1
$\begingroup$

I'm trying to understand hydrodynamics of relativistic CFTs. A paper I'm referring to is this article published in PRL by Itzhak Fouxon and Yaron Oz in 2008.

The paper states that hydrodynamics applies whenever the correlation length of the fluid is much smaller than the characteristic length scale of variations of the fields. They derive an Euler equation from the zeroth-order gradient expansion of the conservation of the stress-energy-momentum (SEM) tensor of the CFT, and state that the SEM tensor only depends on the temperature $T$. The energy density $e$ is argued to scale as $T^4$ in $3+1$ dimensions from dimensional arguments.

I'm having trouble understanding this argument. From the perspective of mass dimension, I can see that $e$ (if interpreted as the Hamiltonian density) should have mass dimension $4$. In that case, why does temperature have mass dimension $1$?

PS: I also don't understand what is meant by the correlation length of the fluid. How should one define this? The correlation length corresponding to correlation functions diverges at a CFT due to scale invariance.

$\endgroup$
3
  • 1
    $\begingroup$ For the first part of the question: temperature has units of energy (not energy density) in units where the Boltzmann constant $k$ is $1$, which you can see (for example) from the Boltzmann factor $\sim e^{- E / kT}$. Energy has mass dimension $1$ in units with $\hbar=c=1$ (eg, $E=mc^2$). Energy $E$ is related to energy density $\rho$ and volume $V$ by $E=\rho V$; since $E$ has mass dimension $1$ and $V$ has mass dimension $-3$, $\rho$ must have mass dimension $4$. $\endgroup$
    – Andrew
    Commented Jul 25 at 23:24
  • $\begingroup$ For the second part, I agree I don't know what the correlation length could mean in a CFT because naively there shouldn't be any length scales. Since I don't have any insight on this part, I'm just leaving this as a comment. $\endgroup$
    – Andrew
    Commented Jul 25 at 23:25
  • $\begingroup$ Ah thank you, this explains the dimensional analysis! $\endgroup$ Commented Jul 25 at 23:56

0

Your Answer

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