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According to the Newtonian gravitational potential:

$$\phi\left(r\right)=-\dfrac{GM}{r}.\tag{1}$$

We can find the minimum possible point in a gravitational well as the one for which the value of the potential is lower (in $r\rightarrow 0$ in this case). How can you do the same in General Relativity? Is it needed to use the Ricci tensor instead of the usual gravitational potential or is it more than sufficient to use the latter? Do you lose relevant information using eq. (1) rather than the usual field equations?

Qmechanic
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Antoniou
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  • Are you asking if general relativity makes all the same predictions as newton's model of gravity? If so, the answer is of course not - why would GR even exist as a theory if it was just a vastly more complex formulation of the exact same thing. There is a limit where Newtonian gravity is approximately valid. But otherwise even the fact that the coordinates ignore that we're talking about a curved manifold proves this can't be right. – AXensen May 30 '23 at 07:55
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    You may find the diagram in this answer helpful: https://physics.stackexchange.com/a/644480/123208 – PM 2Ring May 30 '23 at 08:14
  • @AXensen of course I am not asking if GR makes the same predictions than Newton's laws. It is clear by reading the post. – Antoniou May 30 '23 at 09:26
  • @AntonioPeña Let me put it this way... Assume it's true GR a mass gave rise to a potential $GM/r$. If it was further true that that potential could be added linearly to the potentials created by other masses. And if you finally add the assumption that objects live in a 3d space and distances between them can be given by a coordinate $r=\sqrt{dx^2+dy^2+dz^2}$ - this is the entire basis of Newtonian mechanics. Nothing other than Newtonian gravity would be possible given all these assumptions. But GR is differnent from newtonian mechanics in all three of these ways. – AXensen May 30 '23 at 10:09

1 Answers1

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One can find the effective potential a particle experiences for instance in the most frequent case, the Schwarzschild case. Outside of a non-rotating massive object the effective potential can be found by computing the 4-momentum square in that metric ($r_s$ is the Schwarzschild radius):

$$(mc)^2 = p_\mu p^\mu = m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau} = m^2 \left[ \left(1-\frac{r_s}{r}\right)c^2\dot{t}^2 -\frac{\dot{r}^2}{1-\frac{r_s}{r}} -r^2 \dot{\varphi}^2\right] \tag{1}$$

The motion is governed by 2 constants of motion $E$ and $L_z$.

EDIT:

This can be seen considering the expression $m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau}$ as Lagrangian (more precisely said the Lagrangian multiplied by $-2m$):

$$-2mL = m^2 \frac{dx_\mu}{d\tau}\frac{dx^\mu}{d\tau}$$

In order to find the canonical momenta the derivatives with respect to $\dot{t}$, $\dot{r}$ and $\dot{\varphi}$ are taken. Most interesting are $\dot{t}$ and $\dot{\varphi}$ since $t$ and $\varphi$ are cyclic variables ($r$ is not a cyclic variable since the Lagrangian depends explicitly on $r$):

Then we get (where energy $E$ and angular momentum $L_z$ appear as constants of motion scaled by $2m$ as the Lagrangian):

$$const =\frac{\partial (-2mL)}{\partial \dot{t} } =2m^2c^2 \dot{t} \left(1-\frac{r_s}{r}\right) =2mE$$

and $$const =\frac{\partial (-2m L)}{\partial \dot{\varphi} } =-2m^2c^2 r^2\dot{\varphi} =-2mL_z$$

which turn into:

EDIT END

$$\dot{t} =\frac{E}{mc^2(1-\frac{r_s}{r})} \quad\text{and}\quad \dot{\varphi} =\frac{L_z}{mr^2}$$

After a bit of algebra one gets from (1):

$$ \frac{E^2}{(mc^2)^2} = 1 + \frac{\dot{r}^2}{c^2} +V_{eff}$$

where the effective potential equals:

$$ V_{eff}= -\frac{r_s}{r} + \frac{L_z^2}{m^2c^2}(\frac{1}{r^2} - \frac{r_s}{r^3})$$

The minima and maxima of this potential provide you the requested information.

If the solution of Einstein's field equations (EFE) is already known as here assumed the task is rather easy. Of course to find the rotational- symmetric EFE's solution outside a non-rotating massive object one has to solve the EFE's in vacuum, i.e.

$$ R_{\mu\nu}=0 \quad\text{with the corresponding symmetry conditions} $$

where $R_{\mu\nu}$ is the Ricci tensor.

Frederic Thomas
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  • But two such solutions cannot be added together linearly, correct? Even the theorem that gives rise to an $r$ coordinate is no longer valid when there's two bodies. So while it's useful to point out that the effective potential is not $1/r$ anymore, one should also point out that the whole idea of linearly adding the potential due to different bodies and interpreting $r$ as the distance to those bodies from a particular point in cartesian space breaks down (in a more general case with more than one body). – AXensen May 30 '23 at 08:50
  • How does it go from $\left(mc\right )^2$ in the first definition to $mc^2$ in the $t$ dot definition? How are you solving for both? And why the polar coordinate doesn't go to zero after $d\tau$? – Antoniou May 30 '23 at 09:13
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    The whole expression can be also considered as Lagrangian. Then $x:= t$ and $x:=\varphi$ are cyclic which leads directly to conversed momenta $p_{x} = \frac{d}{dt}\frac{dL}{d\dot{x}}$. $\dot{t} = \frac{dt}{d\tau}$ and $d\tau$ is the proper time. So $\dot{t}$ is dimensionless. $E/(mc^2)$ too. $\frac{d\varphi}{d\tau}\neq 0$ because the particle (e.g. moon around earth) is moving (circulating) in (proper) time. – Frederic Thomas May 30 '23 at 10:42