Does the gravitational field have energy like the electric field ?

[Vanadium 50]

I define potential energy as (positive) energy stored in a field.

Which is not how it's defined in Halliday and Resnick, Sears, or anyone else outside of granpa-land. If I lift a rock, it gains potential energy mgh. The field strength g remains constant, so there is no change in the energy stored in the field.

you arent using the definition of potential energy that I have clearly and repeatedly stated that you must use if the Lagrangian is to be conserved.

No, he's using the definition of potential energy used in places outside granpa-land, places like textbooks.

Have you ever read Alice in Wonderland?

`And only one for birthday presents, you know. There's glory for you!'

`I don't know what you mean by "glory,"' Alice said.

Humpty Dumpty smiled contemptuously. `Of course you don't -- till I tell you. I meant "there's a nice knock-down argument for you!"'

`But "glory" doesn't mean "a nice knock-down argument,"' Alice objected.

`When I use a word,' Humpty Dumpty said in rather a scornful tone, `it means just what I choose it to mean -- neither more nor less.'

If you want to communicate effectively, you can't redefine words to mean whatever strikes your fancy at the moment. Otherwise it's Alice in Granpa-land.

 

[Bergstein]

Now I think that the gravitational field doesn't have energy at all, because the gravity is just the reflection of the structure of space-time, it's nonsense to talk about the Lagrangian and the Hamiltonian of gravitational field.what't your opinion?

 

[non-hermitian]

It's possible to define the energy stored in the gravitational field for Newtonian gravity using a similar approach to the way we define energy stored in the electric field. For the electric field, we start from Coulomb's Law (in CGS units) for the force on a point charge q_{1} which arises from another point charge q_{2}:

\vec{F}_{1} = \frac{q_{1} q_{2}}{{\Vert \vec{r}_1 - \vec{r}_{2} \Vert}^3} (\vec{r}_{1} - \vec{r}_{2})

For a bunch of point charges labeled q_{i}, where i goes from 1 to N, the force on q_{1} adds linearly, so we get

\vec{F}_{1} = \sum^{N}_{j=2} \frac{q_{1} q_{j}}{{\Vert \vec{r}_1 - \vec{r}_{j} \Vert}^3} (\vec{r}_{1} - \vec{r}_{j})

and in general, the force acting on the ith charge is given by

\vec{F}_{i} = \sum^{N}_{j \neq i} \frac{q_{i} q_{j}}{{\Vert \vec{r}_i - \vec{r}_{j} \Vert}^3} (\vec{r}_{i} - \vec{r}_{j})

We define a potential energy function for each charge q_{i}, using \vec{F}_{i} = - q_{i} \vec{\nabla} \Phi_{i} for some scalar potential functions \Phi_{i}, and a vector field \vec{E}_{i} = \frac{1}{q_{i}} \vec{F}_{i} corresponding to the electric field that each charge sees. Note that the potential energy function and electric fields are different for each charge using this method, since we assume that point charges do not see their own fields. This is not strictly true (if I remember correctly, point charges interacting with their own fields are what causes the radiation reaction force for accelerating charges), but the problem will go away when we take the continuum limit and radiation reaction can be ignored here anyway. We find that the potential energy functions have a nice expression

\Phi_{i} = \sum_{j \neq i}^{N} \frac{q_{j}}{\Vert \vec{r}_{i} - \vec{r}_{j} \Vert}

This allows us to calculate the amount of energy necessary to assemble a collection of point charges by bringing them in one at a time from infinity, that is, the potential energy of the system. This is given by

E_{potential} = \frac{1}{2} \sum^{N}_{i=1} q_{i} \Phi_{i} = \frac{1}{2} \sum^{N}_{i \neq j} \frac{q_{i} q_{j}}{\Vert \vec{r}_{i} - \vec{r}_{j} \Vert}

The real electric field is defined using the idea of a test charge. We imagine adding a small charge Q to the system at \vec{r} and define the electric field \vec{E} and electric potential \Phi as the electric field and scalar potential seen by our test charge. This gives the well-known formulae

\vec{E}(\vec{r}) = \sum^{N}_{j=1} \frac{q_{j}}{{\Vert \vec{r} - \vec{r}_{j} \Vert}^3} (\vec{r} - \vec{r}_{j})

and

\Phi(\vec{r}) = \sum_{j =1}^{N} \frac{q_{j}}{\Vert \vec{r} - \vec{r}_{j} \Vert}

With these formulae in hand, we can verify that Gauss' law holds,

- \nabla^2 \Phi = \vec{\nabla} \cdot \vec{E} = 4 \pi \rho(\vec{r})

where we have defined the charge density \rho(\vec{r}) = \sum^{N}_{j = 1} q_{j} \delta^{(3)}(\vec{r} - \vec{r}_{j}) and made use of the identity \nabla^2 \left( \frac{1}{r} \right) = - 4 \pi \delta^{(3)} (\vec{r}). When we take the continuum limit, we no longer have to worry about particles interacting with their own fields, which allows us to write

E_{potential} = \frac{1}{2} \int_{space} \rho(\vec{r}) \Phi(\vec{r}) d^3r = - \frac{1}{8 \pi} \int_{space} \Phi(\vec{r}) \nabla^2 \Phi(\vec{r}) d^3r

We can do the three-dimensional equivalent of integrating by parts using Green's theorem to get

E_{potential} = \frac{1}{8 \pi} \int_{space} \vec{\nabla} \Phi(\vec{r}) \cdot \vec{\nabla} \Phi(\vec{r}) d^3r + \text{boundary term}

and because we are integrating over all space, our boundary term vanishes for localized charge distributions. Using -\vec{\nabla} \Phi = \vec{E}, we discover

E_{potential} = \frac{1}{8 \pi} \int_{space} \vec{E} \cdot \vec{E} d^3r

which motivates the assertion that the density of energy stored in the electric field is given by \frac{1}{8 \pi} \vec{E} \cdot \vec{E}. The same derivation applies to the Newtonian gravitational field mutatis mutandis. In this case, Gauss' law must be changed to

- \nabla^2 \Phi_{g} = \vec{\nabla} \cdot \vec{E}_{g} = - 4 \pi G \rho_{g}

where \Phi_{g} is the gravitational potential, \vec{E}_{g} is the gravitational analogue of an electric field, and \rho_{g} is the mass density. There is an extra minus sign as compared to before because gravity is attractive between two positive masses, whereas two positive charges repel. Nonetheless, we can still define

E_{potential} = \frac{1}{2} \int_{space} \rho_{g}(\vec{r}) \Phi_{g}(\vec{r}) d^3r

This time, however, when we apply Green's theorem, we are short one minus sign as compared to last time, and we discover in the end

E_{potential} = - \frac{1}{8 \pi G} \int_{space} \vec{E}_{g} \cdot \vec{E}_{g} d^3r

motivating us to define gravitational energy density as -\frac{1}{8 \pi G} \vec{E}_{g} \cdot \vec{E}_{g}. This negative energy density is disturbing- what could it possibly mean? Fortunately for us, the question is only academic, because real gravity is described by general relativity. General relativity has its own problems with defining gravitational energy density since gravitational effects are contained in the spacetime metric instead of in a field living on the spacetime manifold, but there are clever constructions one can make to get around this difficulty. As far as Newtonian gravity goes, you really do get a negative energy density if you try to do things in the same way you would in electromagnetism. I've found a paper with more information about this, as well as a way to get around this by including a self-interaction of the gravitational field, here:

www.iop.org/EJ/article/0143-0807/28/6/016/ejp7_6_016.pdf[/URL]

 

[Dale]

Hi non-hermetian,

Welcome to PF, and excellent first post!

 

[egust]

I don't understand why people are so perturbed by a negative energy density. I always understood that a negative energy (compared to zero energy at infinite separation) indicated a bound system. Since gravitational forces are never repulsive, it follows that all systems of gravitationally interacting particles have a net negative potential energy. It's certainly not a problem that requires general relativity to solve.

I would also like to point out that applying the energy density concept to systems of point particles is rather delicate, since all of the integrals diverge formally.