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"mass" of a sphere of light - how pressure causes gravity

This rather old topic has come up again in another thread. I thought I would make an attempt to address it as simply as possible.

First, let's talk about an operational way to measure mass. Suppose we have a spherically symmetric body. In Newtonian theory, we know that we can find the mass of the body by applying Gauss's law - we integrate the gravitational force on a "test mass" over a spherical surface, and we will alway get

$$\frac{GM}{r^2} 4 \pi r^2 = 4 \pi G M$$

independent of the radius r. Furthermore, the gravitational effect of a spherical shell of matter surrounding a probe is zero.

This sort of intergal can be applied in GR, as well as Newtonian theory, if one is willing to make a few very small adjustments. The biggest adjustment needed is that one has to include a factor related to the "time dilation" at the radius r. An alternative way of doing this is to measure the force via a string to an observer at infinity, rather than measuring the force directly with a local observer. This is discussed (in rather advanced terms) in Wald's "General Relativity" on pg 288 - though they do explicitly mention the "string" idea.

For the purposes of this thread, I'm going to ignore the details of how one deals with the time dilation factor, and simply assume that the fields and potentials are such that we never have experimentially significant time dilation, thus that we can ignore this factor. This basically places a modest constraint on the size of our experiment. The gravitational time dilation factor due to the entire mass of the Earth is, for instance, less than 1 part in a billion at its surface.

This "Gauss-law intergal" means of measuring mass is operationally equivalent to GR's defintion of Komar mass. The way we have described it applies to only spherically symmetric distributions of matter, though the concept can be generalized to any static (i.e. unchanging with time) metric.

This particular defintion of mass (Komar mass) is one of a very few in General Relativity that can be localized, i.e. where we can say not only how much mass a system has, but where exactly that mass is located.

Now, let us consider a very strong mirrored container, containing a "photon gas", i.e. randomly oriented high-frequency photons (electromagnetic radiation) bouncing around inside it.

We can use our concept of Komar mass to find, in theory, the amount of mass enclosed in a spherical shell of the gas. by lowering an accelerometer into the sphere and measuring the acceleration due to the gravity of the apparatus at various positions. (Fortuantely, we have an accelerometer that won't melt, and is also senstive enough to measure the minute accelerations accurately, this being a thought experiment). The gravitational field that we measure with our acceleromter will be given by Gauss's law and the amount of mass enclosed in the sphere of the appropriate radius, allowing us to measure the enclosed mass as a function of distance.

When we do this, we find that the "mass" of the ball of light is larger than we expect. This mass, ("Komar mass") will be about 2x as big as E/c^2! A more detailed analysis shows that, in geometric units, the Komar mass of the photon gas in a volume element dV is $(\rho + 3P)dV$. Here [/rho] is the energy density of the photon gas, and P is the pressure of the gas. It is a property of photon gasses that P=$\rho/3$.

At this point, we do another experiment. We take our very strong container, and we generate our photon gas inside it by detonating a nuclerar bomb or an anti-matter device to generate large quantites of x-rays, without adding or subtracting any matter or energy from the sphere.

We measure the gravitational field outside the sphere to see if it changes - if it does, the mass of our system has changed. We find that the gravitational field, and hence the mass, of the system has not changed, when our bomb detonates.

Now we do a third experiment. We take our empty shell, and we essentially "pump" a photon gas into it. We measure the amount of gas we "pump" into the shell (by measuring its energy), and the mass of the entire system. We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2.

Thus we conclude that the shell itself is getting lighter as we "inflate" it with photon gas.

The explanation for this odd behavior is that pressure causes gravity, and hence contributes to mass. The positive pressure of our photon gas causes the photon gas to "weigh more". The tension in our massive shell, needed to hold it together, causes it to "weigh less".

The total contribution of the pressure terms to the mass of the entire closed system is zero, but it changes the apparent distribution of mass, as measured by our gravity probes.
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 Recognitions: Gold Member Science Advisor that is an intriguing thought experiment and cluster of ideas. it would not have occurred to me to notice that the shell is getting less massive due to the negative of pressure yet that is right, so it must be I remember one time trying to estimate the fraction of the mass of the sun due to the pressure at the core, but dont think I ever carried it through. maybe someone knows. it is fascinating to reflect that one of the reasons we orbit around the sun as we do is not just the mass of the sun but the fact that it is compressed and has pressure. thanks for a thread with educational and entertainment value
 For a universe in tension - the stress energy is not detectable because there is no gauge against which it can be measured - if the photon gas can be considered as displacing the vacuum tension, the effect could be explained as a property of the volume rather than the shell

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"mass" of a sphere of light - how pressure causes gravity

 Quote by yogi For a universe in tension - the stress energy is not detectable because there is no gauge against which it can be measured - if the photon gas can be considered as displacing the vacuum tension, the effect could be explained as a property of the volume rather than the shell
There is another problem with attempting to apply the concept of Komar mass to the universe. The Komar mass is defined in GR only for a _static_ metric. Because the universe is expanding, it's metric is not static - it's changing with time. Thus the universe, as a whole, does not have a Komar mass (at least in standard cosmologies).
 What I meant to convey is the idea that w/i the shell, there exists a negative stress energy consequent to expansion - it is the same everywhere throughout the universe - but in the particular case of the sphere injected with a photon gas, the negative stress energy E/c^2 is displaced w/i the enclosed volume - ergo, the effective mass increases by E/c^2 rather than (2E/c^2). A surface to volume transform via the divergence theorem would relate the energy difference to the properties of space rather than the construct of the shell. just a thought...probably of no value.
 pervect says " We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2. Thus we conclude that the shell itself is getting lighter as we "inflate" it with photon gas." Is this the photons momentum you are referring to? If so, why would the shell become lighter due to a high momentum of photons infused? Are you seeing it as momentum is a variable that one, relatively seen, can apply inside a 'enclosed system' at a 'whim'. Then you see momentum as something not 'locked' to the 'particles/photons' exhibiting it? Or is it something else you refer to as you also write 'that the total mass of the system increases only by E/c^2.' That doesn't fit those thought experiments where you've placed photons with momentum inside a 'reflective' box. I'm so totally confused here:)

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 Quote by Yor_on pervect says " We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2. Thus we conclude that the shell itself is getting lighter as we "inflate" it with photon gas." Is this the photons momentum you are referring to? If so, why would the shell become lighter due to a high momentum of photons infused? Are you seeing it as momentum is a variable that one, relatively seen, can apply inside a 'enclosed system' at a 'whim'. Then you see momentum as something not 'locked' to the 'particles/photons' exhibiting it? Or is it something else you refer to as you also write 'that the total mass of the system increases only by E/c^2.' That doesn't fit those thought experiments where you've placed photons with momentum inside a 'reflective' box. I'm so totally confused here:)
Firstly, don't expect a reply from pervect, as you're responding to a post that's over two years old but pervect retired from these forums some time ago.

Secondly, your questions seem to show a lack of understanding of some of the basic principles, so it's a bit difficult to know how to explain the details, but I'll have a go.

Basically, if you have an empty reflective shell and fill it with photons, they will keep bouncing off the inside, creating some pressure, because each photon has momentum given by its energy divided by c, in the direction of travel, and the bounce changes the direction, creating an impulse.

If you imagine a plane slicing the shell in half, then if the overall situation is static, the force created by the photons pushing the halves apart within the shell must be balanced by the force holding the halves of the shell together at the edge. If you divide that force by the area, you get the pressure. There is therefore negative pressure in the shell (probably quite high if the shell is thin) and a lower positive pressure inside it. (Technically, since I'm only referring to pressure in one plane here I should be calling it "normal stress", but "pressure" is close enough).

The formula for the Komar energy says that the integral of pressure with respect to volume counts as being like energy for gravitational purposes. Since pressure is force per area, this integral is the simply same as the total force times the perpendicular distance over which it operates. When you add this up, it's like slicing up the system into thin slices and adding up the forces in each slice times the thickness of the slice. Since the system is static, the pulling and pushing forces in each slice are equal. This means that if you add up the pulling forces in the shell, you get a negative total and if you add up the pushing forces you get a positive total, and if you add the two together they cancel out, so the net gravitational effect is just that of the photons.
 You definitely sound as you know 'your thing'. And you're quite correct in stating that i don't know very much about this:) But regretfully, I still don't get how Pervert thought here? "We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2." He starts with a 'empty' shell That must mean no energy at all? Fills it with a photon gas of 'twice as high mass' as compared to E/c^2. Would that refer to the 'pressure' of the gas being released into the shell? Arbitrarily chosen pressure? But I still get that as the 'force' of the momentum of the photons. If you have a highly reflective chamber the momentum should be retained as 'mass' if one weighted the system as I understand it. What I don't get is the difference between masses here? The mass of something under Arbitrarily high pressure will naturally be compared to the volume, infinitely small volume will give infinitely high pressure for example. But as long as this isn't that, and you have a 'closed system' the volume shouldn't matter I thought? The rest of your explanation I think i got:) Thanks.

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 Quote by Yor_on You definitely sound as you know 'your thing'. And you're quite correct in stating that i don't know very much about this:) But regretfully, I still don't get how Pervert thought here? "We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2." He starts with a 'empty' shell That must mean no energy at all? Fills it with a photon gas of 'twice as high mass' as compared to E/c^2. Would that refer to the 'pressure' of the gas being released into the shell? Arbitrarily chosen pressure? But I still get that as the 'force' of the momentum of the photons. If you have a highly reflective chamber the momentum should be retained as 'mass' if one weighted the system as I understand it. What I don't get is the difference between masses here? The mass of something under Arbitrarily high pressure will naturally be compared to the volume, infinitely small volume will give infinitely high pressure for example. But as long as this isn't that, and you have a 'closed system' the volume shouldn't matter I thought? The rest of your explanation I think i got:) Thanks.
He's pumping in some photons with total energy E, so as the reflective shell is giving them an average momentum of zero, that is also the effective total rest energy, and the equivalent mass is E/c^2. However, according to the Komar energy formula, their gravitational effect also depends on their pressure, which can be assumed to match the normal force per area on the inside of the shell.

I don't recall personally exactly how to work out the pressure (it only involves the geometry of how the photons bounce on average and how frequently they do it, which depends on the radius), but Pervect states that the result is that the pressure summed for all three directions and integrated over the volume also comes to E/c^2, so the gravitational effect of the photon gas's energy plus that of its pressure adds up to 2E/c^2. Although the pressure depends on the radius of the shell, so does the volume of the shell, and these factors cancel so the product of the two is independent of the radius.

From the "slicing" model in my previous response, the gravitational effect of the negative pressure in the surface of the shell is automatically equal to that of the positive pressure inside (at least assuming negligible additional pressure in the center due to the gravity of the photon gas pulling it inwards), so the overall gravitational effect is merely that of mass E/c^2, the effective rest mass of the photon gas.
 Thanks for your patience again :) I looked up the 'Molecular Definition of Pressure'. http://www.grc.nasa.gov/WWW/K-12/airplane/pressure.html " From the kinetic theory of gases, a gas is composed of a large number of molecules that are very small relative to the distance between molecules. The molecules of a gas are in constant, random motion and frequently collide with each other and with the walls of any container. The molecules possess the physical properties of mass, momentum, and energy. The momentum of a single molecule is the product of its mass and velocity, while the kinetic energy is one half the mass times the square of the velocity. As the gas molecules collide with the walls of a container, as shown on the left of the figure, the molecules impart momentum to the walls, producing a force perpendicular to the wall. The sum of the forces of all the molecules striking the wall divided by the area of the wall is defined to be the pressure. The pressure of a gas is then a measure of the average linear momentum of the moving molecules of a gas. The pressure acts perpendicular (normal) to the wall; the tangential (shear) component of the force is related to the viscosity of the gas. " And momentum is defined as an objects 'mass' as well as its velocity. The energy is naturally there:) And being in a pressurized 'colliding' gas there will be difficulties saying where the momentum points but it is to me the same as the 'pressure'? I can understand that pressure might be a easier way to count on a 'gas' of something, but I don't see how it can be separated from momentum and so give a 'added' result? To me they seem the same? http://www.grc.nasa.gov/WWW/K-12/airplane/conmo.html

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 Quote by Yor_on Thanks for your patience again :) I looked up the 'Molecular Definition of Pressure'. http://www.grc.nasa.gov/WWW/K-12/airplane/pressure.html " From the kinetic theory of gases, a gas is composed of a large number of molecules that are very small relative to the distance between molecules. The molecules of a gas are in constant, random motion and frequently collide with each other and with the walls of any container. The molecules possess the physical properties of mass, momentum, and energy. The momentum of a single molecule is the product of its mass and velocity, while the kinetic energy is one half the mass times the square of the velocity. As the gas molecules collide with the walls of a container, as shown on the left of the figure, the molecules impart momentum to the walls, producing a force perpendicular to the wall. The sum of the forces of all the molecules striking the wall divided by the area of the wall is defined to be the pressure. The pressure of a gas is then a measure of the average linear momentum of the moving molecules of a gas. The pressure acts perpendicular (normal) to the wall; the tangential (shear) component of the force is related to the viscosity of the gas. " And momentum is defined as an objects 'mass' as well as its velocity. The energy is naturally there:) And being in a pressurized 'colliding' gas there will be difficulties saying where the momentum points but it is to me the same as the 'pressure'? I can understand that pressure might be a easier way to count on a 'gas' of something, but I don't see how it can be separated from momentum and so give a 'added' result? To me they seem the same? http://www.grc.nasa.gov/WWW/K-12/airplane/conmo.html
Since photons are moving at c, we have to use the Special Relativity definition of momentum, which is $Ev/c^2$ where E is the energy (of the photon in this case) and v is its velocity, which in this case has magnitude c.

For each individual photon, the energy is actually all effectively kinetic energy; it doesn't have any rest mass so it doesn't have any rest energy. However, when you take a collection of photons which are not all moving in the same direction, then the momentum vectors cancel out, reducing the total momentum.

The standard Special Relativity relationship between the overall energy (E), magnitude of momentum (p) and effective overall rest mass (m) of a system is as follows:

$$E^2 = (pc)^2 + (mc^2)^2$$

This means that when the overall momentum partly cancels out, the collection of photons appears to have some rest mass, and when the overall average momentum cancels to zero, the effective rest mass is $E/c^2$, showing the same relationship between total energy and mass as for an ordinary massive particle.

Each time a photon is reflected, its momentum component in the direction perpendicular to the shell is reversed, so the change in momentum is twice that amount. The travel time of a photon across the shell is determined by the size of the shell, and that is the maximum time between collisions, so every photon must hit each side at least that often. From that, one can calculate the average change in momentum per unit time and hence (by definition) the average force, and one can divide it by the area to determine the pressure.

I don't off hand know the simplest way to calculate that average force for a spherical shell, and I don't have the patience to work out the integral from first principles, but if pervect says that the integral of the pressure times the volume is the same as the total energy, I'm happy to take pervect's word for it.

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 Quote by Yor_on pervect says " We find that when we pump a photon gas into the shell, the mass (Komar mass) of the photon gas is twice as high as one would expect from the relation E/c^2, but that the total mass of the system increases only by E/c^2. Thus we conclude that the shell itself is getting lighter as we "inflate" it with photon gas."Is this the photons momentum you are referring to? If so, why would the shell become lighter due to a high momentum of photons infused? Are you seeing it as momentum is a variable that one, relatively seen, can apply inside a 'enclosed system' at a 'whim'. Then you see momentum as something not 'locked' to the 'particles/photons' exhibiting it? …
 Quote by Yor_on … And momentum is defined as an objects 'mass' as well as its velocity. The energy is naturally there:) And being in a pressurized 'colliding' gas there will be difficulties saying where the momentum points but it is to me the same as the 'pressure'? …
Hi Yor_on!

No, momentum is not defined as an object's mass.

Momentum is a vector. Mass, like energy, is a scalar (one ordinary number).

Energy can be considered a form of mass (and vice versa ), but momentum cannot.

(But yes, pressure is more-or-less the same as momentum. Pressure, as you have found, is ultimately caused by the change in momentum when particles bounce off each other.)
 … I can understand that pressure might be a easier way to count on a 'gas' of something, but I don't see how it can be separated from momentum and so give a 'added' result? To me they seem the same?
In GR (general relativity), gravity is determined by mass and pressure.

(The pressure can't be "separated" from the momentum, but it can be "separated" from the mass.)

The Komar mass is a sort of "black box" mass … if you couldn't see what was causing the gravity of something, you'd have to measure the gravity, and calculate the mass from that … the Komar mass is what you get.

(This is what pervect calls "an operational way to measure mass".)

Since (as I said) gravity is determined by mass and pressure, that means that Komar mass is determined by mass and pressure.

So when we observe that shell from a distance, since we can't see inside it, it is easiest to measure its Komar mass.

If we inject more photons into the shell, obviously the mass, and the Komar mass, go up … but so does the pressure of the photons … pervect was asking, if gravity = mass + pressure, and pressure is increasing, doesn't that mean our caclulations are wrong, and he answered himself by saying that the tension in the shell increases to balance out the pressure of the photons … since tension is negative pressure (are you happy with that?), the total pressure increase is zero, and we get back (in this particular case) to gravity = mass.
 First of all Jonathan, thanks for your tries to explain. And I'm entirely serious at that. I'm rather boneheaded at times:) Perhaps I'm slowly seeing the 'light' here. But being as bad as I am at math, I'm still wondering. So on the other hand, don't get any hopes up:) If I get it right the idea hinges on lights momentums vectors canceling each other paths inside a closed system. What makes the most sense to me, if so, would probably be to see photons as waves, and 'then allowing for them to quench each other depending on velocity's. But if one does so, where does the momentum disappear too? :) It is after all an added 'force'. And how do you get it out as 'invariant mass' as that is what it will be in the end? Or do you mean it is 'rest mass' but not 'invariant mass'. ----- Tiny-Tim. Thanks for explaining Komar mass. And yes, I'll go with your description of 'negative tension'. I think I see how you think there, this is a discussion about 'boundaries', well sort of:) You could treat the inside of this shell as one 'boundary' and the shell itself as another 'boundary'. In 'reality' though, isn't there only one 'boundary', but with two forces acting on it. One force that we could say is 'known' within reason:) (inside/Komar mass). And the other is in fact 'unknown' ('outside'). If so, it responds to the first 'force' with an equal 'tension/pressure' thereby negating the first force. But we can't define its strength as we can with the 'inside force/Komar mass'. Am I correct? So Komar mass is what fills a 'black box' then? But I think there actually is mass defined to momentum? The invariant kind as I understand it. But maybe you're thinking of photons?

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 Quote by Yor_on Tiny-Tim. Thanks for explaining Komar mass. And yes, I'll go with your description of 'negative tension'. I think I see how you think there, this is a discussion about 'boundaries', well sort of:) You could treat the inside of this shell as one 'boundary' and the shell itself as another 'boundary'. In 'reality' though, isn't there only one 'boundary', but with two forces acting on it. One force that we could say is 'known' within reason:) (inside/Komar mass). And the other is in fact 'unknown' ('outside').
Hi Yor_on!

I don't see where you get "boundaries" and force from …

I don't see any boundaries as being involved … the gravitational effect (or the Komar mass) is the sum of all the little bits of mass (or energy) and pressure …

that is, the sum (∫∫∫), over the whole volume, of the effects of the mass (or energy) density and "pressure density" of each tiny bit of volume.

The "boundaries" have nothing to do with it (it's not a ∫∫ sum, for example).

And the only place force comes into it is in defining the pressure (or tension) …

although I agree that the pressure force from the photons has to balance the tension force of the shell, that's only to stop the shell from expanding or collapsing, and doesn't directly affect the gravity or the Komar mass.
 If so, it responds to the first 'force' with an equal 'tension/pressure' thereby negating the first force. But we can't define its strength as we can with the 'inside force/Komar mass'.
The first sentence, I agree with, but that's just good ol' Newton's second law. The second sentence, sorry, I don't follow.
 So Komar mass is what fills a 'black box' then? But I think there actually is mass defined to momentum? The invariant kind as I understand it. But maybe you're thinking of photons?
No, "black box" is just a convenient way of saying "let's ignore the details of which bit of gravity comes from where, and just calculate a total sum".

As I said above, forget any mass-momentum connection. And forget momentum itself … gravity comes from mass (or energy) and pressure:
 Quote by pervect the Komar mass of the photon gas in a volume element dV is $(\rho + 3P)dV$. Here [/rho] is the energy density of the photon gas, and P is the pressure of the gas.
(and "invariant mass" - another name for "rest mass" - is independent of velocity)
 Maybe I'm using the wrong definitions Tim? But when you define a 'system' you automatically impose a boundary on it it seems to me? Just by defining it you have separated an area from 'something else'. And when one discuss the 'shell' and its 'negative pressure' one do seem to discuss a 'boundary'? A 'boundary' between 'states', one known or at least defined, which then represent the inside/Komar mass. The other one defined by its tension/resistance towards the first, but in itself representing a unknown 'quantity' if you get my drift. Or you can't see it that way then? ---------- Thinking about it again it seems not that smart does it :) It is after all a 'shell' and answers to a 'pressure from the inside and nothing else in this system.. ---------------- You say " although I agree that the pressure force from the photons has to balance the tension force of the shell, that's only to stop the shell from expanding or collapsing, and doesn't directly affect the gravity or the Komar mass. " Could you clarify how you think here? To me the photons creates the mass. Or if you like , the mess:) The pressure comes from compressing photons into a gas. (I have some difficulties seeing a 'photon gas', as they can be superimposed on each other but I'm trying here:)

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