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.

  • $\begingroup$ $\nabla\phi\, - \,\dot\phi$ 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.