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HerrCaZZini
2009-Jun-26, 08:03 AM
Photons have no rest mass, but have a relativistic mass.
Energy = hbar*omega
and Massa = Energy / cē
But it has no volume; it's a point.
Then its radius is smaller then the schwarzschild radius.
So its a black hole?

And a photon can't jump off another blackhole because the speed of light is the limit of velocity. But they can sit on eachother i've been told.

Can anybody make this a bit clear for me?

grant hutchison
2009-Jun-26, 10:56 AM
Photons have no rest mass, but have a relativistic mass.We're into the overlap between quantum mechanics and GR, where current theory doesn't work. But it's worth pointing out that rest mass is what determines black hole status: no rest mass, no black hole.

Grant Hutchison

astromark
2009-Jun-26, 11:06 AM
Can anybody make this a bit clear for me?.... Yes I think I can.:)
Are photons black holes? .... No.

Lets try and be really sencable about this,.
All that is, is. While all that is not, is not. This fits the is not .
No you do not seem convinced? then lets try, The range of all mater.
Elements and compounds alike all seem to be some place on the spectrum. What is the spectrum ? How can we test it ? Visable light is found to occupie a small fraction of this 'spectrum thing... :) Omitted energy as photons recieved. Where is this going ? Its not a 'Black Hole'... I would suggest a quick search in Wickapedia will reveal some truths of what a Photon might be. It is not a black hole. Does a photon have gravity? Do you find them clumping together... No, you seem to be of on a tangent of some ill placed logic. Where did you get this ? put it back there.:)
A Black Hole is the quite inappropriate name of a collapsed super giant star. That is often found to be at the hart of most if not all galaxies. A matter magnet. A area of massive mater density. not a ball of photons., Where gravity has risen to become the major force... and welcome.

HerrCaZZini
2009-Jun-26, 12:31 PM
It was an idea of my own, because i'm following to internet courses,one about astronomy and one about particle physics. Apparently i forgot that relativity and quantum theory don't stick together. I'll need to wait few years before the string theorists have done there job.

Thanks


Can anybody make this a bit clear for me?.... Yes I think I can.:)
Are photons black holes? .... No.

Lets try and be really sencable about this,.
All that is, is. While all that is not, is not. This fits the is not .
No you do not seem convinced? then lets try, The range of all mater.
Elements and compounds alike all seem to be some place on the spectrum. What is the spectrum ? How can we test it ? Visable light is found to occupie a small fraction of this 'spectrum thing... :) Omitted energy as photons recieved. Where is this going ? Its not a 'Black Hole'... I would suggest a quick search in Wickapedia will reveal some truths of what a Photon might be. It is not a black hole. Does a photon have gravity? Do you find them clumping together... No, you seem to be of on a tangent of some ill placed logic. Where did you get this ? put it back there.:)
A Black Hole is the quite inappropriate name of a collapsed super giant star. That is often found to be at the hart of most if not all galaxies. A matter magnet. A area of massive mater density. not a ball of photons., Where gravity has risen to become the major force... and welcome.

AntiGravity
2009-Jun-30, 05:37 PM
Really fun idea though- never noticed the similarities before

EnigmaPower
2009-Jun-30, 05:44 PM
If photons were black holes how could there be light?

nauthiz
2009-Jun-30, 06:12 PM
We're into the overlap between quantum mechanics and GR, where current theory doesn't work. But it's worth pointing out that rest mass is what determines black hole status: no rest mass, no black hole.

We could, though, modify the question to be about other elementary particles, many of which have rest mass and are still considered point particles - electrons, for example.

That said, we can't say for sure that their masses really are concentrated in a mathematical point. We call them point particles because that's how they're treated in the Standard Model, but that's just a treatment we use because it works. String theory treats the fundamental particles as non-point masses, and folks seem to think that treatment also works fairly well. Unfortunately we don't have the ability to experimentally distinguish between the two, because at the scales we're currently able to detect there's no way to distinguish between tiny but non-point masses of the size predicted by string theory and bonafide point masses.

Either way it's certainly possible for light to get out of an electron even though we call it a point mass.

iquestor
2009-Jun-30, 06:26 PM
I dont see how a massless body could have relativistic mass, but no rest mass. I know that relativistic mass depends on the total energy of the system, but there would seem to be impossible. Could someone who is educated in this field explain if possible?



Interesting topic. :)

DrRocket
2009-Jun-30, 06:53 PM
We're into the overlap between quantum mechanics and GR, where current theory doesn't work. But it's worth pointing out that rest mass is what determines black hole status: no rest mass, no black hole.

Grant Hutchison

Why do you say that ?

John Archibald Wheeler studied gravitational structures generated solely by electromagnetic and gravitational waves, called geons. In his book Geons, Back HOles & Quantum Foam he notes that one possibility for a geon composed of gravitational waves is to collapse into a black hole -- mass without mass forming a black hole in his terminology.

That would be an example of a black hole forming from zero rest mass.

Note that these are theoretical studies and there is no experimental evidence for geons.

EnigmaPower
2009-Jun-30, 06:59 PM
Why do you say that ?

John Archibald Wheeler studied gravitational structures generated solely by electromagnetic and gravitational waves, called geons. In his book Geons, Back HOles & Quantum Foam he notes that one possibility for a geon composed of gravitational waves is to collapse into a black hole -- mass without mass forming a black hole in his terminology.

That would be an example of a black hole forming from zero rest mass.

Note that these are theoretical studies and there is no experimental evidence for geons.
Isn't a black hole formed from the collapse of a star? Without that an object could be black hole like but not a black hole. I thought that book was an autobiography not a scientific text.

DrRocket
2009-Jun-30, 08:11 PM
Isn't a black hole formed from the collapse of a star? Without that an object could be black hole like but not a black hole. I thought that book was an autobiography not a scientific text.

The book is an autobiography of a physicist, and therefore talks about his work. I found it a convenient reference. You can also "google" geons, but that article does not talk about the collapse into black holes, while Wheeler's book does. If you would like the full treatment of the theory of black holes then perhaps you might prefer Chandrasekhar's The Mathematical Theory of Black Holes or Gravitation by Misner, Thorne and Wheeler, but while they are quite good treatments of the basic theory they don't bring out the importance of the "energy" side of mass-energy as forcefully as does the theoretical construction of a black hole from gravity itself as with a geon.

A black hole is a region of space in which the curvature of space-time is such that neither light nor anything else can go beyond a surface called the "event horizon". It is caused by a concentration of mass-energy sufficient to create that extreme curvature. The most common source of the requisite mass-energy density is a collapsed star, but that is not the only possibility. It is, for instance, theoretically possible to create a "mini" black hole from the collision of sub-atomic particles of sufficient energy --- as may be possible in the LHC. Such small black holes are expected to evaporate very quickly due to Hawking radiation.

The point is that all forms of mass-energy go into the stress-energy tensor (also called the momentum tensor) that determines the curvature, not just rest mass.

EnigmaPower
2009-Jun-30, 08:15 PM
A black hole is a region of space in which the curvature of space-time is such that neither light nor anything else can go beyond a surface called the "event horizon". It is caused by a concentration of mass-energy sufficient to create that extreme curvature. The most common source of the requisite mass-energy density is a collapsed star, but that is not the only possibility. It is, for instance, theoretically possible to create a "mini" black hole from the collision of sub-atomic particles of sufficient energy --- as may be possible in the LHC. Such small black holes are expected to evaporate very quickly due to Hawking radiation.

Which are formed in space by the collapse of stars. To say otherwise is mere speculation.

grant hutchison
2009-Jun-30, 08:55 PM
Why do you say that ?Well, I say it because an event horizon is a property of global spacetime, not frame-dependent. Otherwise I could, for instance, turn the Earth into a black hole simply by choosing to observe it from a frame in which its "relativistic mass" was high enough to make it into a black hole. It would be a black hole in one frame and not in another.


John Archibald Wheeler studied gravitational structures generated solely by electromagnetic and gravitational waves, called geons. In his book Geons, Back HOles & Quantum Foam he notes that one possibility for a geon composed of gravitational waves is to collapse into a black hole -- mass without mass forming a black hole in his terminology.Sounds like a kugelblitz, by another name.
I believe the difference is that in the case of a kugelblitz the energy (gravitational waves, photons) must be focussed from different directions to produce the required event horizon. So there is no frame, in any state of motion, from which an observer can measure a mass-energy concentration that is less than what is required for a black hole. An observer who is stationary relative to the centre of momentum of all the incoming radiation sees the minimal mass-energy concentration (the "rest mass" of that converging system of radiation), and that's sufficient to form a black hole.
But I certainly can't offer you a detailed GR justification for that. Might be wrong.

Grant Hutchison

pzkpfw
2009-Jun-30, 08:58 PM
Which are formed in space by the collapse of stars. To say otherwise is mere speculation.

Collapse of stars is not currently thought to be the only way black holes could exist in our Universe.

Black holes themselves are not so far from when they were "mere speculation".

I believe you are being just a little bit nit-picky again.

mugaliens
2009-Jun-30, 08:59 PM
We're into the overlap between quantum mechanics and GR, where current theory doesn't work. But it's worth pointing out that rest mass is what determines black hole status: no rest mass, no black hole.

Grant Hutchison

To piggyback, they don't have a relativistic mass, per se', but a relativistic momentum commensurate with their energy which translates into the potential to curve spacetime.

EnigmaPower
2009-Jun-30, 09:02 PM
Collapse of stars is not currently thought to be the only way black holes could exist in our Universe.

Black holes themselves are not so far from when they were "mere speculation".

I believe you are being just a little bit nit-picky again.

Nope. I said black hole like objects. As far as I have read, all the other ways are theoretical and never have occurred in practice.

DrRocket
2009-Jun-30, 10:35 PM
Which are formed in space by the collapse of stars. To say otherwise is mere speculation.

Black holes are justified on the basis of the general theory of relativity. That certainly does not require them to be formed from collapsed stars.

Nobody has directly seen any black hole, although there is evidence pointing to effects that would accompany one. So to say that black holes are formed by collapsed stars and only by collapsed stars is simply more speculation.

DrRocket
2009-Jun-30, 10:41 PM
Well, I say it because an event horizon is a property of global spacetime, not frame-dependent. Otherwise I could, for instance, turn the Earth into a black hole simply by choosing to observe it from a frame in which its "relativistic mass" was high enough to make it into a black hole. It would be a black hole in one frame and not in another.


Grant Hutchison

Why do you say that black holes are an invariant of space-time? Curvature and "gravity" are not. A black hole is nothing but a manifestation of extreme local curvature. An observer in an "accelerating reference frame" sees curvature that another observer may not see for instance, that is one result of the principle of equivalence.

You may also have problems picking your "reference frame" since GR does not admit global reference frames. That is one effect of curvature.

In any case the stress-energy tensor that determines curvature includes far more than just rest mass. It includes energy, stress, .....

grant hutchison
2009-Jun-30, 11:10 PM
Why do you say that black holes are an invariant of space-time?I say that the black hole's event horizon is a property of global spacetime because it is defined as a surface beyond which light escapes to infinity, and inside which it does not:
... the event horizon is a global property; it depends on constructing null geodesics to determine whether they reach infinity or not, which cannot be judged from the local behaviour.

Black Holes: An Introduction Raine & Thomas


The horizon only marks the boundary between where photons can get out and where they cannot, but that is a property of the large scale structure of the spacetime, not something that can be sensed locally.

Gravity: From the Ground Up Schutz
That seems to be a common enough textbook definition.
Something that arises from a global web of null geodesics doesn't seem to me like something that can be switched on and off by a bit of local manoeuvring.


A black hole is nothing but a manifestation of extreme local curvature. An observer in an "accelerating reference frame" sees curvature that another observer may not see for instance, that is one result of the principle of equivalence.Sure. And in that case we have a Rindler horizon, which is relevant to our observer, but not to the Universe at large.
But are you suggesting that one observer can experience (say) a neutron star, while another sees a black hole? Surely such observers would be messing with a global property, since the light from the neutron star is escaping to infinity, while the light from the black hole is not.

Grant Hutchison

DrRocket
2009-Jul-01, 12:17 AM
I say that the black hole's event horizon is a property of global spacetime because it is defined as a surface beyond which light escapes to infinity, and inside which it does not:That seems to be a common enough textbook definition.
Something that arises from a global web of null geodesics doesn't seem to me like something that can be switched on and off by a bit of local manoeuvring.

But gravity is indistinguishable from acceleration, and acceleration is clearly something that is subject to "a bit of local maneuvering".

This textbook definition appears to make the assumption that you are working in a fixed reference frame. The surface will change with length contraction along the direction of motion. "Infinity" here seems not very well defined as well. The usual definition is that nothing that crosses the event horizon from the outside ever crosses over again -- a hypersurface of no return.

Null geodesics themselves are dependent on the observer. They are determined by the curvature which is in turn determined by the stress-energy tensor which is dependent on the observer.


Sure. And in that case we have a Rindler horizon, which is relevant to our observer, but not to the Universe at large.

When you throw in the notion of strong curvature the very assumptions that go into defining what is meant by "the Universe at large" go out the window. Those assumptions are that the universe is homogeneous and isotropic, which it most clearly is not on small scales. In particular when you are talking about highly curved regions of space-time that assumption is strikingly invalid.

If I am missing something here I would like to know what it is.



But are you suggesting that one observer can experience (say) a neutron star, while another sees a black hole? Surely such observers would be messing with a global property, since the light from the neutron star is escaping to infinity, while the light from the black hole is not.

Grant Hutchison

There is no meaning to "infinity" in this context. I had not thought about this question much, but I think it does mean that one oberver might see a neutron star and another a black hole. That does seem a bit weird, but stranger things can happen in GR -- black holes are pretty weird themselves. It is possible that there are effects due to length contraction that make this not happen -- I dunno. Length contraction would certainly eliminate the solutions that depend on spherical symmetry in most cases.

But the main thing is that when you throw in high curvature, inhomogeneity, and lack of isotropy it is not very clear what a "global property" really is. Perhaps the topology might be thought of as global, but the geometry will vary with the curvature which varies with mass-energy distribution which is not a global property invariant of the observer.

publius
2009-Jul-01, 12:42 AM
A black hole is an invariant property. That is the state of gravitational collapse to a singularity is an invariant property. It will either collapse or it won't. If the mass doesn't collapse in the rest frame, it won't collapse in any frame.

Imagine something whizzing by the earth at 99.999999999999999% c. The earth is very massive, and presents a quite a spike in the components of the stress energy tensor. But from the POV of that frame, there are gravitomagnetic and momentum terms that act against the "Coulomb" part of the field, and that observer concludes that the earth will not collapse into a singularity.

In the rest frame, the horizon is a sphere, but to a moving frame, we'd expect in to suffer a Lorentz contraction in the direction of motion and no longer be spherical (the real solution is a bit more complicated since we have real gravity but it's close). So while the moving earth might be smaller than the rest frame Schwarzschild radius in that direction, it's still larger than the transformed radius itself.

-Richard

publius
2009-Jul-01, 03:47 AM
Now, on the really subtle part that makes my head hurt, and this is this business of what event horizons are, and Rindler vs Schwarschild horizons.

But before we get into that -- Equivalence Principle and "gravity being indistinguishable from acceleration". Well, depends on what you mean by gravity. The modern view is that gravity is the (invariant) Riemann curvature of space-time. And note the curvature correlates with the derivatives of the Newtonian g-field, not the field itself. IOW, gravity is in the tides. Now, even that gets subtle. The Rindler family experiences gradients in proper acceleration to remain mutually stationary (defining the "time-like congruence" which defines spatial hyperslices where the family can consider themselves stationary). But free-falling observers experience no tides in the absence of curvature.

A horizon is null surface, ds = 0 upon it. Being null, it means it's a type of light like boundary. Consider the Rindler Horizon. To inertial observers in flat space-time, the Rindler Horizon is just a plane moving at light speed chasing the Rindler family of accelerating observers. To the Rindler family, that horizon is stationary, but not all agree it is null.

To the inertial observers, any signal sent behind the horizon will never reach the Rindler family because it will never catch up. But other than that, it's unremarkable, just a light-like surface defined as z - ct = 0 (for acceleration in the z direction). That light-like surface is unique for each Rindler family. So it does have a sort of global footing, although otherwise unremarkable.

From the standpoint of the inertial observers, light does go to space-like infinity in time-like infinity, but nonetheless will never catch the Rindler family if launched behind their horizon.

From the Equivalence Principle, we would expect the Schwarzschild horizon to share many similiarities with the Rindler Horizon. And it does. A free faller would see that horizon as a light like surface in some appropriate coordinates. But the curvature, the real gravity, does change the global picture. And that free faller hits the singularity and his coordinates come to and end for him at some point. He has no time-like infinity.

And if we plot the null geodesics inside the horizon, we'll see that they also turn around and hit the singularity.

The Schwarzschild vacuum is a invariant geometric structure (although a very idealized one), which has various properties. The point of no return character of the Schwarzschild horizon does have invariant meaning. On one side all world lines terminate at the singularity. No escape. On the other, world lines can escape

-Richard

grant hutchison
2009-Jul-01, 08:30 AM
But the main thing is that when you throw in high curvature, inhomogeneity, and lack of isotropy it is not very clear what a "global property" really is. Perhaps the topology might be thought of as global, but the geometry will vary with the curvature which varies with mass-energy distribution which is not a global property invariant of the observer.Well, hopefully Richard has provided enough of the GR background to convince you that I'm not making this up. :)
To my understanding, this is a topological issue. Spacetime is threaded by the null surface of the black hole event horizon (or not threaded, if no black hole exists). Coordinatize as you will, you can't get rid of that null surface if it exists, or produce such a surface if one doesn't exist. All observers have to deal with the presence (or absence) of the black hole.

Grant Hutchison

DrRocket
2009-Jul-01, 07:17 PM
A black hole is an invariant property. That is the state of gravitational collapse to a singularity is an invariant property. It will either collapse or it won't. If the mass doesn't collapse in the rest frame, it won't collapse in any frame.

Imagine something whizzing by the earth at 99.999999999999999% c. The earth is very massive, and presents a quite a spike in the components of the stress energy tensor. But from the POV of that frame, there are gravitomagnetic and momentum terms that act against the "Coulomb" part of the field, and that observer concludes that the earth will not collapse into a singularity.

In the rest frame, the horizon is a sphere, but to a moving frame, we'd expect in to suffer a Lorentz contraction in the direction of motion and no longer be spherical (the real solution is a bit more complicated since we have real gravity but it's close). So while the moving earth might be smaller than the rest frame Schwarzschild radius in that direction, it's still larger than the transformed radius itself.

-Richard

OK I think I see basically where you are heading. But a couple of questions.

First, the singularity of a black hole is not the essential feature. If you go so far as to adopt Einstein-Cartan theory then there is no singularity for instance. But the point is that the defining feature is an event horizon.

So, given that the stress-energy tensor most certainly is dependent on the observer, and it includes quite a bit more than rest mass, how does one read off from that tensor, or equivalently the curvature tensor, than there does or does not exist an event horizon ?

If the existence of the horizon is an invariant, then the off-diagonal terms in the tensor must overcome the diagonal terms that include the relativistic mass.

This clearly has nothing whatever to do with any real or imagined "large-scale structure" of space-time, or the stress-energy pseudotensor that includes the self-interaction of gravity with itself.

publius
2009-Jul-01, 08:11 PM
Ah, but the stress-energy tensor is invariant! That's the meaning of tensor according to the accepted language. A psuedo-tensor, like the
Landau-Lifshi-tz (hyphen to avoid rather stupid profanity filter behavior) "gravitational energy" thing is one that is not invariant under the rules.

The components of that tensor do vary with coordinates, but the tensor itself, as a whole, as a geometric entity is invariant. The EFE equates the Einstein Tensor (operation involving the Ricci curvature) with the stress-energy tensor at an event. That is an invariant statement.

The components of the tensor, the coordinate dependent things, are the energy, energy current, momentum, and currents of momentum. That all shifts around from frame to frame, but the tensor itself is invariant.

The best way to think of the stress-energy tensor is as follows. It is usually denoted by the symbol T, with components T_uv. The component T_uv is the flux of the u component of momentum in the v direction, with the understanding that the 0th, or time-like components are intepreted a little differently. The 0th component of momentum is "energy" and the flux in the
0th direction is "density".

Thus T_00 is the flux of 0th component of momentum in the 0th direction. Which translates to energy density. T_0i ( i = 1, 2, 3) is the flux of the 0th component in the three spatial directions. Thus T_0i is the (mass)energy current.

A little reflection will convince us that T must be symmetric. So, T_i0, the momentum densities (the density of the x, y, and z components of momentum if using cartesian spatial coordinates) must equal T_i0.

T_ij is the regular 3-stress, or pressure, tensor. We normally think of pressure as a scalar, but it turns out, in the general case, in must be a rank-2 tensor, as pressure, at it's root, is all about momentum flux, and the flux of a vector requires a rank-2 tensor to specify. Again using cartesians, T_23, or T_yz, would be the flux of the yth component of momentum in the z direction. T_ii, the diagonals, correspond to the familiar scalar pressure if all are equal. If not, I think the scalar pressure can be taken to be 1/3 the trace, or sum of the diagonals.

At any rate, the stress-energy tensor of GR merges the space and time parts all together, putting energy and momentum into one rank-2 4-tensor.

You can see that T_0v, the top row, looks like the 4-momentum vector. And it is, although as a "density" not integrated total.

Indeed, for a point particle, or something close to it, the stress energy tensor would be simply an "outer product" or tensor product "PV" where P is the 4-momentum and V is the 4-velocity (with the appropriate factors of c and all that to get the units right).

Now, that would seem to imply there would be only 4 degrees of freedom. However, there are more than that. For a non-interacting dust, the pressure in the rest frame is not a function of density. However, interacting matter has an equation of state, where the pressure depends on the density, and this gives us all 10 degrees of freedom in the rank-2 tensor.

-Richard

DrRocket
2009-Jul-02, 01:16 AM
Ah, but the stress-energy tensor is invariant! That's the meaning of tensor according to the accepted language. A psuedo-tensor, like the
Landau-Lifshi-tz (hyphen to avoid rather stupid profanity filter behavior) "gravitational energy" thing is one that is not invariant under the rules.



I know that a tensor in the language of GR is invariant, more or less by definition and so the stress-energy tensor must be. The question is how one reads off the existence of an even horizon from that tensor.

Or maybe it is equivalent to seeing why the stress energy tensor is indeed an tensor and therefore invariant since it is not intuitively clear from physical considerations that it should be -- the mass-energy and momentum terms being so clearly dependent on the observer. The tensorial property would show that there must be a way, in principle, to read off the existence of an event horizon, but unfortunately it does not make it clear how to do that. It may well be that there is no known way to actually do this, since the black hole calculations seem to rely on rather special coordinate systems and transformations.

DrRocket
2009-Jul-02, 01:20 AM
A little reflection will convince us that T must be symmetric. So, T_i0, the momentum densities (the density of the x, y, and z components of momentum if using cartesian spatial coordinates) must equal T_i0.

Richard

A little reflection ?

This is not at all obvious. In fact in the Einstein-Cartan theory it is not symmetric. The symmetry is a direct consequence of the assumption that the geometry is torsion-free. The non-symmetry and the abandonment of the assumption (it is purely an assumption) that the torsion tensor is zero is what distinguishes the two theories.

publius
2009-Jul-02, 02:22 AM
A little reflection ?

This is not at all obvious.

So you won't let me get away with that. :)

One "little reflection" is to say that T must look like some PV, the tensor product of two vectors. Such a product is obviously symmetric, although that doesn't include the equation of state effects that give the general T more degrees of freedom that a simple PV (if interested, look up the derivation and application of the regular 3-stress tensor, or the Maxwell Stress Tensor, the EM version -- for the case of a non-interacting dust fluid you can easily the stress tensor takes the form of the tensor product m*vv, where v is the point velocity vector).

Second, which admittedly is more than a little reflection, is that conservation of angular momentum requires a symmetric T. The two statements are equivalent.

In E-C, we add a spin tensor to the mix. Conservation of angular momentum there inolves the sum of T and the spin tensor, which allows "spin orbit" coupling between the two.

That does allow that the metric tensor itself can become non-symmetric, which has "weird" implications.

-Richard

DrRocket
2009-Jul-02, 02:57 AM
So you won't let me get away with that. :)

One "little reflection" is to say that T must look like some PV, the tensor product of two vectors. Such a product is obviously symmetric, although that doesn't include the equation of state effects that give the general T more degrees of freedom that a simple PV (if interested, look up the derivation and application of the regular 3-stress tensor, or the Maxwell Stress Tensor, the EM version -- for the case of a non-interacting dust fluid you can easily the stress tensor takes the form of the tensor product m*vv, where v is the point velocity vector).

Second, which admittedly is more than a little reflection, is that conservation of angular momentum requires a symmetric T. The two statements are equivalent.

In E-C, we add a spin tensor to the mix. Conservation of angular momentum there inolves the sum of T and the spin tensor, which allows "spin orbit" coupling between the two.

That does allow that the metric tensor itself can become non-symmetric, which has "weird" implications.

-Richard

It is not that you can't get away with that. It is that it is simply an assumption. It is not derivable from first principles, or else there would be problems with Einstein-Cartan, and there are not.

The symmetry of the stress-energy tensor in GR comes from the same sort of heuristic argument that is used to justify the symmetry of the stress tensor in continuum mechanics. There the underlying assumption is that one is not admiting "couples" in the stress tensor and hence to prevent rotation you have a symmetric tensor. That basic argument is used in Misner, Thorne and Wheeler to justify a symmetric stress-energy tensor.

There is nothing wrong with the assumption, at least nothing clearly wrong. As you say it avoids some weirdness. But it also results in the singularities that are associated with black holes and the big bang. So perhaps the trade-off is in where you are willing to live with weirdness. I dunno.

It is useful to recognize that it is, however, an assumption and not an inevitable consequence of a piece of physics. If it were due to inevitable physics then E-C would not be a viable theory, but it is. As far as I can tell the reason that E-C is not better known is that it is quite a bit more involved with respect to the mathematics and it is indistinguishable from GR on the basis of experiments that are feasible with current measurement technology. But it serves to point out that there are assumptions in GR that are assumptions and nothing more, and that the singularities that occur in GR may well be a consequence of some of those assumptions. It also serves to point that the singularities in GR that are often attributed to a break-down in the theory requiring a unification with quantum mechanics for resolution may have their origin in other issues. It gives me some incentive to learn more about both GR and E-C.

DrRocket
2009-Jul-02, 03:06 AM
In E-C, we add a spin tensor to the mix. Conservation of angular momentum there inolves the sum of T and the spin tensor, which allows "spin orbit" coupling between the two.

That does allow that the metric tensor itself can become non-symmetric, which has "weird" implications.

-Richard

From the point-of-view of a mathematician, you don't add the spin tensor to the mix, you simply don't rule it out.

There is a fundamental theorem in Riemannian geometry ("Riemannian" here includes what is often called pseudo-Riemannian which allows non-degenerate metrics and not just postive-definite ones), which says that given a metric there is a unique connection (the Levi-Civita connection) that preserves inner products and is torsion-free. The connection is the thing that gives you covariant differentiation and parallel transport and ties everything together. It is that uniqueness of the connection (if "connection" is foreign then think of Christoffel symbols) that reduces everything to calculations based on the metric. The assumption that things are torsion free is what lets you invoke all of the machinery of Riemannian geometry, and that is some important and powerful machinery.

I am not an expert in this stuff, but I am trying to learn it.

Radical Yellow Duck
2009-Jul-06, 08:11 PM
Perhaps I am dipping my foot into a pond much deeper than I expect but the question and its answer seem kind of clear to me. I make the following assumptions.

1 That gravity is a function or attribute of matter.
2 That one of the claims of the theory of relativity (special or general, I am not sure which) is that as the speed of an object approaches the speed of light, ( c ) it mass increases. And at the speed of light, its mass would be infinite.
3 Photons travel at the speed of light in a vacuum.
4 Photons have no measurable mass.
5 Photons are affected by gravity, but that gravity is not affected by photons.
6 One definition of a black hole is a super massive collapsed star, in which the escape velocity from the star exceeds the speed of light. A second definition would be ordinary matter compressed to a size where the escape velocity would exceed the speed of light. For a mass the size of the sun this would be about 3 km and for a mass the size of the earth about 9mm. In both definitions it is an amount of matter compressed into a small volume.
7 It is my understanding that once an object obtains the density of a black hole, no further information from the object can be gathered except for its mass, charge and angular momentum.
8 Photons can display properties outside of mass, charge and angular momentum.

Given the above assumptions, it would seem that photons cannot be black holes.

Black holes have mass, photons do not.

Black holes are made of matter. Photons are not.

Photons can travel at the speed of light. Matter cannot unless pushed by infinite force.

Infinite force is a quantity of force that does not exist in a finite universe.

Ergo, photons are not black holes.

Perhaps my assumptions or definitions are incorrect. But if they are correct then my conclusion seems to stand.

WayneFrancis
2009-Jul-08, 05:26 AM
Perhaps I am dipping my foot into a pond much deeper than I expect but the question and its answer seem kind of clear to me. I make the following assumptions.

1 That gravity is a function or attribute of matter.


Actually gravity is a function of E=Mc2
Photons have no rest mass but do add to the mass of an object. IE a box with photons in it has a higher mass then an empty box.



2 That one of the claims of the theory of relativity (special or general, I am not sure which) is that as the speed of an object approaches the speed of light, ( c ) it mass increases. And at the speed of light, its mass would be infinite.


Correct.



3 Photons travel at the speed of light in a vacuum.


More accurately is Photons travel at c in a vacuum and speed is 299,792,458m/s

Photons always travel at the speed of light. The speed of light varies depending on the medium the photons travel through.



4 Photons have no measurable mass.


Photons have no rest mass. They have mass via the equation E=Mc2



5 Photons are affected by gravity, but that gravity is not affected by photons.


Photons do have "gravity" it is just a very small amount and dependent on their wave length.



6 One definition of a black hole is a super massive collapsed star, in which the escape velocity from the star exceeds the speed of light. A second definition would be ordinary matter compressed to a size where the escape velocity would exceed the speed of light. For a mass the size of the sun this would be about 3 km and for a mass the size of the earth about 9mm. In both definitions it is an amount of matter compressed into a small volume.


A black hole is any area which has enough mass within the Schwarzschild radius of that area to cause an escape velocity of ≥ c. The LHC will produce energy levels high enough to cause black holes, the Schwarzschild radius of them will be so small that even if they didn't evaporate away they would pose no danger as they would not encounter any other mass to increase in size all that often.



7 It is my understanding that once an object obtains the density of a black hole, no further information from the object can be gathered except for its mass, charge and angular momentum.


A black hole has no hair.



8 Photons can display properties outside of mass, charge and angular momentum.


True.




Given the above assumptions, it would seem that photons cannot be black holes.

Black holes have mass, photons do not.


Like I stated above not exactly true.



Black holes are made of matter. Photons are not.


You can make a black hole from pure energy.



Photons can travel at the speed of light. Matter cannot unless pushed by infinite force.


Technically this doesn't have any bearing on a black hole unless you get into advanced theories of black holes and what happens at the event horizon.



Infinite force is a quantity of force that does not exist in a finite universe.


If the universe is finite or infinite is a debatable subject but doesn't matter, no pun intended, as you couldn't get matter to c in an infinite universe either.



Ergo, photons are not black holes.

Perhaps my assumptions or definitions are incorrect. But if they are correct then my conclusion seems to stand.

You got me thinking. What is the "size" of a photon? The quantum packet has to have some "size" because it has a wave length. Even gamma rays have a frequency that is much bigger then the Schwarzschild radius. I'd suspect many orders of magnitude larger too.

DrRocket
2009-Jul-08, 05:16 PM
A black hole has no hair.


This might appear to resolve the black hole information paradox in favor of the original position of Hawking that all information entering a black hole is lost, and will be destroyed when a black hole eventually evaporates due to Hawking radiation.

But it is not that simple. Hawking has conceded a bet, and reversed his position. Not everyone involved agrees.

The concession is based on notions from string theory. You can find a discussion in Leonard Susskind's book The Black Hole War.

That argument may or may not convince you, and it does not convince everyone. But it does serve to show that the problem is not as cut-and-dried as "black holes have no hair".

WayneFrancis
2009-Jul-09, 04:35 AM
This might appear to resolve the black hole information paradox in favor of the original position of Hawking that all information entering a black hole is lost, and will be destroyed when a black hole eventually evaporates due to Hawking radiation.

But it is not that simple. Hawking has conceded a bet, and reversed his position. Not everyone involved agrees.

The concession is based on notions from string theory. You can find a discussion in Leonard Susskind's book The Black Hole War.

That argument may or may not convince you, and it does not convince everyone. But it does serve to show that the problem is not as cut-and-dried as "black holes have no hair".

I'm open to the idea that information may not be lost but for now I gravitate to information being lost when it enters into a black hole.

I don't consider hawking radiation as an example of information coming out but would be happy to be shown where my logic falls down. In theory there could be monopoles but we don't see them so for now I won't worry about them.

grant hutchison
2009-Jul-09, 11:24 AM
I don't consider hawking radiation as an example of information coming out but would be happy to be shown where my logic falls down.It's that word "information", which means many things to different people.
Susskind is concerned about the physical reversibility of processes: being able to time-reverse events and get out what you put in. This is a basic principle for classical physics, and for the wavefunction in quantum mechanics. In principle, then, we can time-reverse the explosion of an atomic bomb, and retrieve the original mechanism. So the information about the bomb is still present in the Universe, albeit in a rather thermalized form.

An external observer never sees an object actually cross the event horizon of a black hole; in fact, that observer is never simultaneous with any events that befall the object along its worldline inside the event horizon. In Susskind's take on the problem, the external observer finds that infalling objects are thermalized at the event horizon and re-emitted as Hawking radiation: in principle, their information is retrievable if you wait for the hole to evaporate and collect all the emitted radiation.
Infalling observers of course don't experience this process, and they progress smoothly towards extinction at the singularity, along with all information they contain. But external observers and infallers are unable to compare notes on what "really" happens to the information.

Grant Hutchison