Distance traveled winding around sphere

1. Nov 18, 2013

scorpius1782

1. The problem statement, all variables and given/known data
I have a sphere that I want to travel from the north pole to the south pole. The route I take is winding around the sphere (instead of the obvious shortest path). My path is dictated by:

$\phi(t)=wt$
$z(t)=R-v_z t$ in the negative z direction per N to S

The form of the final answer is given!

$d_{tot}=\int_?^? A\sqrt{1+B^2sin^n(\theta)}d\theta$

A and B are some constants to find.

2. Relevant equations

dl for spherical:$dl=dr\hat{r}+rd\theta \hat{\theta}+r sin(\theta)d\phi \hat{\phi}$

3. The attempt at a solution

I know that R is constant since we're flying around at the same altitude. I believe the limits for $\phi$ are 0 to 2$\pi$ and $\theta$ 0 to $\pi$. Clearly according to the given integral I don't need to worry abut phi and R anyway though.

I thought that I may take $z(t)=R-v_z t$ and convert the z coordinate to spherical so that $Rcos(\phi)=R-v_z t$ but that hasn't really lead me anywhere near that integral so far.

Just not sure how to start. Anyone care to help point me in the right direction?

2. Nov 18, 2013

Simon Bridge

If $z=R\cos\phi$, then that makes $\phi$ the angle from the +z axis, so the limits for $\phi$ cannot be $0$ to $2\pi$.

Remember, you want to express the equations in $t$, that you have, in terms of equations $\theta$ and the rest. To do that, you need to understand what they are.

3. Nov 18, 2013

scorpius1782

I'm at a complete loss as to what I should do. It seems to me that I need a way to get rid of the z component if I'm to be able to do this in spherical. And, I don't understand the final integral at all. Somehow they removed the phi dependence for the distance and reduced it to a single integral instead of a double as I would expect.

4. Nov 18, 2013

Simon Bridge

OK - perhaps I was not clear: is $\phi$ the angle from the +z axis or not?
If yes - then you should adjust those limits.
If no - then it must be $z=R\cos\theta = R-v_zt$ right?

5. Nov 18, 2013

scorpius1782

$\theta$ should be the angle from the z axis. $\phi$ is the angle from the x axis.

6. Nov 19, 2013

Simon Bridge

Then $R\cos\theta = R-v_zt$ - hint: solve for t and sub into the equation for $\phi$ Or maybe the other way around - gets you $\theta$ in terms of $\phi$.

7. Nov 19, 2013

scorpius1782

I've done the algebra. Going to have to sleep on it to decide what I should do with it next. Not sure at all where the root or powers are coming from.

8. Nov 19, 2013

LCKurtz

Remember that the scalar arc length element $ds$ is the square root of the sum of the squares of the components of $dl$. You can get an easy equation for $\phi$ in terms of $\theta$ as already suggested. Then express $ds$ in terms of $d\theta$ and you will see.

9. Nov 19, 2013

scorpius1782

Okay, since $ds=\sqrt{1+(\frac{d\phi}{d\theta})^2}$

in spherical $dl= \hat{r}dr+rd\theta \hat{\theta}+rsin(\theta)d\phi \hat{\phi}$

I think: $ds=rsin(\theta)\sqrt{1+(\frac{d}{d\theta}\frac{wR}{v_z}(1-cos(\theta))^2}$
$=Rsin(\theta)\sqrt{1+(\frac{wR}{v_z}sin(\theta))^2}$

$A=Rsin(θ)$
$B=(\frac{wR}{v_z})^2$
n=2

the limits would be 0 to $\pi$

Last edited: Nov 19, 2013
10. Nov 19, 2013

LCKurtz

But $\theta$ is not a constant.

Given that $dr$ is zero, what do you get for $ds^2$ if you take the sum of the squares of the components of $dl$?

Last edited: Nov 19, 2013
11. Nov 19, 2013

scorpius1782

I don't know what to put into dl.
dropping hats
$dl=dr+rd\theta+rsin(\theta)d\phi$
I only have 1 equation now

$\phi(t)=\frac {Rcos(\theta)-R}{v_z}$

So I have no equation for r (but dr=0 so okay) or theta. I'm left with only $\phi(t)$

The only reason I just plugged it into the $\frac{d\phi}{d\theta}$ is because it's the only derivative I can do. What could I use for $\frac{d\theta}{d\theta}$?

12. Nov 19, 2013

LCKurtz

Please answer my question before continuing. You don't just "drop the hats".

Last edited: Nov 19, 2013
13. Nov 19, 2013

scorpius1782

$ds^2=0^2\hat{r}+r^2d\theta^2\hat{\theta}+r^2sin^2(\theta)d\phi^2\hat{\phi}$

I was "dropping hats" because latex becomes cumbersome to type after a while.

14. Nov 19, 2013

LCKurtz

$ds^2$ is a scalar. You just want the components squared. If I asked you for the length^2 of a vector $a\vec i + b\vec j + c\vec k$ would you leave the $\vec i,\vec j, \vec k$ in the answer? Fix that and lets see what you have for $ds$.

15. Nov 19, 2013

scorpius1782

$ds^2=r^2d\theta^2+r^2sin^2(\theta)d\phi^2$
$ds^2=r^2(d\theta^2+sin^2(\theta) d\phi^2)$

Since I have no function for $\theta$ (but it goes from 0 to $\pi$) can I go ahead and say that:

$ds^2=r^2(\pi^2+sin^2(\theta) d\phi^2)$

16. Nov 19, 2013

LCKurtz

That's good.
No. But now you do have$$ds=\sqrt{r^2(d\theta^2+sin^2(\theta) d\phi^2)}$$and thats what you want to integrate. You know $r = R$ is constant, and from the previous discussion I think you may have figured out how to get $\phi$ in terms of $\theta$ so you can get $d\phi$ in terms of $d\theta$ and you should be good to go. I will revisit this thread later today to see how you come out.

17. Nov 19, 2013

scorpius1782

So, am I just suppose to put in

$d\phi=\frac {Rcos(\theta)-R}{v_z} d\theta$ and then do algebra to bring out $d\theta$

18. Nov 19, 2013

LCKurtz

I can't follow stuff when you just write some equation down like that where I don't know where you got it. What is the equation giving the relation between $\phi$ and $\theta$? Show what it is and how you get it. After you have that equation you can show how you calculate $d\phi$ from that equation.

19. Nov 19, 2013

scorpius1782

$\phi(t)=wt$
from
$z(t)=R-v_z t$ where $z=Rcos(\theta)$
so $Rcos(\theta)=R-v_z t$
solving for 't' and plugging back into $\phi(t)=wt$
$\phi(t)=w\frac {Rcos(\theta)-R}{v_z}$

Kept forgetting to put 'w' in there.

20. Nov 20, 2013

LCKurtz

You have a sign mistake on the last step, probably from not writing down the steps. Fix that and take differentials to get $d\phi$ in terms of $d\theta$. Then let's see what you get for your integral.