# Relativistic correction to the gravitational force?

1. Apr 8, 2015

### jk22

I considered a free falling object in the Schwarzschild metric at rest at infinity and found as approximation the acceleration :

$$\frac{d^2 r}{dt^2}\approx -\frac{GM}{r^2}+\frac{3G^2M^2}{r^3c^2}$$

This would mean if there are no sign error in some sense that the force becomes repulsive at small distances ?

Last edited: Apr 8, 2015
2. Apr 8, 2015

### A.T.

If t is measured by a clock at infinity, this reduces the coordinate acceleration compared to a value measured with local clocks. Also, if r is the Schwarzschild coordinate, not the actual physical distance, this also reduces the coordinate acceleration, compared to a value measured with local radial rulers.

If your object doesn't start at rest, but is a photon send down at c, you will have this "repulsion" right from the start, not just at small distance to the mass.

Last edited: Apr 8, 2015
3. Apr 8, 2015

### Orodruin

Staff Emeritus
I suggest trying to find the logical inconsistency in these statements:

4. Apr 8, 2015

### jk22

Indeed the acceleration depends on the energy.

The test particle accelerates down towards the mass and at small distances feels repulsion.

Is it this that the nasa uses to propel satellites around a planet since the classical work cancels out ?

5. Apr 8, 2015

### Orodruin

Staff Emeritus
No. Your expression is only valid for $GM/(rc^2) \ll 1$ and you are trying to apply it for a completely different situation.

6. Apr 8, 2015

### jk22

If i don't do this approximation then the acceleration were

$$\frac{-\frac{GM}{r^2}+\frac{4G^2M^2}{r^3c^2}}{\sqrt{1-\frac{2GM}{rc^2}}}$$

So i made an error in the approximation indeed.

7. Apr 8, 2015

### pervect

Staff Emeritus

It's not particularly clear that $\frac{d^2 r}{dt^2}$ should be interpreted as a "force" in any event, this interpretation seems to be based on mixing Newtonian ideas with GR formula. Letting momentum = m dr/dt just doesn't lead to very sensible results, but I believe that the idea that the force is m d^2r / dt^2 is based on the idea that the momentum p = m dr/dt , m is constant, giving dp/dt = m d^2 r / dt^2. (I don't see any other justification for saying that d^2r / dt^2 is related to a "force").

I'm not sure where the equation for d^2r/dt^2 came from and I haven't tried to verify it, but it is true that dr/dt goes to zero at the event horizon. Then it's clear that some object fallign into a black hole has some maximum value of dr/dt outside the event horizon (it is zero at infinity and zero at the event horizon but greater than zero everywhere else, implying a maximum value). This implies that the the quantity m does dr/dt decreases from it's maxximum value. While it's true that dr/dt decreases, it's not particularly sensible to regard this decrease as due to a repulsive force in my opinion.

Looking at some other ways to approach the question of the force on an object as a function of its radial coordinate:

The proper acceleration of an object at rest at some Schwarzschild radius r is infinite as you approach the event horizon. The exact formula is (see http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=641329740 and convert from geometric to standard units - the result is also in Wald's General Relativity) is
$$a = \frac{GM}{r^2 \sqrt{1 - r_s / r}}$$
where r is the Scwarzschild coordinate and $r_s = 2GM / c^2$ is the Schwarzschild radius.

If you specify the "force at infinity" - the force on the end (at infinity) of a massless string suspending an object of mass m (m is also measured at infinity) which has been lowered by said massless string to a schwarzschild cordinate of r and held there by the string, you get:

F = GMm / r^2

which is finite at the event horizon (and closely related to the idea of the surface gravity of a black hole, which has a rather technical wiki writep). The above formula leads to the integral of the acceleration-at-infinty multipled by the surface area at radius r being a a constant for a static black hole as described in the Komar mass article in the wiki.

8. Apr 8, 2015

### Staff: Mentor

Can you show how you are obtaining this result?​

9. Apr 11, 2015

### jk22

In fact i found a mistake. I start from the radial equation of motion in proper time :

$$\ddot{r}=-\frac{GM}{r}$$

and the relation proper time to coordinate time is $$d\tau=\sqrt{1-\frac{GM}{rc^2}}dt$$

I did the following : multiplying by $$\dot{r}$$ and integrating

$$1/2 \dot{r}^2=\frac{GM}{r}+C$$

Hence I have $$\dot{r}$$ in function of $$r$$ then I replaced the derivative towards proper time by derivative towards coordinate time, set C=0 and derivate once :

$$\frac{d^2 r}{dt^2}=-\frac{GM}{r^2}+4\frac{G^2M^2}{r^3c^2}$$ equation in coordinate time (?)

I should be able to verify this in a textbook but I haven't one.

Last edited: Apr 11, 2015
10. Apr 11, 2015

### pervect

Staff Emeritus
This doesn't seem to agree at first glance with the results form "Orbits in Strongly Curved spacetime", http://www.fourmilab.ch/gravitation/orbits/, which I know agrees with my textbook (MTW's Gravitation).

The formula there use two constants of motion, ~E which represents the energy-at-infinity, and ~L, which represents the angular momentum. It also implicitly assumes gemoetric units, so that G=c=1.

If we take ~E = 1 and ~L=0, which represents the simple case of a free fall from infinity with no angular momentum, so that $\frac{dr}{d\tau} = 0$ at r = infinity we get the fairly well known results:

$\frac{dr}{d\tau} = -c \sqrt{2GM / c^2 r}$ and $\frac{dt}{d\tau} = 1/(1 - 2GM/ c^2 r)$
The original used geometric units, I've put the needed factors of G and c in by hand. Hopefully I didn't make any mistakes there, the units seem to come out ok.

The equation for $\frac{dt}{d\tau}$ in particular , seems to disagree directly with the assumptions in your post.

Last edited: Apr 11, 2015
11. Apr 11, 2015

### bcrowell

Staff Emeritus
For something more rigorous that has sort of a similar flavor, see the discussion of the precession of Mercury's orbit in section 6.2.6 of my GR book, http://www.lightandmatter.com/genrel/ . There is a much more detailed treatment of orbits in Exploring Black Holes, by Taylor and Wheeler.

There is an effective potential, and it does have an additional relativistic term with a higher power of r in it. However, the details are all different from the result you got. The sign of the lowest-order relativistic correction term is opposite to yours, the exponent is different, the term vanishes when the angular momentum is zero, and the whole thing is actually a squared energy.

The problem of describing free fall into a black hole with zero angular momentum seems like an interesting special case. Let's see what happens. Following the treatment and notation in my book, we have for the equation of motion:

$$\dot{r}^2 = E^2-(1-2m/r)(1+L^2/r^2)$$

This is in relativistic units (c=1, G=1), L is the angular momentum, and the dot means differentiation with respect to *proper* time s, not Schwarzschild coordinate time. The interpretation of the constant E is discussed in the book. In the special case of axial free fall, L=0, and I believe E=1 if the free fall starts from rest and far away, so this simplifies:

$$\dot{r}^2 = 2m/r$$

Separating variables and integrating gives

$$r = ks^{2/3}$$

where $k=(3\sqrt{m/2})^{2/3}$ and s=0 is taken to be at impact with the singularity.

So unless I've made a mistake here (which is entirely possible), the exact relativistic motion is unexpectedly simple to characterize as the function r(s).

12. Apr 11, 2015

### Staff: Mentor

Where are you getting this equation from?

13. Apr 11, 2015

### Staff: Mentor

I agree.

This appears to agree with the calculation in MTW (equation 25.38).

14. Apr 11, 2015

### pervect

Staff Emeritus
That's is (not very surprisingly) the same as the fourmilab link I quoted earlier and my textbook as well, but without the unit fixes. In retrospect, I think it's better to follow the text more closely (as you did), and explain in detail how to "fix up" the units.

Putting the units back in $\dot{r} = \frac{dr}{ds}$ where s is proper time, I get $\dot{r}^2$= c^2 * a dimensionless quantity, which can be written as $r_s / r$, $r_s$ being 2GM/c^2, the Schwarzschild radius. So the dimensions turn out right, and the only additions are factors of G and c which are equal to 1 in geometric units, so this should be the right ungeometrized result.

I also recall r being proportional to $\tau^\frac{2}{3}$ where $\tau$ ranges from -infinty to zero. (And you square the negative $\tau$ before taking the cube root).

As far as justifying the result to someone unfamiliar with GR, I would suggest that a more productive route than focusing on "forces" as the OP apparently does would be to start with a force-free notion of physics, such as the Lagrangian formulation or using Hamilton's principle, http://en.wikipedia.org/wiki/Hamilton's_principle. In a relativistic context, the Lagrangian can be taken as being proportional to proper time, so Hamilton's principle is a principle of extremal aging, often (over) simplified as the principle of maximal aging.

I'm not aware of any really great treatment of forces in GR, not even a complete discussion of the relation between forces and Christoffel symbols. (And I wouldn't care to try to explain Christoffel symbols in a simple, undergraduate approach). Focusing on a force-free approach seems to me to be by far the easiest way to learn some basics about dynamics in GR.

15. Apr 11, 2015

### bcrowell

Staff Emeritus
I agree that force is not a very useful concept in relativity, but even more so when it comes to gravitational force. The moral of the equivalence principle is that gravity isn't a force.

I took the OP as being in the spirit of just trying to calculate something to see if it would work. Nothing wrong with that.

Taylor and Wheeler's Exploring Black Holes manages to do this sort of thing without introducing the full machinery of Lagrangian or Hamiltonian mechanics, which could be an insuperable barrier for many people. My own treatment is modeled on theirs. The Lagrangian technique really is being introduced from scratch, but not in full generality.

A nice outcome of the present calculation is that it shows pretty explicitly that an infalling observer does reach the singularity in finite proper time, so that in that sense the black hole singularity is a testable scientific prediction.