# Deriving equations of motion in spherical coordinates

1. Sep 15, 2013

### Emspak

1. The problem statement, all variables and given/known data

OK, we've been asked to derive the equations of motion in spherical coordinates. According to the assignment, we should end up with this:

$$\bf \vec{v} \rm = \frac{d \bf \vec{r} \rm}{dt} = \dot{r} \bf \hat{r} \rm + r \dot{\theta}\hat{\boldsymbol \theta} \rm + r \dot{\phi}\sin \theta \bf \hat{\boldsymbol\phi}\rm$$

In this case θ is the angle from the z axis and phi is the angle in the x-y plane.

Now, if I take it that position $$\bf \vec{r} \rm = r \bf \hat{r}$$ and say $$\bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\ \hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\ \hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi\\$$

now maybe I am making this more complex than it is. And maybe it's just a notation problem (I really hate the dot notation sometimes because I feel it obscures things, but I need to know it, I know).

If we assume that when r changes, $\phi$ and $\theta$ and their unit vectors stay the same, then we can safely say that $\frac{d \hat{\boldsymbol\phi}}{dr} = 0$ and $\frac{d \hat{\boldsymbol\theta}}{d r} = 0.$ (someone please tell me if i am wrong).

If we do the same thing with changing θ and $\varphi$ though, the result is different. hen we change θ, r has to change because it changes direction, and when we change $\varphi$ r has to change because it changes direction in that case also.

When I take the derivative of $\hat{r}$ with respect to $\vartheta$, I get the following:

$$\frac{d \bf \hat{r}}{d\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta$$

which as it happens also is equal to $\hat{\boldsymbol{\theta}}$

Now, if I look at $\bf \vec{r} \rm = r \bf \hat{r}$ and take the derivative w/r/t time, I should get $\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm$

I notice that this happens (and some of this is just seeing the notation):
$$\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm = \dot{r} \bf \hat{r} \rm + r \hat{\boldsymbol \theta} \rm$$

ANd I feel like I am almost there. But I am having trouble making that last step. I am getting a bit frustrated because I can't seem to make the differentiation work the way it does in the text and I haven't found a derivation online that matches up with anything I have seen in class. Again, maybe it's just the notation used.

Any help is most appreciated. Thanks.

EDIT: unit vector phi expression corrected.

Last edited: Sep 15, 2013
2. Sep 15, 2013

### vela

Staff Emeritus
Just look at your expressions for the unit vectors. $r$ doesn't appear at all, so changing it leaves the unit vectors unchanged.

You mean $\hat{r}$, not $r$, right?

Note that you calculated $d\hat{r}/d\theta$ in one case and $d\hat{r}/dt$ in the other.

3. Sep 15, 2013

### CAF123

The unit vector for $\hat{\phi}$ is incorrect, check this.

Yes, that's correct.
You have that $\dot{\hat{r}} = \hat{\theta}$ which is incorrect. You forgot the chain rule.

4. Sep 15, 2013

### pasmith

In which case $\mathbf{r} = r(\sin\theta \cos \phi \hat x + \sin\theta \sin\phi \hat y + \cos\theta \hat z)$, and
$$\hat r = \sin\theta \cos\phi \hat x + \sin\theta \sin\phi \hat y + \cos \theta \hat z \\ \hat \theta = \cos\theta \cos\phi \hat x + \cos\theta \sin\phi \hat y - \sin\theta \hat z \\ \hat \phi = -\sin\phi \hat x + \cos\phi \hat y$$

Those derivatives should be partial.

No; if you fix $r$ then you are confined to a sphere of radius $r$. $\hat r$ is always normal to the sphere and $\hat \theta$ and $\hat \phi$ are tangent to the sphere. All three unit vectors are functions of $\theta$ and $\phi$ and independent of $r$. (Actually, as you see from above, $\hat \phi$ depends only on $\phi$).

$$\mathbf{r} = r\hat r$$
and differentiate with respect to time:
$$\dot{\mathbf{r}} = \dot r \hat r + r \dot {\hat r}$$
Now calculate $\dot{\hat r}$ from the definition using the product rule and the chain rule. Then collect terms involving $\dot \theta$ and $\dot \phi$. You should then get the given answer.

5. Sep 15, 2013

### Emspak

So when I look at the change in the unit vector with respect to time, I have to consider it as $\vec{r}=r(\theta + d\theta, \phi+ d\phi)\hat{r}$. Is that the case? So I could rewrite this as $d\hat{r} = d\theta \hat{\theta} + d \phi hat{\phi}$

6. Sep 15, 2013

### Emspak

So when I look at the change in the unit vector with respect to time, I have to consider it as $\vec{r}=r(\theta + d\theta, \phi+ d\phi)\hat{r}$. Is that the case? So I could rewrite this as $d\hat{r} = d\theta \hat{\theta} + d \phi \hat{\phi}$

I was thinking that in two dimensions I did this:

$d\hat{r} = d\theta \hat{\theta}$

and multiplied through by $\frac{d\theta}{dt}$ and got $\frac{d\hat{r}}{d\theta} = \hat{\theta}\frac{d \theta}{dt}$

7. Sep 15, 2013

### vela

Staff Emeritus
That doesn't work units-wise, does it?

8. Sep 15, 2013

### CAF123

I do not quite follow. What you want to find is $\dot{\hat{r}}$. Since $\hat{r} = \hat{r}(\theta, \phi)$, the differential is as follows:$$\frac{d \hat{r}}{dt} = \frac{\partial \hat{r}}{\partial \theta} \frac{d \theta}{dt} + \frac{\partial \hat{r}}{\partial \phi} \frac{d \phi}{dt},$$
Sub this into $\dot{\vec{r}}$ and this should give you your answer.