω,T') where T' = T'(v). Depending on the direction (sign) of v we have T'(v) > T or T'(v) < T.
For T' > T the power spectrum u(ω,T') > u(ω,T) for all frequencies ω.
For T' < T the power spectrum u(ω,T') < u(ω,T) for all frequencies ω.
Therefore when the observer approaches the emitter (sun) the emitter will look brighter for every frequency ω.

 

[tionis]

but I made the same mistake as Kip Thorne;

Huh? Are you saying Kip is WRONG!?

 

[tom.stoer]

Are you saying Kip is WRONG!?

Yes, b/c the second part of

The sun emits infrared radiation, which will get shifted into the visible part of the spectrum ... However, these emissions are weaker than the sun's optical emissions, so the sun will get dimmer ...

is wrong.

For T' > T we have u(ω,T') > u(ω,T) for every single ω.

 

[tionis]

Lala

Please, be serious for a minute. Is Kip wrong?