Farsight versus Einstein, part 1
Once again,
Farsight has stated an important difference between FGR (
Farsight general relativity) and the general theory of relativity put forth by Einstein:
When spacetime is flat, no gravitational field is present.
When spacetime is flat, no gravitational field is present.
You can make it as simple as you like, but it will remain wrong. There simply isn't a direct analog of "gravitational field" in GR, but the closest is the connection, not the curvature. Of course that's meaningless to you, since you have no clue what those words even mean, let alone what the difference is.
Huff puff, sol really doesn't like it when I explain the gravitational field with reference to Einstein. He behaves like a witchdoctor faced with a pharmacist, all outraged, saying "you don't understand it". Tough. I do.
Farsight can huff and puff all he likes, but he's arguing with Einstein here. From the English translation of "The Foundation of the General Theory of Relativity":
Albert Einstein said:
C. THEORY OF THE GRAVITATIONAL FIELD
§ 13. Equations of Motion of a Material Point in the Gravitational Field. Expression for the Field-components of Gravitation
A freely movable body not subjected to external forces moves, according to the special theory of relativity, in a straight line and uniformly. This is also the case, according to the general theory of relativity, for a part of four-dimensional space in which the system of co-ordinates K
0, may be, and is, so chosen that they have the special constant values given in (4).
Einstein's formula (4) is the 4x4 matrix of coefficients for the Minkowski metric in the familiar coordinate system for flat spacetime. Einstein continues by talking about the gravitational field that appears when you use a different coordinate system to describe exactly the same locally flat spacetime:
Albert Einstein said:
If we consider precisely this movement from any chosen system of co-ordinates K1, the body, observed from K1, moves, according to the considerations in § 2, in a gravitational field....
If the Γτμν vanish, then the point moves uniformly in a straight line. These quantities therefore condition the deviation of the motion from uniformity. They are the components of the gravitational field.
According to Einstein, therefore, the 64 coordinate-dependent Christoffel symbols "are the components of the gravitational field". (Because the Christoffel symbols are symmetric in their lower indices, not all of those components are independent.)
Let's do an example. As this example shows, Einstein's "components of the gravitational field" can be nonzero even in flat spacetime, contrary to
Farsight's repeated base
less assertions.
Let's use
the metric form posted by sol invictus. In that coordinate system, the four nonzero components of the covariant metric tensor are
[latex]
\[
\begin{align*}
g_{00} &= - (x^1 - r_0) \\
g_{11} &= \frac{1}{x^1 - r_0} \\
g_{22} &= 1 \\
g_{33} &= 1
\end{align*}
\]
[/latex]
(I'm using MTW notational conventions, and I'm writing the r coordinate as x
1 to prepare for the index and Einstein summation notations used below.)
To calculate the 64 components of the gravitational field, we'll need the components of the contravariant form as well:
[latex]
\[
\begin{align*}
g^{00} &= - \frac{1}{(x^1 - r_0)} \\
g^{11} &= (x^1 - r_0) \\
g^{22} &= 1 \\
g^{33} &= 1
\end{align*}
\]
[/latex]
Einstein's equation (45) states his formula for the components of the gravitational field. Expanding Einstein's notation into modern notation, Einstein's equation becomes
[latex]
\[
\begin{align*}
\Gamma^\lambda_{\mu\nu}
&= \frac{1}{2} g^{\lambda\alpha} \left(
\partial_{\nu} g_{\alpha\mu}
+ \partial_{\mu} g_{\alpha\nu}
- \partial_{\alpha} g_{\mu\nu} \right)
\end{align*}
\]
[/latex]
If I've done that calculation correctly, only 4 of the 64 Christoffel symbols are nonzero:
[latex]
\[
\begin{align*}
\Gamma^0_{01} &= \Gamma^0_{10} = \frac{1}{2 (x^1 - r_0)} \\
\Gamma^1_{00} &= \frac{1}{2} (x^1 - r_0) \\
\Gamma^1_{11} &= - \frac{1}{2 (x^1 - r_0)}
\end{align*}
\]
[/latex]
With those Christoffel symbols in hand, it's straightforwardly tedious to calculate the 256 components of the Riemann curvature tensor via Einstein's equation (43). In modern notation, that equation becomes
[latex]
\[
\begin{align*}
R^{\alpha}_{\beta\gamma\delta} &=
\partial_{\gamma} \Gamma^{\alpha}_{\beta\delta}
- \partial_{\delta} \Gamma^{\alpha}_{\beta\gamma}
+ \Gamma^{\alpha}_{\kappa\gamma}
\Gamma^{\kappa}_{\beta\delta}
- \Gamma^{\alpha}_{\kappa\delta}
\Gamma^{\kappa}_{\beta\gamma}
\end{align*}
\]
[/latex]
That calculation is greatly simplified by the following observations. As can be seen from the Christoffel symbols calculated above, the partial derivatives of the first two terms are zero unless the subscript on the derivative operator is 1. Similarly, every component of the Riemann tensor that has a 2 or a 3 as an index will be zero; that observation reduces the number of components from 256 to 16, even before we take advantage of symmetries.
I calculated the components of the Riemann tensor by hand, and found all of its 256 components are zero.
That means
sol invictus has indeed given us an example of flat spacetime (because the Riemann curvature tensor is zero) that has a nonzero gravitational field (because four of the Christoffel symbols, which Einstein regards as the components of the gravitational field, are nonzero).
In his response to
sol invictus, quoted above,
Farsight said he understands all this stuff. That's great. I invite
Farsight to check my calculations.
The bottom line is that Einstein disagreed with
Farsight's repeated bare assertion that flat spacetime implies a zero gravitational field.
).
