4
$\begingroup$

The first law of black hole mechanics (let's simplify by considering a uncharged and non-rotating black hole) can be written as

$$\delta M = T \delta S$$

If I use the definition of Hawking temperature as $T=\frac{\kappa}{2 \pi}$ ($\kappa$=surface gravity) and entropy as $S=\frac{A}{4}$ ($A$=horizon area) then I can write this as

$$\delta M = \frac{\kappa \delta A}{8 \pi G}$$

Now this makes sense since the mass is essentially a conserved charge which I should get by some kind of Noether-like procedure where the conserved current is the stress tensor and if the Einstein-Hilbert term in my Lagrangian is $\mathcal{L}=-\frac{1}{16 \pi G}R$, then the stress-tensor (and thus $M$) will automatically come with the $8 \pi G$ to balance things out.

My question is to do with what happens if I use units in which the Einstein-Hilbert term of the Lagrangian is $\mathcal{L}=-\frac{1}{2}R$. This will rescale the stress-tensor (and consequently $M$) by $8 \pi G$ and so the factor of $\frac{1}{8 \pi G}$ on the right hand side needs to be removed. At first glance it seems quite straightforward: I can send $S \rightarrow A$ and $T \rightarrow \kappa$ so that my first law now reads $\delta M = \kappa \delta A$ like I want it to.

However, doesn't this affect dimensional analysis? Why have I never seen entropy or temperature defined like this in the literature? It seems a bit strange...

Improve this question
$\endgroup$
10
  • $\begingroup$ If you're rescaling properly (i.e., propagating this everywhere), then it shouldn't affect dimensional analysis. $\endgroup$
    – Kyle Kanos
    Commented Jan 28, 2016 at 20:57
  • 1
    $\begingroup$ @KyleKanos I don't understand what you mean by propagating this everywhere? Also, are you referring to my proposed rescaling of $S$ by $4G$ and $T$ by $2 \pi$ or to something else? $\endgroup$
    – user11128
    Commented Jan 28, 2016 at 20:59
  • 1
    $\begingroup$ @KyleKanos Sorry, I'm afraid I don't follow. Can you confirm what you mean by propagating everywhere? To me this means everywhere in the spacetime but it seems rather obvious; I don't know why we would want to do a rescaling just in some small region? Or am I missing your point? $\endgroup$
    – user11128
    Commented Jan 28, 2016 at 21:06
  • 1
    $\begingroup$ @KyleKanos But the thing that can't change is the 1st Law. I can preserve its structure by rescaling $S$ by $2 \pi$ and $T$ by $4G$. This is of course much less natural (and would lead to different values of temperature and entropy which are physical observables!!!) but I see no reason not to do it this way unless there is some reason from dimensional analysis telling me not to? $\endgroup$
    – user11128
    Commented Jan 28, 2016 at 21:20
  • 1
    $\begingroup$ Sorry to drag it out but I still don't get it. If the original Lagrangian (which defines the unit system) has the $8 \pi G$ included then it appears naturally in mass and in the product of $\frac{1}{4G}$ and $\frac{1}{2 \pi}$ in the definitions of entropy and temperature. If i start with a different Lagrangian (different unit system) that doesn't have an $8 \pi G$ in it then neither will the stress tensor and neither will the definition of mass. Therefore I need to remove the $8 \pi G$ from the product of $TS$. But how do I know how these are defined in the new unit system? $\endgroup$
    – user11128
    Commented Jan 28, 2016 at 21:39

0

Reset to default

Your Answer

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