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Jonathan Scott
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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.
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.
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