# GR - Find the orbit of the planet

• quasar987

Homework Helper
Gold Member

## Homework Statement

Find the orbit of a planet in a grav. field corresponding to the metric

$$d\tau^2=(1+\alpha^2(x^2+y^2))dt^2-(dx^2+dy^2+dz^2)$$

in the Newtonian limit with the initial conditions at t=0, x=R, dx/dt=$\beta$,0=y=z=dy/dt=dz/dt where alpha and beta are cositive constants.

## Homework Equations

The Levi-Civita connexion and the geodesic equations

## The Attempt at a Solution

I calculates the connexion coefficients and wrote down the 4 geodesic equations with the proper time as the parameter:

$$\frac{d^2t}{d\tau^2}+\frac{2\alpha^2x}{(1+\alpha^2(x^2+y^2))} \frac{dt}{d\tau}\frac{dx}{d\tau}+\frac{2\alpha^2y}{(1+\alpha^2(x^2+y^2))}\frac{dt}{d\tau}\frac{dy}{d\tau}=0$$

$$\frac{d^2x}{d\tau^2}+\alpha^2x\left(\frac{dt}{d\tau}\right)^2=0$$

$$\frac{d^2y}{d\tau^2}+\alpha^2y\left(\frac{dt}{d\tau}\right)^2=0$$

$$\frac{d^2z}{d\tau^2}=0$$

I can solve the 4th equation of course:

$$z(\tau)=a\tau +b$$

and if I understand correctly, the Newtonian limit means that

$$\frac{dx^i}{d\tau}<<\frac{dt}{d\tau}$$

But I don't see how that can be applied here to simplify the equations.

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## Homework Statement

Find the orbit of a planet in a grav. field corresponding to the metric

$$d\tau^2=(1+\alpha^2(x^2+y^2))dt^2-(dx^2+dy^2+dz^2)$$

in the Newtonian limit with the initial conditions at t=0, x=R, dx/dt=$\beta$,0=y=z=dy/dt=dz/dt where alpha and beta are cositive constants.

## Homework Equations

The Levi-Civita connexion and the geodesic equations

## The Attempt at a Solution

I calculates the connexion coefficients and wrote down the 4 geodesic equations with the proper time as the parameter:

$$\frac{d^2t}{d\tau^2}+\frac{2\alpha^2x}{(1+\alpha^2(x^2+y^2))} \frac{dt}{d\tau}\frac{dx}{d\tau}+\frac{2\alpha^2y}{(1+\alpha^2(x^2+y^2))}\frac{dt}{d\tau}\frac{dy}{d\tau}=0$$

$$\frac{d^2x}{d\tau^2}+\alpha^2x\left(\frac{dt}{d\tau}\right)^2=0$$

$$\frac{d^2y}{d\tau^2}+\alpha^2y\left(\frac{dt}{d\tau}\right)^2=0$$

$$\frac{d^2z}{d\tau^2}=0$$

I can solve the 4th equation of course:

$$z(\tau)=a\tau +b$$

and if I understand correctly, the Newtonian limit means that

$$\frac{dx^i}{d\tau}<<\frac{dt}{d\tau}$$

But I don't see how that can be applied here to simplify the equations.

In the Newtonian limit you can set t = tau so $\frac{dt}{d \tau} = 1$.

And, of course, $\frac{d^2t}{d \tau^2} =0$.
So the equations become trivial to solve.

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Are you sure?

$$d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

And here, $g_{ii}=-1$, $g_{00}=1+\alpha^2(x^2+y^2)$ so

$$d\tau^2=(g_{00}-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2)dt^2$$

and in the small speed limit, this only reduces to

$$d\tau^2\approx g_{00}dt^2$$

which is not just dtau=dt because g00 is not just 1.

Are you sure?

$$d\tau^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

And here, $g_{ii}=-1$, $g_{00}=1+\alpha^2(x^2+y^2)$ so

$$d\tau^2=(g_{00}-(dx/dt)^2-(dy/dt)^2-(dz/dt)^2)dt^2$$

and in the small speed limit, this only reduces to

$$d\tau^2\approx g_{00}dt^2$$

which is not just dtau=dt because g00 is not just 1.

You are right. Sorry, I had misread the spacetime interval and thought that the (1+alpha^2(x^2+y^2)) term went with the spatial part. So you have an expression for dt/dtau to replace in all the equations. Using the chain rule you can also get an expression for the second derivative d^2 t/dtau^2.

Sorry for my mistake.

Use the weak-field approximation, which gives a straight Newtonian potential of -
$$- \frac{1}{2}c^2( 1 + g_{00})$$

You can derive this by assuming the

$$g_{ij} = \eta_{ij} + h_{ij}$$ where $$\eta$$ is the flat space metric and h
is a small perturbation.

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