Objects turning into a black hole near the speed of light?

In summary, the conversation discusses the Theory of Relativity and its proposal of equivalence between mass and energy. It is suggested that any form of energy can have the same effect as mass on spacetime, potentially leading to the creation of a black hole. However, the concept of relativistic mass and its relationship to black holes is debated, with consideration given to the dynamics of space-time and the conditions necessary for a black hole to form. The conversation also raises questions about the effects of extreme speed on spacetime and the appearance of non-accelerated objects.
  • #1
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The Theory of Relativity proposes equivalence between mass and energy. Well... since they are "equal", when mass disturbes the spacetime which causes a forcefield that we call gravity, energy should have the same effect on spacetime. Putting a huge amount of mass in a very little place in space, we have a black hole. After this little explanation my question ;):

Does any Kind of energy have the same effect? Thus heating up some gas to a 10^xxxxxxxxxxxx Temperatur would -in the end, theoretically- cause a black hole beeing created? Another example would be a gigantic magnetic field. Which causes such a large density of energy in a certain place that spacetime would collapse? Now my last and maybe most controverse example:
Is it possible to accelerate an object on such a high speed that his kinetic energy (from the frame of reference of the observer) would cause the object to let spacetime around it collapse into a black hole?
I thought maybe this effect would not appear since time slows time in the frame ofthe accelerated object. Thus making its spacetime more "spread out". And working against the black-hole thing.
And last but not least: if you place the observer in the object which is Now flying veryvery close to the speed of light. wouldn't everything else which hasn't been accelerated look like a black hole? I'm just wondering and would love to get some answers of people who really know this stuff.
 
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  • #2
yeus said:
The Theory of Relativity proposes equivalence between mass and energy. Well... since they are "equal", when mass disturbes the spacetime which causes a forcefield that we call gravity, energy should have the same effect on spacetime. Putting a huge amount of mass in a very little place in space, we have a black hole. After this little explanation my question ;):
Does any Kind of energy have the same effect? Thus heating up some gas to a 10^xxxxxxxxxxxx Temperatur would -in the end, theoretically- cause a black hole beeing created?
Sure. So long as all the matter is confined within a small enough region of space.
Another example would be a gigantic magnetic field. Which causes such a large density of energy in a certain place that spacetime would collapse?
I don't think so in this sense - If a star with a magnetic field collapses into a black hole then the magnetic field becomes detached. The final black hole has no magnetic field.

Now my last and maybe most controverse example:
Is it possible to accelerate an object on such a high speed that his kinetic energy (from the frame of reference of the observer) would cause the object to let spacetime around it collapse into a black hole?
No. It wouldn't.
And last but not least: if you place the observer in the object which is Now flying veryvery close to the speed of light. wouldn't everything else which hasn't been accelerated look like a black hole?
No.

Pete
 
  • #3
yeus said:
And last but not least: if you place the observer in the object which is Now flying veryvery close to the speed of light. wouldn't everything else which hasn't been accelerated look like a black hole? I'm just wondering and would love to get some answers of people who really know this stuff.
This question is addressed in the Usenet physics FAQ, in the section If you go too fast do you become a black hole?:
In fact objects do not have any increased tendency to form black holes due to their extra energy of motion. In a frame of reference stationary with respect to the object, it has only rest mass energy and will not form a black hole unless its rest mass is sufficient. If it is not a black hole in one reference frame, then it cannot be a black hole in any other reference frame.

In part the misunderstanding arises because of the use of the concept of relativistic mass in the equation E = mc^2. Relativistic mass, which increases with the velocity and kinetic energy of an object, cannot be blindly substituted into formulae such as the one that gives the radius for a black hole in terms of its mass. One way to avoid this is to not speak about relativistic mass and think only in terms of invariant rest mass (see Relativity FAQ Does mass change with velocity?).

The statement that "If enough mass is squeezed into a sufficiently small space it will form a black hole" is rather vague. Crudely speaking we would say that if an amount of mass, M is contained within a sphere of radius 2GM/c^2 (the Schwarzschild radius) then it must be a black hole. But this is based on a particular static solution to the Einstein field equations of general relativity, and ignores momentum and angular momentum as well as the dynamics of space-time itself. In general relativity, gravity does not simply couple to mass as it does in the Newtonian theory of gravity. It also couples to momentum and momentum flow; the gravitational field is even coupled to itself. It is actually quite difficult to define the correct conditions for a black hole to form. Hawking and Penrose proved a number of useful singularities theorems about the formation of black holes, and from astrophysics we know that the theorems should apply to sufficiently massive stars when they reach the end of their life and collapse into a small volume.
 
  • #4
yeus said:
Is it possible to accelerate an object on such a high speed that his kinetic energy (from the frame of reference of the observer) would cause the object to let spacetime around it collapse into a black hole?
I thought maybe this effect would not appear since time slows time in the frame ofthe accelerated object. Thus making its spacetime more "spread out". And working against the black-hole thing.
And last but not least: if you place the observer in the object which is Now flying veryvery close to the speed of light. wouldn't everything else which hasn't been accelerated look like a black hole? I'm just wondering and would love to get some answers of people who really know this stuff.


Take a look at the sci.physics.faq on the topic at the following link:

If I go too fast, do I turn into a black hole?

[add] Oh, I see Jesse beat me to it.

The answer to this part of your question is that you do not turn into a black hole if you go too fast.

I think some of your other ideas could work in theory though (they probably won't actually happen, they are more in the line of thought experiments). The formulas you'll see for the critical density of a black hole assume that one is in the center of mass frame of the system - changing the velocity has no effect on whether or not an object becomes a black hole.

Temperature is a form of energy, so if you could somehow heat an object up hot enough AND keep it confined to a small areaa (a rather difficult job, because it will tend to expand), you could probably in principle form a black hole.

Even in the center of mass frame, the condtions for forming a black hole aren't definitely known, but there is a conjecture, known as the "hoop conjecture", which basically says that an object of mass M has to have a circumference of less than 4*pi*GM/c^2 in all directions in order to form a black hole. See for instance

http://scienceworld.wolfram.com/physics/HoopConjecture.html
 
  • #5
Wait, are you people saying that if object posseses so much relavistic mass in one area of space, a black hole will open? This is BS, I have aksed that same questio here before and somebody told me I didnt know what I was talking about, and that it does NOTwork that way.

I don't know what to believe anymore. I though physicsforums was well educated, now it seems that people just say things to seem kewl, or they don't really know what they are talking about.
 
  • #6
eNathan said:
Wait, are you people saying that if object posseses so much relavistic mass in one area of space, a black hole will open? This is BS, I have aksed that same questio here before and somebody told me I didnt know what I was talking about, and that it does NOTwork that way.
Whomever told you that sure didn't know what they were talking about. Its well known that a star can collapse to form a black hole. Don't think of it as something "opening" though. That way of phrasing it gives me the willies! :bugeye:

Pete
 
  • #7
pmb_phy said:
Whomever told you that sure didn't know what they were talking about. Its well known that a star can collapse to form a black hole. Don't think of it as something "opening" though. That way of phrasing it gives me the willies! :bugeye:

Pete

Well, that situation was reffering to relavistic mass, not rest mass. :yuck:
 
  • #8
eNathan said:
Well, that situation was reffering to relavistic mass, not rest mass. :yuck:
That doesn't tell me why he thinks the equivalence principle is not consistent with the scalar theory. Brans and Dicke didn't create a theory without it that's for sure (and for the reasons I mentioned). Remember that the time component of the energy-momentum tensor is the mass of the system. It remains as such even when the system is composed of photons with random directions etc.

Pete
 
  • #9
eNathan said:
Wait, are you people saying that if object posseses so much relavistic mass in one area of space, a black hole will open?
No, everyone who has answered yeus' question so far said the exact opposite, that an increase in relativistic mass without an increase in rest mass will not cause a black hole to form. Read the responses again.
 
  • #10
pmb_phy said:
Whomever told you that sure didn't know what they were talking about. Its well known that a star can collapse to form a black hole.
eNathan was talking only about an increase in relativistic mass due to increased velocity causing a black hole to form, without any change in mass or density in the object's rest frame--you yourself agreed this couldn't happen in your original response to yeus.
 
  • #11
JesseM said:
eNathan was talking only about an increase in relativistic mass due to increased velocity causing a black hole to form, without any change in mass or density in the object's rest frame--you yourself agreed this couldn't happen in your original response to yeus.
It appeared to me that in his follow up to me that he was now referring to the rest frame - I have to admit that it was a bit unclear and I should have asked for clarity. I took his comment "in one area of space" to be quite general and as such apply to the rest frame as well, as in pack the matter into smaller region of space and you get a black hole.

To be honest, this is getting confusing to me.

Pete
 
  • #12
In Gravitation, by MTW, one of the more whimsical results is a theoretical limit on horspower, about 10^54 h.p., after which point black hole formation is unavoidable.
 
  • #13
Imagine a massive object bouncing around inside a perfect box (ie perfectly elastic, and as rigid as relativity allows).

As the speed of the object increases inside the box, does the external gravity field of the box increase?
 
  • #14
PeteSF said:
As the speed of the object increases inside the box, does the external gravity field of the box increase?
Why would the speed of the object increase? Wouldn't that violate conservation of momentum?
 
  • #15
Good point!

Let me think...

Let's say we have two identical objects in two identical boxes. The object in one box is bouncing around much faster than the object in the other box. Does the box with the fast object have more gravity?
 
  • #16
PeteSF said:
Good point!

Let me think...

Let's say we have two identical objects in two identical boxes. The object in one box is bouncing around much faster than the object in the other box. Does the box with the fast object have more gravity?
I'm not sure...on the one hand, between collisions you can always find a frame where the object in the second box is moving slower, but maybe the constant changes in direction after each collision would break the symmetry...pmb_phy said that increasing temperature could cause a black hole to form, and increasing the temperature of a box filled with gas molecules just means increasing the average speed of all the molecules bouncing around between walls. Maybe someone who understands GR better can give a more definite answer.
 
  • #17
I'm trying to approach it using SR and Newtonian gravity by considering conservation of momentum as the box falls toward another massive object... but I'm a bit too lazy to follow it through rigorously.

My unreliable intuition at the moment is that the fast moving object-in-box results in a high average gravity field... and if I knew any GR I might think about the object-box collisions perhaps producing gravity waves with the same result?
 
  • #18
JesseM said:
...pmb_phy said that increasing temperature could cause a black hole to form, ...
Yes. That is correct.
..and increasing the temperature of a box filled with gas molecules just means increasing the average speed of all the molecules bouncing around between walls.
These are not too different things. They are identically the same thing. Increasing the average speed of particles in a box is synonymous with increasing the temperature of the gas and thus increasing the active gravitational mass of the gas. Increase if far enough and the result will be a black hole (theoritically possible. Too difficult in practice of course.

Note that the gravitational field of a single particle will change with the change of the speed of the particle. However that does not mean a black hole will form.

Pete
 
  • #19
Can I please just get a straight answer here.

1. Does relavistic mass create gravity?

2. Is it as easy to calculate as just plugging in the relavistic mass variable into the universal gravity equation?
 
  • #20
eNathan said:
Can I please just get a straight answer here.

1. Does relavistic mass create gravity?

2. Is it as easy to calculate as just plugging in the relavistic mass variable into the universal gravity equation?
No to both, at least not if you're talking about increasing an object's velocity in a uniform way (increasing the velocity of all the particles in a single direction). As long as the object doesn't look any different in its own rest frame, it will be no more likely to form a black hole.
 
  • #21
pmb_phy said:
Increasing the average speed of particles in a box is synonymous with increasing the temperature of the gas and thus increasing the active gravitational mass of the gas. Increase if far enough and the result will be a black hole (theoritically possible. Too difficult in practice of course.
OK, so I assume this is true even if you have a box containing only a single particle? If so, what is the crucial difference between increasing the velocity of a particle bouncing around in a box and increasing the velocity of a particle flying in a straight line through empty space? Is it just because the particle in the box has to accelerate rapidly to change the direction of its velocity vector each time it hits a wall? Is a black hole likely to form only at the moments and locations that the particle hits the wall?
 
  • #22
JesseM said:
OK, so I assume this is true even if you have a box containing only a single particle? If so, what is the crucial difference between increasing the velocity of a particle bouncing around in a box and increasing the velocity of a particle flying in a straight line through empty space? Is it just because the particle in the box has to accelerate rapidly to change the direction of its velocity vector each time it hits a wall? Is a black hole likely to form only at the moments and locations that the particle hits the wall?
The mass-energy will never remain within a small enough region to form a black hole and that is crucial.

eNathan said:
Can I please just get a straight answer here.

1. Does relavistic mass create gravity?
Yes. Relativistic is the source of gravity just as charge is the source of an EM field. Mass in one frame becomes momentum and pressure in another frame just as charge becomes current in another frame.
2. Is it as easy to calculate as just plugging in the relavistic mass variable into the universal gravity equation?
Nothing is easy in GR! :tongue:

If you're thinking about plugging rel-mass into Newton's equation of the gravitational field then the answer is no.

I while back I calculated the gravitational field of a moving body. The calculation is here

http://www.geocities.com/physics_world/gr/moving_body.htm

JesseM said:
No to both, at least not if you're talking about increasing an object's velocity in a uniform way (increasing the velocity of all the particles in a single direction). As long as the object doesn't look any different in its own rest frame, it will be no more likely to form a black hole.
That question was not about whether a body forms a black hole.

Keep in mind that although a single moving body will never become a black hole in a moving frame if its not one it its rest frame it does not follow, and is incorrect to say, that the gravitational field of the body does not increase. The fact is that it does increase.

Pete
 
  • #23
PeteSF said:
Imagine a massive object bouncing around inside a perfect box (ie perfectly elastic, and as rigid as relativity allows).

As the speed of the object increases inside the box, does the external gravity field of the box increase?

If you have a whole bunch of tiny objects bouncing around inside a box (i.e. atoms in an ideal gas), the gravitational field of the box increases when you heat up the gas (make the atoms move faster).

Furthermore, the increase in the mass of the box is equal to the energy it takes to heat up the atoms / c^2.

The best reference I'm aware of is http://arxiv.org/abs/gr-qc/9909014.

One fine point - one might wonder why the increase in the mass of the box is E/c^2 since pressure causes gravity, and the box is under pressure. The answer to this riddle is that (assuming weak fields and small boxes) the intergal of the pressure of the gas over the volume of the box is equal to the intergal of the tension in the container over the volume of the container, so the two terms cancel out. The easiest case to analyze is a spherical container (there's still a few tricks to it, though, I wound up asking some mech E types for how to calculate the stress).
 
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  • #24
whow! ;) I come back a day later and find 21 answers to my question -actually not bad- *g* thanks guys. Well... except for one:

yeus said:
Another example would be a gigantic magnetic field. Which causes such a large density of energy in a certain place that spacetime would collapse?

pmb_phy said:
I don't think so in this sense - If a star with a magnetic field collapses into a black hole then the magnetic field becomes detached. The final black hole has no magnetic field.

i'm not satisfied with this answer, yet. I mean: We don't know what's "in" a black hole, what a black hole consists of. So would just a great density of energy which consists of a field of any kind, which is capable of "conserving" energy theoretically make a black hole possible?

cheers tom
 
  • #25
eNathan said:
Can I please just get a straight answer here.

1. Does relavistic mass create gravity?

Sort of. It's really the stress-energy tensor that creates gravity in GR. This is too complicated to explain on a popular level, so often an oversimplification is used where instead of dealing with tensors (which scare people), the problem is approximated/oversimplified to the single most important component of the tensor, which is the energy density.

<snip>

I got a bit technical in my last response, I'll just say that it is an oversimplification to say that relativistic mass creates gravity, it's not really wrong, just oversimplified.

2. Is it as easy to calculate as just plugging in the relavistic mass variable into the universal gravity equation?

NO! This definitely will not work.
 
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  • #26
pmb_phy said:
Keep in mind that although a single moving body will never become a black hole in a moving frame if its not one it its rest frame it does not follow, and is incorrect to say, that the gravitational field of the body does not increase. The fact is that it does increase.
Not as seen by an observer at rest relative to the center of mass of the object though, which is why it is no more likely to form a black hole--that's what I thought the question was about. If eNathan was asking about the gravity experienced by an observer who sees the object's center of mass moving at velocity v relative to himself, then I assume you're correct that he will experience a greater field the larger v is.
 
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  • #27
JesseM said:
OK, so I assume this is true even if you have a box containing only a single particle? If so, what is the crucial difference between increasing the velocity of a particle bouncing around in a box and increasing the velocity of a particle flying in a straight line through empty space? Is it just because the particle in the box has to accelerate rapidly to change the direction of its velocity vector each time it hits a wall? Is a black hole likely to form only at the moments and locations that the particle hits the wall?
pmb_phy said:
The mass-energy will never remain within a small enough region to form a black hole and that is crucial.
Wait, are you saying that a single particle bouncing around in a box will never form a black hole no matter how great its speed, but multiple particles bouncing around in a box can form a black hole if their average speed is increased high enough? That doesn't make much sense to me--what would the threshold be? Anyway, if a single particle is moving fast enough, why wouldn't the energy of the collision with the wall be high enough to form a black hole?
 
  • #28
pervect said:
One fine point - one might wonder why the increase in the mass of the box is E/c^2 since pressure causes gravity, and the box is under pressure. The answer to this riddle is that (assuming weak fields and small boxes) the intergal of the pressure of the gas over the volume of the box is equal to the intergal of the tension in the container over the volume of the container, so the two terms cancel out. The easiest case to analyze is a spherical container (there's still a few tricks to it, though, I wound up asking some mech E types for how to calculate the stress).
I'm going to have to disagree with you here pervect. While this is true for a box it is not true for a star, which has no containing wall, for which there is non-zero pressure inside.

I mentioned a while back, and many times since, that "mass" is the ratio of momentum over velocity. Use that definition to find the mass of something like a rod which is under stress and moving along the length of the rod. The mass of the rod will then be a function of [itex]\gamma m_0[/itex] as well as the stress (i.e. pressure) in the rod. (see http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm). Notice that [itex]\gamma m_0[/itex] is not the mass in this case since the rod is not isolated from that which causes the stress.

Now apply this kind of thinking for a gas with no walls. It appears to me (a hunch/guess) that to get the result I just add one p/c term for each spatial dimension. This will reduce the equation to

[tex]\rho_{active} = \rho + 3p/c[/tex]

If you take a look at Einstein's field equations in the week field limit then you'll see that this rho on the left, i.e. what I called "mass" above" will be a function of the pressure in the star.

Pete
 
  • #29
JesseM said:
Wait, are you saying that a single particle bouncing around in a box will never form a black hole no matter how great its speed, but multiple particles bouncing around in a box can form a black hole if their average speed is increased high enough? That doesn't make much sense to me--what would the threshold be? Anyway, if a single particle is moving fast enough, why wouldn't the energy of the collision with the wall be high enough to form a black hole?
I never used the term "particle" since that would be troublesome. A particle has zero radius and is therefore a black hole.

No matter how fast that object moves there will always be a frame in which that single object is at rest and does not fit in the given small region. For two particles not moving in the same direction this argument does not apply though so that's why two objects can clash to form a black hole but a single one can't.

Pete
 
  • #30
pmb_phy said:
I never used the term "particle" since that would be troublesome. A particle has zero radius and is therefore a black hole.

No matter how fast that object moves there will always be a frame in which that single object is at rest and does not fit in the given small region. For two particles not moving in the same direction this argument does not apply though so that's why two objects can clash to form a black hole but a single one can't.
Well, doesn't the wall of the container count as an "object"?
 
  • #31
JesseM said:
Well, doesn't the wall of the container count as an "object"?
Yes. And if you want to take that into account then you don't have a single particle system.

Pete
 
  • #32
pmb_phy said:
I'm going to have to disagree with you here pervect. While this is true for a box it is not true for a star, which has no containing wall, for which there is non-zero pressure inside.

I agree that this formula is not true for a star, or any system that has significant _gravitationally induced_ pressures.

That's why I made a point of saying:

(assuming weak fields and small boxes)

When the box is small, though, the tension in the box walls contributes a negative term to the mass of the box which exactly cancels out the positive terms due to the pressure. In short, the pressure terms in the stress-energy tensor T11, T22, and T33 do not contribute to the mass of the box - the intergal over the entire volume is zero. This is why the mass of the box is E/c^2.


I suppose I should have added that specifically the assumption that the box is an isolated system to avoid your "stressed rod" example, but that was inherent in the original problem definition.

A rod under external stress is not a closed system. E/c^2 works only for a closed system

Right. But it does work for a closed system.

Now apply this kind of thinking for a gas with no walls. It appears to me (a hunch/guess) that to get the result I just add one p/c term for each spatial dimension. This will reduce the equation to

[tex]\rho_{active} = \rho + 3p/c[/tex]

If you take a look at Einstein's field equations in the week field limit then you'll see that this rho on the left, i.e. what I called "mass" above" will be a function of the pressure in the star.

Pete

The answer that I have confidence in for finding the mass of a large isolated, static box (static star) with significant gravitational self-energy via a volume intergal I've posted earlier, and is from Wald:

[tex]
M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV
[/tex]

Here Tab is the stress-energy tensor, T is the trace of the stress-energy tensor, gab is the metric tensor, na is the unit future normal to the volume element, and [tex] \mbox{\xi^b}[/tex] is the Killing vector representing the time translation symmetry of the static system, sutiably normalized so that [tex]|\xi^a \xi_a|[/tex] equals unity at infinity.

Now if we assume that [tex]n^a[/tex] and [tex]\xi^b[/tex] both point in the time direction, which is denoted by the superscript or subscript zero, we can simplify this result a lot -- i.e. we assume that the only non-zero comoonents of n and [tex]\xi[/tex] are [tex]n^0[/tex] and [tex]\xi^0[/tex], and we know that the former has the value of 1. Then we write

[tex]
M = 2\int_{\Sigma} (T_{00} - \frac{1}{2}T g_{00} )\xi^0 dV
[/tex]

[correction]
T is just [tex]T^0_0 + T^1_1+T^2_2 + T^3_3 = g^{00} T_{00}+ g^{11} T_{11} + g^{22} T_{22} + g^{33} T_{33}[/tex]

and we wind up with

[tex]
M = \int_{\Sigma} ((2-g_{00}g^{00}) T_{00} - (g^{11} T_{11}+g^{22} T_{22}+g^{33} T_{33})g_{00}) \xi^0 dV
[/tex]

Unfortunately, last I checked, [tex]\xi[/tex] which is a nice unit vector outside the star isn't necessarily unity inside the star :-(, so you have to solve Killings equation to find [tex]xi^0[/tex] :-(

[add]
I suppose I should add that when g_00 = 1 and g_11 = g_22 = g_33 =-1 (flat spacetime with a +--- metric), which has the timelike Killing vector [tex]\xi^0 = 1[/tex] the above formula reduces to

[tex]
M = \int_{Sigma} (T_{00}+T_{11}+T_{22}+T_{33})dV
[/tex]

which is the same result given in Carlip's paper
http://arxiv.org/abs/gr-qc/9909014

for weak fields (nearly flat space-time).
 
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  • #33
pervect said:
When the box is small, though, the tension in the box walls contributes a negative term to the mass of the box which exactly cancels out the positive terms due to the pressure.
I think you missed my point. You're referring to the entire gas-box system whereas I am referring only to the gas. Also it makes no difference as to the size of the box as to whether F/A is a tension or a pressure term.
In short, the pressure terms in the stress-energy tensor T11, T22, and T33 do not contribute to the mass of the box - the intergal over the entire volume is zero. This is why the mass of the box is E/c^2.
The mass of the box is not E/c^2. It is the total mass of the mass-box system that is E/c^2.
I suppose I should have added that specifically the assumption that the box is an isolated system to avoid your "stressed rod" example, but that was inherent in the original problem definition.
Recall the statement that I was responding to "one might wonder why the increase in the mass of the box is E/c^2 since pressure causes gravity, and the box is under pressure."
Right. But it does work for a closed system.
your point being?
The answer that I have confidence in for finding the mass of a large isolated, static box (static star) with significant gravitational self-energy via a volume intergal I've posted earlier, and is from Wald...
The purpose of my weak field example was to help the layman since it is easier to follow. It more readily shows the relationship of the inertial mass of the gas to the active gravitational mass of the gas. If you have Schutz's new book then look up his derivation as to why the inertial mass of a gas (not box-gas, just the gas) is a function of the pressure of the gas.

(I've been through the rest in the past too manyh times to bother re-addressing it again here - plus I'm getting lazy :biggrin: )

Pete
 
  • #34
pmb_phy said:
I think you missed my point. You're referring to the entire gas-box system whereas I am referring only to the gas.

If you measure the "mass of the gas" by measuring the mass of the box empty, and the mass of the box full (which is both a logical means of measuring the mass of the gas AND the means specified by the original poster), you must take account the contribution of the stresses in the box.

The box was unstressed in the first measurement, and stressed in the second. This makes a difference to the calculation.

When you do this procedure, you get E/c^2 for the mass of the gas, at least for small boxes, which leads us to the second point.

Also it makes no difference as to the size of the box as to whether F/A is a tension or a pressure term.

I'm not even sure what you are trying to say here. (After I can figure out what you are saying, I can start to think about whether or not I agree with it and what your basis for saying it is).

My second point is this - gravity is nonlinear, and with a large enough box this is an important effect. This means that the mass of a gravitationally bound system is not equal to the sum of the masses of its parts. It's also, unfortunately, a rather advanced topic.

With a small enough box, the non-linearity is unimportant. Since this is by far the easiest case to analyze, and interesting in its own right, I think it's the one to address. This is the case that I calculated.

I'd also like to point out that even in our sun, the contribution of the pressure terms to the mass is negligible, so one can have a pretty darn big box without the non-linear effects of gravitation becoming imporant.

Clearly, the approximations I made are good when the box is small enough that the pressure at the center of the box is not significantly different than the pressure at the edge of the box because of the box's self-gravity.

In practice, the approximations are even better than that, because the pressure terms are extremely small anyway, thus a small error in a small term becomes an extremely small error in the final result. The magnitude of the term in standard units will be 3*pressure*volume/c^2, and c^2 is a very big number.
 

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