How to Calculate Christoffel Symbols in Spherical Coordinates?

[latentcorpse]

The metric of Euclidean \mathbb{R}^3 in spherical coordinates is ds^2=dr^2+r^2(d \theta^2 + \sin^2{\theta} d \phi^2).
I am asked to calculate the Christoffel components \Gamma^{\sigma}{}_{\mu \nu} in this coordinate system.

i'm not too sure how to go about this.

it talks about ds^2 being the metric but normally the metric is of the form g_{ab} i.e. a 2-form but ds^2 isn't a 2-form. are these metrics different or do i make g_{\mu \nu}=ds^2 \omega_{\mu} \omega_{\nu} where \omega_i is a 1 form?

i think I'm missing some key point here...

 

[gabbagabbahey]

The metric is often defined according to the equation ds^2=g_{ab}dx^adx^b...In this case, you have x^a\in\{r,\theta,\phi\}...so what are the components of g_{ab}?

 

[latentcorpse]

g_{ab}=\left[ \begin {array}{ccc} 1&amp;0&amp;0 \\ 0&amp;r&amp;0<br /> \\ 0&amp;0&amp; {\sin}{\theta}\end {array} \right]

i didn't know how else to right it.

would that work? because say
g_{33}=g_{\phi \phi} = \sin^2{\theta} dx^\phi dx^\phi

 

[gabbagabbahey]

Don't you mean:

g_{ab}=\begin{pmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; r^2 &amp; 0 \\ 0 &amp; 0 &amp; \sin^2\theta \end{pmatrix}

 

[latentcorpse]

yeah sorry. okay so that would work out for the formula ds^2=g_{ab} dx^a dx^b

now i guees I'm supposed to use 3.1.30 in Wald:

\Gamma^{\sigma}{}_{\mu \nu}=\frac{1}{2} \sum_{\rho} g^{\sigma \rho} \left( \frac{\partial g_{\nu \rho}}{\partial x^{\mu}} + \frac{\partial g_{\mu \rho}}{\partial x^{\nu}} - \frac{\partial g_{\mu \nu}}{\partial x^{\sigma}} \right)

im confused about how this sum is going to work though.
\sigma,\nu,\mu \in \{ r, \theta, \phi \} and so they aren't fixed variables...which is confusing
also what values does \sigma take?

 

[gabbagabbahey]

\sigma,\nu,\mu \in \{ r, \theta, \phi \} and so they aren't fixed variables...which is confusing
also what values does \sigma take?

No, all of \mu,\nu,\sigma,\rho\in\{1,2,3\}, with x^{1}=r, x^{2}=\theta and x^{3}=\phi

Also, your last term in \Gamma^{\sigma}{}_{\mu\nu} has a typo.

So, for example,

\Gamma^{1}{}_{23}=\frac{1}{2} \sum_{\rho} g^{1\rho} \left( \frac{\partial g_{3\rho}}{\partial \theta} + \frac{\partial g_{2\rho}}{\partial \phi} - \frac{\partial g_{23}}{\partial x^{\rho}} \right)=\frac{1}{2} g^{11} \left( \frac{\partial g_{31}}{\partial \theta} + \frac{\partial g_{21}}{\partial \phi} - \frac{\partial g_{23}}{\partial r} \right)=0<br />

 

[latentcorpse]

ahh i think i get it. the sum reduces to just the \rho=1 term because g_{12}=g_{13}=0 which zeroes the whole expression in the cases of \rho=2 or \rho=3.

so \Gamma^{\sigma}{}_{\mu \nu} will have 3^3=27 copmonents, correct? i can't write my final answer as a matrix can i?
i'd just have to write them out explicitly as:
\Gamma^1_{11}= ...
[tiex]\Gamma^1_{12}= ...[/itex]
etc.

that doesn't look very concise though?

 

[gabbagabbahey]

It will look more concise once you realize just how many of those 27 components are zero (Also, you should keep in mind that g^{ab} is the inverse of g_{ab} when doing your calculations)\

As a matter of convention, \Gamma^1{}_{23} is often written as \Gamma^r_{\theta\phi} and so on; which may be what was confusing you earlier.

 

[gabbagabbahey]

Also, my earlier matrix contains a typo, it should be:

g_{ab}=\begin{pmatrix}1 &amp; 0 &amp; 0 \\ 0 &amp; r^2 &amp; 0 \\ 0 &amp; 0 &amp; r^2\sin^2\theta \end{pmatrix}

 

[latentcorpse]

why is g^{ab} the inverse of g_{ab} and how will that be useful?

also when u say \sigma \in \{ 1,2,3 \} and x^1=r, x^2= \theta, x^3 = \phi
this means that if \sigma=1 then \sigma=r and that's why we can write \Gamma^{1}{}_{23}=\Gamma^r{}_{\theta \phi}
doesn't that imply that 1=r rather than x^1=r?

 

[gabbagabbahey]

why is g^{ab} the inverse of g_{ab} and how will that be useful?

Because of its definition; g^{ab}g_{ac}=\delta^{b}{}_{c}...which tells you that multiplying the matrix g^{ab} with the matrix g_{ac} produces the identity matrix...i.e. g^{ab} is the inverse of g_{ab}.

It's useful, because you will need to know the components of g^{ab} to compute the Christoffel symbols; and you can get those components just by taking the inverse of g_{ab}

also when u say \sigma \in \{ 1,2,3 \} and x^1=r, x^2= \theta, x^3 = \phi
this means that if \sigma=1 then \sigma=r and that's why we can write \Gamma^{1}{}_{23}=\Gamma^r{}_{\theta \phi}
doesn't that imply that 1=r rather than x^1=r?

That's why that notation is often confusing; nevertheless, it is still the convention...

 

[latentcorpse]

so because 1=r and x^1=r, don't you mean to write that \sigma \in \{ x^1,x^2,x^3 \}?

i used maple to quickly get g^{ab}=\left[ \begin {array}{ccc} 1&amp;0&amp;0\\ \noalign{\medskip}0&amp;{r}^{-2}&amp;0<br /> \\ \noalign{\medskip}0&amp;0&amp;{\frac {1}{{r}^{2}{\sin}^{2}\theta}}<br /> \end {array} \right]

 

[latentcorpse]

scratch that above post.

i couldn't think of any quick way to do it so i just did all 27 calculations and found the non zero terms are :

\Gamma^1{}_{22}=-r
\Gamma^1{}_{33}=-r \sin^2{\theta}
\Gamma^2{}_{12}=r^3
\Gamma^2{}_{21}=r^3
\Gamma^2{}_{33}=-\frac{1}{2}r^4 \sin{2 \theta}
\Gamma^3{}_{13}=r^3 \sin^4{\theta}
\Gamma^3{}_{23}=r^4 \sin^3{\theta} \cos{\theta}
\Gamma^3{}_{31}=r^3 \sin^4{\theta}
\Gamma^3{}_{32}=r^4 \sin^3{\theta} \cos{\theta}

i'm not sure if there's a pattern i was supposed to spot so i could save myself some time in working out the copmonents or what?

anyway, when it asks for the components of the Christoffel symbol, do i just leave it as a list of the non zero ones like i have done above or am i missing how to write the whole thing neatly as a matrix or something?

 

[gabbagabbahey]

so because 1=r and x^1=r, don't you mean to write that \sigma \in \{ x^1,x^2,x^3 \}?

No, \sigma \in \{ 1,2,3 \} with x^1=r, x^2=\theta and x^3=\phi...so the Christoffel symbols should be labeled \Gamma^1{}_{23} etc... But, by convention they are often labeled \Gamma^r_{\theta\phi} etc...it's sloppy notation to do this, but nevertheless, convention is convention.

 

[gabbagabbahey]

\Gamma^2{}_{12}=r^3
\Gamma^2{}_{21}=r^3
\Gamma^2{}_{33}=-\frac{1}{2}r^4 \sin{2 \theta}

You seem to be missing a factor of 1/r^4

\Gamma^3{}_{13}=r^3 \sin^4{\theta}
\Gamma^3{}_{23}=r^4 \sin^3{\theta} \cos{\theta}
\Gamma^3{}_{31}=r^3 \sin^4{\theta}
\Gamma^3{}_{32}=r^4 \sin^3{\theta} \cos{\theta}

And for these you are missing a factor of 1/(r^4\sin^4\theta)

i'm not sure if there's a pattern i was supposed to spot so i could save myself some time in working out the copmonents or what?

You could save a little time by remembering that the Christoffel symbols are symmetric in the bottom pair of indices...so you only have to calculate 18 of them.

anyway, when it asks for the components of the Christoffel symbol, do i just leave it as a list of the non zero ones like i have done above or am i missing how to write the whole thing neatly as a matrix or something?

You can't really write them as a single matrix, but they are often written as a set of 3 matrices in the form:

\Gamma^r=\begin{pmatrix}0 &amp; 0 &amp; 0 \\ 0 &amp; -r &amp; 0 \\ 0 &amp; 0 &amp; -r\sin^2\theta\end{pmatrix}

\Gamma^\theta=\begin{pmatrix}0 &amp; \frac{1}{r} &amp; 0 \\ \frac{1}{r} &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; -\sin\theta\cos\theta\end{pmatrix}

\Gamma^\phi=\begin{pmatrix}0 &amp; 0 &amp; \frac{1}{r} \\ 0 &amp; 0 &amp; \cot\theta \\ \frac{1}{r} &amp; \cot\theta &amp; 0\end{pmatrix}

Again, this is somewhat sloppy notation, but is still fairly common in the literature.

 

[latentcorpse]

are they ok then?

 

[gabbagabbahey]

See my edited post above^^^

 

[latentcorpse]

i don't see how I am missing those factors. take for example

\Gamma^2{}_{12}=\frac{1}{2} g^{22} \left( \frac{\partial g_{22}}{\partial r} + \frac{\partial g_{12}}{\partial \theta} - \frac{\partial g_{12}}{\partial \theta} \right) = \frac{1}{2} r^2 \frac{\partial}{\partial r} \left(r^2 \right) = r^3

i can't see where I'm missing this factor?

 

[gabbagabbahey]

You seem to have used g^{ab}=g_{ab} instead of using the inverse matrix you calculated...

\Gamma^2{}_{12}=\frac{1}{2} g^{22} \left( \frac{\partial g_{22}}{\partial r} + \frac{\partial g_{12}}{\partial \theta} - \frac{\partial g_{12}}{\partial \theta} \right) = \frac{1}{2} \frac{1}{r^2} \frac{\partial}{\partial r} \left(r^2 \right) = \frac{1}{r}

 

[latentcorpse]

lol. I'm an idiot sometimes...thanks.

the next bit asks me to write out the copmonents of the geodesic equaiton in this coordinate system and verify the solutions correspond ot straight lines in Cartesian coordinates.

so i guess the eqn tehy're referring to is 3.3.5.

\frac{d^2 x^{\mu}}{dt^2} + \sum_{\sigma, \nu} \Gamma^{\mu}{}_{\sigma \nu} \frac{d x^{\sigma}}{dt} \frac{dx^{\nu}}{dt}=0

am i required here to solve 17 different differential equations?
even if the Christoffelsymbol is zero there will still be that first term equal to zero so none of the equaitons are going to be trivial. or do i just solve for one term of the Christoffel symbol and show the solutino is a straight line?

 

[gabbagabbahey]

Once you sum over \sigma and \nu you will only have 3 ODEs (one for each value of \mu)...

 

[latentcorpse]

ok so i get my three equations as:

\frac{d^2 r}{dt^2}-r \frac{d^2 \theta}{dt^2} - r \sin^2{\theta} \frac{d^2 \phi}{dt^2}=0

\frac{d^2 \theta}{dt^2} + \frac{2}{r} \frac{dr}{dt} \frac{d \theta}{dt} - \sin{\theta} \cos{\theta} \frac{d^2 \phi}{dt^2}=0

\frac{d^2 \phi}{dt^2} + \frac{2}{r} \frac{dr}{dt} \frac{d \phi}{dt} + 2 \cot{\theta} \frac{d \theta}{dt} \frac{d \phi}{dt}=0

do they look correct? how on Earth do i solve them?

 

[gabbagabbahey]

Assuming you meant \frac{d^2 \theta}{dt^2} +\frac{2}{r} \frac{dr}{dt} \frac{d \theta}{dt} - \sin{\theta} \cos{\theta} \frac{d^2 \phi}{dt^2}=0 for the 2nd ODE, then yes those look right...Now, do you really need to solve them, in order to verify that the solutions are straight lines in Cartesian coordinates?...What is the general (parametrized in terms of t) form of a straight line in Cartesian coords?...What is it when you convert to Spherical coords?...Does that form satisfy the ODEs? Could there be any other solutions?

 

[latentcorpse]

in cartesian isn't it just going to be

\vec{r}(t)=\vec{a}+\vec{b}t with \vec{a}, \vec{b} \in \mathbb{R}^3.

i'm not sure how to convert that to sphericals though? have i used the wrong form above?

 

[gabbagabbahey]

No, the general parametrized form of a line in 3D, is ax(t)+by(t)+cz(t)=0

 

[latentcorpse]

\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}?

 

[gabbagabbahey]

Did you see my previous post?

 

[latentcorpse]

ok. so if that's the cartesian form i need to convert it to spherical polars:

ar(t) \sin{\theta (t) } \cos{\phi (t) } + br(t) \sin{\theta (t)} \sin{\phi (t)} + c r(t) \cos{\theta (t)}=0

but i don't see where to substitute that into the geodesic eqn. or am i wanting to show

\left( \frac{d^2 x^{\mu}}{dt^2} + \sum_{\sigma, \nu} \Gamma^{\mu}{}_{\sigma \nu} \frac{d x^{\sigma}}{dt} \frac{d x^{\nu}}{dt} \right) \left( ar(t) \sin{\theta (t)} \cos{\phi (t)} + br(t) \sin{\theta (t)} \sin{\phi (t)} + c r(t) \cos{\theta (t)} \right) = 0?

 

[gabbagabbahey]

ok. so if that's the cartesian form i need to convert it to spherical polars:

ar(t) \sin{\theta (t) } \cos{\phi (t) } + br(t) \sin{\theta (t)} \sin{\phi (t)} + c r(t) \cos{\theta (t)}=0

Okay, now take the derivative of the equation with respect to time twice; do some jiggling around and see what you can come up with...

 

[latentcorpse]

sorry. i don't understand what's going on here.

am i substituting that equation into the geodesic equations?

if so what is r theta and phi?

or am i just multiplying the geodesic eqn with the straight line eqn?

 

[gabbagabbahey]

You want to show that r(t), \theta(t) and \phi(t) satisfy each of your 3 ODEs, so long as they also satisfy your straight line equation..

 

[latentcorpse]

ok. so i did the double derivative with time by hand (is there a way to write a simple maple code to do this for me? i tried but couldn't get it to work) and got:

a \ddot{r} \sin{\theta} \cos{\phi} + a \dot{r} \dot{\theta} \cos{\theta} \cos{\phi} - a \dot{r} \dot{\phi} \sin{\theta} \sin{\phi}

+ a \dot{r} \dot{\theta} \cos{\theta} \cos{\phi} + a r \ddot{\theta} \cos{\theta} \cos{\phi}

- a r \dot{\theta}^2 \sin{\theta} \cos{\phi} - a r \dot{\theta} \dot{\phi} \cos{\theta} \sin{\phi} - a \dot{r} \dot{\phi} \sin{\theta} \sin{\phi} - a r \ddot{\phi} \sin{\theta} \sin{\phi} - a r \dot{\theta} \dot{\phi} \cos{\theta} \sin{\phi}

- a r \dot{\phi}^2 \sin{\theta} \cos{\phi} + b \ddot{r} \sin{\theta} \sin{\phi} + b \dot{r} \dot{\theta} \cos{\theta} \sin{\phi} + b \dot{r} \dot{\phi} \sin{\theta} \cos{\phi} + b \dot{r} \dot{\theta} \cos{\theta} \sin{\phi} + b r \ddot{\theta} \cos{\theta} \sin{\phi}

- b r \dot{\theta}^2 \sin{\theta} \sin{\phi} + b r \dot{\theta} \dot{\phi} \cos{\theta} \cos{\phi} + b \dot{r} \dot{\phi} \sin{\theta} \cos{\phi} + b r \ddot{\phi} \sin{\theta} \cos{\phi} + b r \dot{\theta} \dot{\phi} \cos{\theta} \cos{\phi}

- b r \dot{\phi}^2 \sin{\theta} \ubs{\phi} + c \ddot{r} \cos{\theta{ - c \dot{r} \dot{\theta} \sin{\theta} - c \dot{r} \dot{\theta} \sin{\theta} - c r \ddot{\theta} \sin{\theta} - cr \dot{\theta}^2 \cos{\theta} = 0

 

[latentcorpse]

hmm. i don't know why that all posted on one line? can you read it by just clicking on the code?

anyway, i can group some terms together and what not but I'm still not sure what I am doing...surely i want to take the second derivative of r, theta and phi not of the straight line eqn?

 

[latentcorpse]

sorry, i sorted out the LaTeX in post 32. see my previous 2 posts.

 

[latentcorpse]

basically i don't understand why differentitating the straight line eqn wrt time twice helps us to show they also satisfy the geodesic eqn?

 

[gabbagabbahey]

Sorry, the form of the line equation you want to use is the form you posted in #24...You then have x(t)=a_x+b_xt, y(t)=a_y+b_yt, and z(t)=a_z+b_zt in Cartesians...what are these equations in Spherical coords?

 

[latentcorpse]

ok. i would get

r(t) \sin{\theta(t)} \cos{\phi(t)}=a_x + b_x t
r(t) \sin{\theta(t)} \sin{\phi(t)}=a_y + b_y t
r(t) \cos{\theta(t)}=a_z + b_z t

surely i need to rearrange these to get
r=something
theta=something and
phi=something?

 

[dx]

In the first post, you said that the metric is a 2-form. It is not. A 2-form is a covariant anti-symmetric tensor of rank 2, while a metric is always a symmetric tensor.

 

[gabbagabbahey]

surely i need to rearrange these to get
r=something
theta=something and
phi=something?

That shouldn't be too hard to do...the inverse relations between \{r,\theta,\phi\} and \{x,y,z\} are fairly well known

 

[latentcorpse]

ok. so i take
r(t)=\sqrt{x(t)^2+y(t)^2+z(t)^2}
\theta(t)=\tan^{-1}{\frac{y}{x}}
\phi(t)=\cos^{-1}{\frac{z}{r}}

and plug these into the 3 geodesic ODE's, yes?

also could somebody please remind me of the difference between a two form and a rank 2 covariant tensor?
thanks.

 

[dx]

also could somebody please remind me of the difference between a two form and a rank 2 covariant tensor?
thanks.

A 2-form is a rank 2 covariant tensor that is also antisymmetric.

 

[latentcorpse]

so

r(t)=\sqrt{a_x^2+a_y^2+a_z^2+(b_x^2+b_y^2+b_z^2)t^2}

\theta(t)=\tan^{-1}{\frac{a_y+b_yt}{a_x+b_xt}}

\phi(t)=\cos^{-1}{\frac{a_z+b_zt}{r}}

are these what i should sub into the geodesic eqns?

 

[gabbagabbahey]

Sounds like a plan to me...

 

[latentcorpse]

y:=sqrt((a+b*t)^2+(c+d*t)^2+(e+f*t)^2);
r:=diff(y,t,t);
s:=arctan((c+d*t)/(a+b*t));
q:=simplify(y*diff(s,t,t));
l:=arccos((e+f*t)/y);
p:=simplify(y*sin^2(s)*diff(l,t,t));
simplify(r-q+p);

i tried the above MAPLE code because this calculation was getting tediously long by hand but i can't seem to get 0 out as an answer...any ideas where I'm going wrong?

 

[gabbagabbahey]

\arctan(y/x)\neq\tan^{-1}(y/x)

 

[latentcorpse]

huh? why not?
i can't use tan^(-1) as legitmate maple code - it interprets that as 1/tan(...)

 

[gabbagabbahey]

\arctan is single valued, while the inverse tangent is multivalued (same thing for \cos^{-1})...Does Maple have an "inverse" command?

 

[latentcorpse]

apparently arctan is the inverse of tan according to maple? do you use a different program to maple?
i was all for doing it by hand before i got halfway thorugh the first derivative and realized the whole calculation was going to take about 10 pages, somewhere in which i was bound to make a tedious mistake whilst differentiating. as one of my levturers recently said, most physicists just use maple (or i guess some similar program) for these types of calculations.

 

[gabbagabbahey]

It shouldn't be too hard to do by hand...For starters, \dot{x}=b_x, \dot{y}=b_y
and \dot{z}=b_z; so


\dot{r}=\frac{x\dot{x}+y\dot{y}+z\dot{z}}{\sqrt{x^2+y^2+z^2}}=\frac{b_xx+b_yy+b_zz}{r}

\implies \ddot{r}=\frac{b_x^2+b_y^2+b_z^2}{r}-\frac{b_xx+b_yy+b_zz}{r^2}\dot{r}=\frac{b^2}{r}-\frac{(b_xx+b_yy+b_zz)^2}{r^3}

 

[latentcorpse]

\theta=\tan^{-1}{\frac{y}{x}}

\dot{\theta} = \frac{1}{1+(\frac{y}{x})^2} \left( \dot{y} x^{-1} - y x^{-2} \dot{x} \right)

\ddot{\theta} = \frac{\left( \ddot{y} x^{-1} - \dot{y} x^{-2} \dot{x} \right)}{\left(1 + \left(\frac{y}{x} \right)^2 \right)} - \frac{2y}{x} \frac{ \left( \dot{y} x^{-1} - y x^{-2} \dot{x} \right)^2}{ \left( 1 + \left( \frac{y}{x} \right)^2 \right)^2}

is that ok?