How Does the Four-Way Vector Operation Relate to Potential Energy in Gravity?

[Jonathan Scott]

When comparing Newtonian and GR views of gravity, I came across a vector expression in the Newtonian form which happens to integrate to the total potential energy of a system of masses, even in the case of dynamic situations: $-\mathbf{x}\cdot\rho \, \mathbf{g}$, where $\mathbf{x}$ is position (relative to any arbitrary origin), $\rho$ is the local mass density (or energy density in mass units in the relativistic case) and $\mathbf{g}$ is the Newtonian gravitational field at that position.
Using Gauss's law$$-\nabla\cdot\mathbf{g} = 4\pi G\rho$$we get$$\rho = -\frac{\nabla\cdot\mathbf{g}}{4\pi G}$$so$$-\mathbf{x}\cdot\rho \, \mathbf{g} = \mathbf{x}\cdot\frac{\nabla\cdot\mathbf{g}}{4\pi G} \, \mathbf{g} = \mathbf{x}\cdot\nabla\left(\frac{g^2}{8\pi G}\right )$$The quantity in parentheses is the classical expression for the energy density in the gravitational field, so I was interested in the possibility of the above potential energy expression being part of the derivative of an expression which also includes the field energy, roughly as follows:$$\nabla \cdot \left ( \mathbf{x}\cdot \left (\frac{g^2}{8\pi G} \right ) \right ) = \frac{g^2}{8\pi G} + \mathbf{x}\cdot\nabla\left(\frac{g^2}{8\pi G}\right )$$There is an obvious problem with this, which is that the dot operator in the expression on the left of this equation is operating between a vector and a scalar, which isn't meaningful. If the dot is simply omitted, the result is incorrect, because in three dimensions $\nabla \mathbf(x) = 3$. For this equation to work, the correct expression for the numerator of the expression should be a vector in which component $i$ is given by $x_i g_i g_i$, so the left hand side could be written as follows:$$\frac{\partial}{\partial x_i} \left ( \frac{x_i g_i g_i}{8\pi G} \right )$$This is effectively a sort of four-way scalar product, but I'm not aware of any vector notation for this, and I don't even recall anything physically meaningful with four indices being in step in tensors either. Does anyone know of an established vector notation which covers this case, or of any other case where an expression of this form seems to be meaningful?
Edited to add:
I see I've assumed that $\nabla ( g^2 ) = 2 (\nabla \cdot \mathbf{g}) \mathbf{g}$ without checking it, which may invalidate the "meaningful" aspect if it's wrong, but I'm too sleepy to check it tonight.

 

[Orodruin]

I see I've assumed that $\nabla ( g^2 ) = 2 (\nabla \cdot \mathbf{g}) \mathbf{g}$ without checking it, which may invalidate the "meaningful" aspect if it's wrong, but I'm too sleepy to check it tonight.

It is wrong. If it were true it would mean that the magnitude of a vector field would necessarily increase in the direction of the field itself.

Also see commandment 5 in https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

 

[Jonathan Scott]

It is wrong. If it were true it would mean that the magnitude of a vector field would necessarily increase in the direction of the field itself.

OK, thanks, I'll have to find the correct version of that when I'm more awake and less under the influence of Christmas. I thought I remembered that some simplification like that was valid for a vector field which was the gradient of a scalar potential, but I should have checked it.

Also see commandment 5 in https://www.physicsforums.com/insights/the-10-commandments-of-index-expressions-and-tensor-calculus/

I'm well aware that in tensor calculus notation one can only use two instances of an index (and that usually implies summation). That was the reason I was puzzled (and I was aware that one possible reason was that I had got something wrong). I was using four instances only to explain what I meant.

 

[strangerep]

$$-\mathbf{x}\cdot\rho \, \mathbf{g} = \mathbf{x}\cdot\frac{\nabla\cdot\mathbf{g}}{4\pi G} \, \mathbf{g} = \mathbf{x}\cdot\nabla\left(\frac{g^2}{8\pi G}\right )$$

Since $\rho$ is a scalar, this eqn could be better understood if written as:$$-\Big(\mathbf{x}\cdot\mathbf{g}\Big) \rho ~=~ \Big(\mathbf{x}\cdot \mathbf{g}\Big) ~ \frac{\nabla\cdot\mathbf{g}}{4\pi G} ~\ne~ \mathbf{x}\cdot\nabla \left(\frac{g^2}{8\pi G} \right) ~.$$

 

[Jonathan Scott]

It's easy enough to see that those expressions are not equal, despite some similarity, when I'm awake.$$((\nabla \cdot \mathbf{g}) \, \mathbf{g})_i = (\nabla_j g_j) g_i$$ but $$(\nabla (g^2/2))_i = (\nabla_i g_j) g_j $$
But I'm fairly sure I've seen some result (probably in tensors) relating force density to energy in the field (both in Newtonian gravity and in electrostatics) which included a valid relationship which was quite similar to this, probably including additional terms. I've had so little time for physics in the last couple of years (since having to take on several other people's jobs in addition to my own) that I guess I need to do some revision before I ask any more questions.

 

[vanhees71]

Newtonian gravity indeed can be described by the equation
$$\Delta \phi=4 \pi G \rho, \quad \vec{g}=-\vec{\nabla} \phi,$$
where $\phi$ is a scalar potential for the gravitational field $\vec{g}$. It's physical meaning is that a test particle with mass $m$ feels the gravitational force
$$\vec{F}=m \vec{g}.$$
It can be obtained as (one possible) non-relativistic limit of Einstein's field equations, as is explained in almost all introductory textbooks on general relativity.