0
$\begingroup$

When Einstein developed his field equations for general relativity, he attempted to apply them to the entire universe. He found the universe had to expand, which at the time was believed not to be the case. So he added the cosmological constant to his equations to stop the universe from expanding. Later he learnt, from Edwin Hubble, that the universe was expanding, so he removed his cosmological constant, calling it his greatest blunder (though, in a way, you could say his initial prediction, that the universe was expanding, was his greatest success).

In the meantime, Schwarszchild had developed his solution to Einstein's equations, for a single star in otherwise empty space. It was his solution which prediced the existence of black holes. Now I am pretty sure that Schwarszchild developed his solution while Einstein's equations still retained the cosmological constant. Does this mean the cosmological constant is required to stop Schwarszchild's solution from also expanding?

$\endgroup$

3 Answers 3

4
$\begingroup$

The Schwarzschild solution sets the cosmological constant to zero.

The expanding universe is a feature of when you model a universe with a homogenous and isotropic matter content. The Schwarzschild solution models a universe in vacuum, which does not expand. Hence, the cosmological constant is unnecessary.

Usually, the Schwarzschild solution is used to model isolated bodies, such as planets, stars, and black holes. Since we are working at very small scales (when compared to cosmological scales), the cosmological constant doesn't really contribute that much. Hence, we can neglect it, and we very often do.

There are generalizations of the Schwarzschild solution to consider a cosmological constant. These are known as Schwarzschild–de Sitter and Schwarzschild–Anti-de-Sitter spacetimes.

$\endgroup$
2
  • 1
    $\begingroup$ Great answer! Let me also add that the cosmological constant is expected to have absolutely no impact on dense and small scales systems (as e.g. planetary systems, black holes etc) such that the Schwarzschild metric would correctly describe such system without a need for a $\Lambda$ addition. Possibly more informations in this outreach video: youtube.com/watch?v=bUHZ2k9DYHY. $\endgroup$ Commented May 14 at 16:49
  • 1
    $\begingroup$ @LéoVacher Super fair! I added a paragraph commenting on this $\endgroup$ Commented May 14 at 16:51
1
$\begingroup$

The original published version of the GR field equation in 1915 didn't contain a cosmological constant (Λ) term. Schwarzschild found his solution in 1916. Einstein proposed adding the Λ term in 1917.

As Nickolas Alves said, there is a generalization of the Schwarzschild solution to nonzero Λ, but the original solution assumed no (or equivalently zero) Λ.

It's not really relevant to this question, but your description of Einstein's reason for adding Λ may not be accurate. See Did Einstein really invent the cosmological constant to make the universe static in his 1917 paper?

$\endgroup$
-1
$\begingroup$

When Einstein used GR with the Hubble constant, he found the age of the universe to be about 1.5e9 years, which at the time they thought was about the age of the Earth (Einstein, "The Meaning of Relativity", 1953, pg. 120). So this must be wrong. So the cosmo constant was introduced to get a better result [this is wrong it had already been introduced, but to fix this same issue].

Apparently now, theories consider a cosmological scalar, so it can vary over space-time.

Edit1: Maybe I'm misinterpreting his words? From page 120:

"From the measured value of $h$ [Hubble constant] we get for the time of the existence of the world [he means universe] up to now 1.5e9 years. This age is about the same as that which one has obtained from the disintegration of uranium for the firm crust of the earth. This is a paradoxical result, which for more than one reason has aroused doubts as to the validity of the theory."

This is taking the Hubble constant into account, even though, as pointed out on the bottom of page 127, the cosmo constant was introduced before the Hubble's expansion was discovered.

He then shows that a curvature can't fix the cosmological problem on page 127:

"One further obtains, for the time from the start of the expansion up to the present, a value of the order of magnitude of 10^9 years. The brevity of this time does not concur with the theories on the developments of fixed stars.

"(4) The latter result is not changed by the introduction of spatial curvature; nor is it changed by the consideration of the random motion of stars and systems of stars with respect to each other."

It's interesting how he points out The Big Bang, on page 129:

"The[se] theoretical doubts are based on the fact that for the time of the beginning of the expansion the metric becomes singular and the density, $\rho$, becomes infinite."

And Dark Matter, on page 130:

"This is the case because we can hardly form an opinion on how large a fraction of $\rho$ is given by astronomically unobservable (not radiating) masses."

I thought that the cosmo constant was introduced to fix the cosmological problem, that the theory predicts a very short lifetime for the universe comparable to the age of the earth, and way shorter than we think it takes to form stars. Seems he rejected the cosmo constant, page 127:

"The introduction of the "cosmological member" into the equations of gravity, though possible from the point of view of relativity, is to be rejected from the point of view of logical economy."

But then, from page 132 at the end of the appendix for the 2nd edition:

"In this case I see no reasonable solution."

My understanding is that not only is the cosmo constant back in vogue [dark energy], but that it has been upgraded to a cosmo scalar, providing even more degrees of freedom to match theory with observation. I could be wrong on many things here, but if so, I'd like to learn where.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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