Are Intensity and Magnitude independent of the radius of a star?

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Summary:

I’m a bit confused about how different equations relate to each-other, and it seems contradictory?
For a star..
Apparent Magnitude = -2.5log10 I K
And I = I0/d^2
So in terms of I0...
Apparent Magnitude = -2.5log10 (I0/d^2)
And the Stefan-Boltzmann law says:
Energy Flux = Sigma(T^4)
In my reading it says that Intensity is the energy emitted per unit of area per unit of time. It says the same about Energy Flux, so are they the same?
If they are then could you say...
Apparent Magnitude = -2.5log10 [(Sigma T^4)/d^2] K
So could you say the Magnitude of a star is determined by Distance and Temperature and not on the Luminosity which is dependent on its radius?
Luminosity = (4piR^2) (Sigma T^4)
If that was the case then you’d be able to measure the distance to stars knowing only it’s apparent magnitude and temperature (from its peak wavelength) which doesn’t sound right?
 

Answers and Replies

  • #2
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This doesn’t seem right... maybe my source meant to say intensity is the same thing as luminosity and not energy flux per unit surface area? Idk it’s confusing stuff to learn without worrying about bad sources.
 
  • #3
Vanadium 50
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So could you say the Magnitude of a star is determined by Distance and Temperature and not on the Luminosity which is dependent on its radius?
Well, you could always say it, but it would be wrong.

Magnitude depends on luminosity. Bright stars are easier to see. The nearest star to Earth is roughly the same temperature as Betelgeuse, but Betelgeuse is the 10th brightest star in the sky, but Proxima Centauri is invisible.
 
  • #4
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Well, you could always say it, but it would be wrong.

Magnitude depends on luminosity. Bright stars are easier to see. The nearest star to Earth is roughly the same temperature as Betelgeuse, but Betelgeuse is the 10th brightest star in the sky, but Proxima Centauri is invisible.
So I guess intensity is the same as luminosity which does depend on radius? Or was the source wrong to include intensity in the apparent Magnitude equation?
But then what about the part of the star’s energy that is emitted in the opposite direction (since it is a sphere)? Wouldn’t that be part of its luminosity but not it’s apparent magnitude? Or does the magnitude equation take this into account?
 
  • #5
Ibix
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Imagine a solar eclipse. As the sun vanishes behind the moon, what happens to the magnitude? If it depends solely on surface temperature and distance then it is unaffected and the sky does not dim.

The Stefan-Boltzmann law gives you the power emitted per unit area of the surface of a black body. But how bright a star appears depends on the total power it emits, not the power emitted by one square meter of its surface. That is, the total power emitted is ##L=4\pi R_*^2\sigma T^4##, where ##R_*## is the radius of the star. By conservation of energy, the total power crossing a spherical surface of radius ##d## centered on the star must be the power emitted by the star, ##L##. So the total power per unit area received by a detector at distance ##d## is $$\begin{eqnarray*}
I&=&L/4\pi d^2\\
&=&(R_*/d)^2\sigma T^4
\end{eqnarray*}$$That's the intensity that feeds in to the magnitude. Although it does have the same units as the Stefan-Boltzmann emission, intensity is the power received per unit area, not the power emitted per unit area. They're only the same thing if your detector is in contact with the black body.
 
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Oh thanks! So if I understand you correctly, the Stefan-Boltzmann equation gives you the power emitted per square meter on the surface of the star, while I is the power received per square meter on a larger sphere surrounding the Star with a Radius equal to the distance between the center of the star and the observer?
 
  • #7
Ibix
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Yes.

Note that the notional larger sphere only comes in because it's easy to deduce the total power crossing that sphere and calculate its area, so you can just say that ##I## is the power received per square meter at whatever distance you are from the star.
 
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Oh thanks!
And in the apparent magnitude equation:
Apparent Magnitude = -2.5log10(I k)
Is k the Boltzmann constant?
1.380649×10−23?
I plugged in the sun’s Intensity (1360 w/m^2) and only got -7.84 instead of -26.74
Not sure why it came up with the wrong answer
And furthermore, what about the difference between total bolometric radiation emitted and visible light? Is this already accounted for in the magnitude equation when going from Intensity?
Or is the non-visible light negligible?
 
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  • #9
stefan r
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Oh thanks!
And in the apparent magnitude equation:
Apparent Magnitude = -2.5log10(I k)
Is k the Boltzmann constant?
1.380649×10−23?
I plugged in the sun’s Intensity (1360 w/m^2) and only got -7.84 instead of -26.74
Not sure why it came up with the wrong answer
And furthermore, what about the difference between total bolometric radiation emitted and visible light? Is this already accounted for in the magnitude equation when going from Intensity?
Or is the non-visible light negligible?
The scale is set by the magnitude if seen from 10 parsecs. The Sun has absolute magnitude around 5 depending on which color. The Sun is 1 astronomical unit away so it is 2 million times closer. The apparent brightness should be 4 trillion times higher. The 2.5 log of 4 trillion is around 31.5 so the Sun's magnitude fits at -27.

The Boltzman constant is important in the gas law.

Radiation and temperature is better understood while playing with hot objects IMO. Just be careful! You need to separate the perception of heat coming from the air and the sensation of infra-red radiation. If you cup your hand around a candle from the side most of the heat is radiation. Likewise with a fire in a fireplace. The draft is carrying the hot air up the chimney. An electric stove top range will glow brighter as it gets hotter. At higher temperatures the metal glows brighter, gets whiter and radiates more heat. At more extreme temperatures the light gets bluer but you only get that with iron when welding and that radiation can cause blindness. A larger stovetop range radiates more heat even if the two ranges are glowing with the same color of red. That relationship is linear so if the range has 4x the surface it will radiate 4x the heat. The intensity at a distance is squared so a range with 4x heat output should feel the same as the little one where your hand is twice as close (assuming your hand is still outside of the hot air).

Snuggling in the winter might be a safer experiment. A bigger body warms the bed faster. :) This won't pass peer review since there is no control for conduction and convection of heat. You can use a hot water bottle instead of an ape but the research might be less fun. You have to convert body temperature into degrees K. A person's blood is around 309K but might be 305K on the pajama surface. A 77C water bottle would be 350K. The water bottle has 1.14x temperature. Radiation is proportional to temperature to the fourth so the bottle should be radiating 1.73 x as strongly. Since the skin has much larger surface than most water bottles it should be radiating much more. It is wrong to conclude the ape is better for bed warming because the cold sheets are still radiating some. If the sleeping bag stars at 295K the skin is only 10 degree warmer but the bottle has a 55 degree margin. However, if you are trying to watch the Leonid meteor shower the bag temperature may be closer to 270K so a large 305K ape makes a more significant difference. You may only need heat on your hands or feet. Using someone's skin to warm your hands may decrease their enthusiasm and/or willingness to wait for the Leonid meteors. Leo does not rise until late at night on November 16th staying warm is critical.

Temperature of a surface increases luminosity (or intensity) x4
Distance decreases intensity x-2
Surface area directly effects luminosity. Surface area itself proportional to radius squared or volume2/3.
 

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