Exploring the Concept of Energy in General Relativity

[sweet springs]

Let me confirm some calculation based on post #30

Along the geodesic
$$u^0=\frac{1}{\sqrt{1-\frac{2M}{r_{max}}}}$$
$$u^0u_0=(1-\frac{2M}{r})\frac{1}{1-\frac{2M}{r_{max}}}>1$$
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. Say$$x^0=t,x^1=r$$,

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?

 

[sweet springs]

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.

In Newton mechanics the ball problem is solved by applying equation of motion. It is also solved by conservation of energy law intorducing mgh.
They are equivalent.
In GR the ball problem is solved by geodesic. It is also solved by "conservation of energy law introducing gravitational energy".
I am sure about the geodesic way, the former. I would rather doubt whether the latter is just an interpretation for easy applications, introducing virtual and non-localized gravitational energy and law of conservation of energy.

 

[PeterDonis]

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

You are confusing yourself by making an ambiguous statement. It is true that, at every point on the geodesic, $g_{ab} \xi^a u^b$ reduces in the coordinates we are using to $g_{00} \xi^0 u^0 = g_{00} u^0$, because, as you note, $\xi^a = \left( 1, 0, 0, 0 \right)$ everywhere. But that is a special property of these coordinates; it does not mean that you can always ignore all of the components of 4-velocity except $u^0$.

Also, both $g_{00}$ and $u^0$ change as the particle falls; they do not have the same values everywhere that they do at the top of the trajectory. Their changes offset each other so that the constant of the motion remains constant. But you are not just trying to show that the constant of the motion remains constant; you are trying to show how the individual "pieces" of that constant change as the object falls, so that, in the Newtonian approximation, the sum

$$
\frac{1}{2} m v^2 + m g h = E
$$

remains constant. To do that, you need to know how $u^0$ changes as the ball falls (you already know how $g_{00}$ changes because we know the metric in these coordinates). Solving the geodesic equation is the usual way of figuring that out.

It is true that, since we know how $g_{00}$ changes, we can use the constant of the motion to obtain an equation for how $u^0$ changes as well. That amounts to a shortcut that avoids having to solve the full geodesic equation for this case (and if you don't want to take the further shortcut I suggested in post #30, using the Newtonian approximation in which the relationship between $v$ and $h$ is known). But you still have to be careful, because to justify $E$ above being constant in the Newtonian approximation, you need to know how $u^0$ in Schwarzschild coordinates is related to the speed $v$ that appears there. You can't use SR formulas for that.

They are equivalent.

They also are in GR. The "constant of the motion" method I am describing is exact. (Of course you can use it to derive approximations like the Newtonian approximation I gave in post #30, but you don't have to; the general equation for the constant of the motion is exact.)

 

[sweet springs]

Thanks PeterDenis. Your explanation is too kind to me as a beginner so in short the formula in post#38 for my baseball case

so product gμνuμξνg_{\mu\nu}u^\mu \xi^\nu reduces to g00u0ξ0=g00u0g_{00}u^0 \xi^0=g_{00}u^0=its value at the top, all along the geodesic.

$$u^0=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{g_{00}}=\frac{\sqrt{1-\frac{2M}{r_{max}}}}{1-\frac{2M}{r}}$$ is wrong?

 

[sweet springs]

But you still have to be careful, because to justify EE above being constant in the Newtonian approximation, you need to know how u0u^0 in Schwarzschild coordinates is related to the speed vv that appears there. You can't use SR formulas for that.

Normality of 4-velosity, say x^1=r,
$$u^0 u_0+u^1 u_1 =1$$
stands also in GR? If so,
$$u^1u_1=1-u^0u_0=1-\frac{1-\frac{2M}{r_{max}}}{{1-\frac{2M}{r}}}$$
from post#38&#53 tells us about speed, doesn't it ? if it's all right, using $g_{11}$,
$$|u^1|^2=\frac{2M}{r}-\frac{2M}{r_{max}}$$

 

[sweet springs]

Think of it as a correction factor for gravitational time dilation, which is well defined in such spacetimes.

Our constant $\sqrt{1-2M/r_{max}}=\sqrt{g_{00}}$ at $r=r_{max}$. It is same as the correction factor at $r=r_{max}$. Why this top position in trajectory was chosen to give the value ?

 

[stevendaryl]

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

The equation that is most analogous to $\frac{1}{2} m v^2 + mgh =$ Constant in General Relativity is the geodesic equation.

For Schwarzschild coordinates in a spherically symmetric, time-independent spacetime, we can get the geodesics using an effective "Lagrangian" that looks like this:

$L = \frac{m}{2} [A(r) c^2 \dot{t}^2 - \frac{1}{A(r)} \dot{r}^2 - r^2 \dot{\theta}^2 - r^2 sin^2(\theta) \dot{\phi}^2]$

where $\dot{X}$ means derivative with respect to proper time, $\tau$, and where $A(r) = 1 - \frac{2GM}{c^2 r}$

The factor $m/2$, where $m$ is the mass of the test particle, is irrelevant for the equations of motion, but it gives $L$ the dimensions of energy, and makes it look a little more like the nonrelativistic case of $L = \frac{m}{2} v^2$.

The geodesics can be obtained from the Euler-Lagrange equations from this effective Lagrangian. However, we can get them much quicker using invariants. The following quantities are constants of the motion:

$p_t \equiv \frac{\partial L}{\partial \dot{t}} = m c^2 A(r) \dot{t}$
$K^2 \equiv m^2 r^4 (\dot{\theta}^2 + sin^2(\theta) \dot{\phi}^2)$

And finally, $L$ itself is a constant of the motion (which happens to equal $\frac{1}{2} mc^2$).

So plugging these constants into the Lagrangian equation gives:

$\frac{m}{2} [\frac{(p^t)^2}{m^2 c^2 A(r)} - \frac{1}{A(r)} \dot{r}^2 - \frac{K^2}{m^2 r^2}] = \frac{1}{2} mc^2$

That's the equivalent of Newton's equation $\frac{m}{2} \dot{r}^2 - \frac{GMm}{r} = E$

 

[stevendaryl]

$\frac{m}{2} [\frac{(p^t)^2}{m^2 c^2 A(r)} - \frac{1}{A(r)} \dot{r}^2 - \frac{K^2}{m^2 r^2}] = \frac{1}{2} mc^2$

That's the equivalent of Newton's equation $\frac{m}{2} \dot{r}^2 - \frac{GMm}{r} = E$

There are two different candidates for what might correspond to the energy of Newtonian physics. First, you could say that $p^t \equiv mc^2 \frac{dt}{d\tau}$ is sort of an energy. In the limit as $r \rightarrow \infty$, this goes to $\gamma mc^2$.

Second, using the effective Lagrangian above, we could form the corresponding Hamiltonian, which is a conserved quantity. However, as I said, as a conserved quantity, it's kind of boring, since its value is just $\frac{1}{2} mc^2$. So it's independent of the motion of the test particle.

 

[sweet springs]

m2[(pt)2m2c2A(r)−1A(r)˙r2−K2m2r2]=12mc2

Thanks. I checked that your equation is reduced to
$$\frac{\dot{r}^2}{2}-\frac{M}{r}=C$$ constant
with K=0, limitation in radial motion, and c,G=1. It is very interesting because familiar kinetic energy like term v^2 appears here though SR says it is mere approximation.

 

[stevendaryl]

Thanks. I checked that your equation is reduced to
$$\frac{\dot{r}^2}{2}-\frac{M}{r}=C$$ constant
with K=0, limitation in radial motion, and c,G=1. It is very interesting because familiar kinetic energy like term v^2 appears here though SR says it is mere approximation.

Yes, it is interesting. You would expect the same equation to be approximately true when $\dot{r} \ll c$ and $r \gg \frac{2GM}{c^2}$, but the fact that it's exactly true is surprising. (But the equality between the classical and GR cases is only exact when the angular momentum is zero.)

 

[sweet springs]

First, you could say that pt≡mc2dtdτp^t \equiv mc^2 \frac{dt}{d\tau} is sort of an energy. In the limit as r→∞r \rightarrow \infty, this goes to γmc2\gamma mc^2.

This is called "energy at infinity" in previous posts. For the basebll pop up case it is
$$m\sqrt{1-2M/r_{max}}$$
m, energy of still ball in SR, is reduced by the factor $\sqrt{1-\frac{2M}{r_{max}}}$.

m, energy of still ball in SR, is distributed at the position of the ball in SR.
I wonder how the reduced energy distribute in GR systm. Again at the position of the ball ?

 

[PeterDonis]

is wrong?

It's ok as far as it goes. The errors in your previous posts came after that equation.

Normality of 4-velosity

The 4-velocity is a unit vector in GR, yes.

 

[PeterDonis]

tells us about speed, doesn't it ?

It's true that $u^1$ tells you something about speed; but it's not equal to speed. (Also, you need to be careful how you're defining "speed". Speed relative to what?)

 

[Dale]

Why this top position in trajectory was chosen to give the value ?

Because at the top v=0 so it is easier to calculate. Remember, the energy is a constant over the whole trajectory, so we can pick any point to calculate it. We may as well pick a point that makes the math easier

 

[sweet springs]

Second, using the effective Lagrangian above, we could form the corresponding Hamiltonian, which is a conserved quantity. However, as I said, as a conserved quantity, it's kind of boring, since its value is just 12mc2\frac{1}{2} mc^2. So it's independent of the motion of the test particle.

It's independent of the motion and the position. Even $\gamma$ at infinity is disregarded. It is equal to rest mass in local Minkowsky space where the body is at rest. I am afraid that might be all that GR says about energy rigorously.

 

[PeterDonis]

I am afraid that might be all that GR says about energy rigorously.

No, it isn't. An object's energy at infinity is not in general the same as its rest mass. That is obvious from the equation for energy at infinity that we have already derived.

Also, as I have already pointed out, the term "energy" has multiple meanings. Several of them have been covered in this thread, and energy at infinity is only one of them.

At this point I am closing the thread. I think you need to consider more specifically what concept of energy you want to ask about, so that future threads can have a more focused and useful discussion.