Distance traveled winding around sphere

[scorpius1782]

Homework Statement

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.

Homework Equations

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

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?

 

[Simon Bridge]

...I believe the limits for ϕ are 0 to 2π ... convert the z coordinate to spherical so that $R\cos\phi=R-v_zt$

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.

 

[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.

 

[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?

 

[scorpius1782]

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

 

[Simon Bridge]

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

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 $\phi$.

 

[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.

 

[LCKurtz]

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.

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.

 

[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 $\pi$

 

[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.

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

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

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

I don't follow that.

$A=Rsin(θ)$

But $\theta$ is not a constant.

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

the limits would be 0 to $\pi$

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

 

[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 $\frac{d\theta}{d\theta}$?

 

[LCKurtz]

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

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

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 $\frac{d\theta}{d\theta}$?

 

[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.

 

[LCKurtz]

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

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

$ds^2$ 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 let's see what you have for $ds$.

 

[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:

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

 

[LCKurtz]

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

That's good.

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

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

No. But now you do have$$
ds=\sqrt{r^2(d\theta^2+sin^2(\theta) d\phi^2)}$$and that's 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.

 

[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 $d\theta$

 

[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.

 

[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.

 

[LCKurtz]

$\phi(t)=wt$
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.

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.

 

[bowlbase]

Yeah, sorry I copied that from the wrong part of my notes. It gets a little cluttered after a while of trying different things.

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

1)$\frac{\partial \phi}{\partial \theta}=\frac{wR}{v_z}sin(\theta) d\theta$

2)$(\frac{\partial \phi}{\partial \theta})^2=\frac{w^2R^2}{v_z^2}sin^2(\theta)d\theta^2$??

I'm not sure about d$\theta$.

$ds=R\sqrt{d\theta^2+sin^2(\theta)(\frac{wR}{v_z})^2sin^2(\theta) d\theta^2}$

$ds=Rd\theta\sqrt{1+(\frac{wR}{v_z})^2sin^4(\theta)}$

I'm not sure if I am suppose to square my derivative or leave it as is. For instance, should I be plugging 2 in for the integral or should I square the d##\theta in 1? The annotation isn't clear to me.

 

[scorpius1782]

I don't know why it had me logged in as someone else. University computer. Guess I'm not the only one here at the U of I looking for help!

 

[LCKurtz]

$ds=Rd\theta\sqrt{1+(\frac{wR}{v_z})^2sin^4(\theta)}$

So the length is$$
d_{tot}=\int_?^?R\sqrt{1+\left(\frac{wR}{v_z}\right)^2\sin^4(\theta)}~d\theta$$Looks like you are ready to answer the original question.

 

[scorpius1782]

Thanks for being patient with me. I really appreciate the help.