Stars and visible light

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If, for argument sake, you had a spaceship that could go thousands of times the speed of light what would happen to the electromagnetic radiation being emitted from the stars you head towards and away from?

Would the stars that you are travelling towards get duller and duller the faster you go until you are seeing infrared waves?

And at what speed would you no longer be able to see them?

And at what speed would the stars that you are travelling away from no longer be visible?
 

Answers and Replies

2,538
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you are getting at the Doppler effect of light I assume, if you traveling towards the stars the wavelength of the incoming light would get shorter .
 
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Yeah, I thought that as a stars main emmision is within the visible light spectrum (correct me if I'm wrong) as you accelerate, the light from the stars travelling toward you would fade as infrared becomes visible and violet becomes invisible. At a certain speed the wave length would become so short that all radiation would become invisible, the last light you would see would be violet (which would be whatever the longest wavelength wave a star produces)?
 
2,538
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yes that seems right , as you travel towards the star , the infrared would turn to visible and the violet would go towards ultraviolet and out of the visible range, all the light will be blue-shifted , And if you want to calculate it , it should just be standard Doppler effect equations for an approaching observer.
 
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what is the longest wavelength a typical star produces and at what speed would it disappear from visiblity as you speed up?
 
2,538
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I think stars can emit radio waves ,
heres the formula to calculate it . j=sqrt((1-q)/(1+q))*k
where j is the final shifted wave-lenght , and q is the speed of the obsever divided by the speed of light and k is the origianl wavellenght.
 
2,538
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k if i did my calculations right you would haft to go like 99% the speed of light
to shift light from infrared to green light . or more precisely 10^(-5) meters to
520*10^(-9) meters .
After some algebra the formula i used was
v= (1-x)/(1+x) *c
where v is the velocity
and x is the final wavelength divide by the initial wavelength and then quantity squared
and c is the speed of light.
 
diazona
Homework Helper
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A star basically emits blackbody radiation. There's a plot on Wikipedia (click the link) that shows how the radiation intensity of an ideal blackbody depends on wavelength for various temperatures, and there's also a formula for it, http://en.wikipedia.org/wiki/Planck's_Law]Planck's[/PLAIN] [Broken] Law:
[tex]I(\lambda, T) = \frac{2hc^2}{\lambda^5}\biggl(\frac{1}{e^{hc/kT\lambda} - 1}\biggr)[/tex]
So there really is no minimum wavelength emitted by the star, it's just that the energy at a particular wavelength gets gradually less and less as the wavelength gets longer and longer.

However, stars aren't perfectly ideal blackbodies of course, so in practice there could be some maximum wavelength to the radiation they emit. Certainly there is some point at which the energy becomes so low it might as well be zero for any practical purpose. That would depend on how sensitive of a detector you're using.

Not all stars emit most of their radiation as visible light. As you can tell from the graph on Wikipedia, the wavelength of peak emission depends on the temperature; the hotter the star, the lower the wavelength. The peak wavelength is only within the visible range for stars with surface temperatures between about 3900K and 7600K, but there do exist stars that are hotter and stars that are colder.

Notice that as you go to very long wavelengths, the blackbody radiation curve becomes nearly flat. This is the part of the curve you would be seeing as you travel at high speed toward a star, so when you get going to a high fraction of the speed of light, all the stars would appear white. As you go faster and faster, their intensity would fade away as more and more of the radiation shifts to shorter-wavelength forms.
 
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I've been working it out; this is what I found;

Most common star 'M' class-

@ 810,000km/h an 'M' class stars luminosity would double as 800nm waves become visible.

@ 5,130,000km/h an 'M' class stars luminosity would half as 2000nm waves become visible
(Halved from the luminosity of that a stationary person would observe).

@ 16,000 x c, radio waves (5,000,000,000nm) would become visible but are emitted in such low amounts I don't think you'd see anything.

In short it appears as you accelerate stars would become brighter but by the time you reach just a few percent of c they would lose all luminosity.
 
DaveC426913
Gold Member
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Lot of misconceptions and bad physics bouncing around here.

If, for argument sake, you had a spaceship that could go thousands of times the speed of light
This is tantamount to "If the laws of physics were not the laws of physics, what would we see?"

And the answer is just as sensical: "You'd see pink unicorns."


@ 810,000km/h an 'M' class stars luminosity would double as 800nm waves become visible.

@ 5,130,000km/h an 'M' class stars luminosity would half as 2000nm waves become visible
(Halved from the luminosity of that a stationary person would observe).
@ 16,000 x c, radio waves (5,000,000,000nm) would become visible but are emitted in such low amounts I don't think you'd see anything.
This is not how you calculate relativistic velocites. Read up on the Lorentz transform.


If you'd tried to do this calc using classical Newtonian math (which is what you seem to be trying to do) and done the math correctly, you should have gotten non-sensical values.

As you approach c, the very longest wavelengths will shorten and approach zero. At a point arbitrarily close to c, an arbitrarily long wavelength will be shortened to arbitrarily close to zero. At c, your wavelength should have been exactly zero.

So, if you tried to calculate a wavelength while traveliing faster than c, you should have gotten a negative value!


In short it appears as you accelerate stars would become brighter but by the time you reach just a few percent of c they would lose all luminosity.
This is better, good. (It is also in direct contradiction to your other calculations, but that's OK, they were all wrong.)
 
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