People often say that the cosmological constant is too small. $\Lambda=10^{-120}$ in Planck units.

Can we set $\Lambda=\hbar=c=1$ ?

If so what would this give for $G$, the gravitational constant in these units?

  • Well, if $x/y$ is very small, that's true in any system of units. You can set $x = 1$ or $y = 1$, but it won't change anything. – knzhou Dec 8 at 20:41
  • This seems like a reasonable question to me, so I don’t understand the downvotes. Unit systems aren’t particularly interesting, but this is at least a valid one. – G. Smith Dec 8 at 23:01
up vote 3 down vote accepted

Yes, you can do this if you want, because $\Lambda$, $\hbar$, and $c$ have independent dimensions, just like $G$, $\hbar$, and $c$ do. If you do, $G$ will be the dimensionless number $10^{-60}$, so you haven't fixed the "too small" problem... you've just transferred it from $\Lambda$ to $G$.

The cosmological constant is an energy density (i.e., it has the units of energy per unit volume). The Planck mass is $m_P=\sqrt{\hbar c/G}$ and the Planck length is $l_P=\sqrt{\hbar G/c^3}$. From these one finds that the Planck energy density is $\epsilon_P=c^7/\hbar G^2$. Saying that the cosmological constant is $10^{-120}$ in Planck units means that $\Lambda/\epsilon_P=10^{-120}$ or $\Lambda\hbar G/c^7=10^{-120}$. Thus in "Lambda units" where $\Lambda=\hbar=c=1$, we have $G=10^{-60}$.

One also finds that the Lambda-units of length, time, and mass are $$l_\Lambda=10^{-30}l_P=1.6\times 10^{-65}\;\text{m},$$ $$t_\Lambda=10^{-30}t_P=5.4\times 10^{-74}\;\text{s},$$ $$m_\Lambda=10^{30}m_P=2.2\times 10^{22}\;\text{kg}.$$

  • Yes, these would be strange units. With $m_\Lambda$ being close to the mass of the moon! Useful for astronomy? There are some theories where $G$ is not a constant. So these might be reasonable alternative units. In these units the radius of the universe would be $10^{91}$ approximately. – zooby Dec 9 at 15:23

Can we set the cosmological constant to one?

You could set it to one. But what would that achieve? If the cosmological constant is 'too small' then setting it to one would make everything else too large!

  • According to another answer it would make the length and time units too small! – zooby Dec 9 at 15:24

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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