Schwarzschild metric and BH mass

[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 referred 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.

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 referred 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 K^a = [1,0,0,0] and K_a = [(-1 + 2m/r),0,0,0] , so the conserved quantities are K^a E_a and K_a E^a, 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 K_a;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 referred 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</sup>

 

[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 separate 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 Schwarzschild 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 g_uv 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 referred 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. 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</sup>

 

[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]v^a

where v^a is the 4-velocity of the observer.</sub>

<sub>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) v^a = (1, 0, 0, 0)

E = -p0[/sup]

Pete</sub>

 

[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 P_a does not depend on x^a 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 P_a does not depend on x^a 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 T_ab, we should be able to integrate over some volume \Sigma as follows

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

where n^a is a unit vector normal to the volume element dV, the so-called "unit future", and K^b 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 K^b 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 T_ab

So therefore there is some sort of flaw in viewing T_ab 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 x^a is clyclic then the canonical momentum conjugate to x^a 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 = r_s. 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 = r_s.

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.

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.

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 Metric...


'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 traveling 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 equations, what is the radius and wavelength for a 'Massless' Photonic-Schwarzschild BH?

Based upon the Orion1 equations, what is the energy magnitude for this single photon?
[/color]

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

The Orion1 solution descibes a classical GR 'massless' Photonic-Schwarzschild BH composed of a single photon traveling at luminous velocity.
[/color]

 

[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 traveling 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 equations, what is the radius and wavelength for a 'Massless' Photonic-Schwarzschild BH?

Based upon the Orion1 equations, what is the energy magnitude for this single photon?
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if a black hole is made from photons, would it be massless and move at the speed of light?
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The Orion1 solution descibes a classical GR 'massless' Photonic-Schwarzschild BH composed of a single photon traveling at luminous velocity.
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No it doesn't. This is yet another example why "relativistic mass" is a bad concept.

 

[pervect]

'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. (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.