How to Calculate Christoffel Symbols in Spherical Coordinates?

[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?

 

[gabbagabbahey]

Shouldn't you have $\phi=\tan^{-1}(y/x)$?

 

[latentcorpse]

ok so

$$\ddot{\phi}=\frac{\ddot{y} x^{-1} -2 \dot{y} x^{-2} \dot{x} - y x^{-2} \dot{x} +2y x^{-3} \dot{x}^2}{1+ \left( \frac{y}{x} \right)^2} - \frac{2y}{x} \frac{ \left( \dot{y} x^{-1} - y x^{-2} \dot{x} \right)^2}{\left( 1+ \left( \frac{y}{x} \right)^2 \right)^2}$$

 

[gabbagabbahey]

That doesn't look quite right to me...I'd start by simplifying $\dot{\phi}$ as much as possible, and then converting it to spherical coordinates...what do you get when you do that?

 

[latentcorpse]

$$\dot{\phi}=\frac{1}{1+ \left( \frac{y}{x} \right)^2 } \left( \dot{y} x^{-1} - y x^{-2} \dot{x}} \right)=\frac{1}{1+ \left( \frac{a_y+b_yt}{a_x+b_xt} \right)^2} \left( \frac{b_y}{a_x+b_xt}-\frac{b_x(a_y+b_yt)}{(a_x+b_xt)^2} \right)$$
that ok?

 

[gabbagabbahey]

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

I'd multiply both the numerator and denominator by $x^2$, sub in $\dot{x}=b_x$ and $\dot{y}=b_y$ and then write $x$ and $y$ in terms of $r$, $\theta$ and $\phi$...

 

[latentcorpse]

$$\dot{\phi} = \frac{r \sin{\theta}}{1+r^2 \sin^2{\theta} \sin^2{\phi}} \left( b_y \cos{\phi} - b_x \sin{\phi} \right)$$

is that going to help? shouldn't i take the second derivative before i switch to spherical coordinates?

 

[gabbagabbahey]

$$\dot{\phi} = \frac{r \sin{\theta}}{1+r^2 \sin^2{\theta} \sin^2{\phi}} \left( b_y \cos{\phi} - b_x \sin{\phi} \right)$$

is that going to help? shouldn't i take the second derivative before i switch to spherical coordinates?

It would help, if you did it properly

Try again, you have an error...and don't be afraid to simplify your result as much as possible.

 

[latentcorpse]

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

$$\ddot{\phi} = \frac{ \left( \ddot{y} x + \dot{y} \dot{x} - \dot{y} \dot{x} - y \ddot{x} \right) \left( 1+ y^2 \right) - 2 y \dot{y} \left( \dot{y} x - y \dot{x} \right) }{ \left(1+y^2 \right)^2 }$$ note the first term in the numerator will dissappear because $$\ddot{y}=\ddot{x}=0$$ and $$\dot{y}\dot{x}-\dot{x}\dot{y}=0$$.

$$\ddot{\phi} = - \frac{2r \sin{\theta} \sin{\phi} b_y \left( b_y r \sin{\theta} \cos{\phi} - b_x r \sin{\theta} \sin{\phi} \right) }{ \left(1+r^2 \sin^2{\theta} \sin^2{\phi} \right)^2}$$
$$\ddot{\phi}=-\frac{2 b_y r^2 \sin^2{\theta} \sin{\phi} \left( b_y \cos{\phi} - b_x \sin{\phi} \right)}{ \left(1+r^2 \sin^2{\theta} \sin^2{\phi} \right)^2}$$

i can't see how to simplify that further...

 

[gabbagabbahey]

When you multiply $1+\left(\frac{y}{x}\right)^2$ by $x^2$, should you get $x^2+y^2$?

 

[latentcorpse]

$$\dot{\phi}=\frac{\dot{y}x-y \dot{x}}{x^2+y^2}$$

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

$$\ddot{\phi} = \frac{-2 \left( b_x r \sin{\theta} \cos{\phi} + b_y r \sin{\theta} \sin{\phi} \right) \left( b_y r \sin{\theta} \cos{\phi} - b_x r \sin{\theta} \sin{\phi} \right) }{ \left( r^2 \sin^2{\theta} \cos^2{\phi} + r^2 \sin^2{\theta} \sin^2{\phi} \right)^2}$$

$$\ddot{\phi}=\frac{-2r^2 \sin^2{\theta} \left(b_x \cos{\phi} + b_y \sin{\phi} \right) \left( b_y \cos{\phi} - b_x \sin{\phi} \right)}{r^4 \sin^4{\theta}}$$

$$\ddot{\phi} = \frac{-2 \left( b_x b_y \cos^2{\phi} + b_y^2 \sin{\phi} \cos{\phi} - b_x^2 \sin{\phi} \cos{\phi} - b_x b_y sin^2{\phi} \right)}{r^2 \sin^2{\theta}}$$

hopefully that looks better? i was hoping the numerator would simplify nicer though.

 

[gabbagabbahey]

Your $\LaTeX$ contains a few typos, but yes, that's correct.

Also, it is worth writing $\dot{\phi}$ and $\dot{r}$ in spherical coordinates;

$$\dot{r}=b_x\sin\theta\cos\phi+ b_y\sin\theta\sin\phi+b_z\cos\theta$$

$$\dot{\phi}=\frac{b_y\cos\phi-b_x\sin\phi}{r\sin\theta}$$

These will come in handy when calculating $\ddot{\theta}$ (Since $\dot{\theta}$ will be a function of $r$, $\theta$ and $\phi$; knowing these derivatives allows you to find $\ddot{\theta}$ quickly, using the chain rule)

 

[Phrak]

I recall that a straight line in cartesian coordinates is $$\ \frac{d^2 x^i}{d \lambda^2} = 0 \ ,$$
where the curve is parametric in lambda. The parameter can be time.

$$\frac{d^2 x^i}{d t^2} = 0$$

 

[gabbagabbahey]

I recall that a straight line in cartesian coordinates is $$\ \frac{d^2 x^i}{d \lambda^2} = 0 \ ,$$
where the curve is parametric in lambda. The parameter can be time.

$$\frac{d^2 x^i}{d t^2} = 0$$

This is true; which is why the solution is of the form $x^i=a_i+b_it$ in Cartesians.

 

[latentcorpse]

ok. so finally $\ddot{\theta}$

$$\theta=\cos^{-1} \left( \frac{z}{r} \right)$$

$$\dot{\theta} = - \frac{1}{\sqrt{1 - \left( \frac{z}{r} \right)^2 }} \left( \dot{z} r^{-1} - z r^{-2} \dot{r} \right)$$

$$\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)$$

is that ok for $\dot{\theta}$?

 

[gabbagabbahey]

Looks fine so far; now express it in terms of spherical coordinates and $b_x$, $b_y$ and $b_z$...

 

[latentcorpse]

$$\dot{\theta}=-\frac{1}{r^2 \sqrt{1-\cos^2{\theta}}} \left( \dot{z}r - z \dot{r} \right) = -\frac{1}{r^2 \sin{\theta}} \left( b_z r - \left( a_z + b_z t \right) \left(b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$

looking ok?

 

[gabbagabbahey]

I'd use $z=r\cos\theta$ instead of $z=a_z+b_zt$...

 

[latentcorpse]

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z - \cos{\theta} \right) \left( b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$
that's ok?

should i take the derivative of the above or should i take the derivative of something shorter like:

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left( \dot{z} r - z \dot{r} \right)$$?

this gives

$$\ddot{\theta}=\frac{ \left( \ddot{z} r + \dot{z} \dot{r} - \dot{z} \dot{r} - z \ddot{r} \right) r \sin{\theta} - \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left( \dot{z} r - z \dot{r} \right) }{r^2 \sin^2{\theta}}$$

$$\ddot{\theta}=\frac{-zr \ddot{r} \sin{\theta}- \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left( \dot{z} r - z \dot{r} \right)}{r^2 \sin^2{\theta}}$$

on the right lines?

 

[gabbagabbahey]

$$\dot{\theta}=-\frac{1}{r \sin{\theta}} \left(b_z - \cos{\theta} \right) \left( b_x \sin{\theta} \cos{\phi} + b_y \sin{\theta} \sin{\phi} + b_z \cos{\theta} \right)$$
that's ok?

Do $b_z$ and $\cos\theta$ have the same units? If not, the appearance of the factor $b_z-\cos\theta$ should be a dead giveaway that you've made an error somewhere...(It's always a good idea to check the units of your expression at each step of a complicated calculation, it will help cut down on the number of error you make)

 

[latentcorpse]

you said i was fine at $$\dot{\theta}=-\frac{1}{\sqrt{r^4-z^2r^2}} \left(\dot{z}r-z \dot{r} \right)$$

so i need to change the coords here

$$\dot{\theta}=-\frac{1}{\sqrt{r^4-r^4 \cos^2{\theta}}} \left(b_z r - r \cos{\theta} \dot{r} \right) = -\frac{1}{r \sin{\theta}} \left( b_z - \cos{\theta} \dot{r} \right)$$
ok that should be better

now

$$\ddot{\theta}=- \left( \frac{ \left( - \ddot{r} \cos{\theta} + \dot{r} \dot{\theta} \sin{\theta} \right) r \sin{\theta} - \left( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta} \right) \left(b_z - \dot{r} \cos{\theta} \right)}{r^2 \sin^2{\theta}} \right)$$

is that ok? it doesn't look like it's going to get much simpler : i can bring the minus at the front in and cancel the $r \sin{\theta}$ in the first term i guess...