Talk:Radiation pressure

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"In heaviest stars radiation pressure is the dominant pressure component." Can anyone explain why in heavy stars radiation pressure has a greater effect than gas pressure?

Temperatures in heavy stars are higher than in lighter stars, and radiation pressure grows with the fourth power of temperature whereas gas pressure grows linearly with temperature. Thus in a star with 10 times the core temperature of the Sun, the gas pressure is tenfold, but the radiation pressure is 10,000-fold. - Andre Engels 22:53, 22 Jan 2005 (UTC)

I know that this sort of question comes up from time to time when people discuss solar sails, but I'd like to ask it here for the possibility of getting different kinds of reply. It is always stated that the radiation pressure is doubled if the incident radiation is entirely reflected, but if the momentum change is transferred entirely to the object the radiation is incident on, then conservation of energy is violated. My question, then, is: "Is the solution simply that we cannot have perfect reflection?" or is there something more complicated going on? -MarkHudson 08:09, 29 July 2005 (UTC)

I suspect the answer goes like this: In the reflection process, the photon is first absorbed by the reflective surface, typically by kicking a bound electron to an excited energy level. The electron later decays back to its ground state, releasing the photon back. Let us assume a perfect reflector for the sake of argument. Initially, the reflector is at rest (in our frame of reference) and the photon incoming, with a certain energy and momentum. It is absorbed, transferring both to the reflector. The now moving reflector emits the photon back the way it originally came. That photon will be, in our frame, very slightly red-shifted, decreasing its energy and momentum. The new photon energy and momentum, added to the reflector's increase in energy and momentum, should equal the photon's original energy and momentum.

Anyone care to do the math and prove this one way or another?

Urhixidur 13:57, 2005 July 29 (UTC)

Thanks for your reply. I've been having a think about this, and am finding that being away from physics for just 4 years has entirely addled my brain. Dammit. I'll see if I can come up with something... Not even sure if I need to bother with relativistic velocity and momentum equations, seeing as the velocities involved will be so small be could probably just use the usual

p 2 / 2 m

kind of approximations. But then we could probably do away with the red shifting and everything and we'd end up back with the "the pressure is doubled" statement! I'll see how far I get before getting into a mess with too many factors of

γ

-MarkHudson 09:16, 4 August 2005 (UTC)

[edit] Commentary in article

I'm moving this commentary out of the article and onto the talk page:

NOTE:

This page can be confusing to the uninformed. It gives the total flux density of 1370W/sq. M and then tells us down a ways "that the pressure against a surface exposed in a space traversed by radiation uniformly in all directions is equal to 1/3 the total radiant energy per unit volume within that space."

Now, I have one term shy of a BS in physics and a BS in science education with some post grad work. I look at this and think it must mean that over 400W/m^2 is transferred to pressure on the exposed surface near earth's orbit! Of course the calculations that result from this are insane. For instance a 1M cube of aluminum would accelerate AWAY from the sun at over half a meter/sec^2 or some such thing!! Wow...who needs rockets? :-)

Of course if a person happens to know what a Pascal is and they calculate the pressure on this object from 4.6 uPascal....then that's 4.6 u-newtons to oppose a force of 4 newtons produced by the suns gravity on the object roughly. So the solar wind force is really only a bit over 1 millionth of the force on the object from gravity according to my calculations. -Bob Weigel / Sound Doctorin'

I don't know enough physics to know how to deal with this, so I'll just leave it here. :)

-- Michael Kelly 09:57, 12 June 2006 (UTC)

[edit] Energy density incorrect?

I have a problem with the statement that the energy density for a black body is $σ T 4 / 3 c$ . It should be $4σ T 4 / c$ which is the radiation constant

The energy density of a black body per unit frequency is (see Planck's law of black body radiation or Hyperphysics)

$u_\nu=\frac{8\pi h}{c^3}\,\frac{\nu^3}{e^{h\nu/kt}-1}$