# A Cosmological constant in the semiclassical limit of quantum gravity

1. Jun 17, 2017

### Afonso Campos

Why is it the case that, in a semiclassical description of the Einstein-Hilbert action, the cosmological constant is small in Planck units?

Why does this mean that

$$\ell \gg G$$

for $\Lambda = - 1/\ell^{2}$?

2. Jun 17, 2017

### Staff: Mentor

Nobody knows. This is an open area of research.

3. Jun 17, 2017

### Afonso Campos

Ah! Right!

For the second part, we have, in Planck units,

$\Lambda \ll M_{P}^{4}$

$-1/\ell^{2} \ll M_{P}^{4}$

$-1/\ell^{2} \ll 1/G^{2}_{N}$

$1/\ell^{2} \gg 1/G^{2}_{N}$

$G^{2}_{N} \gg \ell^{2}$

$\ell^{2} \ll G^{2}_{N}$

$\ell \ll G_{N}$

What have I done wrong here?

4. Jun 17, 2017

### Staff: Mentor

Why did you have $\Lambda = - 1 / \ell^2$ with a minus sign?

5. Jun 17, 2017

### Afonso Campos

Let's say that we have $AdS$ spacetime.

6. Jun 17, 2017

### Staff: Mentor

Then your derivation is only valid for $AdS$ spacetime. Do you have a reference that claims that $\ell \gg G$ for $AdS$ spacetime?

7. Jun 17, 2017

### Afonso Campos

It is in page 2 of this paper - https://arxiv.org/pdf/hep-th/9712251.pdf.

See equation (2.2) and the text above the equation.

8. Jun 17, 2017

### Staff: Mentor

Hm. I think their statement that $\ell \gg G$ is only meant as a heuristic, because the two don't even have the same units; the units of $G$ are length squared (inverse mass squared), while the units of $\ell$ are, of course, length. So really it should be $\ell^2 \gg G$.

Regarding your derivation, you should not be including the minus sign regardless of the sign of $\Lambda$, since the inequality only refers to relative magnitudes, not signs. In other words, the strictly correct way of writing the above inequality (with units corrected) is $\vert \ell \vert^2 \gg \vert G \vert$. Or, if you write it in terms of $\Lambda$ and $M_P$, it is $\vert \Lambda \vert \ll \vert M_P \vert^2$. (Note the exponent, btw; it's the Planck mass squared, since the units of $\Lambda$ are mass squared, or inverse length squared.) The fact that the sign of $\Lambda$ is negative in $AdS$ doesn't change any of the above--its magnitude is still small compared to the magnitude of $M_P$, which is the necessary requirement for the semiclassical approximation.

9. Jun 17, 2017

### Afonso Campos

I find that the reviews of important results in section 2 and section 3 of the paper could have better written.

Thanks for pointing this out!