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I know that the cosmological constant was developed as an addition to the Einstein Field Equation as an anti-gravity force: $$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} + \Lambda g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu}$$

so that he could get a static universe. But later it was discarded. And now it is related to dark energy.

But what exactly does the cosmological constant mean. Is it a measure of the amount of dark energy? If so, how can we measure it? Please note that I want to know the physical meaning of the constant in our universe (but some mathematical exposure can be helpful anyway)

NOTE: The answer given here: What does the 'cosmological constant' represent? is more inclined towards the fact that the cosmological constant was developed as a tool to keep the universe static, or that it represents curvature where $T_{\mu \nu} = 0$, but what I am looking for is what do we think of the cosmological constant today? I am not asking how or why it crops up in the EFE, or why it was developed to make sense of a solution to the EFE, but simply what it really is, physically?

PNS
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  • The cosmological constant term should be $\lambda g_{\mu\nu}$. –  May 18 '20 at 13:23
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    Thanks for noting. Fixed it. – PNS May 18 '20 at 13:25
  • Not exactly. That one seems to be more about the mathematical need for the constant. But I want to know what it physically represents. Maybe dark energy.. or something else? – PNS May 18 '20 at 13:29
  • I disagree. The linked question specifically asks about what physical aspect of the universe we are referring to when we use the cosmological constant. –  May 18 '20 at 14:05
  • @Dvij D.C, not exactly.. the first two answers I read were mainly about $\Lambda$ as a means for a particular solution of EFE. But the answer by The_Symphatizer is the one I was looking for. What exactly is the cosmological constant, not how they factor into the solutions of the EFE. – PNS May 18 '20 at 14:23
  • Let us continue this discussion in chat. –  May 18 '20 at 14:24

1 Answers1

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The cosmological constant is a measure of the (uniform) energy density associated with space: it is how much "intrinsic energy" each parcel of space has, just matter has an energy associated with it due to its mass. Because of how the equation is typically written, it is not directly given in units of energy density, however

$$\rho_0 = \frac{c^2 \Lambda}{8\pi G}$$

is in units of energy density. For the empirically measured value of $\Lambda \approx 1.1056 \times 10^{-52}\ \mathrm{m^{-2}}$, $\rho_0 \approx 5.9238 \times 10^{-27}\ \mathrm{J/m^3}$ or if you like, $5.9238\ \mathrm{J/Gm^3}$. Thus, each cubic gigameter (a cube of roughly similar volume to our Sun) of space holds a bit less than six joules of inherent energy simply due to its being space itself.

The_Sympathizer
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  • Thanks for the answer +1. But can you explain how we got the $1.1056 * 10^{-52}$? – PNS May 18 '20 at 13:37
  • @PNS: Since the theory goes that the dark energy powers the expansion of the universe, measuring the expansion measures $\Lambda$. And that is done by analyzing redshift distributions of distant galaxies, and other techniques. – The_Sympathizer May 18 '20 at 13:38
  • Ok..got it..thanks – PNS May 18 '20 at 13:39
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    This is one possible interpretation of the cosmological constant. The Einstein equations are, as such, agnostic about how to interpret it. You can interpret it also as the curvature of spacetime in the absence of all matter/energy. –  May 18 '20 at 14:07
  • It is unclear from your answer why the energy of space itself should be repulsive while the energy contained in space is attractive. One could claim that the ADM mass of a black hole also is the energy of the curved space, but in this case it is attractive. Your answer makes a statement without explaining it and does not help with understanding the physical nature of the constant. In contrast, the interpretation given above by @DvijD.C. is very clear. I am not saying it is better (although it may be), but only that yours is not explained to be equally clear. – safesphere May 18 '20 at 17:27
  • In addition neither your answer nor comment explains how exactly the constant is measured. For example, measuring it based on the non-accelerated space expansion would be somewhat different from measuring based on the supernova luminosity data. You also do not mention how the sign of the constant affects it’s physical interpretation or whether or not it’s existence contradicts the equivalence principle, on which general relativity is based. Finally it may worth mentioning whether or not the constant is unique to the Friedman cosmology that would not match observations without this tweak. – safesphere May 18 '20 at 17:42
  • Hi so does the value of $\Lambda \approx 1.1056 \times 10^{-52}\ \mathrm{m^{-2}}$, inferred that the universe is expanding at a rate of 1.10x10^(-52) m^(‐2) ? – Tivity Aug 22 '20 at 13:22
  • The units of energy density for $\rho_0$ in the accepted answer are actually $kg/m^3$ not $J/m^3$. – Mr Anderson Mar 22 '22 at 11:40