# What is the correct ratio of Newtonian to General Relativistic gravitational effects for Sun + single planet orbital system

For a hypothetical orbital system (Sun + single planet) the Newtonian model and the General Relativity (GR) model produce different expressions for the gravitational effect of the Sun on the planet. This is well known.

The ratio between the Newtonian and GR effects is expressed in different ways by different writers.

I am having trouble reconciling two such expressions of the Newtonian:GR ratio.

Firstly Walter (2008) (equation 12.7.6, page 482) presents the following expression for the equation of motion produced from the GR model

$$\frac{d^2\,u}{d\theta^2} + u = \frac{GM}{h^2} + \frac{3GM}{c^2}u^2$$ where $u= (1/r)$, $h=vr$, $G$ is the universal constant of gravitation, $M$ is mass of the Sun, $c$ is speed of light. Here the term $GM/h^2$ is the ordinary Newtonian term and the term $3GMu^2/c^2$ is the additional term introduced by GR.

From this Walter derives the approximate ratio between Newtonian and GR effects as $(1)$ to $(1 + 3v^2/c^2)$ where $v = \sqrt{GM/r}$ is the orbital speed of the planet in a circular orbit (with distance $r$ = $a$, the semi-major axis).

Secondly, an alternative presentation (referring to the so-called Schwartzchild solution) is given by Goldstein in Classical Mechanics (3rd Edition) pages 536-538. The GR potential $V_{GR}$ is given by $$V = -\frac{GMm}{r} -\frac{b}{r^3}$$ where $m$ is the target body mass and $b$ is a constant (Goldstein uses $h$ instead of $b$, see below, but I have already used $h$ to mean something different in Walter above,)

differentiating the potential with respect to distance $r$ to give force we derive $$F_{GR} = \frac{GMm}{r^2} + \frac{3b}{r^4}$$

Now Goldstein defines the constant $b$ thus :- $$b = \frac{k\,l^2}{m^2c^2} \qquad\text{ Goldstein eqtn [12.48]}$$ where $$k=GMm$$ and $$l^2 = mka(1-e^2) \qquad\text{ Goldstein eqtn[12.50]}$$

So

$$b = \frac{GMm\, m\, GMm\, a(1-e^2)}{m^2c^2} = \frac{GMm \, L^2}{c^2} = \frac{GMm \,m^2 v_c^2 a^2}{c^2}$$

So the GR force equation becomes $$F_{GR} = \frac{GMm}{r^2} + \frac{3GMm m^2 v_c^2 a^2}{r^4c^2}$$ substituting $a$ by $r$ we get $$F_{GR} = \frac{GMm}{r^2} + \frac{3GMm m^2 v_c^2}{r^2c^2} = \frac{GMm}{r^2} \left( 1 + \frac{3 v_c^2 \, m^2}{c^2} \right)$$

So the Newtonian:GR ratio derived from Goldstein is the same as the ratio derived by Walter except that the former has the additional term of $m^2$ in the numerator. Even if we tried to fudge this numerically by invoking a unit mass target, it would still be dimensionally incorrect.

So what is the correct ratio? 

UPDATE ---------------------------------------------------------------------

In the refactoring of $b$ I used angular momentum $L$ when I should have used specific angular momentum $\mathfrak{l}$. After correction the extra $m^2$ disappears. Goldstein agrees with Walter. My thanks to Stan Liou for illumination.

Corrected analysis:- $$b = \frac{GMm\, m\, GMm\, a(1-e^2)}{m^2c^2} = \frac{GMm \, \mathfrak{l}^2}{c^2} = \frac{GMm \,v_c^2 a^2}{c^2}$$

So the GR force equation becomes $$F_{GR} = \frac{GMm}{r^2} + \frac{3GMm v_c^2 a^2}{r^4c^2}$$ substituting $a$ by $r$ we get $$F_{GR} = \frac{GMm}{r^2} + \frac{3GMm v_c^2}{r^2c^2} = \frac{GMm}{r^2} \left( 1 + \frac{3 v_c^2}{c^2} \right)$$

So the correct ratio of Newtonian to GR gravitational force is:- $$F_{Newtonian}:F_{GR} \approx 1: \left( 1 + \frac{3 v_c^2}{c^2} \right)$$

NOTES

This ratio is approximate and only applies in the "low velocity, weak field" sub-domain of the GR model.

Goldstein also emphasises that the GR effect is not a velocity effect (presumably as in the velocity of the target body through any kind of aether or flux).

Coincidentally (in the same sub-domain e.g. Mercury orbitting the Sun) a modified Newtonian radial force of magnitude $f=GMm/r^2 * [1 + 3v_t^2/c^2]$, where $v_t$ is the instantaneous transverse velocity of a small target planet, produces non-Newtonian apsidal rotation ("perihelion precession") of the same magnitude (within 1%) as GR.

Goldstein needs to be read with care. Here he uses $l$ to denote angular momentum elsewhere (e.g. eqtn [1.7]) he uses $L$. He often refers to $V$ as "potential" when he is clearly referring to "potential energy" (e.g. eqtn [3.49]).

• There is no one specific "ratio". If that's all there was to general relativity, GR would be easy. GR is not "easy". The relationship posted in this question is perhaps the simplest of simple linearizations of general relativity. – David Hammen Oct 16 '14 at 11:53
• David is of course correct in that this kind of comparison only makes sense for slow orbits in a weak-field approximation, though fortunately that's also the context of this question. It may be noted that for the specific case of Schwarzschild spacetime, orbits are exactly described by the effective potential; the approximation comes in when one treats the radial coordinate and proper time as if they were Newtonian, which is invalid in more general situations. – Stan Liou Oct 16 '14 at 14:46
• David & Stan: Thanks. Yes I was aware but have added clarification at end of the Question. – steveOw Oct 17 '14 at 15:26

Orbits in Schwarzschild spacetime can be described by the effective potential $$V_\text{eff} = -\frac{GM}{r} + \frac{\mathfrak{l}^2}{2r^2} - \frac{GM\mathfrak{l}^2}{c^2r^3}\text{,}$$ where $\mathfrak{l} = r^2\dot{\phi}$ is the specific angular momentum of the orbit, which is a conserved quantity. The first two terms match the form of the Newtonian effective potential, except that here we're referring to the Schwarzschild radial coordinate $r$ and proper time of the orbiting particle, instead of radial distance and coordinate time. The first term being the usual gravitational potential and the second being the centrifugal potential, so Goldstein's $V$ does make some sense as a gravitational potential energy term instead.

Therefore, $l^2 = mka(1-e^2)$ with $k = GMm$ means that $l$ is the angular momentum, $l = m\mathfrak{l}$, and $$\frac{GM\mathfrak{l}^2}{c^2r^3}m = GMm\frac{l^2}{m^2}\frac{1}{c^2r^3} = \underbrace{\frac{kl^2}{m^2c^3}}_b\frac{1}{r^3}\text{,}$$ just as Goldstein says. If we differentiate $\frac{1}{2}m\dot{r}^2 + mV_\text{eff} = \mathcal{E}$ with respect to proper time, then $$\begin{eqnarray*}m\ddot{r} - \frac{l^2}{mr^3} &=& -\frac{k}{r^2} - \frac{3b}{r^4}\\ &=& -\frac{k}{r^2}\left(1 + 3\frac{l^2}{m^2c^2}\frac{1}{r^2}\right)\\ &=& -\frac{k}{r^2}\left(1 + 3\frac{m(GMm)a(1-e^2)}{m^2c^2}\frac{1}{r^2}\right)\\ &=& -\frac{k}{r^2}\left(1 + 3\frac{v_c^2}{c^2}\frac{a(1-e^2)}{r}\right)\text{.} \end{eqnarray*}$$ This is dimensionally correct, as both $v_c/c$ and $a/r$ are dimensionless, while $$\frac{l^2}{mr^3} = \frac{k}{r^2}\frac{a(1-e^2)}{r}\text{.}$$ The left-hand side has the Newtonian form.

• $\ell$ has dimension $MD^2/T$ due to inclusion of $k=GMm$ in Eqtn 12.50. Maybe Eqtn.12.48 for $b$ should have $m^4$ in denominator rather than $m^2$? – steveOw Oct 15 '14 at 20:44
• Aha! I see the error is mine. I confused angular momentum with specific angular momentum (no mass). Goldstein agrees with Walter. His $\ell^2$ is OK as the second $m$ comes from the term $k=GMm$. – steveOw Oct 15 '14 at 21:18
• @steveOw yeah, I misread how the variables were defined too. Eh! – Stan Liou Oct 15 '14 at 21:24
• I think your last equation but one should have (a^2/r^2) instead of (a/r) assuming the previous equation is correct. Either way, any presence of r in this term disproves my thesis...which I am now re-evaluating....I may ask a separate question to clarify my thoughts. – steveOw Nov 7 '14 at 16:30
• @steveOw Since $l^2/m^2 = GMa(1-e^2)$ as introduced in the beginning of the second paragraph (cf. also here but with $m\ll M$), $a/r$ is correct. But you're very welcome to ask any follow-up questions. – Stan Liou Nov 7 '14 at 17:43

The expression that is used by NASA/JPL to approximate the relativistic effects on the orbit of a single planet plus the Sun in our solar system is known as the "post-Newtonian expansion" and looks like:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\left(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2}\right)\hat{r} +\frac{4GM}{r^2}\left(\hat{r}\cdot \hat{v}\right)\frac{v^2}{c^2}\hat{v}$$

If there are many planets the expression becomes more complex. You can compare this to the classical Newtonian acceleration:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}$$

I am not a big fan of this approximation but it is what is mostly used.

For a pure circular orbit in Schwarzschild coordinates you get the same orbital velocity in GR (in coordinate time) as classically.

I you are dropping an object from rest the initial acceleration in GR (in coordinate time) is the same as classically.

Generally if you want to know if GR or classical Newtonian gravitation results in more acceleration you have to decide if you are interested in the result in "coordinate time" or in "proper time" and the fraction will also vary depending on in what direction the planet is moving as compared to the Sun.