The reciprocal of the Planck length, $\Lambda=1/l_P$, is used as a high-frequency cutoff in the particle-physics estimation of the vacuum energy.

For example in "the cosmological constant problem" Steven Weinberg says that summing the zero-point energies of all normal modes of some field of mass $m$ up to a wave number cutoff $\Lambda >> m$ yields a vacuum energy density (with $\hbar=c=1$) $$\langle\rho\rangle=\int_0^\Lambda \frac{4 \pi k^2 dk}{(2\pi)^3}\frac{1}{2}\sqrt{k^2+m^2}\simeq \frac{\Lambda^4}{16\pi^2}.$$ He goes on to say that "if we believe general relativity up to the Planck scale" then $$\langle\rho\rangle \approx \frac{1}{16\pi^2}\frac{1}{l_P^4}.$$ It seems that in this calculation the Planck length $l_P$ is taken to be the size of the smallest interval of space that can be described by general relativity.

But the FRW metric implies that the length of any interval of space, that can be described by general relativity, expands with the scale factor $a(t)$.

Therefore should the Planck length actually be a proper length so that $$l_P=a(t)\ l_{P0}$$ where $l_{P0}$ is a constant representing the Planck length at the present time $t_0$?

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    $\begingroup$ "if we believe general relativity up to the Planck scale", then whatever happens at that scale need not be explained by GR, and by extension, by FRW. Concepts as "proper length", "scale factor", or even "space-time" need not be well-defined at scales comparable to the Planck length. You can certainly take $l_P=a(t)\ l_{P0}$ as a definition, but its meaning is far from clear to me. What do you mean by this expression? $\endgroup$ Aug 13, 2017 at 11:16
  • $\begingroup$ If we assume that QFT and GR are correct for spatial intervals up to and including the Planck length then I think that the expanding FRW metric implies that the size of that minimum spatial interval, i.e. the Planck length, must expand with the scale factor. This assumption would affect cosmology through modified Friedmann equations with $G \propto a^2$. $\endgroup$ Aug 13, 2017 at 11:42
  • $\begingroup$ So you are assuming that both QFT and GR are correct up to $\ell_P$ but not beyond? And this even in spite of the multiple reasons we have not to trust neither of them up to those scales? This seems to be a big leap of faith to me, a pretty unjustified one. [BTW, FWIW, I didn't downvote you] $\endgroup$ Aug 13, 2017 at 11:45
  • $\begingroup$ Sorry I mean that QFT and GR are both valid for a spatial interval whose size is greater than or equal to the Planck length $l_P=a(t)l_{P0}$. $\endgroup$ Aug 13, 2017 at 11:50
  • $\begingroup$ Planck length is just a scale of distance. Like 1 meter. Space expands by the scale factor $a(t)$, but does the measure of 1 meter change? $\endgroup$
    – VacuuM
    Apr 4, 2018 at 10:22

1 Answer 1


The calculation that Weinberg presents is of the energy density. The integrand contains $\sqrt{k^2~+~m^2},$ which is a scalar. The invariant momentum interval $m^2~=~E^2~-~k^2$ is involved. The integration is cut off at the $\Lambda~=~1/\ell_p$. In a formal sense the mathematics is scalar, and so we can say this energy is an invariant.

There are some confusions though about what is meant by the Planck scale. It is not that spacetime stops at this scale. The Fermi satellite data on a range of frequencies of light from very distant Burstars indicates no dispersion. The experimental data indicates space is smooth to within $\ell_p/50$. What the Planck scale reflects is the inability to isolate a qubit on a scale smaller than $\ell_p$. It is not so much that physics ends there.

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    $\begingroup$ How does this answer the question of whether or not the Planck length is a proper length? This answer seems to contain some generalities about the Planck length that don't actually answer the question posed. Also, energy density is certainly not a scalar (but rather the 00-component of stress-energy), so $\Lambda$ shouldn't be a scalar in the sense of GR, either. $\endgroup$
    – ACuriousMind
    Aug 13, 2017 at 12:40
  • $\begingroup$ The Einstein field equations are $R_{ab}~-~\frac{1}{2}Rg_{ab}$ $+~\Lambda g_{ab}~=~8\pi kT_{ab}$. the cosmological constant is then a constant curvature term, similar to Ricci scalar. You are wrong. $\endgroup$ Aug 13, 2017 at 15:09
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    $\begingroup$ The $\Lambda$ here is not the cosmological constant, but an energy cutoff, and the Einstein field equations have nothing to do with this. $\endgroup$
    – ACuriousMind
    Aug 13, 2017 at 15:49
  • $\begingroup$ With inflationary cosmology it does. the CC $\Lambda$ is the cut off at or near Planck scale. Beside the integral is evaluated in the lab frame. $\endgroup$ Aug 13, 2017 at 15:53

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