B How is energy conserved in General Relativity?

[sweet springs]

I would like to understand better about the conservation of energy in GR.

Let us think of infinitesimal vacuum volume dr\ sin\theta d\theta d\phi around the star in center.
Light emitted from the star hit the bottom surface, r, of the volume. Say violet light photons hit the area 1 photon/1 second of the bottom local time.
Ligth escapes the volume at the top surface, r+dr, with red-shifted color with the rate less than 1 photon/1 second of the top local time, I guess.

In coming energy and out going energy should be equal in this stationary case. How can I cofirm it for the case?

 

[Cryo]

In coming energy and out going energy should be equal in this stationary case. How can I cofirm it for the case?

I am not an expert in GR, so I appologize in advance if I say something wrong, but why should the energy be conserved here? Energy of a particle is just the zero-th component of its four-momentum, here you clearly have a situation where the manifold is not trivial, so components of a four-vector will change if you parallel-transport it around the space (I am guessing), so why would you expect the energy conservation to apply?

Having said that. Looking at the Schwartzschild metric, it has no explicit time dependence, so, I guess, the relevant Lagrangian density will also not have explicit time dependence, and therefore there should be some quantity that is conserved as a result of this. But will it correspond to what we usually call energy?

 

[sweet springs]

Thanks Cryo.
I have understood conservation of energy as denial of perpetual motion machine that can produce as much energy as we wish.
If GR denies conservation of energy, I am interested in how it denies, e.g. do perpetual motion machines revive? (though I do not expect so.)

 

[Cryo]

I have understood conservation of energy as denial of perpetual motion

That sounds rather abstract. How would you actually calculate the energy?

 

[sweet springs]

Rather abstract, I admit. I am not sure how I calculate the energy.

I am not sure also about definition of conservation.

a.\ T^{\mu\nu}_{,\nu}=0 , where "," means ordinary coordinate derivative, holds in classical and SR physices but doses not hold in GR
b.\ T^{\mu\nu}_{:\nu}=0, where ":" means covariant derivative, holds in general.
Which is the formula of "conservation" in definition ?

If it is a., energy is not coserved in GR. Based on our wish that something else of energy should be "conserved" non tensor t is invented , i.e.
(T^{\mu\nu}+t^{\mu\nu}){,\nu}=0

 

[Dale]

Which is the formula of "conservation" in definition ?

It is b.

 

[sweet springs]

Thanks Dale.

 

[sweet springs]

I have studied GR by old text Dirac 1975 section 24 of which says,
"In flat spacetime equation (24.5) ( my b.) would become (my a.) and would then give conservation of energy and momentum. In curved space the conservation of energy and momentum is in only approxomation. The error is to be ascribed to the gravitational field working on the matter and having itself some energy and momentum."

I find the text available on the web at https://www.google.com/url?sa=t&rct...lativity.pdf&usg=AOvVaw108LkvdoRz-qsnip_6hwIP

This old interpretaiton was the reason why I thought that definition of conservation might be a.

Best

 

[sweet springs]

In elementary physics of motion in gravitational field we learn the consercvation of energy
\frac{1}{2}mv^2+mgh=Constant
introducing gravitational potential energy.

After learning GR I think GR could give reinterpretaion of the conservation relation, not using potential energy that we cannot ascribe where it is because it does not have tensor property, but using curved spacetime or covariant derivative forluma of conservation relation. I would like to know it if there already is.

Best

 

[Dale]

I don’t know why it would be described as an approximation. Your b is exact in GR.

 

[sweet springs]

Thanks again for reconfirming that b. stands in GR and it is called conservation relation, and a. is not called conservtion relation in GR and does not hold.

 

[sweet springs]

Light emitted from the star hit the bottom surface, r, of the volume. Say violet light photons hit the area 1 photon/1 second of the bottom local time.
Ligth escapes the volume at the top surface, r+dr, with red-shifted color with the rate less than 1 photon/1 second of the top local time, I guess.

In elementary physics of motion in gravitational field we learn the consercvation of energy

12mv2+mgh=Constant​

\frac{1}{2}mv^2+mgh=Constant
introducing gravitational potential energy.

In these cases of motion in gravitational field, the formula of covariant derivative divergence of stress energy tensor

T^{\mu\nu}_{:\nu}=T^{\mu\nu}_{,\nu}+\Gamma^{\mu}_{\nu\alpha}T^{\nu\alpha}+\Gamma^{\nu}_{\nu\alpha}T^{\mu\alpha}=0,
is applicable.

The second and the third term in RHS are proportional to T in rough saying.
Christoffel symbol Gammma should correspond to traditional g, gravitationla accerelation constant.

Gravitation force in hypothetical flat space in classical mechanics and SR is to be reinterpreted as the part of covariant divergence term in curved space.

 

[Matterwave]

In elementary physics of motion in gravitational field we learn the consercvation of energy
\frac{1}{2}mv^2+mgh=Constant
introducing gravitational potential energy.

After learning GR I think GR could give reinterpretaion of the conservation relation, not using potential energy that we cannot ascribe where it is because it does not have tensor property, but using curved spacetime or covariant derivative forluma of conservation relation. I would like to know it if there already is.

Best

Unfortunately, moving to GR, you lose such a neat and convenient definition of "gravitational potential energy". In GR, such a concept is simply not well defined. A potential energy has to do with (conservative) forces - in GR, gravity is no longer a force so it becomes much harder to define its potential energy.

 

[PeterDonis]

moving to GR, you lose such a neat and convenient definition of "gravitational potential energy". In GR, such a concept is simply not well defined

This is true in general, but there is a class of spacetimes, stationary spacetimes, in which there is a well-defined gravitational potential energy. The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.

 

[PeterDonis]

After learning GR I think GR could give reinterpretaion of the conservation relation

There is; the general concept you are looking for here is called "energy at infinity" and is a constant of the motion for free-falling test objects in a stationary spacetime.

Note, however, that this has nothing to do with your question about conservation laws for the stress-energy tensor; that is a whole different subject.

 

[Matterwave]

This is true in general, but there is a class of spacetimes, stationary spacetimes, in which there is a well-defined gravitational potential energy. The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.

Could you remind me how the gravitational potential energy is defined in stationary space times? I do not recall a general definition. All I could say is given a time-like Killing field $\xi^a$, there is a conserved quantity $\xi^a u_a$ for geodesics $u^a$...but how that conserved quantity relates to a gravitational potential energy is escaping me at the moment.

EDIT: Specifically, I know one could associate a "total energy" (including gravitational) to the quantity $E=-g_{ab}\xi^a u^b$ and this quantity is conserved along geodesics. But how to deconstruct that into part Gravitational Potential energy and part "other" energy (kinetic energy/rest energy) is not apparent to me.

 

[PeterDonis]

given a time-like Killing field $\xi^a$, there is a conserved quantity $\xi^a u_a$ for geodesics $u^a$

Yes. This is the energy at infinity, which can be thought of as the total of kinetic energy + potential energy. (More precisely, it's the energy per unit mass at infinity, assuming $u^a$ is a 4-velocity and not a 4-momentum.) The potential energy (more precisely, the potential energy per unit mass) is just $V = \vert \xi^a \vert = \sqrt{ g_{ab} \xi^a \xi^b}$, i.e., the norm of the Killing field.

 

[PeterDonis]

The potential energy (more precisely, the potential energy per unit mass) is just $V = \vert \xi^a \vert = \sqrt{ g_{ab} \xi^a \xi^b}$, i.e., the norm of the Killing field.

Actually, this definition makes the potential energy at infinity equal to $1$ for asymptotically flat spacetimes; the usual convention is for it to be $0$, so that potential energy at any finite radius is negative. So $V = \vert \xi^a \vert - 1$ is the more usual definition.

 

[sweet springs]

The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.

Thanks Peter.　I wonder where is or in other words distribution of gravitation energy as for this well-defined potential energy for my scenario.

 

[Matterwave]

Actually, this definition makes the potential energy at infinity equal to $1$ for asymptotically flat spacetimes; the usual convention is for it to be $0$, so that potential energy at any finite radius is negative. So $V = \vert \xi^a \vert - 1$ is the more usual definition.

That's an interesting definition. I don't think I can recall seeing it anywhere, but admittedly it has been a while since I've done too much in GR. Could you give some motivating background for why it's a good definition, or perhaps a source that talks about it?

 

[PeterDonis]

Could you give some motivating background for why it's a good definition, or perhaps a source that talks about it?

I think Wald discusses it, but it's been a while since I looked at the sources for this.

As a quick heuristic motivation for the definition, consider that $\vert \xi^a \vert$ is the "redshift factor". So, for example, it gives the factor by which the energy of a photon emitted by an observer at rest at a finite radius is redshifted when it is received by an observer at infinity. Then consider Einstein's thought experiment for why there must be gravitational redshift, with the two observers being the ones just mentioned: the observer at infinity takes an object at rest relative to him and drops it; it free-falls down to the observer at finite radius, who takes it, converts it entirely into photons, and emits them back up to the observer at infinity. You will see that the "redshift factor" must also give the potential energy change between the observer at infinity and the observer at the finite radius.

 

[Matterwave]

I think Wald discusses it, but it's been a while since I looked at the sources for this.

As a quick heuristic motivation for the definition, consider that $\vert \xi^a \vert$ is the "redshift factor". So, for example, it gives the factor by which the energy of a photon emitted by an observer at rest at a finite radius is redshifted when it is received by an observer at infinity. Then consider Einstein's thought experiment for why there must be gravitational redshift, with the two observers being the ones just mentioned: the observer at infinity takes an object at rest relative to him and drops it; it free-falls down to the observer at finite radius, who takes it, converts it entirely into photons, and emits them back up to the observer at infinity. You will see that the "redshift factor" must also give the potential energy change between the observer at infinity and the observer at the finite radius.

I don't recall seeing this in Wald, but I will have a look. Thanks for these posts, they've been insightful. :)

 

[sweet springs]

Re: post#5,#6, #8 and #10

It is b.

Landau-Lifshitz old text says in chapter of energy momentum pseudotensor, that \int T^k_i\sqrt{-g}dS_k is conserved in condition that \frac{\partial \sqrt{-g}T^k_i}{\partial x^k}=0 so the integral is not conserved in case covariant derivative of T equals zero. How should I conciliate one ane the other?

 

[Dale]

I tend to avoid the pseudo tensor entirely. It isn’t a real tensor and I would rather stick with true tensors.

The conservation law I identified above applies for the actual stress energy tensor. If you use the pseudotensor then you should not expect it to apply.

 

[sweet springs]

Thnaks　Dale.

The formula of differention,
T^{\mu\nu}_{:\nu}=0
,is a equation of continuity (of energy-momentum).

The formula of 3d volume integration
\int(t^0_\mu+Y^0_\mu)\sqrt{-g}dx^1dx^2dx^3 Dirac(31.4)
, is a equation of conservation of energy momentum of the system where no flow in 3d surface boundary.

Differential thus local continuation of T is not enough and it needs t to expalin integral thus global conservation of energy momentum. Am I right? The formula of continuation and the formula of conservtion do not have to be the same one ?

 

[PAllen]

The (integral conservation) equations Einstein, Landau-Lifschitz, and presumably Dirac, specify for the energy momentum pseudo-tensor are exact rigorous equations. They hold in any coordinates. The issue is that the pseudo-tensor is not covariant, nor tensor density or any other standard geometric object. Further, interpreting it as 'energy including gravity' in some given coordinates, typically leads to implausible results. Thus, the majority view is that the pseudotensor does not meaningfully localize energy, and the conservation laws are not meaningful. This majority view is further bolstered by Noether's theorems, which imply there cannot be well formed conservation laws in GR.

However, there are experts who disagree with this view. Nakanishi (google his papers) argues that the pseudotensor produces physically plausible results in harmonic coordinates, and this plus the fact that conservation equations hold in any coordinates, rescues the relevance of the pseudotensor.

 

[sweet springs]

Thnaks PAllen

Thus, the majority view is that the pseudotensor does not meaningfully localize energy, and the conservation laws are not meaningful. This majority view is further bolstered by Noether's theorems, which imply there cannot be well formed conservation laws in GR.

Noether's theorem tells us conservation laws cannot stand in GR. It's great.

Back to my post#9, I will be satisfied better if someone show me how conservation of energy in gravitational field
\frac{1}{2}mv^2+mgh=const.
in high school physics, should be reinterpreted in GR rigorously.

 

[PeterDonis]

I will be satisfied better if someone show me how conservation of energy in gravitational field

$$
\frac{1}{2}mv^2+mgh=const.
$$

in high school physics, should be reinterpreted in GR rigorously.

I already answered that in post #15: the energy at infinity is a constant of geodesic motion in GR, for spacetimes in which it is well-defined. The Newtonian formula you give is an approximation to the GR energy at infinity for the case where the range of height $h$ is small compared to the range over which $g$ changes, and for which $v$ is much less than $c$. (The approximation also requires subtracting off the rest energy $mc^2$, which is part of the energy at infinity in GR, and adjusting the "zero point" of potential energy to a finite height.)

 

[sweet springs]

Post #15

There is; the general concept you are looking for here is called "energy at infinity" and is a constant of the motion for free-falling test objects in a stationary spacetime.

Let me take an example to understand your explanation. On the ground I pop up the baseball of mass m. The ball reaches height h and falls back into my glove. In this case of high school physics class, what, how much and where is "energy at infinity" and "potential energy" in rigorous GR?

 

[PeterDonis]

On the ground I pop up the baseball of mass m. The ball reaches height h and falls back into my glove. In this case of high school physics class, what, how much and where is "energy at infinity" and "potential energy" in rigorous GR?

Remember that the formula for energy at infinity (per unit mass) is $\xi^a u_a = g_{ab} \xi^a u^b$, where $\xi^a$ is the timelike Killing vector field and $u^b$ is the 4-velocity. We can just multiply by the rest mass $m$ to get the energy total.

We will assume that the Earth is not rotating, so we can use the simple values from the Schwarzschild metric. Then, in standard Schwarzschild coordinates, we have

$$
\xi^a = \left( 1, 0, 0, 0 \right)
$$

The simplest place to calculate the energy at infinity is at the top of the ball's trajectory, where it is momentarily at rest. Then its 4-velocity in Schwarzschild coordinates is

$$
u^b = \left( \frac{1}{\sqrt{1 - 2M / r}}, 0, 0, 0 \right)
$$

The energy at infinity is then simply

$$
E = m g_{ab} \xi^a u^b = m \left( 1 - \frac{2M}{r} \right) \frac{1}{\sqrt{1 - 2M / r}} = m \sqrt{1 - \frac{2M}{r}}
$$

where $r$ here is the radial coordinate of the ball at the top of its trajectory. For this case we can approximate $r = R + H$, where $R$ is the radius of the Earth and $H$ is the maximum height the ball reaches. So we have

$$
E = m \sqrt{1 - \frac{2M}{R + H}}
$$

Since $H << R$ and $M << R$ (note that we are using units in which $G = c = 1$), we can further approximate this as

$$
E = m \sqrt{1 - \frac{2M}{R} \left( 1 - \frac{H}{R} \right)} = m \left( 1 - \frac{M}{R} \left( 1 - \frac{H}{R} \right) \right) = m \left( 1 - \frac{M}{R} + \frac{M}{R^2} H \right)
$$

Now we simply observe that $g = M / R^2$ near the Earth's surface, and subtract off the constant $m \left( 1 - M/R \right)$, to obtain $E = m g H$. Note that, since the ball is at rest at this point, the total energy is equal to the potential energy.

From here, since we are in a regime where Newtonian gravity is a good approximation, finding the general formula $E = \frac{1}{2} m v^2 + m g h$ is straightforward, since $E$ is a constant of the motion and we already know how $v$ and $h$ are related, given that the ball is at rest with $v =0$ at $h = H$.

 

[sweet springs]

Thanks PeterDonis

The energy at infinity is then simply

m\sqrt{1-\frac{2M}{r}}
Why this energy is called "at infinity"? The ball is bound on the Earth and does not go to infinity.

 

[PeterDonis]

Why this energy is called "at infinity"?

History, mostly. Many physicists thought of this constant of the motion as "the energy an observer at infinity would observe the object to have". There are certain scenarios that can motivate this interpretation (for example, imagining the ball being moved from rest at infinity to some particular state at a finite height, and measuring the energy the ball has to give up). But the name is not intended to imply that it is only valid for objects that move to or from infinity.

 

[sweet springs]

Many physicists thought of this constant of the motion as "the energy an observer at infinity would observe the object to have".

In cases that the ball move to or from infinity an observer at infinity could observe energy of their ball in his local Minkowsky spacetime. His local Minkowsky spacetime and the Schwartzshild spacetime coincide at infinity.

In ohter cases that the ball is bound around the Earth, it is not sure how an observer at infinity would observe the distant motion, e.g. what kind of reference system he use for events in such a distant place.

 

[Dale]

what kind of reference system he use for events in such a distant place.

Actually, for this quantity you don’t need a specific reference frame. It is an invariant quantity. All you need is the timelike Killing vector field. Think of it as a correction factor for gravitational time dilation, which is well defined in such spacetimes.

 

[PAllen]

And the timelike killing field implies time translation symmetry, so a robust conservation of energy is expected per Noether.

 

[sweet springs]

Let me confirm some calculation based on post #30

Along the geodesic
u^0=\frac{1}{\sqrt{1-\frac{2M}{r_{max}}}}
u^0u_0=(1-\frac{2M}{r})\frac{1}{1-\frac{2M}{r_{max}}}&gt;1
So square of velocity with minus signature is,say x^1=r, u^1u_1=1-u_0u^0=\frac{\frac{2M}{r}-\frac{2M}{r_{max}}}{1-\frac{2M}{r_{max}}}

In GR is energy mu^0u_0 or "energy at infinity" that includes "gravitational potential energy" ?

 

[PeterDonis]

Along the geodesic
$$
u^0 = \frac{1}{\sqrt{1 - \frac{2M}{r_{max}}}}
$$

No, this is not true everywhere along the geodesic. It is only true that the point at the top of the trajectory, where the ball is momentarily at rest.

To get an equation for the components of $u^a$ that is valid everywhere, you would need to solve the geodesic equation in Schwarzschild coordinates.

In GR is energy
$$
mu^0u_0
$$
or "energy at infinity" that includes "gravitational potential energy" ?

"Energy" is a term that can have multiple meanings. Energy at infinity is a constant of geodesic motion, so it's the best way of addressing the question you originally asked, which is what the GR counterpart is of the Newtonian energy equation you gave.

As far as $m u^0 u_0$, that quantity does not correspond to any definition of "energy" I have seen, so I don't know why you bring it up.

 

[sweet springs]

Thanks. I correct miscalculatin in post#36. Sayx^0=t,x^1=r,

Along the geodesic

g_{ab}\xi^au^b=g_{00}\xi^0u^0=\sqrt{1-\frac{2M}{r_{max}}}
u^0=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{g_{00}\xi^0}=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}

Except the top of trajectory the ball has velocity whose square with minus signature is
u_1u^1=1-u^0u_0

Applying the formula of energy of free particle in SR,
E=p^0=mu^0=m\frac{\sqrt{1-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}
E^2=m^2u^0u_0=m^2\frac{1-\frac{2M}{r_{max}}}{1-\frac{2M}{r}}\approx m^2[1+2g(H-z)] where z is height from the ground.
E\approx m+mg(H-z)

Your check comments are appreciable.

 

[PeterDonis]

Applying the formula of energy of free particle in SR

SR formulas are not valid here. We're using curvilinear coordinates in curved spacetime.

 

[sweet springs]

SR formulas are not valid here. We're using curvilinear coordinates in curved spacetime.

In SR, the energy-momentum of particle is mu^i. If It does not apply to GR, what is energy-momentum of particle in GR? The concept of energy-momentum of particle in SR does not survive in GR?

From the last formula of post#38 is written as
|p^0|+mgz \approx m+mgH=const.

We usually interpret it that
a. this constant is total energy that consists of matter energy |p^0| and gravitational energy mgz. Energy conserves.

But where is gravitational energy mgz stored? (See post#19) In a particle? on geodesic? field around the Earth and the ball? This gravitational energy as part of RHS of Einstein equation does curve spacetime nearby ? I do not think so.
I would rather say
b. energy is just |p^0|. There is no other energy. |p^0| varies along the geodesic so energy does not conserve.

However, ref. post#35, Noether theorem allows the product of Killing vector and 4-velocity along the geodesic is constant. This constant should be defined as energy as its name "energy at infinity" suggests ? I will appreciate your teachings.

 

[PeterDonis]

what is energy-momentum of particle in GR?

The energy-momentum 4-vector of a particle is $m u^a$, where $u^a$ is the 4-velocity. This is the same as in SR. But the energy-momentum 4-vector is not the same as "energy".

From the last formula of post#38

I have already told you that your use of SR formulas in post #38 is wrong. Garbage in, garbage out.

I have also already told you how to derive a general formula, exactly valid in GR, for the energy at infinity in terms of the 4-momentum components for the entire trajectory (instead of just the point of maximum height): you need to solve the geodesic equation in Schwarzschild coordinates. That is what you need to do, instead of waving your hands with wrong formulas.

where is gravitational energy mgz stored?

Nowhere. Gravitational potential energy cannot be localized.

Noether theorem allows the product of Killing vector and 4-velocity along the geodesic is constant. This constant should be defined as energy as its name "energy at infinity" suggests ?

You can certainly call the energy at infinity the "energy" of the particle. Nothing is stopping you. But then you need to stop trying to say that other things are "energy".

If you want to define something else as "energy", you need to explain what it is you are defining as "energy" and why such a definition is meaningful. In doing so, you cannot rely on definitions that work in SR, since we are not dealing with SR here.

 

[sweet springs]

The energy-momentum 4-vector of a particle is muam u^a, where uau^a is the 4-velocity. This is the same as in SR. But the energy-momentum 4-vector is not the same as "energy".

Thanks. 0-component of energy-momentum 4-vector of particle
p^0=mu^0
is energy in SR. Is it same in GR?

 

[sweet springs]

I have already told you that your use of SR formulas in post #38 is wrong. Garbage in, garbage out.

Your post #37 to my post #36 helped me to write correction post #38. I am sad that post #38 is wrong again. I am still want to make correction. So,

Your check comments are appreciable.

.

 

[sweet springs]

Actually,

you need to solve the geodesic equation in Schwarzschild coordinates. That is what you need to do, instead of waving your hands with wrong formulas.

I thought in my post #38, without solving the equation the conserved quantity as mentioned in

All I could say is given a time-like Killing field ξa\xi^a, there is a conserved quantity ξaua\xi^a u_a for geodesics uau^a.

works. Is this quantitiy is conserved along a geodesic ?

 

[sweet springs]

Nowhere. Gravitational potential energy cannot be localized.

I wonder whether the existence of not-localized entity is real and healthy one.
Maybe I am wrong as Einstein was in EPR paradox.

 

[sweet springs]

|p0|+mgz≈m+mgH=const.

Notion |p^0| in post#40 was ambiguous and incorrect. It shoud be read as or replaced by \sqrt{p^0p_0}.

 

[PeterDonis]

Is it same in GR?

The best "B" level answer I can give is "no".

Is this quantitiy is conserved along a geodesic ?

Yes, but the equations you are waving your hands and writing down by guessing don't express that conservation law. To express it correctly the way you want to, in terms of the 4-velocity components of the baseball at an arbitrary point on its trajectory, you need to solve the geodesic equation in Schwarzschild coordinates.

I wonder whether the existence of not-localized entity is real and healthy one.

We are talking about classical physics here, not QM. The fact that gravitational potential energy cannot be localized was already known in Newtonian physics; it is not something new that GR introduced. There is no issue with it in classical physics.

Also, while the concept of "gravitational potential energy" is useful, it is not necessary. You can solve the equations and make correct predictions without using it at all. So if you have a problem with non-localized entities, you can just discard gravitational potential energy and make predictions without it.

It shoud be read as or replaced by

$$
\sqrt{p^0 p_0}
$$

Which still has nothing to do with anything we're discussing, so I don't know why you keep bringing it up. Stop guessing and do what I've asked you to do several times now.

 

[sweet springs]

The best "B" level answer I can give is "no".

Thanks.　　Though form of energy momentum tensor $T^{ik}$, e.g. for dust $\rho u^i u^k$, is familiar in RHS of Einstein equation, you say energhy of particle $p^0=mu^0$ in SR has nothing to to with in GR. Quite amazing. There is a lot of things for me to learn GR.

 

[PAllen]

Thanks.　　Though form of energy momentum tensor $T^{ik}$, e.g. for dust $\rho u^i u^k$, is familiar in RHS of Einstein equation, you say energhy of particle $p^0=mu^0$ in SR has nothing to to with in GR. Quite amazing. There is a lot of things for me to learn GR.

Well, not quite nothing. Given 4 velocity, which in GR is the absolute derivative along path by proper time (i.e. connection coefficients involved), then given a local timelike basis vector, the mass times 4-velocity contracted (via the metric) with the timelike basis vector will be the energy measured by a local device whose 4 velocity is the same as that timelike basis vector. More generally, with correct covariant definitions, SR kinematic and dynamic quantities go over into local observables in GR

 

[sweet springs]

Yes, but the equations you are waving your hands and writing down by guessing don't express that conservation law.

My guessing is
\xi^\mu=(1,0,0,0) in everywhere. And there is no non-diagonal element in the metric,
g_{0i}=0 so product $g_{\mu\nu}u^\mu \xi^\nu$ reduces to $g_{00}u^0 \xi^0=g_{00}u^0$=its value at the top, all along the geodesic.
What is wrong with it?