Leonard Suskind gives the following formulation of the energy-momentum tensor in his Stanford lectures on GR (#10, I believe):

$$T_{\mu \nu}=\partial_{\mu}\phi \partial_{\nu}\phi-\frac{1}{2}g_{\mu \nu}\partial_{\sigma}\phi \partial^{\sigma}\phi$$

In an intro to GR book I find this formulation of the same:

$$T^{\mu \nu}=[\rho+\frac{P}{c^2}]u^{\mu}u^{\nu}+g^{\mu \nu}P$$

I'm having trouble seeing how they are describing the same thing. Is $\partial_{\sigma}\phi \partial^{\sigma}\phi$ equal to $P$? In the book equation, $u^{\nu}$ and $u^{\mu}$ are four velocities differentiated w.r.t. time. Where do these appear in Suskind's equation? What happens to the factor of $\frac{1}{2}$? If I understand the lectures correctly, $\rho$ is essentially the $T^{00}$ component of the tensor in the Newtonian limit. I don't see how the second equation reduces to $\rho$ in that limit (slow and flat).

I appreciate your help.

  • $\begingroup$ Observe what happens when you take $u^{\mu} = \partial^{\mu}\phi$ and impose "obvious" fluid-dynamical conditions on $u^{\mu}$... $\endgroup$ Jul 23, 2012 at 3:10
  • $\begingroup$ $u^{\mu}=\frac{dx^{\mu}}{d\tau}$, so $u^{\mu}=\partial^{\mu}\phi$ would mean the gradient of $\phi$ is equal to velocity. Is that what you mean to say? (Unfortunately I'm not versed in "obvious fluid dynamical conditions.") Oh, and what happens to the factor of $\frac{1}{2}$? $\endgroup$
    – ben
    Jul 23, 2012 at 4:51

3 Answers 3


Keep in mind that the stress-energy tensor is described by the matter and energy fields. Each model of matter or field will have different expressions for this tensor

the first one is for a scalar field $\phi$ under the usual $E^2 - c^2 P^2 = m^2 c^4$ relativistic identity defined by the Klein-Gordon equation, the second is something entirely different, it is the energy tensor for a perfect fluid of density $\rho$ and mean pressure $P$ (not sure about the units)

But i think you want to know under which assumptions the scalar klein-gordon field will be approximated by a perfect fluid equation. Can you post a link to the lectures so we can get a little bit more of context?

  • $\begingroup$ Here is the link to the video: youtube.com/… He finishes deriving the equation at about 1:08 $\endgroup$
    – ben
    Jul 23, 2012 at 0:23

I actually attended a lecture recently on a problem similar to this although the stress energy tensor we were handed was slightly different (possibly due to generalization to curved spacetime with metric $ g_{\mu \nu}$ ). We had that: $$T_{\mu \nu}=\partial_{\mu} \phi \partial_{\nu} \phi - \mathcal{L} g_{\mu \nu} $$ The Lagrangian used is that of GR (In units where $c=1$ and a negative signature metric $(+--- )$ is used) $$ \mathcal{L} = R +\frac{1}{2}g^{\mu \nu}\partial_{\mu} \phi \partial_{\nu} \phi -V(\phi) $$ Known as the minimally coupled Lagrangian density since there are no cross terms between R and $\phi$ ($R \phi$ and so on). Substituting this in yields $$T_{\mu \nu}=\partial_{\mu} \phi \partial_{\nu} \phi - (R +\frac{1}{2}g^{\alpha \beta}\partial_{\alpha} \phi \partial_{\beta} \phi -V(\phi)) g_{\mu \nu} $$ We can then define the normalized four velocity as $$ u_{\mu} = \frac{\partial_{\mu}\phi}{\sqrt{g^{\alpha \beta} \partial_{\alpha}\phi \partial_{\beta}\phi}} $$ So that $u_{\mu}u^{\mu} = 1 $.

We then define a time derivative as the component of the field dotted with the four velocity: $$ \dot\phi=u^{\mu} \nabla_{\mu}\phi = u^{\mu} \partial_{\mu}\phi $$ At which point it is normally left to the reader to verify that the stress energy becomes, with this definition $$ T_{\mu \nu} = \left(\frac{1}{2} (\dot\phi)^2 + V(\phi) \right)u_{\mu}u_{\nu} +\left(\frac{1}{2} (\dot\phi)^2 - V(\phi) \right) \left(u_{\mu}u_{\nu} - g_{\mu \nu} \right)$$ Identifying: $$ \begin{eqnarray*} \rho &=& \left(\frac{1}{2} (\dot\phi)^2 + V(\phi) \right) \\ P &=& \left(\frac{1}{2} (\dot\phi)^2 - V(\phi) \right) \end{eqnarray*} $$ Yields the Tensor $$ T_{\mu \nu} = \rho u_{\mu}u_{\nu} + P \left(u_{\mu}u_{\nu} - g_{\mu \nu} \right)= (\rho +P) u_{\mu}u_{\nu} - P g_{\mu \nu} $$ Im not sure how much is relevant, but this is the derivation I was introduced to involving perfect fluids and the scalar field.


0. Caveat Lector: This was done before I drank my morning coffee, so there may be some errors in the reasoning (well, the physical reasoning, the mathematics should be kosher).

1. Perfect Fluid. So we have two stress-energy tensors here. One is the stress energy tensor for a perfect fluid $$\tag{1}T^{\alpha\beta}_{\text{fluid}} = \rho \, u^\alpha \, u^\beta + p \, h^{\alpha\beta}$$ where we have

  1. the worldlines of the fluid's particles have velocity $u^\alpha$
  2. the projection tensor $h_{\alpha\beta} = g_{\alpha\beta} + u_\alpha \, u_\beta$ projects other tensors onto hyperplane elements orthogonal to $u^\alpha$
  3. the matter density is given by the scalar function $\rho$,
  4. the pressure is given by the scalar function $p$.

We'd need extra terms if there were heat flow or shear involved.

2. Scalar Field. Now, we have another distinct stress-energy tensor for a massless scalar field: $$\tag{2}T^{\mu \nu}_{\text{scalar}} =\partial^{\mu}\phi\, \partial^{\nu}\phi-\frac{1}{2}g^{\mu \nu}\partial_{\rho}\phi\,\partial^{\rho}\phi$$ We would use this equation when modeling, e.g., massless pions (or some other massless spin-0 field).

3. Problem: Are these two related?

Now if we take our matter density to be, in the appropriate units, $$\tag{3a} \rho = 1 + \frac{1}{2}\partial_{\rho}\phi\,\partial^{\rho}\phi $$ and the pressure $$\tag{3b} p = \frac{-1}{2}\partial_{\rho}\phi\,\partial^{\rho}\phi $$ then (2) resembles (1). This is after pretending $\partial^{\mu}\phi=u^{\mu}$, which terrifies the original poster (but that's what condensed matter physicists do, so I suppose I could end here content).

Is this kosher?

We should first note if we wanted to take the derivative of some function along the worldline $x^{\mu}(s)$ with respect to the "proper time" (length) $s$ we have $$\tag{4} \frac{\mathrm{d}f}{\mathrm{d}s}=\frac{\mathrm{d}x^{\mu}}{\mathrm{d}s}\frac{\partial f}{\partial x^{\mu}}$$ by the chain rule. For general relativity, we use the "comma-goes-to-semicolon" rule, but for a scalar quantity $f$ we have $$ \nabla_{\mu}f = \partial_{\mu}f.$$ (If this is not obvious, the reader should consider it an exercise to prove it to him or herself.) The punchline: identifying $\partial^{\mu}\phi=u^{\mu}$ is kosher. How?

Observe in Equation (4) the guy in front, the $\mathrm{d}x^{\mu}/\mathrm{d}s$ is just some vector. So in the very, very special case that equations (3a) and (3b) hold, and $\mathrm{d}x^{\mu}/\mathrm{d}s=(1,0,0,0)$, we see that we can indeed recover the first stress-energy tensor as a special case of the scalar field's stress-energy tensor.


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