Calculating Christoffel Symbol with Rotating Metric

[MathematicalPhysicist]

I have the metric:
$$ds^2 = −(1 − \Omega^2 (x^2 + y^2 ))dt^2 + 2\Omega(ydx − xdy)dt + dx^2 + dy^2 + dz^2$$

In order to calculate the Christoffel symbol I need to compute the inverse matrix of the above, is there some computational shortcut besides the joke:" using mathematica"?

Thanks!

 

[Orodruin]

In order to calculate the Christoffel symbol I need to compute the inverse matrix of the above, is there some computational shortcut besides the joke:" using mathematica"?

Never use
$$
\Gamma_{\alpha\beta}^\gamma = \frac{1}{2} g^{\gamma\delta}(\partial_\alpha g_{\delta\beta} + \partial_\beta g_{\alpha\delta} - \partial_\delta g_{\alpha\beta})
$$
to actually compute the Christoffel symbols of the Levi-Civita connection. It is a bookkeeping nightmare and requires you to compute $g^{\gamma\delta}$. Instead, construct and extremise
$$
\mathcal S = \int g_{\alpha\beta} \dot x^\alpha \dot x^\beta ds
$$
and then identify the Christoffel symbols from the resulting Euler-Lagrange equations.

 

[Orodruin]

Addendum: If you got your metric earlier by switching to the rotating coordinate system, you can also find the geodesic equations from starting with the geodesic equations in non-rotating coordinates and making the change of variables. Either way, you will be able to identify the Christoffel symbols directly from the geodesic equations.

 

[MathematicalPhysicist]

Has someone actually done this bookkeeping nightmare?

I actually started calculating the inverse matrix of this 4x4 monster, but got quite frustrated; I know of a solution that goes through the E-L equation.

 

[Orodruin]

If you really want to do it by computing the "monster", you should have a 4x4 matrix on the form
$$
\begin{pmatrix}
A & B & C & 0 \\
B & D & 0 & 0 \\
C & 0 & D & 0 \\
0 & 0 & 0 & D
\end{pmatrix},
$$
which should be pretty straightforward to diagonalise and then invert (make a xy-rotation to put the off-diagonals in one element only, then you will have a 2x2 block diagonal matrix where one block is proportional to identity and the other is a 2x2 matrix that can be easily inverted).

 

[Dale]

is there some computational shortcut besides the joke:" using mathematica"?

What’s wrong with Mathematica?

 

[MathematicalPhysicist]

What’s wrong with Mathematica?

Nothing, but in class format the tutor prefers to do the calculations by hand since in the exam you cannot use mathematica.
:-)

 

[kent davidge]

I wonder if the metric inside the integral in post #2 should be inside a square root.

 

[martinbn]

I wonder if the metric inside the integral in post #2 should be inside a square root.

You don't need the square root. You get the same equations. The root only complicates the calculations.

 

[Orodruin]

I wonder if the metric inside the integral in post #2 should be inside a square root.

No, it should not. If you include the square root you are getting the equation that describes general paths that optimise the proper time. This is not what you want for the geodesic equations. What you want is the differential equation for an affinely parametrised geodesic. It can be shown that this is exactly what you get when you take away the square root.

You don't need the square root. You get the same equations. The root only complicates the calculations.

You do not generally get the same equations. You will get an equation on the form
$$
\nabla_{\dot x} \dot x = \alpha \dot x
$$
instead of an equation on the form
$$
\nabla_{\dot x} \dot x = 0.
$$
The difference is whether or not the geodesic is affinely parametrised, i.e., whether or not you allow for geodesics with tangent vectors that changes length along the geodesic or not (this will generally not change the length of a curve).

 

[martinbn]

Ah, good point.

 

[sweet springs]

Introducing cylindrical coordinate
x=r\ cos\phi, y=r\ sin\phi

Components of metric tensor are
g_{00}=1-\Omega^2 r^2
g_{11}=-1
g_{22}=-r^2
g_{33}=-1
g_{02}=g_{20}=\Omega r
where
x^0=t,x^1=r,x^2=\phi,x^3=z

Forget about trivial $x^3$ then we can easily calculate inverse of 3x3 matrix in due course.

Note: I wrote in my familiar convention of $(+,-,-,-)$.

 

[sweet springs]

PS. Correction of coefficients in the previous post

Introducing cylindrical coordinate
x=r\ cos\phi, y=r\ sin\phi

Components of metric tensor are
g_{00}=1-\Omega^2 r^2
g_{11}=-1
g_{22}=-r^2
g_{33}=-1
g_{02}=g_{20}=\Omega r^2
otherwiswe 0, where
x^0=t,x^1=r,x^2=\phi,x^3=z, in convention ( +,-, -, - )

If we put
x^0=ct,x^1=r,x^2=\phi,x^3=z, $\Omega$ should be replaced by $\frac{\Omega}{c}$ dimension L^-1.Forget about $x^3$ then we can easiy get inverse of 3x3 matrix in ordinary mathematics techniuque.
. Thus

g^{00}=1
g^{11}=-1
g^{22}=-\frac{1-\Omega^2 r^2}{r^2}=-\frac{1}{r^2}+\Omega^2
g^{33}=-1
g^{02}=g^{02}=\Omega
Otherwise 0.

Only $g^{22}$ is a function of r, other components are constant. Calculations of Christoffel symbols are easy. I should appreciate it if OP could check my calculations above.

 

[MathematicalPhysicist]

Never use
$$
\Gamma_{\alpha\beta}^\gamma = \frac{1}{2} g^{\gamma\delta}(\partial_\alpha g_{\delta\beta} + \partial_\beta g_{\alpha\delta} - \partial_\delta g_{\alpha\beta})
$$
to actually compute the Christoffel symbols of the Levi-Civita connection. It is a bookkeeping nightmare and requires you to compute $g^{\gamma\delta}$. Instead, construct and extremise
$$
\mathcal S = \int g_{\alpha\beta} \dot x^\alpha \dot x^\beta ds
$$
and then identify the Christoffel symbols from the resulting Euler-Lagrange equations.

@Orodruin How would you identify Christoffel symbols from Euler-Lagrange equations?

 

[Orodruin]

@Orodruin How would you identify Christoffel symbols from Euler-Lagrange equations?

Solve for $\ddot x^\mu$. The geodesic equations take the form
$$
\ddot x^\mu + \Gamma^\mu_{\nu\gamma} \dot x^\nu \dot x^\gamma = 0.
$$

Example (polar coordinates in $\mathbb R^2$): The line element in polar coordinates is $ds^2 = dr^2 + r^2 d\phi^2$, so the integrand is $\mathscr L = g_{\mu\nu}\dot x^\mu \dot x^\nu = \dot r^2 + r^2 \dot\phi^2$. Consequently, the variation with respect to $r$ leads to
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
\dd{\mathscr L}{r} - \frac{d}{ds} \dd{\mathscr L}{\dot r} =
2r \dot\phi^2 - 2\ddot r \quad \Rightarrow \quad \ddot r - r\dot \phi^2 = 0.
$$
Identifying this with the geodesic equation for $r$ directly let's you identify $\Gamma^r_{\phi\phi} = -r$. Similarly, variation with respect to $\phi$ leads to
$$
\frac{d}{ds}\dd{\mathscr L}{\dot\phi} - \dd{\mathscr L}{\phi} = 2\frac{d}{ds} r^2 \dot\phi = 2r^2\ddot \phi + 4r\dot r\dot\phi = 0
\quad \Rightarrow \quad
\ddot \phi + \frac{2}{r} \dot r \dot \phi = 0.
$$
Identification with the geodesic equation for $\phi$ gives you $\Gamma^\phi_{r\phi} + \Gamma^\phi_{\phi r} = 2/r$, but the Levi-Civita connection is torsion free so $\Gamma^{\phi}_{r\phi} = \Gamma^{\phi}_{\phi r} = 1/r$.

 

[MathematicalPhysicist]

Ah ok, thanks!

 

[MathematicalPhysicist]

I see now that you were correct, it's indeed easier tool to calculate Christoffel symbols!

 

[stevendaryl]

Never use
$$
\Gamma_{\alpha\beta}^\gamma = \frac{1}{2} g^{\gamma\delta}(\partial_\alpha g_{\delta\beta} + \partial_\beta g_{\alpha\delta} - \partial_\delta g_{\alpha\beta})
$$
to actually compute the Christoffel symbols of the Levi-Civita connection. It is a bookkeeping nightmare and requires you to compute $g^{\gamma\delta}$. Instead, construct and extremise
$$
\mathcal S = \int g_{\alpha\beta} \dot x^\alpha \dot x^\beta ds
$$
and then identify the Christoffel symbols from the resulting Euler-Lagrange equations.

Using the Euler-Lagrange equations for the "Lagrangian" $\frac{1}{2} g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu$ (I like to throw in the 1/2 because it makes it look more like a kinetic energy) gives:

$g_{\mu \nu} \ddot{x}^\nu + \frac{1}{2} (2 \partial_\lambda g_{\mu \nu} - \partial_\mu g_{\lambda \nu}) \dot{x}^\lambda \dot{x}^\nu = 0$

That gives us something sort of like a connection $A_{\mu \nu \lambda} = \frac{1}{2} (2 \partial_\lambda g_{\mu \nu} - \partial_\mu g_{\lambda \nu})$

But to get to $\Gamma^\mu_{\nu \lambda}$, it seems that we still need $g^{\mu \nu}$:

1. $A_{\mu (\nu \lambda)} \equiv \frac{1}{2} (A_{\mu \nu \lambda} + A_{\mu \lambda \nu})$
2. $\Gamma^\mu_{\nu \lambda} = g^{\mu \mu'} A_{\mu' (\nu \lambda)}$

I don't see how to avoid calculating $g^{\mu \nu}$.

 

[Orodruin]

Using the Euler-Lagrange equations for the "Lagrangian" $\frac{1}{2} g_{\mu \nu} \dot{x}^\mu \dot{x}^\nu$ (I like to throw in the 1/2 because it makes it look more like a kinetic energy) gives:

$g_{\mu \nu} \ddot{x}^\nu + \frac{1}{2} (2 \partial_\lambda g_{\mu \nu} - \partial_\mu g_{\lambda \nu}) \dot{x}^\lambda \dot{x}^\nu = 0$

That gives us something sort of like a connection $A_{\mu \nu \lambda} = \frac{1}{2} (2 \partial_\lambda g_{\mu \nu} - \partial_\mu g_{\lambda \nu})$

But to get to $\Gamma^\mu_{\nu \lambda}$, it seems that we still need $g^{\mu \nu}$:

1. $A_{\mu (\nu \lambda)} \equiv \frac{1}{2} (A_{\mu \nu \lambda} + A_{\mu \lambda \nu})$
2. $\Gamma^\mu_{\nu \lambda} = g^{\mu \mu'} A_{\mu' (\nu \lambda)}$

I don't see how to avoid calculating $g^{\mu \nu}$.

You just solve for the $\ddot x^\mu$. There is no need to actually compute the inverse. Although that is one way of solving for ##\ddot x^\mu#}, it is usually faster and less bookkeeping to just use the equations themselves and take linear combinations of them.

 

[MathematicalPhysicist]

Solve for $\ddot x^\mu$. The geodesic equations take the form
$$
\ddot x^\mu + \Gamma^\mu_{\nu\gamma} \dot x^\nu \dot x^\gamma = 0.
$$

Example (polar coordinates in $\mathbb R^2$): The line element in polar coordinates is $ds^2 = dr^2 + r^2 d\phi^2$, so the integrand is $\mathscr L = g_{\mu\nu}\dot x^\mu \dot x^\nu = \dot r^2 + r^2 \dot\phi^2$. Consequently, the variation with respect to $r$ leads to
$$
\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}}
\dd{\mathscr L}{r} - \frac{d}{ds} \dd{\mathscr L}{\dot r} =
2r \dot\phi^2 - 2\ddot r \quad \Rightarrow \quad \ddot r - r\dot \phi^2 = 0.
$$
Identifying this with the geodesic equation for $r$ directly let's you identify $\Gamma^r_{\phi\phi} = -r$. Similarly, variation with respect to $\phi$ leads to
$$
\frac{d}{ds}\dd{\mathscr L}{\dot\phi} - \dd{\mathscr L}{\phi} = 2\frac{d}{ds} r^2 \dot\phi = 2r^2\ddot \phi + 4r\dot r\dot\phi = 0
\quad \Rightarrow \quad
\ddot \phi + \frac{2}{r} \dot r \dot \phi = 0.
$$
Identification with the geodesic equation for $\phi$ gives you $\Gamma^\phi_{r\phi} + \Gamma^\phi_{\phi r} = 2/r$, but the Levi-Civita connection is torsion free so $\Gamma^{\phi}_{r\phi} = \Gamma^{\phi}_{\phi r} = 1/r$.

How would you use this technique in this rotating metric in this OP?

I mean I get after calculating E-L equations the equations:
$$\ddot{t}+\dot{t}(2x\dot x +2y \dot y)/(x^2+y^2)+(y\ddot x -x \ddot y)/(\Omega(x^2+y^2))=0$$
$$\ddot{x}+2\Omega \dot y \dot t + \Omega y \ddot t - \Omega^2 x\dot t^2=0$$
$$\ddot{y}-2\Omega \dot x \dot t - \Omega x \ddot t - \Omega^2 y\dot t^2=0$$

I tried plugging the $\ddot x$ and $\ddot y$ back into the eqaution of $\ddot t$, but I get that zero equals zero which doesn't help me.

 

[Orodruin]

Your EL equations are wrong. Double check your arithmetic.

 

[MathematicalPhysicist]

@Orodruin what do you get?

 

[Orodruin]

I am not going to give you the answer. I have pointed you in the direction of your current error, but you need to be able to do the arithmetic correctly. Suffice to say that you should not be getting 0=0. That would mean that your metric is not invertible.

 

[MathematicalPhysicist]

OK, I fixed it now, and I get the correct maple answer.

 

[kent davidge]

You don't need the square root. You get the same equations. The root only complicates the calculations.

You do not generally get the same equations. You will get an equation on the form $$\nabla_{\dot x} \dot x = \alpha \dot x $$
instead of an equation on the form $$\nabla_{\dot x} \dot x = 0.$$
The difference is whether or not the geodesic is affinely parametrised, i.e., whether or not you allow for geodesics with tangent vectors that changes length along the geodesic or not (this will generally not change the length of a curve).

I guess one might define $g_{\mu \nu} d \dot x^\mu d \dot x^\nu = \mathcal{L}^2$ and $\int \sqrt{g_{\mu \nu} d \dot x^\mu d \dot x^\nu}ds$ but the Euler Lagrange equations would have to be checked to see what they give.

 

[Orodruin]

I guess one might define $g_{\mu \nu} d \dot x^\mu d \dot x^\nu = \mathcal{L}^2$ and $\int \sqrt{g_{\mu \nu} d \dot x^\mu d \dot x^\nu}ds$ but the Euler Lagrange equations would have to be checked to see what they give.

This makes no sense. You cannot leave the $d$s and put a dot on the coordinates. It is well known that
$$
\int g_{\mu\nu}\dot x^\mu \dot x^\nu ds
$$
gives the same curves as
$$
\int \sqrt{g_{\mu\nu}\dot x^\mu \dot x^\nu} \, ds
$$
with the difference that they are affinely parametrised.