Deathray 6

I was wondering if anyone could give me a quick explanation on determining these things. I know little about the two except that AM is something like radius x mass x velocity that the S/B law says that a blackbody's energy is proportional to its temperature to the fourth power. I would like some preferrably simple equations to determine these things and what units of measure are used in the final answer to the equation. Also if it isn't too much trouble, could someone also use an example (like the AM of the Sun & Jupiter or the energy of the 2.7 K microwave background). Thanks.

Coragyps

A lot of this is memory, so don't sue me:
angular momentum of, say, Jupiter = Jupiter's mass x radius of its orbit squared x angular velocity, where the velocity needs to be in radians per second. This velocity would be calculated from 2 x pi (3.14159....) /Jupiter's year, expressed in seconds. Mass is 1.9 x 10^27 kg, radius is 7.78 x 10^11 meters; 12 Earth years = approximate Jupiter year.
The Sun is more complicated - a high-side estimate is AM = 0.4 x mass x (radius) x (radius) x angular velocity. (This is high because it assumes that the density of the Sun is homogeneous, and it isn't). Here, mass is 1.99 x 10^30 kg, radius is 1.39 x 10^9 meters, and period is 25 days and 9 hours. Again, 2 x pi/period in seconds gives angular velocity. Units are kg-m^2/sec.

If you do all this, you will find that Jupiter has most of the whole solar system's angular velocity. This is largely because the Sun has been losing mass through the (largely ionized) solar wind for 4.5 billion years now, and the Sun's magnetic field dragging through that wind acts as a brake on its rotation, "robbing" angular momentum and turning it into (heat, I suppose?) References, if I can find them, when I get home tonight.
Black bodies: I don't remember how, but it's a simple little equation, and 4th power sounds familiar.
I hope this helps.

[ March 15, 2002: Message edited by: Coragyps ]</p>

Coragyps

<a href="http://www.edpsciences.org/articles/aa/abs/2001/04/aa8852/aa8852.html" target="_blank">This paper</a> provides experimental measures of how the solar wind "carries off" the sun's angular momentum.

Tim Thompson

Angular Momentum

Angular momentum is the rotational analog for linear momentum. So, where the formula for ordinary, "linear" momentum is (1) {momentum = mass X velocity}, the formula for angular momentum is (2) {angular momentum = moment of inertia X angular velocity}. Or, in more direct symbols:

(1) p = mv
(2) L = Iw

Physicists always use "p" for (linear) momentum, and "l" (or "L") for angular momentum; I don't know why, they just do. In equation (1) "m" is mass and "v" is velocity. In equation (2), "I" represents the "moment of inertia" (a rotational effective mass), and "w" is where a lower case Greek omega would be, which looks very "w" like anyway.

The moment of inertia for a point mass is just its mass (since it has no structure). In the general case, you would do a volume integral of the product of the density and the appropriate distance from the desired axis of symmetry (rotation). So the angular momentum for an orbiting "particle" is L = mvr. In the case of Jupiter that's (1.8986x10^27 kg)(1.307x10^4 m/sec)(7.7857x10^11 m) = 1.932x10^43 kg-m^2/sec (and see <a href="http://www.astro.ufl.edu/~doug/hw2a.html" target="_blank">here</a>, where they get slightly different numbers).

For more general expositions, see "<a href="http://scienceworld.wolfram.com/physics/AngularMomentum.html" target="_blank">Angular Momentum</a>" and "<a href="http://scienceworld.wolfram.com/physics/MomentofInertia.html" target="_blank">moment of inertia</a>", both courtesy of <a href="http://scienceworld.wolfram.com/physics/" target="_blank">Eric Weisstein's World of Physics</a>.

Stefan Boltzmann

The Stefan-Boltzmann law is an integral of the more general Planck Law, over all angles and wavelengths. It's a simple formula:

(3) S = sigma*T^4

Here "S" is the total energy flux, "sigma" is the Stefan-Boltzmann constant (5.67032x10^-8 W m^-2 k^-4; W = Watts where 1 Watt = 1 Joule/sec), so S is in Watts per meter squared. So, if the temperature of something (like the CMBR) is 2.7 K, then its total emitted energy flux is (5.67032x10^-8)(2.7^4) = 3.01x10^-6 Watts per square meter.

The Stefan-Boltzmann formula returns the energy per unit area, emitted by a body of a given temperature. So, you could measure the flux and recover the temperature, or you could measure the temperature, and recover the expected flux. But the sky does not have a surface area in square meters, so you don't get the temperature of the CMBR from the Stefan-Boltzmann Law. Rather, it is derived directly from the shape of the spectrum (energy vs wavelength or frequency), fitted to a Planck Law curve (<a href="http://space.gsfc.nasa.gov/astro/cobe/firas_spectrum.jpg" target="_blank">like this</a>).

See Weisstein's <a href="http://scienceworld.wolfram.com/physics/Stefan-BoltzmannLaw.html" target="_blank">Stefan-Boltzmann Law</a>.

repoman

What is described above is the Ideal SB law, it is the maximum amount of energy that can be absorbed or emitted at a specific temp. You have to multiply the equation by emissivity (e - epsilon) or absorptivity (a -alpha) which are basically the same. This can vary between 0.02 at 300K for polished silver up to 0.98 for black paint. Surprisingly water is 0.96, I thought it would be much lower!
Of course this is why Mirrors are back with silver and they are still used for some thermoses.
The interesting thing is that the emissivity in the tables you may see is an average value of all the wavelengths at the temp listed. But for example if emissivity is listed at say 0.3 but you are shining only green light, the emissivity therefore absorbtivity may be say 0.2. Almost forgot Kirchoff's law -- at the same temp and wavelength absorbtivity and emissivity are the same.


The way my boook puts it is ...
Q = heat flow
Qradiative = e*Sigma(Tsurface^4 - Tsurrounding^4)

IF you were talking about the cosmic background temp of about 3 K then that is the Tsurrounding part of the equation. But I am not sure how far out of the solar system or galaxy have to be to get near that temp.

Also this may be of interest to you about weather:

The reason that cloudy nights stay warmer than clear nights (given the same temp at dusk) has to do with blackbody radiation. Clouds than only be about 0 C, while the upper atmosphere during a cloudless night is really really cold and gives back almost no heat in the two way radiation exchange. In fact it is even possible for a puddle of water to freeze at night when the air temp is a few degrees about zero just from the blackbody radiation lose to the upper atmosphere. The radiative heat loss is faster than the convective heat flow from the air near the puddle in that case.