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I am going through some problems from Sean Carroll's Introduction to GR textbook. In the first chapter, he gives the energy-momentum tensor of a scalar field on spacetime as $$T^{\mu\upsilon}_{scalar} = \eta^{\mu\lambda}\eta^{\upsilon\sigma}\partial_\lambda\phi\partial_\sigma\phi\; - \;\eta^{\mu\upsilon}\left[\frac 12\eta^{\lambda\sigma}\partial_\lambda\phi\partial_\sigma\phi\,+\,V(\phi)\right] $$ where $$\phi(x^\mu)\,:\,(spacetime)\,\to\,R$$ In an exercise question he asks us to express this equation in terms of 3-vector notation, and I do not really know how to do this. I tried to split up the $\partial_\lambda\phi$ terms into their time and space components and act the metric on them, which gave me something like $(\nabla\phi\, - \,\dot\phi)(\nabla\phi\, - \,\dot\phi)$ for the first term of the equation, but that is still a difference of two four-vectors, right? I also realize that the term in square brackets should work out to just being a scalar as well - so that the indices of both terms of the equation agree - but I do not know how to express something like $\eta^{\mu\upsilon}$ in 3-vector notation.

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  • $\begingroup$ $\nabla\phi\, - \,\dot\phi$ doesn’t make sense. The first term is a 3-vector and the second one isn’t. $\endgroup$
    – Ghoster
    Jun 24, 2023 at 17:52
  • $\begingroup$ I separated them according to the sum $\partial_\lambda\phi = \partial_0\phi + \partial_i\phi$. So I actually treated them both as four-vectors, but with the extra dimensions being 0. Am I abusing notation by calling $(\frac{\partial\phi}{\partial t}, 0, 0, 0) = \dot\phi$? $\endgroup$
    – Chidi
    Jun 24, 2023 at 19:31
  • $\begingroup$ $\partial_\lambda\phi$ is not a sum. $\endgroup$
    – Ghoster
    Jun 24, 2023 at 19:55
  • $\begingroup$ It's a dual vector, which I can express as a sum of dual vectors i.e $(\frac{\partial\phi}{\partial t}, \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z}) = (\frac{\partial\phi}{\partial t}, 0, 0, 0) + (0, \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z})$, right? $\endgroup$
    – Chidi
    Jun 24, 2023 at 20:05
  • $\begingroup$ Yes, you can do that. But that’s not the same as $\partial_\lambda\phi=\partial_0\phi+\partial_i\phi$. You can write $\partial_\lambda\phi=(\partial_0\phi,\partial_i\phi)$ to express a four-vector as a temporal component and a three-vector. $\endgroup$
    – Ghoster
    Jun 24, 2023 at 20:14

1 Answer 1

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After the helpful comments made by Ghoster, I went back and looked at the problem differently, and I now believe I am able to solve it. If I make a mistake in answering the question, please let me know.

The key I was missing was considering the result as a series of simultaneous equations for each variable of spacetime (t, x, y, z), rather than looking for a way to break up four-vectors into separate 3-vectors and a time component. I realized you cannot break up $T^{\mu\upsilon}$ that way.

Acting the metric on the appropriate dual vectors in the equation for $T^{\mu\upsilon}$, you get $$T^{\mu\upsilon} = \partial^\mu\phi\partial^\upsilon\phi\; - \;\eta^{\mu\upsilon}\left[\frac 12\left(-(\partial_0\phi)^2+(\partial_x\phi)^2+(\partial_y\phi)^2+(\partial_z\phi)^2\right) + V(\phi)\right]$$ In terms of 3-vector notation, you can express $-(\partial_0\phi)^2+(\partial_x\phi)^2+(\partial_y\phi)^2+(\partial_z\phi)^2$ as $-\dot\phi^2 + |\nabla\phi|^2$, where I have used $\nabla\phi\,\cdot\nabla\phi = |\nabla\phi||\nabla\phi|\cos\theta$ to express the spatial component of the sum. For the relevant components of $T^{\mu\upsilon}$, I matched them to the relevant components of the RHS: $$T^{00} = -|\nabla\phi|^2-V(\phi)$$ $$T^{0i} = T^{i0} = \dot\phi(\partial^i\phi)$$ $$T^{ij} = (\partial^i\phi)(\partial^j\phi)\, -\,\left[-\dot\phi^2 + |\nabla\phi|^2 + V(\phi)\right] $$

I don't know if expressing objects like $\partial^i\phi$ as $\nabla\phi$ make sense in this case, since $T^{0i}$ and others like it are components of a tensor, so this is as far as I got in solving the problem.

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