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## Question

Shouldn't this article give an explanation as to the assumed contradiction that light cannot exert pressure because photons have zero mass? —Preceding unsigned comment added by 213.84.241.121 (talk) 11:44, 3 February 2010 (UTC)

There is no contradiction. While photons have no mass, they most certainly posses momentum, given by p=hν/c. Oz1sej (talk) 08:31, 8 June 2010 (UTC)
Its not that photons have zero mass, its that mass is not a parameter used to describe a photon. If you absolutely insist on assigning a mass to a photon by using the equation p=mv then p=hν/c and the mass of a photon is hν/c^2, a function of frequency. If you want to use E=mc^2 then again, its hν/c^2. But this is not its rest mass. A photon has no rest mass, it cannot ever be at rest, since it travels at the speed of light, which is constant. The idea that the rest mass is mo=m*Sqrt[1-v^2/c^2] and v=c gives a rest mass of zero is pointless. PAR (talk) 03:28, 10 June 2010 (UTC)
Gamma factor of speed of light is infinite. Mass equivalent hν/c^2 of photon divided by infinity is zero. Zero multiplied by infinity is undefined quantity. It can be defined only by additional functional relations and calculated using limits. Mass equivalent of photon is physically inconsistent and it is only mathematical game. Photon phenomenon cannot be understood as particle phenomenon. However, moving entity enclosed in finite volume can be interpreted as particle, if we can interpret a pulse as particle. However, it is more convenient not to interpret a pulse or packet of radiation as particle. Physical particles emit field. Photon does not emit field. Emitors of field have inertial mass. Nonemitors of field does not have inertial mass. Nonemitors of field cannot be detected, because they do not exert influence. Nonemitors of field can exist only as packets of radiation, if we interpret a packet of radiation as particle. However, effective physical action of packets of radiation is basically different from effective physical action of field emitors. Effective physical action is the only way how to determine and classify properties of physical entities. Effective physical action depends on the system beeing affected. Softvision (talk) 10:58, 22 January 2011 (UTC)
This is not correct. If we could have a massless mirrored box with a photon bouncing around inside of it and we put it on a scale, it would weigh g(hν/c^2) where g is the acceleration due to gravity. That box would "emit" a gravitational field. To be more realistic, if we have a black body cavity at temperature T, and weigh it, its weight will be due to the mass of the cavity material at that temperature, plus the "effective weight" of the photons contained inside of it, and its gravitational field will be proportional to that mass. PAR (talk) 14:18, 22 January 2011 (UTC)
You have just postulated the EM photonic origin of gravitational domain. Why to search for Higgs ? On what experimental fact is this claim based ? If this can be true, than the cavity must obtain also inertial mass from the photons. Otherwise, the inertial mass of "massless" cavity is zero, and if it is gravitationally responding, than the acceleration of the cavity in gravitational field is infinite - a = F/m. If you claim, that cavity will obtain the inertial mass from photons encolsed, than you claim the photons will obtain inertial mass by enclosure in cavity, because free photons do not have inertial mass. However, according to E=mc^2, you can suppose, that energy of photons enclosed in cavity are physically acting as inertial mass of cavity. But where is a experimental confirmation, that the cavity will respond on gravitational force. Moreover, if there is gravitational response, than there must be a gravitational influence. There is no evidence of existence of elements, responding on the field without emiting that field. That means the photons enclosed in cavity should be origin of gravity. What is the mechanism which gives the gravitational response to the photons ? Do you really mean this ? Or the gravitational response is overall result of the cavity system ? Can we call photons enclosed in massless cavity a system ? Or we forget the mirroring and photons and start a new view at the moment we enclose the photons ? Gravitational response is bound with inertial mass and gravitational influence. Mathematics is important science, but physics is constrained by reality - experiments, measurement, physical interaction. Reality cannot be postulated. Reality must be observed. Softvision (talk) 22:02, 22 January 2011 (UTC)
Yes, I really mean this. A free photon (theoretically) has inertial and gravitational mass and generates a gravitational field, as does a photon trapped in a mirrored box. I doubt if the gravitational field of a free photon has ever been measured, but I expect that if it were ever found that the photon did not have the expected gravitational field, there would have to be some serious changes made in the present theory of special and general relativity. The general theory of relativity states that inertial mass=gravitational mass, so lets drop the distinction. The idea of a photon bouncing around in a massless box is kind of a bad idea, because conservation of momentum is ignored. Lets take the more realistic case of a massive black body cavity which contains black body radiation, everything at some temperature T and the whole system is at rest, so the total momentum of the system is zero. What I am saying is that the rest mass of the system is equal to the rest mass of the cavity plus the rest mass of the photon gas. The rest mass (m) of the photon gas is given by $m=E/c^2$ where E is the total energy of the photon gas: The sum of $h\nu$ over every photon in the cavity. This mass will be both inertial and gravitational. PAR (talk) 18:02, 23 January 2011 (UTC)
Check out the Poynting vector article which describes the energy flux and momentum of a classical electromagnetic wave. When you say stable oscillation, I assume you mean a plane wave in a box or something like that. The NET momentum will be zero, but there will be radiation pressure on any wall, and the force on one wall will be equal and opposite to the force on the opposite wall. PAR (talk) 04:36, 24 January 2011 (UTC)
Stable oscilation of element in plane wave is, when speed of element is -pi/2 retarded against the E(B) field of wave, due to inertial mass of oscilating element. The effect of wave field is then symmetricaly acting in both directions - the x direction of wave propagation and y direction of wave intensity oscilation. That means the integral effect is neutral and no net momentum increase is provided by the wave to the element. That means, in the case of wave packet, the momentum of element will be same before and after element-packet interaction - zero. Plane wave packet does not create the change of element momentum, due to the symmetry of influence in both x and y directions. Computer simulation confirms this result. Softvision (talk) 20:47, 25 January 2011 (UTC)

I am just guessing here - have you taken into account the radiation from the oscillating particle? The radiation from the oscillating particle will carry momentum and it could be that for the case you have mentioned, the momentum lost by the radiation of the oscillating particle is equal to the amount gained from the external EM field. PAR (talk) 22:32, 25 January 2011 (UTC)

I am not saying, that the phenomenon of radiation pressure does not exists in reality, and I understand the energetical arguments, that lead to quantitative expression of the radiation pressure, and I believe it is consistent with the measurements. I am writing, that the classical EM theory seems to have problems to explain this qantitativly. E is driving the element speed and the speed in B field drives the magnetic force and corresponding oscilation. The net momentum resulting from interaction is strongly dependent on initial conditions. On the other side, in the case of "stable" oscilation, described above, the effects of interaction will be symmetric in x and y directions, creating no momentum asymmetry - the pressure in one direction. I consider the case of "stable" oscilation the natural case, because the wave packets lead to such oscilation, and I consider wave packets are statisticaly prevalent, as a result of natural atomic processes and state transitions. However it seems, that the EM quantification based on Maxwell theory, does not lead to pressure in direction of wave packet propagation. That is the reason why I was trying to find classical EM explanation and derivation of the radiation pressure, and I have realised, that there is a strong symmetry in the case of "stable" oscilation interaction of EM wave and element. If we consider the plane wave (one symmetry) and the fact, that Lorentz force does not recognize the direction of wave propagation (other symmetry), because the E and B vectors are perpendicular to the propagation line, the overall symmetry is inevitable. The Radiation pressure article declares, that such EM quantification or derivation of radiation pressure, the asymmetric effect of interaction, exists. This is in the center of my consideration. Radiation pressure is wave pressure, not the field propagation pressure (like a wind), I guess. Softvision (talk) 23:43, 26 January 2011 (UTC)

—Preceding unsigned comment added by 217.149.104.6 (talk) 15:42, 12 April 2008 (UTC)

"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 $p^{2}/2m$ 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 $\gamma$ -MarkHudson 09:16, 4 August 2005 (UTC)
The radiation pressure being "doubled" is slightly inexact if the measurements are done by an observer in the frame of reference where the reflective surface is initially at rest. Choose the special frame where the reflector is moving towards the oncoming photon with a velocity such that the momenta of the photon and the reflector are equal. Then when they collide they will each rebound with reversed momenta, and the energy will be unchanged. If you translate this result back into the frame where the reflector starts at rest, the momentum transfer will no longer be double the original photon momentum, and the original and reflected photons will each be Doppler-shifted in opposite directions. This effect won't impact an actual sail design, and can probably be ignored. BruceThomson 01:01, 12 November 2007 (UTC)

## 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)

Weigel was not reading carefully. He took the solar flux, which is clearly unidirectional from the Sun, and applied it in a place requiring an omnidirectional flux. As he quoted: "traversed by radiation uniformly in all directions".
Example: Release a reflective sphere in solar space--it will experience pressure (force) on the sunlight side and will accelerate away from the Sun. Release that same sphere in an environment of uniform omnidirectional flux--it will experience compressive pressure on all sides, and will not move. See the section on radiative compression.--SpaceSailor 12:25, 11 August 2013 (UTC)
The flux density is 1370W/sq.m. but to get from there to the energy density per unit volume, you need to divide by the speed of light!
I.e. W (Watt) is Joule per second, and in one second the energy in that light gets distributed over 3e8 meters.
Check: 1370/3e8 = 4.56e-6, and without further checking units, I'll wager that is 4.56μPa, which seems to fit with the numbers in the article!
But this seems to have been clarified in the first paragraph of the article, so this section of talk could really be removed. Tøpholm (talk) 01:23, 5 August 2009 (UTC)

## Energy density incorrect?

I have a problem with the statement that the energy density for a black body is $\sigma T^4/3 c$. It should be $4 \sigma T^4/c$ 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)

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

Integrating this from ν=0 to infinity gives for the energy density

$u=\frac{8k^4\pi^5}{15c^3h^3}\,T^4$

and the Stefan Boltzmann constant is (see Stefan-Boltzmann law and Stefan-Boltzmann constant)

$\sigma=\frac{2\pi^5k^4}{15c^2h^3}\,T^4$

substituting gives $u=4 \sigma T^4/c$. Did I go wrong here or is the article wrong? PAR 00:17, 20 November 2006 (UTC)

The compression section now has the correct value, same as above. SpaceSailor (talk) 02:50, 10 August 2013 (UTC)

## Pressure of Light Can Move Liquid

Neat image. Would be cool to get permission for the image somehow. --Eean 15:04, 29 March 2007 (UTC)

The effect is from heat transfer, not light pressure.--SpaceSailor 17:13, 10 August 2013 (UTC)

## Pressure of the Sun's light on Earth

What is the total amount of light pressure of the Sun light on Earth? —Preceding unsigned comment added by 86.163.165.101 (talk) 12:20, 27 August 2008 (UTC)

Approximately 4.7*10^-6 Pascal. —Preceding unsigned comment added by 86.92.80.224 (talk) 19:23, 11 June 2009 (UTC)
Anyone else can confirm this? As atmosphere pressure is only double this at ~ 9.8×10−6 —Preceding unsigned comment added by Seb-Gibbs (talkcontribs) 19:27, 2 August 2009 (UTC)
As the article states, the pressure at the earth is 4.6*10^-6 Pascal. That does not depend on the size of the object (i.e. the Earth). What you probably want is the force, which you get by multiplying the pressure with the cross section of the Earth. That yields a total force of Pi * (6,371 km)^2 * 4.6*10^-6 Pa = 587,000,000 newtons (i.e. the pressure from a 60 kiloton weight on the surface). Not that impressive when applied to a planet :-)
Oh, and the pressure of the atmosphere at the surface is around 10^5 Pa, ie. 2*10^10 larger. At height ofcourse the atmospheric pressure drops to zero (and as it does, I guess the solar radiation is partly responsible for some of it escaping. Which is why we have almost no free hydrogen and helium in the atmosphere) Tøpholm (talk) 01:44, 5 August 2009 (UTC)
it states that, if the whole light from the sun is absorbed, the pressure is 4.6*10^-6 Pascal. It doesn't say at any point that the earth absorbs all light, it's just a supposition in case it absorbs it all. Does any one know the real pressure? It should be between 4.6*10^-6 (fully absorbed) and 9.15*10^-6 (fully reflected) Pascal. — Preceding unsigned comment added by 79.147.28.155 (talk) 18:18, 2 January 2012 (UTC)

## Radiation Pressure in Defense Systems

Useful for active ablative armor systems, feedback loops offer protection from bullets to explosions, as well rebreather filters can utilize this force given a sufficient power source. However verifiable sources of applied consumer goods as well as concepts are to be calculated and debated. Aditya.m4 (talk) 03:16, 19 January 2010 (UTC)

The primary real world application of radiation pressure is in the two-stage thermonuclear warhead. In that case, the pressure is not slight, it is huge (140 TPa). You should mention it here. https://en.wikipedia.org/wiki/Thermonuclear_weapon Anorlunda (talk) 13:27, 20 July 2014 (UTC)

## elementary explaination

hey i got a web http://www.blazelabs.com/f-g-rpress.asp that explain it quiet well for elementary level —Preceding unsigned comment added by ArielGenesis (talkcontribs) 21:52, 9 June 2010 (UTC)

## Casimir Effect

Including Casimir Effect as it is related... LucianSolaris (talk) 07:53, 9 October 2010 (UTC)

Stupid me, Casimir effect is in an environment lacking electromagnetic radiation. Edit removed.
LucianSolaris (talk) 07:56, 9 October 2010 (UTC)

## Section Theory is insufficient

"It may be shown by electromagnetic theory" is insufficient theoretical reference. Please, add more detailed explanations or derivations, or apropriate references. Especially useful will be direct derivation from Maxwell equations or Lorentz force equation. Softvision (talk) 10:30, 22 January 2011 (UTC)

Derivations from those equations would be out of scope for this article. See those respective articles. I am working on improving the explanations, which are definitely in need of reworking. SpaceSailor 02:48, 2 August 2013 (UTC)

Yes. The article is good and the source is OK, but is the article referring to reflected or absorbed radiation pressure? Is the coefficient 1/3 or 2/3? Probably a combination of sources are needed here, some with simple and some with mathematical explanations.. 172.129.73.71 (talk) 20:12, 8 September 2011 (UTC)BG.

The "theory" section has been reworked, which should take care of the concerns. Emission pressure is now included (Previously missing).--SpaceSailor 00:30, 11 August 2013 (UTC)

## no explanation of WHY --- and way too hard

This article doesn't have any explanation of WHY radiation exerts pressure. How about something that explains this in a sort of intuitive way, or if that is impossible, says that no one understands why it happens, but there is experimental proof that it does. Right now the article is very technical, pitched towards people who have degrees in physics rather than the average reader. It seems to assume that anyone reading it already has a basic understanding of what light pressure is and how it works. Which is what I was trying to find out when I looked it up. Can't you put in a top level section that is easy enough for most people to understand? Right now you are losing about 99% of the population. —Preceding unsigned comment added by 74.38.247.10 (talk) 13:50, 21 May 2011 (UTC)

The revised text uses momentum to explain the existence of pressure, hopefully making it more understandable. SpaceSailor 02:54, 10 August 2013 (UTC)

## Derivation from thermodynamics alone?

The article says "It may be shown by electromagnetic theory, by quantum theory, or by thermodynamics, making no assumptions as to the nature of the radiation, that the pressure against a surface exposed in a space traversed by radiation uniformly in all directions is equal to one third of the total radiant energy per unit volume within that space.[citation needed]". Is it really possible to derive it from thermodynamics alone, without any information about electromagnetic theory? I would be very interested to see the reference if so, but I suspect it's not possible. Planck derived most of the properties of radiation from thermodynamics alone, but he got the starting point of p=U/3V from Maxwell's equations. Nathaniel Virgo (talk) 10:40, 2 January 2012 (UTC)

## Expression for radiation force is incorrect

$F_{SR} = -p_{SR}c_{R}A_{\odot}r_{\oplus\odot}$ is apparently incorrect: for black bodies it yield zero force!

The correct one should be: $F_{SR} = -p_{SR}(1+c_{R})A_{\odot}r_{\oplus\odot}$

The equation was correct, but not useful. The equation says nothing about black bodies. It was removed when the section was rewritten. SpaceSailor 03:11, 10 August 2013 (UTC)

## Article overhaul

I am making an effort to tighten up the article. A lot of things have gotten into it that are out of scope and will be removed. An explanation of momentum is being added under the photon section. SpaceSailor 20:25, 31 July 2013 (UTC)

## Radiation pressure not used for fusion

In the section Laser applications of radiation pressure there's a mention of laser radiation pressure being used for fusion research. The force on the pellet is primarily ablation of the surface. In fact in many cases, radiation pressure is confused with other effects involving the transfer of real mass - see the Crookes Radiometer for example.

I will revise this paragraph, and perhaps add a reference to Effects Commonly Confused With Radiation Pressure. Thomasonline (talk) 15:39, 11 September 2014 (UTC)