# A few questions about Einstein's field equations

1. Jul 22, 2014

### space-time

1. What exactly is the cosmological constant, what is its value (or how do you derive it if it is something that must be derived for various situations) and how do I know when to use it and when not to use it in the Einstein field equations?

I ask this last question because sometimes I see the equations with just:

R$\mu$$\nu$ - $\frac{1}{2}$g$\mu$$\nu$R = 8$\pi$G T$\mu$$\nu$

while other times I see them written as:

R$\mu$$\nu$ - $\frac{1}{2}$g$\mu$$\nu$R + g$\mu$$\nu$$\Lambda$ = [(8$\pi$G)/c4] T$\mu$$\nu$

(Sometimes I also see the equations with or without the c4 term.)

When do I use the cosmological constant, and when do I leave it out? What is the value of the constant, or how do I derive it? Why do people sometimes omit the speed of light term in the equations (though I'd guess that it is probably due to the c=1 convention, but correct me if I am wrong).

2. How do they get that c = 1 convention to make sense? What units are they using? I originally thought that they were saying that c = 1 light second per second, but I later found out that they were still using the SI unit meters.

How is that? c ≈ 3 * 108 m/s. Do the units for c affect the units that you must use throughout the field equations? For the stress energy tensor for an electromagnetic field, I have seen multiple sources use CGS units as opposed to SI units. Does the c = 1 convention have any affect on the types of units that you must use for this tensor or is it your choice on whether you want to use SI or CGS?

3. In multiple sources that I have seen, people have used the (+ - - -) signature for tensors such as the stress energy momentum tensor. For example: In the electromagnetic stress energy momentum tensor, all of the elements that had a 0 as one of the indices were positive while all the purely spatial elements were negative.

Does this mean that if I want to use the (- + + +) signature that all of the elements that are purely spatial will be positive while all of the elements that have a 0 as one of the indices will be negative? I ask this because I am relatively new to deriving tensors in 4D (with time) and the only 4D tensor I ever saw before recently was the Minkowski metric tensor which is all 0 except for the diagonal. Therefore, I never actually got to see a full example of how the sign signature affects all of the elements including the ones that are not on the diagonal.

4. Finally, is it true that when solving the Einstein field equations, you are solving for the metric tensor? If so, how would I go about solving these, considering that all of the tensors on the whole space time curvature side of the equations are derived from the metric tensor? After all, if I don't have the metric tensor, then how would I derive the Cristoffel symbol which is needed to derive the Riemann tensor which is needed to derive the Ricci tensor which is needed to derive the curvature scalar?

Thank you

2. Jul 22, 2014

### Staff: Mentor

The short answer is, it depends on what you're trying to model. See further comments below.

And sometimes you might see them written this way:

$$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = 8 \pi \left( T_{\mu \nu} + T^{vac}_{\mu \nu} \right)$$

where $T^{vac}_{\mu \nu} = - \frac{\Lambda}{8 \pi} g_{\mu \nu}$ is the "vacuum stress-energy". This is pretty much the modern view of what the cosmological constant "is": it's a manifestation of the fact that the vacuum has nonzero stress-energy, which we understand as arising from quantum vacuum fluctuations.

As for when to include it in the equations, as I said above, it depends on what you're trying to model, and how important the effects of vacuum stress-energy will be. The main area in which those effects are known to be important is cosmology: vacuum stress-energy (which is called "dark energy" in this context) is our current best explanation for why the expansion of the universe is accelerating. So our current best-fit cosmological model of the universe includes the cosmological constant in the field equations.

In other applications I'm aware of, such as modeling the solar system, or compact objects like white dwarfs and neutron stars, or binary pulsars, or gravitational waves, there are no observations that indicate a significant effect from vaccuum stress-energy, so we don't include the cosmological constant in the field equations in these models.

This is just a matter of units; the LHS of the equation describes curvature, which has units of inverse length squared, and the RHS describes energy density, which in conventional units has units of energy per volume, or energy per length cubed. The factor of $G / c^4$ converts energy to length, so that energy per length cubed becomes length per length cubed, or inverse length squared, the same units as the curvature on the LHS.

The reason the $G / c^4$ is often left out in the equations is that most relativists prefer to use "natural" units in which $G = c = 1$; in other words, a system in which mass, distance, and time all have units of length. See further comments below.

If distance is in meters, and $c = 1$, then the unit of time is the time it takes light to travel 1 meter (about 3.3 nanoseconds). Or if the distance unit is light seconds, then the time unit is seconds. Or if the distance unit is light-years, then the time unit is years. In other words, $c = 1$ just means the unit of time is the time it takes light to travel the unit of distance. You can do this with any unit of distance you want.

It's your choice what distance unit you want to use; but once you pick a distance unit, the time unit is determined as above. (And if you also use units in which $G = 1$, then the mass unit is also determined.)

Yes, this is often called the "timelike convention", because the timelike component has a + sign and the spacelike ones have a - sign.

Yes, this is often called the "spacelike convention", because the spacelike components have a + sign and the timelike component has a - sign.

The signature only tells you the signs of the components when the tensor is diagonalized (because what it's actually telling you, mathematically, is the signs of the eigenvalues of the tensor when it's viewed as a 4 x 4 matrix).

Yes.

The same way you solve any other second-order partial differential equation, since the LHS of the field equations is just a set of second-order partial differential equations in the metric tensor components (because you obtain the Christoffel symbols, and hence the Riemann tensor, Ricci tensor, and Ricci scalar by taking derivatives of the metric tensor).

3. Jul 22, 2014

### bcrowell

Staff Emeritus
Re units, I think usually the best strategy is this. Do all your calculations in purely symbolic form (no numbers), in units with c=1. Get a final algebraic result. Reinsert the factors of c in this result. (There is always a unique way to do this.) Then plug in numbers in SI units.