What is the significance of the factor 8π in the Einstein Equations?

[NicosM]

Hi!

I would like to ask where the factor of 8π comes from in the Einstein Equations. I understand that it is augmented in order to comply with the Newtonian limit. But why did Einstein decide to place 8π and not for example 8.1π or 3π-3π/e=8.27 or whatever around this value (which would give almost the same correspondence with the experiment)?

Thanks

 

[kevinferreira]

It comes from the Newtonian limit, as you said, and the Newtonian formula for the potential $\Delta\Phi=4\pi\rho$, which gets multiplied by 2 somewhere. This $4\pi$ factor here ultimately comes from the spherical symmetry of the Newtonian gravitation theory.

 

[NicosM]

Thanks for your reply. I understand the 4π factor of the Newtonian formula. Can you tell me where the 2 comes from, in case you know? Thanks again

 

[Ben Niehoff]

Ultimately, the constant in Einstein's equations is arbitrary and can be set to any value with an appropriate choice of units. I typically use

$$R_{\mu\nu} = \frac12 T_{\mu\nu}$$
because I do theoretical physics and I don't care about units. I guess I'm working with $\hbar = 1, c = 1, G = 1/16\pi$.

 

[NicosM]

Thanks for your reply too. Then my question should be restated as why G/h=1/8 ? :)

 

[Ben Niehoff]

Thanks for your reply too. Then my question should be restated as why G/h=1/8 ? :)

It doesn't mean anything at all. G and Planck's constant have different units, so G/h also has units, and can be set to any value we choose by choosing the right system of units.

The only meaningful numbers in physics are dimensionless (ironically enough). The rest is convention.

 

[PAllen]

Isn't it because Einstein enjoyed pie more than Newton?

 

[NicosM]

Actually Archimedes enjoyed that more than both of them :)

 

[rbj]

there is a very good and very old discussion of this $8 \pi$ thing in an old sci.physics.research thread.from John Baez:

The $4 \pi$ comes from the area of the sphere, just like electromagnetism. The extra factor of 2 comes from the fact that $h_{00}$ is -2 times the Newtonian gravitational potential $\Phi$. What's $h_{00}$? Well, $h_{00}$ measures the amount of "time dilation due to the gravitational field". Technically, $h_{00}$ is the difference between the time-time component of the actual spacetime metric and that of flat Minkowski spacetime.

Here's how it works. In the Newtonian limit, Einstein's equations say that

$$\nabla^2 h_{00} = -8 \pi G \rho$$

On the other hand, according to Newtonian gravity we have

$$\nabla^2 \Phi = 4 \pi G \rho$$

These are consistent, thanks to that extra factor of 2.

(The minus sign is related to the fact that we're using a convention where the time-time component of the metric is negative.)

By the way, I think this factor of 2 is secretly the same as the 2 you see in the formula for the Schwarzschild solution: the spacetime metric for a spherical object of mass $M$ is described by a formula involving the quantity $2M$.

I suppose now someone is going to ask where the 2 comes from in the formula $h_{00} = -2 \Phi$.

from Daryl McCullough:

I thought that the source of this 2 is the square-root appearing in the proper time formula:

$$dT = \sqrt{g_{\mu,\nu} dx^{\mu} dx^{\nu}}$$

In weak, time-independent gravitational fields, $g_{\mu,\nu}$ takes the form

$$g_{0,0} = 1 + h$$
$$g_{1,1} = g_{2,2} = g_{3,3} = -1$$
$$g_{\mu,\nu} = 0 \quad \quad (\mu \ne \nu)$$

So using
$$\left(\frac{v}{c}\right)^2 = \left(\frac{dx^1}{dx^0}\right)^2 +\left(\frac{dx^2}{dx^0}\right)^2 +\left(\frac{dx^3}{dx^0}\right)^2$$

the formula for proper time becomes
$$dT = \sqrt{1+h - \frac{v}{c}^2} dt$$

For $A$ and $v$ small, we can expand the square-root to get

$$dT = \left( 1 + \frac{1}{2} h - \frac{1}{2} \left(\frac{v}{c}\right)^2 \right) dt$$

The equations of motion for a test particle are obtained by maximizing the integral of $\left(1 + \frac{1}{2} h - \frac{1}{2} \left(\frac{v}{c}\right)^2 \right) dt$. Since the constant 1 is irrelevant for the equations of motion, this is the same as maximizing the integral of $D dt$, where

$$D = \frac{1}{2} h - \frac{1}{2} \left(\frac{v}{c}\right)^2$$

But, in this limit (small $A$ and small $v$), Newtonian mechanics can be used, so we should also be able to obtain the equations of motion by minimizing the integral of $L dt$, where

$$L = \frac{1}{2} mv^2 - m \Phi$$

(where $\Phi$ is the Newtonian gravitational potential). This suggests that $D$ and $L$ are linearly related. The only way for this to be true is if

$$h = 2 \Phi/c^2$$

i think i transcribed the math into LaTeX accurately.

i found this paper to be very useful:

http://math.ucr.edu/home/baez/einstein/einstein.html][/PLAIN] The Meaning of Einstein's Equation


and the main statement in the paper:

Given a small ball of freely falling test particles initially at rest with respect to each other, the rate at which it begins to shrink is proportional to its volume times: the energy density at the center of the ball, plus the pressure in the $x$ direction at that point, plus the pressure in the $y$ direction, plus the pressure in the $z$ direction.

Baez and Bunn tell us that this statement is equivalent to Einstein's

$$G_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu}$$

the constant of proportionality in the Baez-Bunn statement is, i believe, $\frac{4 \pi G}{c^2}$ not $\frac{8 \pi G}{c^2}$. i believe it is:

$${\ddot V\over V} \Bigr\vert _{t = 0} = - \frac{4 \pi G}{c^2} \left(\rho c^2 + P_x + P_y + P_z \right)$$

this reinforces my opinion that normalizing $4 \pi G = 1$ rather than $8 \pi G = 1$ is more natural in defining the most "natural" units. i think Planck missed it by a factor of $4 \pi$ regarding $G$ and i think that subsequently, they should have normalized $\epsilon_0$ rather than $4 \pi \epsilon_0$ regarding the electric charge in either cgs or in Planck units..

 

[NicosM]

thanks rbj

 

[rbj]

yer welcome. can't say that i can explain it.