# The unit of the cosmological constant

In natural units, it’s known that the unit of the cosmological constant is $$\text{eV}^2$$. I don‘t get why the paper arXiv: 2201.09016, p. 1, it is said the value of $$\Lambda \sim \text{meV}^4$$, this means $$\Lambda \sim (10^6 ~ \text{eV})^4 \sim 10^{24} \text{eV}^4$$, should not the unit be $$\text{eV}^2$$ instead?

• I'm not sure what cosmological constant value should be in the natural unit system. However your mentioned paper in a first page talks about energy scale and not just about cosmological constant, though $\Lambda$ and energy scale may be related. Commented Jun 1, 2022 at 8:49
• What's your source for it being energy squared? In the Friedmann equation $\Lambda$ is a density, so its mass dimension is $1-(-1)\times3=4$.
– J.G.
Commented Jun 1, 2022 at 9:19
• The paper you link says: "One effective solution to the Cosmological Constant Problem could be for a single scalar field coupled to the metric to eat up the large vacuum energy density and produce an effective cosmological constant at the required energy scale ($\sim \mathrm{meV}^4$)". That number is not the cosmological constant. It's the energy scale (with units of energy density) where the scalar field appears as an effective cosmological constant. Commented Jun 4, 2022 at 12:16

Dr. phy asked: "What is the unit of the cosmological constant?"

In SI units it is $$\rm [1/m^2]$$, since in the Friedmann equation $$\rm (\dot{a}/a)^2+...=\Lambda c^2/3$$ with dimensionless $$\rm a=r/r_0=[m/m]=[1]$$, the square of the Hubble parameter $$\rm H^2=(\dot{a}/a)^2=(da/dt/a)^2$$ has units of $$\rm [1/s^2]$$, which is the same as $$\rm \Lambda c^2$$ when $$\rm \Lambda$$ is in $$\rm [1/m^2]$$.

In the covariant field equation $$\Lambda g_{\mu \nu}+...=\rm T_{\mu \nu} 8 \pi G/c^4$$ with $$g_{\rm t t}$$ in $$\rm c^2=[m^2/s^2]$$ and $${\rm T_{t t}}/g_{\rm t t}$$ and $${\rm T^{t t}}/g^{\rm t t}$$ in units of $$\rm [J/m^3]$$ the $$\Lambda$$ also has to be in $$\rm [1/m^2]$$ in order for the units to match.

The energy stress momentum tensor component for the dark energy density is $$\rm T^t_t = c^4 \Lambda/(4 \pi G)$$ so for that to have the units $$\rm [J/m^3]=[kg/m/s^2]$$ you need $$\Lambda$$ in $$\rm [1/m^2$$].

In our ΛCDM universe the value is around $$\rm \Lambda= 10^{-52}/m^2$$. In a de Sitter universe containing no matter and radiation so that the $$+...$$ term vanishes, the relation to the Hubble parameter is simply $$\rm H=c\sqrt{\Lambda/3}$$, and $$\rm H$$ has the units $$\rm [m/m/s]=[1/s]$$ (the frequency of one e-fold).

As you can see in the other answer above, sometimes it is given already multiplied with $$\rm c^2$$, then the numeric value is around $$\rm 10^{-36}/s^2$$. That is the case in sources where it simply says $$\rm H^2=\Lambda/3$$, but in that case they should mention that they set $$\rm c=1$$ and then you also should not see any other $$\rm c$$s in the equation.

So if you read different numbers in different references and they don't mention which constants they omitted, just take the $$\rm \Lambda=10^{-52}/m^2$$ and multiply the number in your source with the constants in question until the number matches the units, then you know for sure by what's the conversion factor between your different references.

with Friedman equation

$$\left(\frac{\dot a}{a}\right)^2+\ldots =\frac{1}{3}\Lambda\,c^2$$

with $$~c=[1]~$$ you obtain that the unit of $$~\Lambda~$$ is $$\left[\frac{1}{s^2}\right]=[eV]^2$$

with $$E=m\,c^2~,c=1\quad\text{the unit of mass and energy is ~[eV]}$$ with $$E^2-p^2\,c^2=m^2\,c^4,c=1\quad\text{the unit of p is ~[eV]}$$ with $$\Delta p\,\Delta L\ge h,h=1\quad \text{the unit of L is ~[eV]^{-1}}$$ with $$c=\frac{[L]}{[t]},c=1\quad\text{the unit of t is ~[eV]^{-1}}$$