MTW Exercise 12.6: Formulating Newtonian Gravity Using Curvature

[jbergman]

Homework Statement

Not homework but looking for some help on a problem.

Relevant Equations

$$ \Phi = \frac{-GM}{|\textbf{r}|} $$
$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

Hello, I am attempting to work through problem 12.6 in MTW which involves formulating Newtonian Gravity using Curvature as opposed to the standard formulation. This is a precursor before standard GR. In it he states that the curvature tensor in this formulation is as follows.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$

For problem 12.6 I am trying to calculate this for the earth with the potential,

$$ \Phi = \frac{-GM}{|\textbf{r}|} $$

And for example, for one of the components I get.

$$R^j_{0k0}=-R^j_{00k} = \frac{\partial^2\Phi}{\partial x^j \partial x^k}$$
$$\frac{\partial^2}{\partial x^j\partial x^k}(-GM/|\textbf{r}|)=
\frac{\partial}{\partial x^k}\left( \frac{GM x^j}{|\textbf{r}|^3}\right) =
\frac{\partial}{\partial x^k}(x^j) \frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^j x^k}{|\textbf{r}|^5}$$
$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^2}{|\textbf{r}|^5}$$

However, I am not convinced this is correct based on p.37 where it looks like,
$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3}$$

However, MTW states that z is in the radial direction which would suggest that $x=0$ and $y=0$. So, it looks like MTW is using a specific position to simplify the problem. Is my understanding correct or did I botch the calculation.

 

[TSny]

I believe you are thinking about it correctly. At the point $p$ where you are finding the components $R^j_{0k0}$, the $z$ axis is assumed to pass through $p$ in the radial direction. So, the derivatives $\frac{\partial^2}{\partial x^j \partial x^k}$ are evaluated at $z = r$ and $x = y = 0$.

 

[jbergman]

I believe you are thinking about it correctly. At the point $p$ where you are finding the components $R^j_{0k0}$, the $z$ axis is assumed to pass through $p$ in the radial direction. So, the derivatives $\frac{\partial^2}{\partial x^j \partial x^k}$ are evaluated at $z = r$ and $x = y = 0$.

@TSny can I ask a follow up? The problem suggests calculating the geodesic deviation in the spaceships coordinates.
However, I'm a bit puzzled by what the fiducial geodesic looks like in that frame.

Wouldn't $\gamma'(t)$ be 0 along that geodesic since the spaceship is always at the origin of this coordinate system? Which in turn kills the curvature term in the geodesic deviation equation.

 

[TSny]

@TSny The problem suggests calculating the geodesic deviation in the spaceships coordinates.
However, I'm a bit puzzled by what the fiducial geodesic looks like in that frame.

Wouldn't $\gamma'(t)$ be 0 along that geodesic since the spaceship is always at the origin of this coordinate system? Which in turn kills the curvature term in the geodesic deviation equation.

The fiducial geodesic is the geodesic followed by the spaceship. The spacetime coordinates of the spaceship in the primed frame at any time $t$ are: $x^{k '} = 0$ and $x^{0'}= t$.

In this problem, you want to find the equations of motion, in the primed frame, of the bag of garbage ejected from the spaceship. The garbage bag also follows a geodesic. This geodesic is "nearby" the geodesic of the spaceship. The motion of the bag relative to the spaceship is determined from the geodesic deviation equation $$\frac{D^2 n^{\alpha '}}{d \lambda^2} + R^{\alpha '} _{\; \beta ' \gamma ' \delta '} \frac{dx^{\beta '}}{d \lambda} n^{\gamma '} \frac{dx^{\delta '}}{d \lambda} = 0.$$
We can let the parameter $\lambda$ be the Newtonian time $t$. The components of the curvature tensor are evaluated on the fiducial geodesic; that is, at the origin of the primed frame. The spatial components $n^{k '}$ of the deviation vector represent the spatial primed coordinates of the garbage bag.

In the geodesic deviation equation, $\large \frac{dx^{\beta '}}{d \lambda}$ is the derivative of the spaceship coordinate $x^{\beta '}$ with respect to $\lambda$. Thus, $$\frac{dx^{\beta '}}{d \lambda} = \frac{dx^{\beta '}}{d t} = \delta_{ \beta ' 0'}.$$ So, in the geodesic deviation equation, there will be some nonzero terms involving components of the curvature tensor.

This problem was discussed in this thread. You might find some of the discussion there to be helpful.

 

[jbergman]

The fiducial geodesic is the geodesic followed by the spaceship. The spacetime coordinates of the spaceship in the primed frame at any time $t$ are: $x^{k '} = 0$ and $x^{0'}= t$.

In this problem, you want to find the equations of motion, in the primed frame, of the bag of garbage ejected from the spaceship. The garbage bag also follows a geodesic. This geodesic is "nearby" the geodesic of the spaceship. The motion of the bag relative to the spaceship is determined from the geodesic deviation equation $$\frac{D^2 n^{\alpha '}}{d \lambda^2} + R^{\alpha '} _{\; \beta ' \gamma ' \delta '} \frac{dx^{\beta '}}{d \lambda} n^{\gamma '} \frac{dx^{\delta '}}{d \lambda} = 0.$$
We can let the parameter $\lambda$ be the Newtonian time $t$. The components of the curvature tensor are evaluated on the fiducial geodesic; that is, at the origin of the primed frame. The spatial components $n^{k '}$ of the deviation vector represent the spatial primed coordinates of the garbage bag.

In the geodesic deviation equation, $\large \frac{dx^{\beta '}}{d \lambda}$ is the derivative of the spaceship coordinate $x^{\beta '}$ with respect to $\lambda$. Thus, $$\frac{dx^{\beta '}}{d \lambda} = \frac{dx^{\beta '}}{d t} = \delta_{ \beta ' 0'}.$$ So, in the geodesic deviation equation, there will be some nonzero terms involving components of the curvature tensor.

This problem was discussed in this thread. You might find some of the discussion there to be helpful.

Very helpful answer. I forgot about the time coordinate. Doh!

 

[jbergman]

Reply @TSny I looked at that thread you linked and it has me a bit puzzled. I get a bit of a mess for the Riemann Curvature Tensor in the spaceships coordinates. Here are my calculations.

First as I posted above, I calculated in the earth coordinates,

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^2}{|\textbf{r}|^5}$$

Then transforming this to the prime coordinates I get,

$$ R^{x'}_{0x'0} = L^{x'}_{x}L^{x}_{x'}R^x_{0x0} +
L^{x'}_{y}L^{y}_{x'}R^y_{0y0} $$
$$ R^{x'}_{0x'0}=
\cos^2 \omega t \left(\frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^2}{|\textbf{r}|^5}\right)
+ \sin^2 \omega t \left(\frac{GM}{|\textbf{r}|^3} -
\frac{3GM y^2}{|\textbf{r}|^5}\right)$$
$$ = \frac{GM}{|\textbf{r}|^3} - \frac{3GM}{|\textbf{r}|^5}(x^2\cos^2 \omega t + y^2\sin^2 \omega t)
$$

Lastly using the relations between the prime and unprimed coordinates,

$$x=r_0\cos \omega t + x' \cos\omega t -y'\sin \omega t$$
$$y=r_0\sin \omega t + x' \sin\omega t +y'\cos \omega t$$

I replace $x$ and $y$ in $R^{'x}_{0x'0}$

$$x^2 = (r_0^2 + x'^2 + 2r_0x')\cos^2\omega t + y'^2 \sin ^2 \omega t -(2x'y' + 2r_0y') \cos \omega t \sin \omega t $$
$$y^2 = (r_0^2 + x'^2 + 2r_0x')\sin^2\omega t + y'^2 \cos ^2 \omega t + (2x'y' + 2r_0y') \cos \omega t \sin \omega t $$

and get,
$$R^{x'}_{0x'0} =
\frac{GM}{|\textbf{r}|^3} - \frac{3GM}{|\textbf{r}|^5}( (r_0^2 + x'^2 + 2r_0x')\cos^4\omega t \sin^4\omega t
+ 2y'^2 \cos^2 \omega t \sin ^2 \omega t + (2x'y' + 2r_0y')\cos \omega t \sin \omega t)$$

Is there some simplification that I am missing?

 

[TSny]

First as I posted above, I calculated in the earth coordinates,

$$R^x_{0x0} = \frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^2}{|\textbf{r}|^5}$$

Then transforming this to the prime coordinates I get,

$$ R^{x'}_{0x'0} = L^{x'}_{x}L^{x}_{x'}R^x_{0x0} +
L^{x'}_{y}L^{y}_{x'}R^y_{0y0} $$
$$ R^{x'}_{0x'0}=
\cos^2 \omega t \left(\frac{GM}{|\textbf{r}|^3} -
\frac{3GM x^2}{|\textbf{r}|^5}\right)
+ \sin^2 \omega t \left(\frac{GM}{|\textbf{r}|^3} -
\frac{3GM y^2}{|\textbf{r}|^5}\right)$$
$$ = \frac{GM}{|\textbf{r}|^3} - \frac{3GM}{|\textbf{r}|^5}(x^2\cos^2 \omega t + y^2\sin^2 \omega t)
$$

In using the geodesic deviation equation, you will only need the primed components of the curvature tensor at points on the fiducial geodesic (i.e., at the location of the spaceship). Convince yourself that this means that you can let $t = 0$, $r = r_0$, $x = r_0$, and $y = 0$ in your equation for $R^{x'}_{\; 0' x' 0'}$.

 

[jbergman]

In using the geodesic deviation equation, you will only need the primed components of the curvature tensor at points on the fiducial geodesic (i.e., at the location of the spaceship). Convince yourself that this means that you can let $t = 0$, $r = r_0$, $x = r_0$, and $y = 0$ in your equation for $R^{x'}_{\; 0' x' 0'}$.

Hmm. I'm struggling to convince myself of this.

It seems to me that if we just want to calculate the accelerations of the deviation at $t=0$ this is true or if we wanted an approximate solution assuming the curvature is roughly constant in a neighborhood.

But if we actually want to solve the system of differential equations for the deviation $n(t)$ then we wouldn't do that because the curvature changes with time.

I was trying to do the latter.

 

[TSny]

At any fixed point in the primed frame, it should be evident that the components of the curvature tensor in the primed frame are time-independent. This is because the spaceship is in a circular orbit around the earth and the primed axes rotate such that the x' axis is always in the radial direction from the earth's center toward the spaceship. In particular, the primed components of the curvature tensor at the origin of the primed frame (at the spaceship) are time-independent.

So, if you evaluate the primed components of the curvature tensor at the location of the ship at $t = 0$ (when the x-axis and the x'-axis are parallel), that will give you the primed components of the curvature tensor at the location of the ship at any other time.

However, your post #6 seems to show that $R^{x'}_{\;0x'0}$ is time-dependent. It took me a while to resolve this.
You wrote

$$ R^{x'}_{0x'0} = L^{x'}_{x}L^{x}_{x'}R^x_{0x0} +
L^{x'}_{y}L^{y}_{x'}R^y_{0y0} $$

There are additional terms that should be included.
$$ R^{x'}_{\;0x'0} = L^{x'}_{x}L^{x}_{x'}R^x_{\;0x0} +
L^{x'}_{y}L^{y}_{x'}R^y_{\;0y0} +L^{x'}_{x}L^{y}_{x'}R^x_{\;0y0} + L^{x'}_{y}L^{x}_{x'}R^y_{\;0x0} $$
Here, $$R^x_{\;0y0} =R^y_{\;0x0} = \frac{\partial ^2 \Phi}{\partial x \partial y} = -\frac{3GM}{r^5} xy$$ This would be zero at $t = 0$ when the spaceship is on the unprimed x-axis (so $y= 0$).

At the spaceship we have $r = r_0$, $x = r_0 \cos \omega t$ and $y = r_0 \sin \omega t$. Then, when I evaluate $R^{x'}_{\;0x'0}$ at the spaceship, I find that all of the time dependence vanishes and I get the expected time-independent result $R^{x'}_{\;0x'0} = \large -\frac{2GM}{r_0^2}$. Similarly, you can show $R^{y'}_{\;0y'0} = \large -\frac{GM}{r_0^2}$ and $R^{x'}_{\;0y'0} = R^{y'}_{\;0x'0} =0$ at any time.

This is all fairly tedious. A simpler approach is to say that at any point of the orbit of the spaceship, you can introduce an unprimed, earth coordinate system such that the x-axis is in the radial direction of the spaceship at that instant. So, the x and x' axes are parallel at this instant. Then work out the primed curvature tensor components at the spaceship for this instant. The result must then hold for all other points of the orbit of the ship.

 

[jbergman]

At any fixed point in the primed frame, it should be evident that the components of the curvature tensor in the primed frame are time-independent. This is because the spaceship is in a circular orbit around the earth and the primed axes rotate such that the x' axis is always in the radial direction from the earth's center toward the spaceship. In particular, the primed components of the curvature tensor at the origin of the primed frame (at the spaceship) are time-independent.

So, if you evaluate the primed components of the curvature tensor at the location of the ship at $t = 0$ (when the x-axis and the x'-axis are parallel), that will give you the primed components of the curvature tensor at the location of the ship at any other time.

However, your post #6 seems to show that $R^{x'}_{\;0x'0}$ is time-dependent. It took me a while to resolve this.
You wroteThere are additional terms that should be included.
$$ R^{x'}_{\;0x'0} = L^{x'}_{x}L^{x}_{x'}R^x_{\;0x0} +
L^{x'}_{y}L^{y}_{x'}R^y_{\;0y0} +L^{x'}_{x}L^{y}_{x'}R^x_{\;0y0} + L^{x'}_{y}L^{x}_{x'}R^y_{\;0x0} $$
Here, $$R^x_{\;0y0} =R^y_{\;0x0} = \frac{\partial ^2 \Phi}{\partial x \partial y} = -\frac{3GM}{r^5} xy$$ This would be zero at $t = 0$ when the spaceship is on the unprimed x-axis (so $y= 0$).

At the spaceship we have $r = r_0$, $x = r_0 \cos \omega t$ and $y = r_0 \sin \omega t$. Then, when I evaluate $R^{x'}_{\;0x'0}$ at the spaceship, I find that all of the time dependence vanishes and I get the expected time-independent result $R^{x'}_{\;0x'0} = \large -\frac{2GM}{r_0^2}$. Similarly, you can show $R^{y'}_{\;0y'0} = \large -\frac{GM}{r_0^2}$ and $R^{x'}_{\;0y'0} = R^{y'}_{\;0x'0} =0$ at any time.

This is all fairly tedious. A simpler approach is to say that at any point of the orbit of the spaceship, you can introduce an unprimed, earth coordinate system such that the x-axis is in the radial direction of the spaceship at that instant. So, the x and x' axes are parallel at this instant. Then work out the primed curvature tensor components at the spaceship for this instant. The result must then hold for all other points of the orbit of the ship.

Thanks. Your physical reasoning makes sense with respect to the primed coordinates respecting the symmetry of the potential hence basically the curvature tensor should be unchanged. I never thought of it like that.

As for the additional terms, good catch. For some reason I was under the impression that all mixed spatial indices terms disappeared but I will have to go and double check why I was under that impression.

Thanks again for the help!