0
$\begingroup$

I am following Carroll Spacetime and geometry.

I want to derive the continuity equation from the conservation of the stress-energy tensor: $$ \partial_t \rho + \nabla \cdot (\rho \vec{v}) = 0. $$

Suppose we assume that the Energy-stress tensor of a perfect fluid is conserved. Then since $u^\mu$ is the 4-velocity, $$ T^{\mu \nu} = (\rho + p)u^\nu u^\mu + \eta^{\mu \nu}p. $$ It follows directly that $$ 0 = \partial_\mu T^{\mu \nu} = (\partial_\mu p + \partial_\mu \rho) u^\mu u^\nu + (p + \rho)u^\nu \partial_\mu u^\mu + (p + \rho) u^\mu \partial_\mu u^\nu + \partial^\nu p \qquad (\text{Eq. 1}) $$ We know that since $u^\nu u_\nu = -1$ we get from the product rule that $u_\nu\partial_\mu u^\nu = (1/2)\partial_\mu(u^\nu u_\nu) = 0$. Multiplying equation 1 and using this identity we get $$ 0 = u_\nu\partial_\mu T^{\mu \nu} = \cdots = - \partial_\mu (u^\mu \rho) - p \partial_\mu u^\mu $$ Assuming that we are in the non-relativistic limit $u^\mu = (1, u^i)$, $|u^i| \ll 1$ and $p \ll \rho$, Carroll states that the continuity equation follows. I however find $$ 0 = - \partial_\mu (u^\mu \rho) - p \partial_\mu u^\mu = - (p + \rho)\partial_\mu u ^\mu - u^\mu \partial_\mu \rho \stackrel{p \ll \rho}{=} - \rho \partial_\mu u^\mu - u^\mu \partial_\mu \rho= - \partial_\mu (u^\mu\rho)\\ = \partial_t v^0 \rho - \nabla \cdot (\rho \vec{v}) = \partial_t\rho - \nabla\cdot(\rho \vec{v}) \neq \partial_t\rho + \nabla\cdot(\rho \vec{v}) $$ This is clearly problematic. May I get a few suggestions to rectify the derivation of the continuity equation?

$\endgroup$
2
$\begingroup$

The error seems to lie near the end of your derivation, when you claim

$$-\partial_{\mu}(u^{\mu}\rho)=\partial_{t}(v^{0}\rho)-\boldsymbol{\nabla}\cdot\left(\boldsymbol{v}\,\rho\right)$$

(Where I am using bold text to refer to vectors now.) This is, in fact, not the case. Recall that $\partial_{\mu}=(\partial_{t},\boldsymbol{\nabla})$, and requires no minus signs, since derivatives are naturally covariant. On the other hand, $u^{\mu}=(u^{0},\textbf{u})$ naturally transforms like coordinate vectors, and thus naturally has an upstairs index. Thus, the real expression would be

$$-\partial_{\mu}\left(u^{\mu}\rho\right)=-\left(\partial_t(v^{0}\rho)+\boldsymbol{\nabla}\cdot(\boldsymbol{v}\,\rho)\right)$$

With this, the continuity equation is immediately in the form you want it.

The lesson is that, when doing these kinds of calculations, it is important to recall how vectors like $\partial$ and $v$ are defined. Do they naturally have upstairs or downstairs indices? You get negatives when you contract two of the same type of vector, but none when you contract a vector and a covector.

I hope this helps!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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