# Basic relativity problem: a limit on the cosmological constant

I am working through a problem in Schutz's A First Course in General Relativity. With respect to Einstein's field equations, $$G^{\alpha\beta}+\Lambda g^{\alpha\beta}=8\pi T^{\alpha\beta},$$ he asks, "find the Newtonian limit and show that one recovers the motion of the planets only if $|\Lambda|$ is very small. Given that the radius of Pluto's orbit is $5.9\times 10^{12}\,\mathrm{m}$, set an upper bound on $|\Lambda|$ from solar system measurements."

The answer he gives is $|\Lambda| < 4\times10^{-35}\mathrm{m}^{-2}$, but I am completely stumped. When I take the Newtonian limit, I get something like $$-\frac{1}{2}\nabla^2\bar{h}^{\alpha\beta} + \Lambda\left(\eta^{\alpha\beta}+h^{\alpha\beta}\right)=8\pi T^{\alpha\beta},$$ but cannot see a way to reduce further, let alone come up with an inequality.

## 1 Answer

Rewrite the Einstein Equation as: $$G^{\alpha\beta}=\frac{8\pi G}{c^2}(T^{\alpha\beta}-\frac{c^4\Lambda}{8\pi G}g^{\alpha\beta})$$ Thus we can absorb the cosmological constant into our stress energy. We use the stress energy tensor of the solar system as that of a single point particle (the Sun), ignoring all the other planets and teapots and such: $T = diag(M_\odot c^2\delta(\vec r),0,0,0)$.

We now examine the time-time component of the Einstein equation: $$G^{00}\approx -2\nabla^2\Phi = \frac{8\pi G}{c^2}(M_\odot c^2 \delta(\vec r)-\frac{c^4\Lambda}{8\pi G})$$ Integrating over a sphere of radius $r$ centered at the Sun, we find the force field: $$\Phi'(r) = -G\frac{M_\odot }{r^2}+\frac{c^2\Lambda}{6}r$$ Thus we see that the Cosmological Constant generates a repulsive force that gets STRONGER the farther away one is from the source of the gravitational field (the Sun).

Since we know that Pluto is bound, the net force on the little guy must point towards the Sun. This translates into $\Phi'(R_P ) < 0$. Rearranging the furniture to suit our needs gets us to: $$\Lambda < \frac{6GM_\odot}{c^2R_P^3}=3\frac{R_{S,\odot}}{R_P^3}$$ In the last line I've written the answer in terms of the Schwarzschild radius of the Sun, $R_{S,\odot}\approx 3km$. We can already see that this means that the order of magnitude is going to be incredibly small. Plugging in everything, we find: $$\Lambda < 4.32\times 10^{-35} m^{-2}$$

• Everything makes sense except one thing: you used the fact that $G^{00}\approx -2\nabla^2\Phi$. Where did this come from? According to Schutz's text, in order to arrive at $G^{00}\approx -2\nabla^2\Phi$, we must assume $\Lambda=0$. But in this problem, we are no longer assuming anything about $\Lambda$. – Doubt Apr 27 '17 at 20:45
• Think about how we're interpreting $\Lambda$. The Einstein tensor is unchanged because of this. – Damian Sowinski Apr 28 '17 at 3:41
• Can you explain more? The result $G^{00}\approx-2\nabla^2\Phi$ is a result of $\mathbf{G}=8\pi\mathbf{T}$, correct? – Doubt May 1 '17 at 22:18
• Yup... and adding the $\Lambda$ changes only ${\bf T}$ since we are interpreting the cosmological constant as a constant stress energy everywhere. – Damian Sowinski May 1 '17 at 22:48
• We absorbed the cosmological constant into the stress energy. That was the whole point, to reinterpret $\Lambda$: $\rho = \rho_m+\rho_\Lambda$ – Damian Sowinski May 9 '17 at 2:15