UPDATE September 19, 2023: Our claim that the geodesic equation is incorrect, is wrong. We ignored the fact that
g_xy / dx
is non-zero at the test mass m. Thus the first derivative, after all, reveals that the metric is curved.
The y coordinate line at P leans to the right, while it leans to the left at Q. This means that the sign of g_xy flips at the middle, and the derivative g_xy / dx is non-zero.
The metric itself can be seen as consisting of first derivatives. Taking a derivative of a component of a metric is, in a sense, a second derivative.
Our analysis of the force F₂ in the first section below is correct, though.
----
The force F₂ appears in the Schwarzschild orbit of a test mass m
● M
|\
| \ β = angle y axis
vs. (M, m)
r
(0, 0) = coordinates of m
• ----> v coordinate velocity
m
\o
| observer
/\
^ y
|
-------> x
We want to determine the y acceleration of the test mass m. We assume the Schwarzschild metric around M.
As m moves to the right in the diagram, a part of its tangential velocity is converted into radial velocity. The observer sees that gravity G M / r² attracts the test mass m toward M. He also sees that tangential momentum measured by him
p_t' = m v / sqrt(1 - r_s / r)
of the test mass is partially converted into radial momentum p_r' measured by him. The observer does not notice anything special in this conversion. Momentum is conserved.
If the observer measures a tangential velocity v_t', then the coordinate tangential velocity is
v_t = sqrt(1 - r_s / r) v_t'.
That is because the clock of the observer ticks slower.
We used this fact above in the formula. If the observer measures a radial velocity v_r', then the coordinate radial velocity is
v_r = (1 - r_s / r) v_r'.
This is because the clock of the observer ticks slower and his ruler is contracted in the radial direction from M.
If we ignore the acceleration
G M / r²
of m toward M, then the observer would see m to move along a path which he thinks is a straight line. Some of the tangential momentum p_t is converted to radial momentum p_r. But in the Schwarzschild coordinates, the same velocity v' which the observer measures in the radial direction is only
sqrt(1 - r_s / r)
times the corresponding velocity v' that the observer measures to the tangential direction. This means that in Schwarzschild coordinates, the path of m bends up, to the positive direction of the y axis. Let us calculate the acceleration. We assume that m starts from coordinates x = 0 and y = 0. If m would move along straight line in the Schwarzschild coordinates, its radial velocity at a location x << r would be
v_r = v sin(β)
= v x / r
But the radial velocity is shrunk by the factor
sqrt(1 - r_s / r) = 1 - 1/2 r_s / r.
Let
x = v t.
The shrinking of the velocity v_r in the time t is
v * v t / r * 1/2 r_s / r,
and the associated acceleration to the positive y direction
1/2 v² / c² * G M / r².
This corresponds to the force F₂ of our previous blog post.
The force F₂ is not present in the geodesic equation acceleration to the y direction
In the formulae above, μ is y. The second term in the geodesic equation (top) is zero because dy / dt is zero for the test mass m. In the first term, α and β can be x or t, because dt / dt and dx / dt are non-zero.
Cross terms in the Schwarzschild metric g at the test mass m are zero.
----
UPDATE September 19, 2023: Here we have an error. The derivative dg_xy / dx is NOT zero!
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Thus, only the terms
∂g_tt / dy
and
∂g_xx / dy
can contribute to the y acceleration. The tangential spatial metric is 1 close to the test mass m, thus
∂g_xx / dy = 0.
The term
∂g_tt / dy
corresponds to the familiar acceleration
G M / r².
Thus, the acceleration corresponding to the force F₂ is missing from the geodesic equation.
Conclusions
The geodesic equation is clearly incorrect. We have to check if anyone has expressed doubts about its correctness. The fact that one tries to derive a second derivative thing, the acceleration, from first derivatives is highly suspicious.
A brief net search did not return anything about the geodesic equation and second derivatives. We have to check the derivations of the geodesic equation in Wikipedia. How would they handle our counterexample?
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