# I Special Relativity Approximation of Gravitation

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1. Sep 13, 2016

### tomdodd4598

Hey there,

I have two questions - the first is about an approximation of a central gravitational force on a particle (of small mass) based on special relativity, and the second is about the legitimacy of a Lagrangian I'm using to calculate the motion of a particle in the Schwarzchild metric.

First of all, I should mention I have not yet properly studied GR, but have simply had some ideas about how to calculate trajectories of particles using different metrics.

1. A couple of sources, including Susskind's theoretical minimum, show that the Lagrangian of a free relativistic particle (in flat spacetime) is: $$L=-m{ c }^{ 2 }\sqrt { 1-\frac { { v }^{ 2 } }{ { c }^{ 2 } } }$$ I noticed that, in spherical coordinates, this is equal to: $$L=-m{ c }^{ 2 }\sqrt { \frac { -{ \eta }_{ \mu \nu }{ \dot { x } }^{ \mu }\dot { x } ^{ \nu } }{ { c }^{ 2 } } }$$ where $$\eta =\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & { r }^{ 2 } & 0 \\ 0 & 0 & 0 & { r }^{ 2 }\sin ^{ 2 }{ \theta } \end{pmatrix}$$ is the metric for flat spacetime.
Now, I can add a potential energy term to this Lagrangian, so could I not simply add a gravitational potential energy term $$L=-m{ c }^{ 2 }\sqrt { \frac { -{ \eta }_{ \mu \nu }{ \dot { x } }^{ \mu }\dot { x } ^{ \nu } }{ { c }^{ 2 } } } +\frac { GMm }{ r }$$ and solve the Euler-Lagrange equations?
When I do this, and set the initial conditions such that the particle is in an elliptical orbit, I actually do get an apsidal precession, but the rate of this precession is much smaller than that I get from what I did below.

2. Having noticed the form of the Lagrangian above, I decided to change the flat metric to the Schwarzchild one, and set θ to be π/2 so that I could simplify the problem: $$\eta \rightarrow g=\begin{pmatrix} -\left( 1-\frac { { r }_{ s } }{ r } \right) & 0 & 0 & 0 \\ 0 & { \left( 1-\frac { { r }_{ s } }{ r } \right) }^{ -1 } & 0 & 0 \\ 0 & 0 & { r }^{ 2 } & 0 \\ 0 & 0 & 0 & { r }^{ 2 } \end{pmatrix}$$ This gives me the Lagrangian: $$L=-m{ c }^{ 2 }\sqrt { \frac { -{ g }_{ \mu \nu }{ \dot { x } }^{ \mu }{ \dot { x } }^{ \nu } }{ { c }^{ 2 } } } =-m{ c }^{ 2 }\sqrt { 1-\frac { { r }_{ s } }{ r } -\frac { { \left( 1-\frac { { r }_{ s } }{ r } \right) }^{ -1 }{ \dot { r } }^{ 2 }+{ r }^{ 2 }{ \dot { \varphi } }^{ 2 } }{ { c }^{ 2 } } }$$ Solving the Euler-Lagrange equations of this Lagrangian gives me an apsidal precession much larger than the one before.

My two questions boil down to this: First, what are the main issues with the SR approximation in part 1, and is the Lagrangian in part 2 (using the Schwarzchild metric) legitimate?
As an extra, perhaps, why do the two Lagrangians both give rise to an apsidal precession, but with different rates of precession?

2. Sep 14, 2016

### pervect

Staff Emeritus
The principle of stationary action says $\int L \, dt$ must be stationary. (Sometimes you'll see this writen less precisely as extremal action, or even more loosely as "least action").

We note that $\sqrt{1-v^2/c^2} dt = d\tau$, where $\tau$ is proper time. Because m is constant for a test particle and c is always constant, we can re-write the action integral as $-m c^2 \int d\tau$, thus the principle of stationary action boils down to the principle of stationary proper time.

When we go from flat space-time of SR to the curved space-time of GR, there isn't any additional "potential" added to the Lagrangian L. We still have $\int d\tau$ being stationary, that principle will give us the correct equations of motion for a force-free test mass in curved space-time.

3. Sep 14, 2016

### vanhees71

The problem with the suggested Lagrangian is that it doesn't describe gravity but the motion in an attractive coulomb potential. It's simply the wrong interaction to describe gravity (abelian massless gauge field rather than a massless spin-2 field for gravitation).

The ingenious insight by Einstein was that the most natural way to obtain a relativistic theory of gravitation is to "geometrize" it by assuming a pseudo-Riemannian manifold, which makes it compatible with the equivalence principle.

4. Sep 14, 2016

### tomdodd4598

Right, I see - thanks. So it's simply a coincidental fact that both give rise to precession.
But is it still a better approximation than the non-relativistic one, and is my Schwarzchild Lagrangian valid?

5. Sep 14, 2016

### Orodruin

Staff Emeritus
Even in classical mechanics most central potentials lead to precession. The Kepler problem and harmonic potentials are very special cases.

6. Sep 14, 2016

### tomdodd4598

I see... so I'm guessing that my approximation has no reason to be any more accurate than the Newtonian one.

7. Sep 15, 2016

### vanhees71

Yes, and you can show that they are the only potentials that lead to closed curves (for the non-relativstic problem only; already with special relativistic kinematics you get a precession).

8. Sep 15, 2016

### haushofer

A while ago we had a topic here about a picture you sometimes encounter when discussing precession, like

Those can be a bit deceiving, because it suggests that precession is purely a geometric phenomenon. As Orodruin indicates, it isn't.

9. Sep 16, 2016

### tomdodd4598

Right, I understand now - thanks everyone!

So, from what's been said above, my Schwarzchild Lagrangian is fine, and my attempted SR-based approximation would be used to describe, as an example, a charged particle orbiting another oppositely charged one (of much greater mass), right?

10. Sep 19, 2016

### Jonathan Scott

If you use your original Lagrangian but are careful to use coordinate values (where the coordinate speed of light varies from the standard value) and work in isotropic coordinates (where the space term in the metric has the same factor in all directions and the "coordinate value of c" doesn't depend on direction) then the Euler-Lagrange equations become equivalent to the geodesic equations.

11. Sep 25, 2016

### tomdodd4598

Aha! Thank you very much, Jonathan. I was wondering why my numerical solutions to the geodesic equations were different to those from the Lagrangian, especially close to the Schwarzchild radius - I'm very glad I came back to this thread because it seems I had accidentally forgotten about what you told me.

However, I am now very confused - if I solve the geodesic equations, not only do I get a totally different trajectory (probably to do with what Jonathan said), but also a precession which occurs in the opposite direction to that which I would expect.

Last edited: Sep 25, 2016
12. Sep 26, 2016

### Jonathan Scott

I'm sure that your approach of attempting to add a potential to an SR Lagrangian will not give results consistent with GR, although I don't have the time and patience to check out the details.

By the "original" Lagrangian, I just mean the SR Lagrangian without any potential term but with the coordinate speed of light $c$ varying as a function of the location as determined by the metric, which also means that the coordinate rest mass $m$ varies (since $mc^2$ varies with dimensions of coordinate energy, which is determined by the time dilation factor). However, this is not a good way to understand GR; it is rather that once you have got to grips with the Schwarzschild solution in isotropic coordinates, this provides a neat way to relate it back to SR and Newtonian theory. Also, be aware that using $c$ and $m$ in this way to mean coordinate values rather than local values is extremely unconventional. In a text book treatment of the GR Lagrangian you will find everything expressed in terms of tensor expressions involving the metric (which effectively converts between local values and coordinate values), but in this simple case of a static metric in isotropic coordinates it is possible to simplify the Lagrangian by using symbols to represent the coordinate speed of light and the coordinate rest mass, and if you simply use $c$ and $m$ for the purpose (with an appropriate warning to the reader!) the result looks exactly the same as the SR Lagrangian.

The Euler-Lagrange equations in this form are equivalent to the geodesic equations in that they determine the same space-time path, but they are not identical in form because they give the rate of change of coordinate momentum $(E/c^2) \, \mathbf{v}$ rather than coordinate velocity $\mathbf{v}$. For a static metric, $E$ is constant but the coordinate speed of light varies with radial position.

You should not expect to be able to compute the perihelion precession from any simplified model as it is dependent on the second-order (post-Newtonian) term in the potential (that is, the time dilation factor in the metric). If the time dilation to post-Newtonian accuracy is $1 - Gm/rc^2 + \beta/2 \, (Gm/rc^2)^2$ then the amount of precession depends on $\beta$. For General Relativity, $\beta = 1$.

13. Sep 26, 2016

### tomdodd4598

I think I understand what you mean. However, my issue is that the equations of motion from the E-L equations of this Lagrangian
$$L=-m{ c }^{ 2 }\sqrt { 1-\frac { { r }_{ s } }{ r } -\frac { { \left( 1-\frac { { r }_{ s } }{ r } \right) }^{ -1 }{ \dot { r } }^{ 2 }+{ r }^{ 2 }{ \dot { \varphi } }^{ 2 } }{ { c }^{ 2 } } },$$
which is the Schwarzchild Lagrangian, are very different to the equations of motion I get from the geodesic equations
$${ \ddot { x } }^{ \mu }={ \Gamma }_{ \alpha \beta }^{ 0 }{ \dot { x } }^{ \alpha }{ \dot { x } }^{ \beta }{ \dot { x } }^{ \mu }-{ \Gamma }_{ \alpha \beta }^{ \mu }{ \dot { x } }^{ \alpha }{ \dot { x } }^{ \beta }.$$
I would have expected that they were the same, but obviously I'm missing something (sorry if it's obvious). Is this because of a similar coordinate based issue that you are talking about? Or perhaps my Schwarzchild Lagrangian is invalid?

14. Sep 26, 2016

### vanhees71

As I said, you original Lagrangian is the motion of a charged relativistic particle in a electrostatic Coulomb field. That's obviously very different from the motion of particle in the gravitational field of a static spherical mass.

The equations of motion of the Lagrangian
$$L=-mc^2 \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}$$
is, by definition, the geodesic equation. It's even the most general form for the most general parameterization of the geodesic. In this case you use the coordinate time as indpenedent variable, and this leads to different-looking equations for the geodesic.

It's much simpler to use the proper time of the (massive) particle. Then you can use the simpler Lagrangian
$$L=\frac{1}{2} g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}.$$
Here, the parameter is automatically proportional to the pseudolength of the curve, because since the Lagrangian doesn't depend explicitly on that parameter, the "Hamiltonian"
$$H=p_{\mu} \dot{x}^{\mu}-L=L$$
is conserved, and thus you can normalize it to 1 (or $c^2$, if you want proper time as the independent variable).

15. Sep 26, 2016

### tomdodd4598

But that's exactly why I am confused. I am using that Lagrangian with the Schwarzchild metric, and I'm getting totally different equations of motion and trajectories to when I use the geodesic equations which involve the Christoffel symbols.

16. Sep 26, 2016

### vanhees71

Yes, but with this Lagrangian you get a different equation, because your parameter can be anything. In the form you wrote it, it's the coordinate time of the Schwarzschild coordinates.

If you use a parameter proportional to the proper time, you get the form with the Christoffel symbols. To see this we set
$$L=-\sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}=-\sqrt{G}.$$
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}}=\frac{\partial L}{\partial x^{\mu}}.$$
Now we have
$$\frac{\partial L}{\partial x^{\mu}}=-\frac{1}{2 \sqrt{G}} \frac{\partial G}{\partial x^{\mu}}$$
and
$$\frac{\mathrm{d}}{\mathrm{d} \lambda} \frac{\partial L}{\partial \dot{x}^{\mu}} =-\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} \lambda} \left (\frac{1}{2 \sqrt{G}} \frac{\partial G}{\partial \dot{x}^{\mu}} \right).$$
Only if $G=\text{const}$ along the trajectory you get the equation with the Christoffel symbols.

Now you used the correct equation for general parameters and then fixed the parameter to the coordinate time $t$. For this choice in the case of the Schwarzschild metric, you don't get the equation with the Christoffel symbols, because the coordinate time is not proportional to the proper time $\tau$.

17. Sep 26, 2016

### tomdodd4598

Right. So is this equation,
$${ \ddot { x } }^{ \mu }={ \Gamma }_{ \alpha \beta }^{ 0 }{ \dot { x } }^{ \alpha }{ \dot { x } }^{ \beta }{ \dot { x } }^{ \mu }-{ \Gamma }_{ \alpha \beta }^{ \mu }{ \dot { x } }^{ \alpha }{ \dot { x } }^{ \beta },$$
where dots represent derivatives by $t$, wrong? Things would make much more sense to me if they were, because I'm still struggling to understand how the equations of motion can be different if my parameter in both cases is $t$.

EDIT: Some possibly interesting and useful information: as the magnitude of the velocity get closer to $c$, the two sets of equations of motion yield increasingly similar trajectories. Is it possible that my geodesic equation, which I got from here, is only valid for massless particles?

Last edited: Sep 26, 2016
18. Sep 27, 2016

### vanhees71

You can just write down the Euler-Lagrange equation from your action. Then you get the correct equations of motion, describing the geodesic in the 4D pseudo-Riemannian spacetime in somewhat unfortunately chosen coordinates. However, sometimes an explicit 1+3 decomposition (choice of "time slices") can be advantegeous to solve the dynamical systems of Einstein equations and the motion of matter and radiation, particularly when it comes to numerics. You can use this as well for massless particles since the coordinate time does not depend on the choice of proper time as the parameter, which becomes meaningless for massless particles. Nevertheless you can also express the null geodesics in terms of a scalar parameter. You can use the quadratic Lagrangian. The only difference to timelike geodesics is that for the trajectory the Lagrangian becomes 0 rather than a value >0.

19. Sep 27, 2016

### tomdodd4598

Unfortunately, I can no longer record my animated graphs on Mathematica due to a horrible update to the application I used to use to record the desktop, but I will try to explain the difference in the motion I get when using the Lagrangian (with the square root) and the (questionable?) geodesic equation I showed in post #17. My initial conditions are always such that the radial velocity is 0.

When using the Lagrangian, aside from issues simply due to tiny inaccuracies in the numerical solution near the Schwarzchild radius, the particle, depending on the initial radial position and angular velocity, either:
falls in towards the event horizon, slowing down as it should,
orbits with an apsidal precession or
gets flung out and escapes, following a path that becomes straighter as it gets further from the origin.
However, there is one big issue which is that if I set the angular velocity to be such that the speed of the particle is equal to that of light (at that radius), then the calculation fails due to infinite expressions appearing. I, of course, also get no solution if the mass is 0.

When using the geodesic in post #17, things sometimes look a little odd. Either:
the particle falls in towards the event horizon, slowing down as it should,
orbits in a strange, skewed path which seems to have negative apsidal precession (only if the initial speed is lower that that of light) or
gets flung out and escapes, but instead of just escaping in a similar way to before (or like in Newtonian gravitation), the particle seems to curve away from the origin and shoot off such that it's velocity vector is almost perfectly in line with the origin.
Unlike before, however, I can set the initial speed to that of light (at that radius) without any issues, and there's no issue with the mass like before.

It's these drastically differing results that's really confusing me right now.

Anyway, something's telling me that I simply need to do a bit more studying to get these things better understood - thanks all for your help :)

Last edited: Sep 27, 2016
20. Sep 27, 2016

### tomdodd4598

I understand you now, and think I have it right. In the line element I set $dτ$ to 0 and made $t$ my parameter by dividing everything by 1 - rs/r, and then used the new metric tensor :)