How is there any pressure at all in isotropic radiation?

[phyTens]

TL;DR Summary

Standard theory on radiation says that pressure in isotropic radiation is given by $P_\nu=\frac 1 3 U_\nu$. However, why is there any pressure at all, since isotropic means that no direction is preferred? What is actually the definition of pressure inside radiation field?

It is a standard result that in a blackbody radiation there is a pressure (at a certain frequency), given by
$$P_\nu=\frac 1 3 U_\nu$$
However, I am quite confused by this result.

Firstly, how do we even define pressure in radiation gas? I would think that this would be the pressure on a small hypothetical surface $dA$. However, since blackbody radiation is isotropic and no direction is preferred, why do we expect the imaginary surface to be blown away by the pressure? Certainly my definition of pressure cannot be correct.

The reason I think that pressure is defined as described above, is that in a book "Astrophysics for Physicists" the author says that "The pressure of the radiation field over a surface is given by the flux of momentum perpendicular to that surface". However, if radiation is isotropic, how can there be any flux of momentum?

Any insight would be greatly appreciated.

 

[Frabjous]

Pressure is always normal to a surface. If you have a thin shell with identical radiation on both sides, the forces on each side will point toward the surface cancelling out.
Gradients in pressure cause motion.

 

[Vanadium 50]

I would say that radiation is at pressure P if a perfectly reflective baloon with pressure P neither expands nor contracts when placed in that radiation area.

Pressure in a gas is also (close to) isotropic.

 

[Ibix]

The way I see it is, if I carve out a little cubical hollow inside the field and orient it according to my coordinate system, what forces do I have to apply to its walls so it doesn't change shape or size? (A variant on the two answers above, I see).

 

[WernerQH]

Even if the radiation field is isotropic, the photons do not contribute equally to the momentum flux. Photons traveling at an angle of 60°, for example, carry only half of the momentum (with respect to normal of the imagined surface $dA$), and their flux is also reduced by a factor $\cos \theta$. The momentum flux is proportional to an integral $$
\int \sin\theta d\theta \quad \left( I_\nu(\theta) \cos \theta \right) \times
\left( {h \nu \over c} \cos \theta \right)
$$ The factor $\cos^2 \theta$ averages to the $1/3$ that you already know.

 

[phyTens]

I understand your explanations (that pressure can be interpreted as hollowing a cavity of some sort and see what force is being applied by the gas), and I would totally agree with you if it weren't for the derivation of the pressure in the aforementioned book:

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The pressure of the radiation field over a surface is given by the flux of momentum perpendicular to that surface. The momentum associated with energy $dE_\nu$ is $dE_\nu/c$ and its component normal to the surface $dA$ is $dE_\nu\cos\theta /c$. By dividing this by $dA\, dt$, we get the momentum flux associated with $dE_\nu$, which is
$$\frac{dE_\nu\cos\theta}{c}\frac{1}{dA\, dt}=\frac{I_\nu}{c}\cos^2\theta\, d\Omega$$
on making use of (2.2). The pressure $P_\nu$ is obtained by integrating this over all directions, i.e.
$$P_\nu=\frac{1}{c}\int I_\nu\cos^2\theta\, d\Omega$$.
If the radiation field is isotropic, then we get
$$P_\nu=\frac{I_\nu}{c}\int \cos^2\theta\, d\Omega=\frac{4\pi}{3}\frac{I_\nu}{c}$$
It follows from (2.5) that
$$U_\nu=4\pi\frac{I_\nu}{c}$$
for isotropic radiation. Combining this with (2.8), we have
$$P_\nu=\frac 1 3 U_\nu$$
for isotropic radiation.

------------
(2.2) there refers to the definition of specific intensity.

As you can see, the author integrates the momentum flux over the whole solid angle, and this is the main reason I am struggling to see how there is any pressure using this integral.

With the definition with the balloon: the pressure by the radiation is evaluated over the hemisphere, not the whole sphere.

With the definition with the cavity: similarily, the pressure by the radiation is integrated only over the hemisphere.

There is clearly a discrepancy.

 

[Vanadium 50]

The pressure of the radiation field over a surface is given by the flux of momentum perpendicular to that surface.

OK, so how does this differ from a gas?

 

[WernerQH]

As you can see, the author integrates the momentum flux over the whole solid angle, and this is the main reason I am struggling to see how there is any pressure using this integral.

To calculate the momentum flux correctly, you need to include also the opposite hemisphere: photons with the opposite (negative) momentum, and with a "negative" flux, because they travel in the opposite direction. There's no flaw in the book-keeping.

There is clearly a discrepancy.

No. If there is a boundary, the radiation field need no longer be isotropic. The total force on the balloon will be zero, whether it's perfectly reflecting or a perfect absorber emitting blackbody radiation. But of course the pressure in the balloon cannot be zero.

 

[Ibix]

With the definition with the cavity: similarily, the pressure by the radiation is integrated only over the hemisphere.

Indeed. That's why you need to apply a force to stop the cavity from changing shape or size.

Look at it another way: consider a really low density gas such that only two particles pass through the elementary area in the time, and both happen to be perpendicular to the surface. One is moving to the right and carries momentum $p$, so its contribution to the momentum flux is $p$. The other is moving to the left and carries momentum $-p$, so its contribution to the momentum flux is $-p\times-1=p$, where the second negative comes from the cosine term. The center of mass of the system hasn't moved, so we're in the rest frame of the gas, but there's a net momentum flux across the surface.

That's a gas but, as Vanadium 50's rhetorical question implies, a similar argument applies to the radiation field.

 

[phyTens]

Thank you for all the answers! I think I understand this much better now. The idea that a particle moving in the opposite direction has negative momentum, has particularly helped.

 

[Ken G]

It's not your fault, in my opinion physics textbooks really drop the ball on the definition of pressure. Many will say pressure is defined by a force per unit area on a surface, but that is only correct if you only consider the force on one side of the surface, and the particles are elastically bouncing off, both special cases that have nothing to do with what pressure actually is. Pressure is just a (vector) momentum flux per unit area through a purely hypothetical surface, a property purely of the motion of the particles that requires no collisions of any kind, so no forces either. (The force people talk about is actually a result of the presence of pressure under special conditions, such as a real barrier that does not need to be there for there to be pressure there. It's a lot like defining the force of gravity to be what a scale reads, which would be a silly way to define gravity.)

Also, it's a tensor that acts like a constant times the identity matrix expressly when the particles are isotropic, which ironically means that pressure only takes its common meaning in the isotropic case, so that's not the weird case, it is the common case!

 

[Frabjous]

It's not your fault, in my opinion physics textbooks really drop the ball on the definition of pressure. Many will say pressure is defined by a force per unit area on a surface, but that is only correct if you only consider the force on one side of the surface, and the particles are elastically bouncing off, both special cases that have nothing to do with what pressure actually is. Pressure is just a (vector) momentum flux per unit area through a purely hypothetical surface, a property purely of the motion of the particles that requires no collisions of any kind, so no forces either. (The force people talk about is actually a result of the presence of pressure under special conditions, such as a real barrier that does not need to be there for there to be pressure there. It's a lot like defining the force of gravity to be what a scale reads, which would be a silly way to define gravity.)

Also, it's a tensor that acts like a constant times the identity matrix expressly when the particles are isotropic, which ironically means that pressure only takes its common meaning in the isotropic case, so that's not the weird case, it is the common case!

This is not necessarily a satisfactory defintion. One has to count particles going in one direction separately from particles going in the opposite direction. At the continuum level, one could have a net momentum flux of zero with a non-zero pressure. What am I missing?

 

[Ken G]

This is not necessarily a satisfactory defintion. One has to count particles going in one direction separately from particles going in the opposite direction. At the continuum level, one could have a net momentum flux of zero with a non-zero pressure. What am I missing?

It is satisfactory, because a (vector) momentum flux already does what you are saying, automatically. So when you take a hypothetical surface, and count the flux, a flux that goes one way has positive sign while the flux the other way has negative sign. But the momentums are also opposite, that's their vector quality, so they add rather than canceling. In the isotropic case, they all give positive contribution when added up this way, so you get a pressure that acts like a scalar because it gives the same positive momentum flux regardless of how you orient the hypothetical surface. It is that positive momentum flux that is what isotropic pressure really is (and when it is not isotropic, it is actually a stress-energy tensor), it has nothing to do with forces but can yield forces under certain conditions (just like gravity can yield a normal force under the conditions of stepping on a scale). Some of those conditions include inserting a barrier, or hollowing out a region as mentioned above, or even adding up a bunch of particles together into a fluid description. (In that last case, it is a pressure gradient, not a pressure, that will act like a force per unit volume on the fluid, i.e., the pressure gradient will look like a force on itself, not a force on some wall that isn't even there.)

 

[Frabjous]

It is satisfactory, because a (vector) momentum flux already does what you are saying, automatically. So when you take a hypothetical surface, and count the flux, a flux that goes one way has positive sign while the flux the other way has negative sign. But the momentums are also opposite, that's their vector quality, so they add rather than canceling. In the isotropic case, they all give positive contribution when added up this way, so you get a pressure that acts like a scalar because it gives the same positive momentum flux regardless of how you orient the hypothetical surface. It is that positive momentum flux that is what pressure really is, it has nothing to do with forces but can yield forces under certain conditions (just like gravity can yield a normal force under the conditions of stepping on a scale).

How does one calculate this for a simple continuum. The is only a single value for momentum at any point in space.

 

[Ken G]

How does one calculate this for a simple continuum. The is only a single value for momentum at any point in space.

The momentums are associated with each particle, not each point. So you pick your hypothetical surface, and count the particles going across, and add their momentums, but particles going one way get a positive count and particles going the other way get a negative count. Since it is their momentums that you are accumulating, a negative count of a negative momentum adds positively to the momentum flux total. That's how vector fluxes always work (see Ibix above).

 

[Frabjous]

The momentums are associated with each particle, not each point. So you pick your hypothetical surface, and count the particles going across, and add their momentums, but particles going one way get a positive count and particles going the other way get a negative count. Since it is their momentums that you are accumulating, a negative count of a negative momentum adds positively to the momentum flux total. That's how vector fluxes always work.

I understand the particle definition. I am missing how the momentum flux definition applies to a continuum.

 

[Ken G]

I understand the particle definition. I do not understand how the momentum flux definition applies to a continuum.

Momentum flux definitions are particle definitions, so I'm not seeing why you are distinguishing them.

 

[Frabjous]

Momentum flux definitions are particle definitions, so I'm not seeing why you are distinguishing them.

It was not clear that your post 11 only applied to particles. Thanks.

 

[Ken G]

Perhaps you are thinking in terms of the radiation intensity, but one can regard that as coming from a photon flux, so it's still a particle definition. But either way, the bottom line is there are two powers of the direction cosine normal to the surface when you do the integral, one because of the flux rate and the other because of the vectors being added. So it's positive either way you look at it.

 

[vanhees71]

Also, it's a tensor that acts like a constant times the identity matrix expressly when the particles are isotropic, which ironically means that pressure only takes its common meaning in the isotropic case, so that's not the weird case, it is the common case!

That's the only correct definition! As any stress also pressure is a tensor. It's defined in a fluid, which by definition is isotropic in the local rest frame of the fluid cell. By definition it's defined in Cartesian coordinates as $\sigma_{jk}=-P\delta_{jk}$.

https://en.wikipedia.org/wiki/Cauchy_stress_tensor

 

[Ken G]

Yes, my point is that isotropy is not the situation where pressure is hard to understand, it is the idealization that simplifies the concept of a momentum flux density.