schwarzschild metric and BH mass

[kurious]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?
Also, if a black hole is made from photons, would it be massless and move at
the speed of light?

[DW]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?
Are you proposing an alternative?
Also, if a black hole is made from photons, would it be massless and move at
the speed of light?
No.

[marlon]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?
Also, if a black hole is made from photons, would it be massless and move at
the speed of light?

I think this is an unvalid question, kurious? The concept of the Schwarzschildradius comes from the use of the schwardzschild metric. It is just a boundary that indicates the infinite-red-translation of the spectra. I am sure you know the story with the speeds exceeding the lightspeed.

A black hole made of fotons ?????

regards
marlon

[pervect]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?
Also, if a black hole is made from photons, would it be massless and move at
the speed of light?

The Scwarazschild metric *probably* isn't a stable solution for the metric inside a black hole - I posted more details on this somewhere, it's from Kip Thorne's popular book. The metric inside a BH is, according to Thorne, most likely something called a BKL singularity.

However, there is just a wee bit of a problem testing this theoretical prediction. There's an even bigger problem reporting the results back :-)

As for your second question, as I recall, two parallel light beams do not attract each other when moving in the same direction. It's possible to form a black hole out of only light, but it requires that the light not all be travelling in the same direction. This results in a black hole that moves at less than 'c'.

[marlon]

Interesting, how can a black hole be formed by using only light ???

regards
marlon

[kurious]

Higgs theory says an object either has mass or is massless.
A black hole made from photons would be indistinguishable from one made from rest masses with the same energy.Is a black hole made from photons massless or does it have mass?

[pmb_phy]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?

There is no evidence of this. Nobody has ever been inside a event horizon to tell us if its correct. And if he was there then we'd have to be inside the event horizon too. While we might find out the answer, we'd pay for it with our lives. Frankly, I, personally am not that anxious to find out. :smile:

Also, if a black hole is made from photons, would it be massless and move at
the speed of light?Interesting question. If there were a system of photons which had a zero total momentum then it would be possible for a black hole to form. All that is required is that the photons be located within the black hole. Since the photons have mass then its possible for a black hole to form. This mass would then be the M in the Schwarzschild metric. But this M would be the E/c2 as measured in the zero momentum frame. Some people like to call that "rest mass".

However, if all the photons were moving in the same direction, then a black hole can't form, even if all the mass of the photons is within the "Schwazchild radius" associated with the total mass of the photons. This is easy to see since if all the photons are moving in the same direction there is no zero momentum frame of reference. If a black hole formed then there would be. A person in that frame would then measure a violation of the conservation of momentum. However you'd have to ask yourself how such a beam of photons could be created in the first place. If the matter from which the photons were emitted fit within Rs to begin with then it would seem to me that it was a black hole to begin with (with photons moving in the opposite direction?) and therefore one can't create such a beam.

Pete

[kurious]

If a large number of pions decay into photon pairs which travel in
opposite directions, and one member of each pair travels to a fixed
point in space,
a black hole would form at the fixed point in space.Since the photon
polarizations are coupled, I could get information about the
microstates in the black hole by measuring the polarization angles of
the photons that are outside the black hole.
And by placing a number of polarizing filters in a line, for each
photon travelling outside the black hole, with one photomultiplier per
photon to detect each photon, I could gain information on the
microstates in the black hole at different periods in time.So I would
know more about a black hole than just its total spin,mass and
charge.Any objections to this?

[pmb_phy]

If a large number of pions decay into photon pairs which travel in opposite directions, and one member of each pair travels to a fixed
point in space, a black hole would form at the fixed point in space.

At what fixed point in space? Why would a black hole form?

Since the photon polarizations are coupled, I could get information about the
microstates in the black hole by measuring the polarization angles of the photons that are outside the black hole. And by placing a number of polarizing filters in a line, for each photon travelling outside the black hole, with one photomultiplier per photon to detect each photon, I could gain information on the microstates in the black hole at different periods in time.So I would know more about a black hole than just its total spin,mass and charge.Any objections to this?
You're speaking of commuinication faster than light using entangled states. There is some controversy about whether this is possible or not. People claim to have done it in fact. But I've seen this discussed in the physics literature. I have no opinion on this otherwise since I haven't studied it in detail.

However, as you've described it, this wouldn't seem to work. All you'll get is a random sequence of polarization states and you'd have no way to interpret them.

There was an article in Scientific American back in 1993 called Faster than Light? which "discusses experiments in quantum optics which they claim shows that two distant events can influence each other faster than any signal could have traveled between them."

Note: I've collected a few articles from the physics literature on FTL communication. If anyone is interested in this topic and would like the references than I can post them. They're from journals such as The American Journal of Physics and Annals of Physics etc. I'm just too lazy to post them otherwise. :smile:

Pete

[da_willem]

The Schwarzschild metric is an exterior solution and is derived as such. So when the empty spacetime field equation holds theoretically the Schwarzschild metric is valid. When the radius of an object is smaller than 2GM/c^2 there is a point where the Schwarzschild line element diverges (g_{11}=-(1-2GM/rc^2)^{-1}). So I guess all bets are off then. In terms of e.g. Eddington-Finkelstein coordinates you can still investigate what happens inside the black hole using the Schwarzschild metric and it predicts (not unreasonable) ingoing null geodesics. But it's more a matter of faith than science because no experiment could be performed to investigate the validity of the Schwarzschild metric inside a black hole.

[selfAdjoint]

AFAIK the black hole is a prediction of the metric, and of theories of stellar collapse. As far as experiment goes, we have never seen an absolute guaranteed black hole. We have inferred that there are black holes at the center of the galaxies but we cannot see them, we only see the violent physics we interpret as coming from a hidden black hole. So the answer to your question is none, just like Hawking radiation and the Beckenstein entropy, and so forth; it's all theory.

[marlon]

What evidence is there that the schwarzschild metric is valid inside a black hole (as opposed to outside the Sun where evidence comes in the form of mercury's perihelion)?
Also, if a black hole is made from photons, would it be massless and move at
the speed of light?


A blach hole must have restmass. two fotons can generate restmass so I think (it is my opinion though) that a blach hole made of fotons can never be massless or move at the speed of light.

Isn't it the restmass only that curves spacetime???

regards
marlon

[DW]

A blach hole must have restmass. two fotons can generate restmass so I think (it is my opinion though) that a blach hole made of fotons can never be massless or move at the speed of light.

Isn't it the restmass only that curves spacetime???

regards
marlon
Its not restmass, its just mass. Restmass is a bad word. And, it is the stress-energy tensor that acts as the source term in Einstein's field equations.

[marlon]

ok,ok,ok,gues, i stand corrected

regards
marlon

[marlon]

So just to get the record straight : it is relativistic mass that curves spacetime, right???

But i have some difficulties accepting the concept of a black hole moving at the speed of light. This is relative. Suppose we can "attach" an observer to the black hole. In his eyes the black hole doesn't move so if it cannot be constructed out of fotons.

Besides, following the QED-feynmann-graphs. Interactions between the fotons (let's say two to make it easy) could occur, generating fermionic-matter-particles like electrons and positrons. The fotons wouldn't last though.


regards
marlon

[pmb_phy]

So just to get the record straight : it is relativistic mass that curves spacetime, right???

Yes. Misner, Thorne and Wheeler call it mass-energy, or sometimes they call it just "mass", as Wald does at times (at least one place in his text that I know of) but its all the same thing. So long as you pay close attention to how I described it above.

But i have some difficulties accepting the concept of a black hole moving at the speed of light.

I don't recall ever saying that a black hole could move at the speed of light.

Pete

[kurious]

If a black hole consists of photons how could the photons survive given that
they must reach the singularity and stop moving?This violates relativity theory
which says a photon cannot be at rest!
And how could fermions which obey fermi-dirac statistics all be present at the singularity? This would be at odds with quantum mechanics!
Black holes are bad news for quantum mechanics and relativity.The fact that they
are bad news for relativity and that relativity theory describes them suggests an internal inconsistency in relativity theory.

[chroot]

kurious:

You are correct. Quantum mechanics and general relativity are at odds with each other. One location where they are not compatible is the "singularity" at the center of a black hole.

In all likelihood, there is no actual singularity at the center of a black hole, just a very dense state of matter. Quantum mechanics prohibits singularities for many reasons. M-theory or another GUT will answer these questions.

- Warren

[zefram_c]

There's a solution of the equations of general relativity called the Aichelburg-Saxl solution, which describes massless black holes moving at the speed of light. There's no experimental evidence that these actually exist, but they're fun to think about nonetheless, since you can use the equations of general relativity to figure out what they would do if they did. – Dr. John Baez
This might be relevant... no time to check now. No idea what they could be made of, but remember that GR is a classical theory.

[pervect]

So just to get the record straight : it is relativistic mass that curves spacetime, right???


Right, modulo the question of wheter or not "relativistic mass" should be called by it's more modern name, "energy".


But i have some difficulties accepting the concept of a black hole moving at the speed of light.


Not a problem, since black holes don't move at the speed of light.

[pmb_phy]

Right, modulo the question of wheter or not "relativistic mass" should be called by it's more modern name, "energy".

Relativistic mass is not another name for energy. Proportionality of two physical quantities does not mean that the two physical quantities have the same physical meaning. They are defined differently. In fact in relativistic electrodynanics, the term energy is used to describe the sum E = Kinetic Energy + Rest Energy + Potential energy = K + E0 + V = mc2 + V. The quantity T = mc2 is inertial energy and the quantity E is "energy" or "total energy". It would be a mistake to confuse E with T. E is proportional to the time component of the generalized 4-momentum while T is proportional to the regular 4-momentum.

Consider also the relationship energy of a photon and its frequency, i.e. E = hf. Since f is proportional to E would you say that frequency is another name for electromagnetic energy? I know I wouldn't.

Pete

[pervect]

It's not just proportionality, it's identity. But I don't think there's much point in arguing the semantics. I refer interested readers to

http://math.ucr.edu/home/baez/physics/Relativity/SR/light_mass.html

for why I like to avoid the term "relativistic mass".

A fuller answer to "what curves spacetime" would be "the stress energy tensor" rather than just "energy", of course.

[Haelfix]

The question of whether a conglomeration of photons can form a blackhole is rather interesting actually, though its vanishingly small even if the background was a vacuum.

In the real world, regular matter and interactions will swamp the source terms making any second order gravitational QFT interactions (with newtonian coupling constants) tiny and negligable.

Now that I think about it, its darn near impossible, b/c we don't live in a vacuum, and thermodynamics will introduce strong cutoff terms in the QFT lagrangians. Not to mention that its unclear to me at least, how a many body problem like that would work out statistically. Clearly you cant just treat it as a classical thermal bath of photons.

[Orion1]

Although the Schwarzschild Metric is a solution in General Relativity, 'static' Schwarzschild BHs with zero angular momentum L = 0, and based upon the violent nature of their improbable formation, cannot exist in the known Universe. The Schwarzschild Metric is a mathematical solution only, not a real solution in the real Universe.

Schwarzschild BHs do not exist.

[DW]

So just to get the record straight : it is relativistic mass that curves spacetime, right???
No. There is no place for relativistic mass in modern relativity. It is the stress-energy tensor that is the source term in Einstein's field equation.

[kurious]

Kurious:
Also relativistic mass leads to odd conclusions such as an object moving
very close to light speed having so much mass it becomes a black hole.


Haelfix:
The question of whether a conglomeration of photons can form a blackhole is rather interesting actually, though its vanishingly small even if the background was a vacuum.

Kurious:
A single photon with enough energy can in principle become a black hole.

[pmb_phy]

pervect - Please understand that I know this is simply a difference of opinion and I can certainly respect your opinion.
It's not just proportionality, it's identity.
This part is not an opinion, it is a incorrect statement. There is a huge difference between identity and equality. When you claim that it is an identitiy, as you are, you're claiming that the term "relativistic mass" is a name which was created to mean the exact same thing as "energy", I.e. a synonym. That is incorrect. Relativistic mass is defined in one way (as the m such that mv is a conserved quantity) and inertial energy, T (which is the notation used in some portions of Goldstein 3rd Ed.), is defined in another way, i.e. T = enegy in the absense of potential energy, V. This is sometimes called free-particle energy (e.g. Jackson 2nd Ed.). It is then proven that T = mc2.

Refering to it as an identity gives the false impression that this is not something that has to be proven. It also gives the impression that it includes potential energy. I had an extremely hard time trying to convince one person that it doesn't contain potential energy of position, but it was a lost cause because people almost always refer to T as "energy" and label it "E".

Yes. I'm quite familiar with that web page. However its arguements are quite poor. That is why I don't choose to abadon relativistic mass and why I think its bad to refer to it as such.

A fuller answer to "what curves spacetime" would be "the stress energy tensor" rather than just "energy", of course.

That view confuses the physical quantities which generate the gravitational field with the mathematical quantity which describes them. But to each his own. But if that is your view then it would be inconsistent to say that charge generates an EM field. So you might want to avoid that in the future should the subject arise. To be consistent with your view then you'd have to say that 4-current generates a EM field. Note: There can be g-fields in the absesnce of matter (e.g. gravitational radiation) and there can be EM fields in the absence of charge (e.g. cosmic background radiation).

Notice how I left out "curves spacetime" in all my responses? That is because a non-vanishing energy-momentum tensor can generate a non-vanishing gravitational field with no spacetime curvature (at points seperate from where the energy-momentum tensor does not vanish).

Pete

[Haelfix]

"Notice how I left out "curves spacetime" in all my responses? That is because a non-vanishing energy-momentum tensor can generate a non-vanishing gravitational field with no spacetime curvature (at points seperate from where the energy-momentum tensor does not vanish)."

What do you mean by curvature? The full Riemann tensor, Gauss curvature, the Ricci scalar.. What exactly? If you're not very careful you will violate the field equations.

All this is moot anyway, in the modern paradigm, everything is in the choice of connection.. Curvature is basically a man made description.

[pmb_phy]

What do you mean by curvature? The full Riemann tensor, Gauss curvature, the Ricci scalar.. What exactly? If you're not very careful you will violate the field equations.

In GR when someone speaks of spacetime curvature they are speaking of the non-vanishing of the Riemann tensor

All this is moot anyway, in the modern paradigm, everything is in the choice of connection.. Curvature is basically a man made description.That doesn't make much sense to me. What do you mean "everything is in the choice of connection"? Any geometry can be described by the metric as well as the affine connection. Affine geometry and metric geometry are two ways of detecting curvature. Curvature is a description of properties of a manifold. It is no more manmade than any other quantity in mathematics.

Gaussian curvature is not defined for any dimension other than two. The dimensions of spacetime is 4 and therefore cannot apply. The Ricci scalar vanishes in vacuum and therefore cannot described the curvature of spacetime outside of matter.

Haelfix - I made the mistake of reading a post by someone I have on my block list and posted a correction to the post. In retrospect I decided it was a bad idea to read and respond to a blocked post/poster so I've deleted that post. Since some of that message might have been read by you in the meantime I'd be more than happy to discuss the content or Gaussian curvature with you and why its not defined in dimensions other than two. If you didn't read it then never mind.

Pete

[DW]

I
Gaussian curvature is not defined for any dimension other than two.
This is a common misconception. Gaussian curvature describes excess radius for a closed path in higher dimensional space just as Riemannian curvature describes the variation in a vector parrallel transported around a closed path in a higher dimentional spacetime. In both cases the path taken can be described as a displacement along a 2d surface, but in both cases the surface is in a higher dimentional imbedding. Your argument is just like claiming that Riemannian spacetime curvature only applies to a 2 dimensional spacetime which is rediculous.

[pervect]

Yes. I'm quite familiar with that web page. However its arguements are quite poor. That is why I don't choose to abadon relativistic mass and why I think its bad to refer to it as such.


I respect your opinion, and am not quoting the webpage to try and "convert" you. I'm just trying to explain my position to other posters in the thread.

But it looks like we still have some issues to resolve when you claim my usage is "incorrect", unfortunately. I don't mind someone calling a spade a spade, or a portable entrenching tool, it's not worth arguing about IMO. When you claim I'm wrong, you more-or-less force a debate on a topic that is starting to become a bit - tedious.


That view confuses the physical quantities which generate the gravitational field with the mathematical quantity which describes them. But to each his own. But if that is your view then it would be inconsistent to say that charge generates an EM field.


The physical quantity that generates the gravitational field is the stress-energy tensor. The physical quantity that generates the EM field is the current density J, a 4-vector.

When I'm speaking less formally, I'll sometimes say that charge generates an EM field, and I will also say that energy generates the gravitational field. In both cases, I'm making the same conceptual simplification - which is really an oversimplification. The simplification is to talk about one component of the tensor and to ignore the others. The simplificaiton can be fully justified when charges (or masses) are not "moving too quickly".

I'd really like to speak less formally sometimes, it's a bit annoying to have to be "on my guard" all the time to be very formal and correct. It's also not conducive to communication with laypeople who may be reading the board, who probably won't appreciate all the formal correctness anyway. I think that using the word "Tensor" tends to scare laypeople :-(. And I don't want to scare people away from such a fasciniting topic as gravity.


To be consistent with your view then you'd have to say that 4-current generates a EM field.


see above

As far as your argument about potential energy, et al. When I say energy, I mean the first component of the energy-momentum 4-vector, whether that 4-vector be the 4-vector of a particle, or an electromagnetic field. This is really standard modern usage. I don't mind you using your own usage, but I *really* wish you'd stop attacking modern usage as being wrong at seemingly every point, seemingly endlessly.

Note that this *is* an identity. Tij is the density of the energy momentum 4-vector per unit volume (with a suitable vector defintion of volume). When we multiply T00 by the volume, we get energy, which is just the first coordinate in the energy-momentum 4-vector.

Also, note that you yourself are making the same simplification, you just call the first component of the energy-momentum 4-vector by a different name than I do.

[Haelfix]

"Affine geometry and metric geometry are two ways of detecting curvature. Curvature is a description of properties of a manifold. It is no more manmade than any other quantity in mathematics."

When people talk about curvature, it is meaningless, unless a *choice* of connection is made. There is no god given choice, although there is a preffered one for a metric (called the metric connection). The Riemann tensor follows from that.

Nowdays, people have reworked gravity in many ways. They often work directly with the connection (actually the space of connections modulo gauge transformations). There are many formalisms for dealing with this.

However the previous post is a little vague.

If your stress energy tensor is non zero, it follows from the field equations that the geometry is nonzero. Sure, you can pick coordinate's such that it vanishes, but thats only a local phenomenon.

[pmb_phy]

When people talk about curvature, it is meaningless, unless a *choice* of connection is made.

Why? All I need to know is the metric and the curvature is well defined. I don't understand why you'd say that. I can give you the metric OR the connection. Either will do.

There is no god given choice, although there is a preffered one for a metric (called the metric connection). The Riemann tensor follows from that.
Please define the term metric connection

If your stress energy tensor is non zero, it follows from the field equations that the geometry is nonzero. Sure, you can pick coordinate's such that it vanishes, but thats only a local phenomenon.
Sorry, but I don't know what you mean when you say "the geometry is nonzero".

If the energy-momentum tensor is non-zero at a point P in spacetime then it follows that the spacetime curvature at P is non-zero. If there is a vacuum at a nearby point Q then it does not follow that the curvature at Q is non-zero.

Pete

[kurious]

pervect:

I'd really like to speak less formally sometimes, it's a bit annoying to have to be "on my guard" all the time to be very formal and correct. It's also not conducive to communication with laypeople who may be reading the board, who probably won't appreciate all the formal correctness anyway

Kurious:

This is all very true.
I am a layperson to some degree but I have learned a lot
from this forum because many people have gone out
of their way to explain advanced concepts with the minimum of
mathematical detail (maths, to me, is the formal side of physics).
And when you think of how many topics on this forum have not
got a shred of experimental evidence to back them (strings, hawking radiation for example), I think many people will be unmotivated to understand the finer details of these topics.

[pmb_phy]

I'd really like to speak less formally sometimes, it's a bit annoying to have to be "on my guard" all the time to be very formal and correct. It's also not conducive to communication with laypeople who may be reading the board, who probably won't appreciate all the formal correctness anyway

My appologies if I contributed to making you feel that you had to be on gaurd. If I was one of the posters who made you feel this way then please PM me an example and I will try to avoid doing so again.


This is all very true.
I am a layperson to some degree but I have learned a lot
from this forum because many people have gone out
of their way to explain advanced concepts with the minimum of
mathematical detail (maths, to me, is the formal side of physics).
And when you think of how many topics on this forum have not
got a shred of experimental evidence to back them (strings, hawking radiation for example), I think many people will be unmotivated to understand the finer details of these topics.I try to avoid the math when possible myself when I'm not 100% sure that the person I'm addressing the post too is not very familiar with the math/physics. However since there are many people reading these same posts then perhaps you can recommend a way to do both.

Thanks

Pete

[kurious]

I think the best way for people to know what level to pitch their replies to posts at is for the people posting questions to declare what level of maths they know.
I have asked questions on sci.physics.research and have received long replies
with a level of maths which frankly I would expect many of the moderators would have to look at a textbook to check out! However this forum is a science forum and
I think it is a good thing that there are experts on here who keep the standard high and who can let you know if you have really understood something about a subject.
I wish more experts would go onto the theory development and hold me and others to account for what we post on there.Sometimes I get the impression on theory development that the blind are leading the blind and that a well placed comment from
someone with detailed knowledge of a subject could stop posters spending weeks going round in circles.
My mathematical knowledge of relativity comes from Schultz's book
"a first course in relativity". I haven't read it all but I think I
can say that I know the basic mathematical principles on which GR is founded.

[pervect]

"Affine geometry and metric geometry are two ways of detecting curvature. Curvature is a description of properties of a manifold. It is no more manmade than any other quantity in mathematics."

When people talk about curvature, it is meaningless, unless a *choice* of connection is made. There is no god given choice, although there is a preffered one for a metric (called the metric connection). The Riemann tensor follows from that.


Does the connection define the derivative operator? The derivative operator defines parallel transport, and while there are many ways in principle that one could parallel transport vectors, only one preserves the dot product of two vectors, g_{ab} u^a u^b - given a metric to define the dot product.

[marlon]

Does the connection define the derivative operator? The derivative operator defines parallel transport, and while there are many ways in principle that one could parallel transport vectors, only one preserves the dot product of two vectors, g_{ab} u^a u^b - given a metric to define the dot product.

Yes, a connection defines the covariant derivative. What is the use of them connections ? Well...

As already stated they are used to study the properties of some manifold. One takes a vector and projects it onto the tangential space. These tangent-vectors are the ones that will be parallel transported. During the transport the vektors follows a certain trajectory called the loop (just like them Wilson loops). Once the loop is completed the tangent-vektor will be in the exact initial point where we started the transport. By comparing how the position of the vektor has changed after the transport (the relation between the original vektor and the vektor that is transported is the connection) one can deduce info about the curvature of the manifold, by just performing LOCAL operations. This is just like checking out the fact that the earth is a sphere by doing "measurements" in our locally flat surrounding space.

Regards
marlon

[marlon]

If a black hole consists of photons how could the photons survive given that
they must reach the singularity and stop moving?This violates relativity theory
which says a photon cannot be at rest!
And how could fermions which obey fermi-dirac statistics all be present at the singularity?

Did you,kurious state that one foton can become a black hole ??? If so, please tell me why you think such a thing.
One foton can never decay into something ......whatever....

Your point about the fermions at the singularity is correct, that is indeed a major flaw of GTR as well as QFT.

[marlon]

"
Nowdays, people have reworked gravity in many ways. They often work directly with the connection (actually the space of connections modulo gauge transformations). There are many formalisms for dealing with this.

.


Haelfix, i have a little question.

When you talk about connections modulo gauge transforms, what exactly do you mean. I would say that in this case the gauge fields are the connections. So that all different metrics and connections make up the configuration space and all possible loops are the state space. This is the picture of LQG, where we can perform quantization of the gauge field because of the diffeomorfism invariance (this is the covariance in GTR) and local gauge invariance (just like in QFT).

Basically in on order to work with a model that does not depend on a specific metric, we take as basis the Wilson loops and all the info of the manifold are in the gauge-fields...

regards
marlon

(ps : this basis of loops is not really used. In stead one uses the spin networks of Penrose)

[kurious]

Marlon:
Did you,kurious state that one foton can become a black hole ??? If so, please tell me why you think such a thing.
One foton can never decay into something ......whatever....

Kurious:
A photon cannot decay into rest masses spontaneously because four momentum would not be conserved.

[marlon]

Marlon:
Did you,kurious state that one foton can become a black hole ??? If so, please tell me why you think such a thing.
One foton can never decay into something ......whatever....

Kurious:
A photon cannot decay into rest masses spontaneously because four momentum would not be conserved.


I know, this is the reason why I was asking this question to you

[pmb_phy]

I respect your opinion, and am not quoting the webpage to try and "convert" you. I'm just trying to explain my position to other posters in the thread.

Okey dokey. I assumed that was the case but I wanted to make sure. A while back I wrote an article to explain/describe my view on this. It was necessary since a precise treatment required too much space to ever place in a post or a thread. It have it on my web page at
http://www.geocities.com/physics_world/
The article is called "On the concept of mass in relativity."

But it looks like we still have some issues to resolve when you claim my usage is "incorrect", unfortunately.

I don't recall saying that your usage is incorrect per se. I said that the statement you made It's not just proportionality, it's identity is incorrect. I don't say these things lightly. In fact it took me many years of studying this one point before I took a rigid position on this point. To be precise, the relativistic mass of a particle, whose 4-momentum is P, is proportional to P0 whereas the energy of the particle is proportional to P0 and as you know, these two quantities are not always equal, especially in GR. This is not an opinion. Any GR text will confirm this point. I.e. every GR text that I've seen refers to P0 as the energy of the particle in all cases and there is nothing I've ever seen which ever indicated that relativistic (aka "inertial") mass is not P0 in all cases. Even Einstein used that in his GR text to some extent.

I don't mind someone calling a spade a spade, or a portable entrenching tool, it's not worth arguing about IMO.

Okay. Then I'll drop it. But I don't post these things because I'm addressing you and only you. I'm posting what I do because many people read this thread and the topic of this thread has to do with mass. I have no way of knowing what people reading this thread do or do not know so I don't see the point in omiting key points. Especially in this case since the website you refered to addressed SR only. It was not intended to speak on the mass in GR and that is the topic in this thread. In any case all of this is discussed in another thread here anyway and all I've ever had to say on this topic is in the article in my website so there's no need for me to continue if nobody is interested. I can see that nobody is so unless requested I will drop this here and now.

When you claim I'm wrong, you more-or-less force a debate on a topic that is starting to become a bit - tedious.

Well this is a discussion forum. It is not a debate forum per se. You're not required to respond, you could simply ask me to prove what I claim. But
I can't find where I used the term "wrong" in this thread. I only used the term "incorrect". I do not believe that what I said about relativistic mass and energy is incorrect. And that is something I can certainly back up quite rigorously. In fact this is reflected in the GR literature. However I know you wish to drop this part of the discussion. By the way, I had several discussions with the person, Don Koks, who maintains that web page you mentioned. He explained to me that the page is meant only to address the topic within the context of SR. It was never meant to breach into GR. That's why it doesn't touch on the parts that I mentioned in GR above.

As far as your argument about potential energy, et al. When I say energy, I mean the first component of the energy-momentum 4-vector, whether that 4-vector be the 4-vector of a particle, or an electromagnetic field. This is really standard modern usage. I don't mind you using your own usage, but I *really* wish you'd stop attacking modern usage as being wrong at seemingly every point, seemingly endlessly.

I'd really wish you didn't claim that I'm "attacking" modern usage. I'm not attacking modern usage. I'm using it. You seem to think that everyone in the relativity community means the exact same thing when they use the term "energy." They do not. In fact it often depends on the context or on the topic. If you take a look at purely SR textbooks then you'll see the term "total energy" used to refer to what you just mentioned. It is not clear to me whether people assume that the terms "energy" and "total energy" may or may not refer to the same thing. However when you look in the relativity sections in texts on electrodynamics "total energy" is not used to refer to the time component of 4-momentum and I can't see why anyone would use the term "energy" to refer to anything but "total energy." Two such EM text that do this are Classical Electrodynamics - 2nd Ed., J.D. Jackson and Classical Field Theory: Electromagnetism and Gravitation, Francs Low. A classical mechanic text which does this exact same thing is Classical Mechanics 3rd Ed., Goldstein, Safko and Poole. These are not obscure texts. They are the staples of the physics community. I'd even go so far as to say that they define modern usage. If people here are studying relativity then these are very important and subtle points to learn. Ignoring them can lead to confusion and is a possible source of error.

Since I cannot know whether everyone knows these different uses then I can't see a reason not to mention it since this is a discussion board and people are here to learn. I wish people had mentioned these differences to me when I was learning relativity. It is for that reason I mention these things.

Please note that I do try to avoid these matters of opinion when I see them, i.e. I don't want to get into whether a spade is called a spade or an etc. When I do see someone say that a spade is that which is squeezed from Florida oranges I will say that's incorrect. If I see someone say that a spade is an automotive tool I will disagree with it. But when the subject of definition does present itself I will post what I think about it and hopefully only once in a particular thread. It is not always meant for you to read but for others who are following along who might want to know.

Pete

[DW]

Okey dokey. I assumed that was the case but I wanted to make sure. A while back I wrote an article to explain/describe my view on this. It was necessary since a precise treatment required too much space to ever place in a post or a thread. It have it on my web page at
http://www.geocities.com/physics_world/
The article is called "On the concept of mass in relativity."

I don't recall saying that your usage is incorrect per se. I said that the statement you made It's not just proportionality, it's identity is incorrect. I don't say these things lightly. In fact it took me many years of studying this one point before I took a rigid position on this point. To be precise, the relativistic mass of a particle, whose 4-momentum is P, is proportional to P0 whereas the energy of the particle is proportional to P0 and as you know, these two quantities are not always equal, especially in GR. This is not an opinion. Any GR text will confirm this point. I.e. every GR text that I've seen refers to P0 as the energy of the particle in all cases and there is nothing I've ever seen which ever indicated that relativistic (aka "inertial") mass is not P0 in all cases. Even Einstein used that in his GR text to some extent.

Okay. Then I'll drop it. But I don't post these things because I'm addressing you and only you. I'm posting what I do because many people read this thread and the topic of this thread has to do with mass. I have no way of knowing what people reading this thread do or do not know so I don't see the point in omiting key points. Especially in this case since the website you refered to addressed SR only. It was not intended to speak on the mass in GR and that is the topic in this thread. In any case all of this is discussed in another thread here anyway and all I've ever had to say on this topic is in the article in my website so there's no need for me to continue if nobody is interested. I can see that nobody is so unless requested I will drop this here and now.

Well this is a discussion forum. It is not a debate forum per se. You're not required to respond, you could simply ask me to prove what I claim. But
I can't find where I used the term "wrong" in this thread. I only used the term "incorrect". I do not believe that what I said about relativistic mass and energy is incorrect. And that is something I can certainly back up quite rigorously. In fact this is reflected in the GR literature. However I know you wish to drop this part of the discussion. By the way, I had several discussions with the person, Don Koks, who maintains that web page you mentioned. He explained to me that the page is meant only to address the topic within the context of SR. It was never meant to breach into GR. That's why it doesn't touch on the parts that I mentioned in GR above.

I'd really wish you didn't claim that I'm "attacking" modern usage. I'm not attacking modern usage. I'm using it. You seem to think that everyone in the relativity community means the exact same thing when they use the term "energy." They do not. In fact it often depends on the context or on the topic. If you take a look at purely SR textbooks then you'll see the term "total energy" used to refer to what you just mentioned. It is not clear to me whether people assume that the terms "energy" and "total energy" may or may not refer to the same thing. However when you look in the relativity sections in texts on electrodynamics "total energy" is not used to refer to the time component of 4-momentum and I can't see why anyone would use the term "energy" to refer to anything but "total energy." Two such EM text that do this are Classical Electrodynamics - 2nd Ed., J.D. Jackson and Classical Field Theory: Electromagnetism and Gravitation, Francs Low. A classical mechanic text which does this exact same thing is Classical Mechanics 3rd Ed., Goldstein, Safko and Poole. These are not obscure texts. They are the staples of the physics community. I'd even go so far as to say that they define modern usage. If people here are studying relativity then these are very important and subtle points to learn. Ignoring them can lead to confusion and is a possible source of error.

Since I cannot know whether everyone knows these different uses then I can't see a reason not to mention it since this is a discussion board and people are here to learn. I wish people had mentioned these differences to me when I was learning relativity. It is for that reason I mention these things.

Pete
You are mixing up the energy which is P^0 with the conserved "energy parameter" which is in some spacial cases equal to P_{0}.

[kurious]

There have been debates on the research section of this forum about energy in GR and
none of them have been conclusive about how to define energy or to conserve it in GR.This just goes to show that there is more work to be done on GR - isn't
quantum gravity supposed to solve all these problems?

[pmb_phy]

There have been debates on the research section of this forum about energy in GR and
none of them have been conclusive about how to define energy or to conserve it in GR.This just goes to show that there is more work to be done on GR - isn't
quantum gravity supposed to solve all these problems?Energy is not a quantity which can be defined. The more basic a quanity the harder it is to define. Or as H.A. Kramers stated it

My own pet notion is that in the world of human thought generally, and in physical science particularly, the most fruitful concepts are those to which it is impossible to attach a well-defined meaning.

Richard Feynman touched on this point in the Feynman lectures.

It is important to realize that in physics today, we have no knowledge of what energy is. We do not have a picture that energy comes in little blobs of a definite amount. It is not that way. However, there are formulas for calculating some numerical quantity, and we add it all together it gives “28” - always the same number. It is an abstract thing in that it does not tell us the mechanism or the reasons for the various formulas.

The inertial energy of a particle (in SR this is defined as total Ienergy - potential energy) is proportional to the time component of a the energy-momentum 1-form and not of the energy-momentum 4-vector. In SR there is no real difference but not in GR. In my opinion it is quite unfortunate that people don't always make this point more prominently in the relativity literature. At least Shutz makes this point clear in his GR text where, for a photon, he defines E as E = -p0. For a particle he defines this as E = -p0/m where m is the particle's proper mass.

To be more precise, on page 190 in his text A first course in general relativity Schutz states

For instance, suppose we have a stationary gravitational field. Then a coordinate system can be found in which the metric components are time indepenant, and p0 is conserved. Therefore p0 (or, really, - p0) is usually called the 'energy' of the particle, without qualifying it with 'in this frame'.

It is -p0 which reduces to the Newtonian formula for energy, i.e. in the Newtonian limit

-p_0 \simeq m_0 + m_0\phi + p^2/2m_0

where m0 is the particle's proper mass and p2/2m0 is the particle's kinetic energy.

Pete

[DW]

Energy is not a quantity which can be defined. The more basic a quanity the harder it is to define. Or as H.A. Kramers stated it

Richard Feynman touched on this point in the Feynman lectures.

The inertial energy of a particle (in SR this is defined as total Ienergy - potential energy) is proportional to the time component of a the energy-momentum 1-form and not of the energy-momentum 4-vector. In SR there is no real difference but not in GR. In my opinion it is quite unfortunate that people don't always make this point more prominently in the relativity literature. At least Shutz makes this point clear in his GR text where, for a photon, he defines E as E = -p0. For a particle he defines this as E = -p0/m where m is the particle's proper mass.

To be more precise, on page 190 in his text A first course in general relativity Schutz states

It is -p0 which reduces to the Newtonian formula for energy, i.e. in the Newtonian limit

-p_0 \simeq m_0 + m_0\phi + p^2/2m_0

where m0 is the particle's proper mass and p2/2m0 is the particle's kinetic energy.

Pete
Instead of using out of date concepts one should use modern relativity according to which this translates to saying that the conserved "energy parameter" in the absence of nongravitational fields is in some special cases equal to the time element of the covariant momentum four-vector which in a Newtonian limit reduces to p_{0} = mc + \frac{1}{2c}mv^{2} + m\frac{\Phi }{c}

[marlon]

hey guys, anyone got some comment on my questions of the previous page ???

[pmb_phy]

hey guys, anyone got some comment on my questions of the previous page ???Sorry but I didn't see a question for us back there that. I saw only one question and that was to Haelfix where you asked him what he meant by something. Was there something in particular that you're refering to?

Pete

[pervect]

OK, I'm feeling a little calmer, now.



To be precise, the relativistic mass of a particle, whose 4-momentum is P, is proportional to P0 whereas the energy of the particle is proportional to P0 and as you know, these two quantities are not always equal, especially in GR.


OK, here you start to lose me. In particular, I've never seen anyone demand that any tensor quantity only be subscripted or superscripted. You'll have to do a lot more convincing if you want me to believe that there is only "one true way" to represent energy. In fact, now that I think about it, you'll have to do a heck of a lot of convicing. One of the main features of tensors is that one can use covariant or contravariant indices, at will.


This is not an opinion. Any GR text will confirm this point. I.e. every GR text that I've seen refers to P0 as the energy of the particle in all cases and there is nothing I've ever seen which ever indicated that relativistic (aka "inertial") mass is not P0 in all cases. Even Einstein used that in his GR text to some extent.


Every text I've read says that one can raise and lower indicies at will, and that a tensor is a tensor regardless of whether it has covariant or contravariant indicies.

Your statement above fails to convince me otherwise.

[pmb_phy]

OK, I'm feeling a little calmer, now.

Please feel free to PM me at anytime if you are irritated by something I've posted. In such cases it is highly likely there is miscommunication between us which is better resolved in PM so we don't mess the board up.

OK, here you start to lose me. In particular, I've never seen anyone demand that any tensor quantity only be subscripted or superscripted.

These are two different entities which have a one-to-one correspondence between the two. One is a 4-vector (contravariant vector) which lies in the tangent space and is normally called the energy-momentum four vector. The other is a 1-form (covariant vector) which lies in the cotanget space. 1-forms and 4-vectors are different animals and no corresponence exists without a metric. I don't know of any notation which makes this distinction other than the abstract index notation. Are you familiar with this? Wald uses it amlost exclusively.

Here it will be useful for me in this post to let P = 4-momentum and let q be the 1-form which is the dual to P. Then q0 is the particle's energy.

Same with tensors of higher rank. e.g. T could be a covariant tensor of rank 2. That means that it maps vectors to real numbers. But the T notation does not tell you what kind of animal the tensor is. It can't tell you if its a tensor which maps 1-forms to reals. But when there is a metric there is a one-to-one correspondence between the covariant tensors and the contravariant tensors then we can forget about the distinction. But we can't forget about the physical meaning, especially in the present example.

You'll have to do a lot more convincing if you want me to believe that there is only "one true way" to represent energy. In fact, now that I think about it, you'll have to do a heck of a lot of convicing. One of the main features of tensors is that one can use covariant or contravariant indices, at will.

Do you know why P0 is refered to as the energy of the particle and P0 is not? Its because P0 has the properties one expects of energy, namely that in a time independant g-field the energy of the particle is a constant of motion. P0 does not have this property.

Your statement above fails to convince me otherwise.Okay.

Am I 100% positive about this? No. Just 90% positive. But when Schutz and others goes out of their way to refer to P0 as energy then I listen very carefully. Could they be wrong or could I misunderstand them? Sure. Of course. I can't seem to determine literature wide how energy is defined in GR especially since nobody ever really defines it in general. But as I said, I will only know everything about 1 year after I pass away. :biggrin:

Pete

[pervect]

These are two different entities which have a one-to-one correspondence between the two. One is a 4-vector (contravariant vector) which lies in the tangent space and is normally called the energy-momentum four vector. The other is a 1-form (covariant vector) which lies in the cotanget space. 1-forms and 4-vectors are different animals and no corresponence exists without a metric.


Very basic stuff so far. Once a metric is defined, though, it's very common to see indices raised an lowered by the metric. It happens all the time.


Here it will be useful for me in this post to let P = 4-momentum and let q be the 1-form which is the dual to P. Then q0 is the particle's energy.


OK, now we are getting to our basic disagreement.

In special relativity, with a flat metric, E^2-P^2 = m^2.

The E here is not q0. The E here is P0. By definition. This equation comes from

g_{ab} E^a E^b = m^2

and the assumption of the flat matric. So the point is, the answer to the following question:

What is the name of the first component of the energy-momentum 4-vector?

is

Energy.

I really *do* want to be able to say in the future that "the first component of the energy - momentum 4-vector is energy" without getiting into a big long argument.


Do you know why P0 is refered to as the energy of the particle and P0 is not? Its because P0 has the properties one expects of energy, namely that in a time independant g-field the energy of the particle is a constant of motion. P0 does not have this property.


I went through a big long derivation with Wald's approach of killing vectors, but MTW says it more simply

pg 655-656

S = -(1-\frac{2M}{r})dt^2+\frac{dr^2}{1-\frac{2M}{r}}+r^2(d \theta ^2 + sin^2\theta d \phi ^2)


The expression [for S] shows that the geometry is unaffected by the translations t->t+\delta t, \phi -> \phi + \delta \phi . Thus the conjugate momenta p0 = -E and p_{\phi} = L are conserved.


But this is a consequence of the particular killing vectors of the Scwarzschild metric.

The killing vectors for the scwarzschild space time are Ka = [1,0,0,0] and Ka = [(-1 + 2m/r),0,0,0] , so the conserved quantities are Ka Ea and Ka Ea, where E is the energy-momentum 4 vector.

We know the above are the killing vectors because they satisfy Killing's equation

\nabla_a K_b + \nabla_b K_a = 0 , i.e. that the covariant derivative \nabla_a K_b must be antisymmetric. This is sometmes written as the requirement that Ka;b be anti-symmetric.

Elsewhere, as has been pointed out, MTW uses P0 as energy, for instance on pg 462. So I don't see any basis for your claims that one must use a particular covariant or contravariant indice for the energy-momentum 4 vector to be able to call the first component of it energy.


Am I 100% positive about this? No. Just 90% positive. But when Schutz and others goes out of their way to refer to P0 as energy then I listen very carefully.


I've seen a lot of good things about Schutz, but I don't have it/haven't read it.

[pmb_phy]

The E here is not q0. The E here is P0. By definition. This equation comes from

In SR q0 = P0 so there is nothing inconsistent about what I said. This fact also demonstrates why I don't think it is correct to refer to P0 as "energy" because in general q0 does not equal P0. People identify P0 in SR because in SR E = mc2 where m = relativistic mass.

I really *do* want to be able to say in the future that "the first component of the energy - momentum 4-vector is energy" without getiting into a big long argument.

Fine by me. I'm not here to force things on people, just to explain/describe things.

But this is a consequence of the particular killing vectors of the Scwarzschild metric.

And you don't wonder why MTW refer to -P0 as energy and not P0. But I don't see what your point is. The energy of a particle does not always need to be conserved. One must only expect the energy of a particle to be conserved when the field is time independant. A g-field is time independant when the components of the metric are not functions of time. This is equivalent to saying that there exists timelike killing vectors.

Elsewhere, as has been pointed out, MTW uses P0 as energy, for instance on pg 462.

The P0 on that page (i.e. Eq. 20.7) is not the time component of a particle in a gravitational field. It is the time component of the total 4-momentum of the source of gravity. As such, what value do you think P0 has?

Can you check something for me? You have Wald's text, correct? If so then please see Eq. (4.3.19) on page 72. Wald writes "..where rho is the mass (i.e. energy)...". That rho looks like relativistic mass density. MTW have this same equation. However I made this mistake earlier where I didn't pay close enough attention to what Wald was saying. Did I mess up here too? Thanks.

Pete

[sal]

Do you know why P0 is refered to as the energy of the particle and P0 is not? Its because P0 has the properties one expects of energy, namely that in a time independant g-field the energy of the particle is a constant of motion. P0 does not have this property.
Pete
In a time-independent gravitational field, I thought we had

P^0 = m \gamma = T + m = energy\ as\ measured\ by\ observer

and

-P_0 = T + V + m = total\ (conserved)\ energy

So they're both "energy", of a sort, but only P0 is conserved.

Is that not correct?

In flat space, they're equal.

[pmb_phy]

In a time-independent gravitational field, I thought we had

P^0 = m \gamma = T + m = energy\ as\ measured\ by\ observer

On what basis are you calling this energy as measured by observer? The equality you speak of is achieved by assuming an inertial frame. If you don't assume an inertial frame then what basis is there for estabilishing that equality?

And how can you call one "energy" if it isn't conserved while the other is?

There is a good reason to say that P0 is proportional to energy when the coordinate system represents and inertial frame. If the frame is not inertial then that reason is not there. Note that if you say that P = m0U then it remains to be shown that P0 is proportional to energy. You can't just say it is or define it that way. Of course you can try to define P[sup]0[/sub] as energy but you are no longer assured that it has the properties that one demands of energy once the frame is not inertial.

Pete

[sal]

In a time-independent gravitational field, I thought we had
P^0 = m \gamma = T + m = energy\ as\ measured\ by\ observer

On what basis are you calling this energy as measured by observer? The equality you speak of is achieved by assuming an inertial frame. If you don't assume an inertial frame then what basis is there for estabilishing that equality?

And how can you call one "energy" if it isn't conserved while the other is?

Taking the last question first, the kinetic energy of a single object is generally not conserved (it varies), but it's still called "energy", albeit with a qualifier. "Kinetic energy", "potential energy", "total energy", and "relativistic energy" are all examples of things people call "energy". All crabs are crustaceans, but not all crustaceans are crabs.

The reason I called it energy as measured by the observer is because that's what I thought it was. Sitting here on Earth, in our totally non-inertial frame, we can measure the kinetic energy of an object in our rest frame (by stopping it and measuring the heat released). If we then drop the object into Einstein's magic gedanken machine that converts the whole thing into photons, and measure the energy of the photons which come out, we have a measure of the mass-energy which was present as well. I thought the sum of those two measurements was P0.

That is why I said I thought P0 = T+m.

I'm not going to endeavor to prove it; rather, I'm going to ask, Is that not the case?

Certainly P0 = m*(dt/dtau), regardless of the frame.

Obviously that kind of energy is not conserved -- if I drop the object 50 feet it has more of it when it gets to the bottom. P0 is conserved in that case.

[pmb_phy]

Taking the last question first, the kinetic energy of a single object is generally not conserved (it varies), but it's still called "energy", albeit with a qualifier.

We're not talking about kinetic energy. We're talking about energy aka total energy. We don't refer to kinetic energy as energy. It would be a very poor way to speak of kinetic energy. It is energy which is conseved, not kinetic energy, potential energy etc.

The reason I called it energy as measured by the observer is because that's what I thought it was. Sitting here on Earth, in our totally non-inertial frame, we can measure the kinetic energy of an object in our rest frame (by stopping it and measuring the heat released). If we then drop the object into Einstein's magic gedanken machine that converts the whole thing into photons, and measure the energy of the photons which come out, we have a measure of the mass-energy which was present as well. I thought the sum of those two measurements was P0.

The term "kinetic energy" is a term which, usually, applies when you can seperate out the potential energy. I.e. you don't want kinetic energy to be a function of the particles position in the gravitational field. You want it to only be a function of speed. Otherwise you'll have to be more precise on what you mean by "kinetic energy".

Since P0 is not a conserved quantity then it will depend on where in the g-field you evaluate it. It'd be enightening to actually work out an example explicitly. Especially in the Newtonian limit.
I'm not going to endeavor to prove it; rather, I'm going to ask, Is that not the case?No. To see this note that dt/d(tau), in a Schwarzchild field, is a function of Phi = gravitational potential. Therefore P0 will also be a function of Phi. So why would T + m be a function of Phi? Since it shouldn't then it follows that P0 is not the sum of kinetic energy and proper energy. If its what you think it is then it should be a function of speed and rest mass only.

Ohanian and Ruffinit touch on this topic in their text Gravitation and Spacetime - 2nd Ed. page 358-359. Note: Eq. (185) is \xi^{\mu}P^{\mu} where \bold \xi is a Killing vector

If the spacetime geometry is time independant, so guv does not depend on x0, then \xi^{\mu} = (b, 0, 0, 0) is the corresponding Killing vector, and the conservation theorem [185] tells us that P0 is constant. This is the law of conservation of energy.
This conservation law also applies to photons moving through curved spacetime. Thus, if the spacetime geometry is time independant, the photon "energy" P0 is constant.

To date I see no reason to question Ohanian (or Schutz) on this point.


In SR I believe that this energy, i.e. E = cP0 should be refered to as ineritial energy and labeled T. Even in SR (inertial frames) P0 is not a conserved quantity in general and I think that the terms "energy" or "total energy" should only refer to conserved energy when a particle is moving in a conservative field. It is E = total energy = kinetic + potential + rest that is conserved. Therefore E = T + m0c[sup]2[sup]. T, defined here, is proportional to the time component of 4-momentum whereas E is proprtional to the time component of the canonical 4-momentum.

Pete

[DW]

We're not talking about kinetic energy. We're talking about energy. We don't refer to kinetic energy as energy. It would be a very poor way to speak of kinetic energy. It is energy which is conseved, not kinetic energy, potential energy etc.

Since P0 is not a conserved quantity then it will depend on where in the g-field you evaluate it. It'd be enightening to actually work out an example explicitly. Especially in the Newtonian limit.
No.

Pete
No, you are not talking about energy. You are talking about the energy parameter and p^{0} is trivial to work out for a Kerr-Newman spacetime of which the Schwarzschild result is a special case given p^{0} = mc(\frac{dt}{d\tau })
and that \frac{dt}{d\tau } is given by the differential time travel equation for arbitrary motion in the spacetime of a charged rotating black hole:
\frac{dt}{d\tau } = \frac{\rho ^{2}(\gamma - \frac{\omega }{c}\frac{l_{z}}{c})}{\Delta - a^{2}sin^{2}\theta + \rho ^{2}\frac{\omega ^{2}}{c^{2}}\varpi ^{2}}
equation 12.3.1 at
http://www.geocities.com/zcphysicsms/chap12.htm#BM12_3
It was derived for geodesic motion but given an energy parameter and angular momentum parameter as functions of proper time, it works for arbitrary motion.

[pmb_phy]

OK, now we are getting to our basic disagreement.

In special relativity, with a flat metric, E^2-P^2 = m^2.

The E here is not q0. The E here is P0. By definition.

I'll skirt the question of whether E = P0 by definition since we've covered this.

Since you seem to respect Wald's text then please see page 69

The energy of the particle as determined by an observer who is present at the event on the particle's world line at which the energy is measured is again

E = -p[sub]a[/sup]va

where va is the 4-velocity of the observer.

This, of course, is a bit different than Ohanian & Rufini since they don't require the observer to be at the event at which the energy is to be determined by the observer.

Choose coordinates in which the observer is at rest. Then (letting c = 1) va = (1, 0, 0, 0)

E = -p[sub]0[/sup]

Pete

[pervect]

And you don't wonder why MTW refer to -P0 as energy and not P0. But I don't see what your point is. The energy of a particle does not always need to be conserved.


Actually, I don't expect the energy or momentum of a particle, as defined by the energy momentum 4-vector of the particle, to be conserved at all as the particle falls towards a larger mass.

Using classical mechanics, I would expect that the kinetic energy plus the potential energy of a falling particle would be conserved.

However, the energy-momentum 4 vector does not include the potential energy.

Therfore we would not expect to see the energy-momentum 4 vector conserved by a falling particle.

You've noticed that in one particular coordinate system, one component of the energy-momentum 4 vector is numerically constant.

This is correct, and useful, but it has apparently misled you into thinking that the energy-momentum-4 vector should include potential energy, as near as I can tell. I say this because I can see no reason that you'd expect energy to be conserved if you didn't include potential energy. You wold expect, instead, that energy would not be conserved as the particle fell.

But we've already discussed that there is no way to localize "potential energy" in a gravitaitonal field in a tensor manner, and I thought you'd agreed on this point.

I don't want to get too far afield, so I'll stop here, and see if I'm guessing correctly that you believe that the energy-momentum 4-vector includes potential energy somehow.

[pmb_phy]

Actually, I don't expect the energy or momentum of a particle, as defined by the energy momentum 4-vector of the particle, to be conserved at all as the particle falls towards a larger mass.

Why not? Just because I can't write the energy as a linear sum of other energy terms?

Therfore we would not expect to see the energy-momentum 4 vector conserved by a falling particle.

Nobody said that the energy-momentum 4 vector should be conserved for a falling particle.

Tell me. Now that you know that if the gravitational field is time independant, what do you think of the fact that the energy of the falling particle is constant? Does that surprise you?

I was speaking about the conservation of one component of a 1-form which has little to do with conservation of 4-momentum. Conservation of the time component of the 1-form which is dual to P does not mean conservation of the spatial components.

You've noticed that in one particular coordinate system, one component of the energy-momentum 4 vector is numerically constant.

Do you understand the physical significance of this? Think of Lagrangian dynamics - What is the requirement in classical (non-relativistic) mechanics for the energy of a particle to be constant? The total energy of a particle is not always constant even in classical non-relativistic mechanics. There are conditions for it to be constant. In general it isn't. Same in GR.

This is correct, and useful, but it has apparently misled you into thinking that the energy-momentum-4 vector should include potential energy, as near as I can tell.

I don't see how you could arrive at such a conclusion. I only said that the time component of a 1-form is the "energy" of the particle whose 4-momentum 4-vector is dual to this 1-form. Where did this "potential" thing come from? It wasn't from me.

I say this because I can see no reason that you'd expect energy to be conserved if you didn't include potential energy.

I expect the energy of a particle to be conserved when I see a field which is time-independant.
You wold expect, instead, that energy would not be conserved as the particle fell.

I don't see it that way. And its good that I don't since I'd be wrong if I did. In some sense P0 it does include potential energy. It is just not well-defined. Not being well defined does not mean that it does not exist or is not meaningful. P0 is simply not a linear sum of terms which has something which you'd call "potential". But P0 is a function of position and that is one of the trademarks of potential energy.

But we've already discussed that there is no way to localize "potential energy" in a gravitaitonal field in a tensor manner, and I thought you'd agreed on this point.

I'm not quite sure what you mean by this and I don't see why you're bringing this up? My comments on E = P0 = "energy" have nothing to do with potential energy. Since I'm not interested in potential energy I have nothing more to say about it in this post.

I don't want to get too far afield, so I'll stop here, and see if I'm guessing correctly that you believe that the energy-momentum 4-vector includes potential energy somehow.I don't recall ever saying that it did. What led you to believe that? I do think of it in some sense to contain potential energy, but not in the way that you think that I was.

As far as conservation goes, there are other ways to think about conservation. One way is to determine whether a quantity is cyclic in a conjugate variable. If Pa does not depend on xa then the momentum conjugate to that variable will be conserved. When a = 0 then P0 = constant = energy. Sound familar? Think of Lagrangian mechanics.

Pete

[pervect]

Tell me. Now that you know that if the gravitational field is time independant, what do you think of the fact that the energy of the falling particle is constant? Does that surprise you?


At first it surprised me slightly, but I realized that it was a reasonably straightforwards result from having a timelike Killing vector - which is equivalent to your description of time-independent Christoffel symbols (fields).


Do you understand the physical significance of this?


I'll have to think aboout this a bit. Mainly I'm concerned as to if this conserved quantity is the same as the other conserved quanties arising from asymptotic flatness. I think the fact that the timlike killing vectors are still timelike at infinity insures that this measure of energy is the same as the Bondi energy, so I'm leaning towards the point of view that you actually do have the energy of the system as E0 if you have a timelike Killing vector.



As far as conservation goes, there are other ways to think about conservation. One way is to determine whether a quantity is cyclic in a conjugate variable. If Pa does not depend on xa then the momentum conjugate to that variable will be conserved. When a = 0 then P0 = constant = energy. Sound familar? Think of Lagrangian mechanics.

Pete

I think you are going way,way,way,way far afield here in your interpretation of the significance of the constancy of E0. Since you haven't derived it from a Lagrangian (the Einstein-Hilbert action), your observations about it being the same as a classical Lagrangian are suspect at the very least. I also think you'll find that people tried for a very long time to get a conserved quantity out of the Einstein-Hilbert action, with *no* successs. Basically you have an interesting and useful result, but it won't generalize.

But at this point I want you to start thinking about the other disagreement we had, about the Riemann at the event horizon, so that we can at least to attempt to start unconfusing some confused newbies.

[pervect]

Oh, I just realized something else fairly important, that's been bothering me for a bit.

One of the things that one needs to do to find that E0 is conserved is to assume that the falling object follows a geodesic. This is true for small objects, but not for extremely heavy ones. Thus, for instance, two black holes orbiting around the common center of mass will not be following geodesics in space time, they will actually be emitting gravitational radiation and departing slightly from geodesic motion as they spiral into each other.

[pervect]

Let me expand a bit on my previous remark, or if you prefer, mention something else that's been bothering me.

Given that E0 is supposed to be the energy we should be able to get the total system energy by integrating E0. If we take the energy density Tab, we should be able to integrate over some volume \Sigma as follows

\int_\Sigma T_{ab} n^a K^b dV

where na is a unit vector normal to the volume element dV, the so-called "unit future", and Kb is a Killing vector

We should be able to do this because
T_{ab} n^a dV gives us the subscripted energy-momentum 4-vector, and multiplication of the subscripted energy-momentum-4 vector by the Killing vector Kb picks out the time component, giving us E0

However, we find that the above is NOT the correct formula for the total energy of a system with a time-like Killing vector given in Wald!

Wald, pg 289, eq. 11.2.10 gives the correct expression as

2 \int_{\Sigma} (T_{ab} - \frac{1}{2}Tg_{ab}) n^a K^b dV

T is I believe the trace of Tab

So therfore there is some sort of flaw in viewing Tab as the energy density of the system, because we can't simply integrate it to get the total energy.

[pmb_phy]

I think you are going way,way,way,way far afield here in your interpretation of the significance of the constancy of E0. Since you haven't derived it from a Lagrangian (the Einstein-Hilbert action), your observations about it being the same as a classical Lagrangian are suspect at the very least.

I wasn't thinking about a Lagrangian. I was thinking only of Lagrange's equations for a particle in a field. If xa is clyclic then the canonical momentum conjugate to xa is zero. That holds in GR too.

But at this point I want you to start thinking about the other disagreement we had, about the Riemann at the event horizon, so that we can at least to attempt to start unconfusing some confused newbies.Only thing they'll be confused about is if they didn't read what I said. I specifically stated that the tidal forces measured by a freely falling observer are finite at the event Horizon. I then said that when you calculate the Riemann tensor in Schwarzschild coordinates then at least one component is infinite at the horizon and that what that means physically is a bit difficult to interpret. Anyone who is a newbie can't confuse the first comment with yours since they are identical. The later is a result of a calculation. The result for one of the components is

R^0_{101} = \frac {r_s}{r^3}\frac{1}{1-r_s/r}

I mean nothing more and nothing less than this component in this coordinate system is infinite at r = rs. I don't know if this means that a person at rest relative to a BH will measure finite or infinite tidal forces on a person who sits closer and closer to the event horizon.

Are you saying that the component above in Scwazchild coordinates is incorrect? Are you saying that this component in these coordinates is finite when r = rs.

Pete

[pmb_phy]

Just a note: Someone once said it is only possible for the energy of a particle to be conserved when there exists a timelike Killing vector. In which case

\xi_{\mu}P^{\mu} = constant

That is a coordinate indepenant way to phrase this. That statement leads to P0 = constant in a particular frame. Another way to say this is to say that if there exists a coordinate system in which the metric is time-independant. This is a coordinate dependant way of saying the same exact thing. Each is correct.

Pete

[selfAdjoint]

But the existence of a time-like Killing vector is by no means guaranteed. It depends on the spacetime.

[pmb_phy]

But the existence of a time-like Killing vector is by no means guaranteed. It depends on the spacetime.
Hence the term when in ..when there exists a timelike Killing vector.

Even in classical mechanics and electrodynamics the energy of a particle in a field isn't always conserved. Its only conserved when the field is conservative, e.g. when the potential is not a function of time.

Pete

[pervect]

Just a note: Someone once said it is only possible for the energy of a particle to be conserved when there exists a timelike Killing vector.


Who was that?

We've been talking a lot about the case where there is a timelike Killing vector, because it's common, useful, and is much easier to deal with.

But asymptotic flatness of the metric is all that's needed to be able to define a conserved energy according to my text (Wald), and the sci.physics.relativity FAQ.

The defintion of asymptotic flatness ala Wald is actually quite involved, though the basic concept is simple. The simple, conceputal version is that the metric coefficients have to look like a Minkowski metric far enough away from the source. So we don't need timelike Killing vectors to define energy, but it's nice when we have them.

[pmb_phy]

Who was that?Steve Carlip

We've been talking a lot about the case where there is a timelike Killing vector, because it's common, useful, and is much easier to deal with.
And harder to understand for those here who are less adept at this crazy math stuff. :smile:

But asymptotic flatness of the metric is all that's needed to be able to define a conserved energy according to my text (Wald), and the sci.physics.relativity FAQ.
The total energy of the gravitating source or for a particle moving in the field the source creates? I was speaking of the later.

Pete

[pervect]

Steve Carlip
And harder to understand for those here who are less adept at this crazy math stuff. :smile:


Yes, but when you actually want to calculate things, Killing vectors are extremely handy. With different clocks ticking at different rates, what constitutes a "time translation" can get confusing without the formalism. With the formalism, this notion is made unambiguous and independent of various coordiante choices (covariant or contravariant coordinates, for instance).


The total energy of the gravitating source or for a particle moving in the field the source creates? I was speaking of the later.
Pete

I was speaking of the former

[Orion1]

'Massless' Photonic-Schwarzschild BH:

Radial solution for sphereically symmetric Schwarzschild metric:
r_b = \frac{2GM_b}{c^2}

Mass-Energy Equivalence principle for Schwarzschild BH:
M_b = \frac{E_b}{c^2} = \frac{r_b c^2}{2G}
E_b = \frac{r_b c^4}{2G}

Energy Equivalence for single photon:
E_p = \frac{\hbar c}{\overline{\lambda_p}}
\overline{\lambda_p} - wavebar (photon wavelength)

General Relativity Mass-Energy Equivalence Principle:
E_p = E_b

\frac{\hbar c}{\overline{\lambda}} = \frac{r_b c^4}{2G}

Radial-Wavebar solution for spherically symmetric 'massless' Schwarzschild BH:
r_b \overline{\lambda} = \frac{2 \hbar G}{c^3}
r_b = \overline{\lambda}
r_b = \sqrt{\frac{2 \hbar G}{c^3}

This solution describes a 'Massless' Photonic-Schwarzschild BH composed of a single photon travelling at luminous velocity.

'Massless' Photonic-Schwarzschild BHs exist as a mathematical solution in General Relativity due to the General Relativity Mass-Energy Equivalence Principle. In Classical GR, BHs can be composed of both mass and/or energy.

Based upon the Orion1 equasions, what is the radius and wavelength for a 'Massless' Photonic-Schwarzschild BH?

Based upon the Orion1 equasions, what is the energy magnitude for this single photon?


if a black hole is made from photons, would it be massless and move at the speed of light?


The Orion1 solution descibes a classical GR 'massless' Photonic-Schwarzschild BH composed of a single photon travelling at luminous velocity.

[DW]

'Massless' Photonic-Schwarzschild BH:

Radial solution for sphereically symmetric Schwarzschild metric:
r_b = \frac{2GM_b}{c^2}

Mass-Energy Equivalence principle for Schwarzschild BH:
M_b = \frac{E_b}{c^2} = \frac{r_b c^2}{2G}
E_b = \frac{r_b c^4}{2G}

Energy Equivalence for single photon:
E_p = \frac{\hbar c}{\overline{\lambda_p}}
\overline{\lambda_p} - wavebar (photon wavelength)

General Relativity Mass-Energy Equivalence Principle:
E_p = E_b

\frac{\hbar c}{\overline{\lambda}} = \frac{r_b c^4}{2G}

Radial-Wavebar solution for spherically symmetric 'massless' Schwarzschild BH:
r_b \overline{\lambda} = \frac{2 \hbar G}{c^3}
r_b = \overline{\lambda}
r_b = \sqrt{\frac{2 \hbar G}{c^3}

This solution describes a 'Massless' Photonic-Schwarzschild BH composed of a single photon travelling at luminous velocity.

'Massless' Photonic-Schwarzschild BHs exist as a mathematical solution in General Relativity due to the General Relativity Mass-Energy Equivalence Principle. In Classical GR, BHs can be composed of both mass and/or energy.

Based upon the Orion1 equasions, what is the radius and wavelength for a 'Massless' Photonic-Schwarzschild BH?

Based upon the Orion1 equasions, what is the energy magnitude for this single photon?


if a black hole is made from photons, would it be massless and move at the speed of light?


The Orion1 solution descibes a classical GR 'massless' Photonic-Schwarzschild BH composed of a single photon travelling at luminous velocity.

No it doesn't. This is yet another example why "relativistic mass" is a bad concept.

[pervect]

[color=blue]
'Massless' Photonic-Schwarzschild BH:

Radial solution for sphereically symmetric Schwarzschild metric:
r_b = \frac{2GM_b}{c^2}

etc etc etc



This random assortment of equations appears to me to be "not even wrong".

[pmb_phy]

This random assortment of equations appears to me to be "not even wrong".
Seems to be that Orion1 is confusing the proper mass, M, which appears in the Schwazchild metric with the relativistic mass of a photon. So the lack of clarity of this distinction has take one more soul. :rofl: (just kidding of course).

Pete

[Orion1]

Classical Radial solution for spherically symmetric Schwarzschild metric:
L = 0 - angular momentum
r_b = \frac{2GM_b}{c^2}

Relativistic Radial solution for spherically symmetric Schwarzschild metric:
L = 0 - angular momentum
n_v = \left( \frac{v}{c} \right) - velocity number
\gamma = \sqrt{(1 - n_v^2)}^{-1}
r_b = \frac{2G \gamma M_b}{c^2}

r_b = \frac{2GM_b}{c^2 \sqrt{(1 - n_v^2)}}
n_v = .998

Based upon the Orion1 solution, what is the relativistic effect on the Schwarzschild metric for a near-luminous velocity Schwarzschild BH with L = 0 angular momentum?

[pervect]

I think I've said about all I need to say about my opinion of Orion's ramblings.

To move onto more positive matters, I will go back to an old issue. It turns out that when pmb was talking about E_0 being conserved, and tying this to a Lagrangian, there is a rigorous justification for this.

The geodesic deviation equations can be derived from a "least action" principle. In this sense, there is a rigorously justifiable "Lagrangian" and a so-called "Super-Hamiltonian" that can be talked about for geodesic motion. This also justifies MTW's remarks about how t^a and E_0 were "conjugate momenta" .

I thought this was sort of interesting, when I ran across it. It doesn't affect of course any of my remarks about the computation of system energy, but it is relevant to conserved quantites for systems following geodesic motion.

[Chronos]

Hmm, it seems a lot simpler to assume the net total energy of particles in a conserved energy field, such as gravity, is always zero.

[pervect]

Hmm, it seems a lot simpler to assume the net total energy of particles in a conserved energy field, such as gravity, is always zero.

Well, if you scroll back over the previous long discussion, you'll see that the conservation of energy in GR isn't, in general, very simple.

There are a few very simple quantites that are conserved in the Schwarzschild metric, though, that are very useful, and easy to explain without trotting out the whole dog_and_pony show. One of these is the covariant value of the energy component of the energy-momentum 4-vector, E0.

While this only works in Schwarzschild coordinates, or other coordinates that have a unit timelike killing vector, it's quite handy when it can be used.