# Deriving equations of motion in spherical coordinates

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 $\phi$ though, the result is different. hen we change θ, r has to change because it changes direction, and when we change $\phi$ $r$ has to change because it changes direction in that case also.

When I take the derivative of $\hat{r}$ with respect to $\theta$, 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. But I am trying to understand where the $\dot{\theta}$ term comes up, and how to get that extra dimension in.

Any help is most appreciated. Thanks.

EDIT: fixed the unit vector phi expression.

• flag on the play. Your "unit" vector in the $\phi$-direction doesn't have length one... I continue reading, but this is troubling to my hat-customs as a former physics major. See page 383 of supermath.info/math231_CurlDivNoncartesian_375_384.pdf – James S. Cook Sep 15 '13 at 20:27
• I fixed the $\hat{\phi}$ expression, but other than that our comment doesn't make a lot of sense to me -- doesn't have length one? What are you referring to? Understand I am not always so intuitive about these things. I'm going with what the prof gives us, so I don't understand what the problem with hat-customs is. – Jesse Sep 15 '13 at 20:33
• When we write $\hat{A}$ in physics-type notation this indicates the vector has length one. This beautiful notation allows us to express any nonzero vector $\vec{A}$ as the product of a magnitude and direction; $\vec{A} = A\hat{A}$. This formula also tells you how to calculate $\hat{A}$. To find $\hat{u}$ for a curvelinear coordinate we can calculate $\nabla u = \langle u_x,u_y,u_z \rangle$ and then normalize it to length one by dividing by $| \nabla u |$. For the spherical radius the gradient already has length one, but for $\phi$ some normalization is needed. – James S. Cook Sep 15 '13 at 20:37
• So what should it look like? Is the book wrong here, or what? Why would my instructor wrote it this way? – Jesse Sep 15 '13 at 20:41
• I think after your edit the spherical unit vectors are fine... I'm just telling you how to derive them via vector calculus. Others prefer to take a geometric approach. That is especially appropriate if this is from a physics course. That said, look at the answer given. Austen's advice is very good. I'm not sure what you mean by differentiation w.r.t. $\theta$. Perhaps, a partial derivative, but mostly, for this problem, I would think about $d/dt$ and keep in mind the spherical frame has a time-dependence but... the cartesian does not. That's the coordinate way to work it out. – James S. Cook Sep 15 '13 at 20:46

## 2 Answers

Suppose $\vec{r} = r \hat{r}$ where $\hat{r} = \cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z}$. Now, this symbol represents the position of a particle at time $t$ where the spherical frame $\hat{r},\hat{\theta},\hat{\phi}$ is co-moving. On the other hand, the cartesian frame $\hat{x},\hat{y},\hat{z}$ is constant across all of three-dimensional space. So, differentiate with respect to time \begin{align} \frac{d}{dt}\vec{r} &= \frac{dr}{dt} \hat{r}+r\frac{d\hat{r}}{dt} \\ &= \dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z} \right] \\ &= \dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \right] \hat{x}+r\frac{d}{dt} \left[\sin \phi \sin \theta \right]\hat{y}+ r\frac{d}{dt} \left[\cos \theta \right]\hat{z} \\ &= \dot{r} \hat{r}+r \left[-\dot{\phi}\sin \phi \sin \theta+\dot{\theta}\cos \phi \cos \theta \right] \hat{x}+ \\ & \qquad +r \left[\dot{\phi}\cos \phi \sin \theta + \dot{\theta}\sin \phi \cos \theta \right]\hat{y}+ r \left[- \dot{\theta}\sin \theta \right]\hat{z} \\ &= \dot{r} \hat{r}+ r\dot{\theta}\left[\cos \phi \cos \theta \hat{x} +\sin \phi \cos \theta \hat{y}-\sin \theta \hat{z}\right] \\ &\qquad + r\sin \theta \dot{\phi} \left[ -\sin \phi \hat{x}+\cos \phi \hat{y}\right] \end{align} Again, the concept here is that the spherical frame rides along with the particle motion and you're trying to express the velocity in terms of the co-moving frame. Identify the $\hat{\phi}$ and $\hat{\theta}$. As Austen was telling you, there is no $\hat{r}$ component in the $\frac{d}{dt} \hat{r}$ term because if there was a change then the length of the unit-vector would evolve... or you can just differentiate $\hat{r} \cdot \hat{r}=1$ with respect to time to see that the time-rate of change for $\hat{r}$ is infact orthogonal to $\hat{r}$ itself.

I would set-aside the questions about differentiating w.r.t. $\phi$ or $\theta$ for a moment until you complete this.

• OK, so when I differentiate this: $\dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z} \right]$ with respect to time I should end up with $\dot{r} \hat{r}+r \left[-\cos \phi \cos \theta \frac{d\theta}{dt} -\sin \phi \sin \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z} \right] – Jesse Sep 15 '13 at 21:38 •$\dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z} \right]$with respect to time I should end up with$\dot{r} \hat{r}+r \left[(-\cos \phi \cos \theta -\sin \phi \sin \theta) \frac{d \theta}{dt}\hat{x}+(\sin \phi \cos \theta + \sin \theta \cos \phi)\frac{d \theta}{dt} \hat{y}- \sin \theta \frac{d \theta}{dt}\hat{z} \right]=\dot{r} \hat{r}+r\dot\theta \left[(-\cos \phi \cos \theta -\sin \phi \sin \theta) \hat{x}+(\sin \phi \cos \theta + \sin \theta \cos \phi)\hat{y}- \sin \theta \hat{z} \right]$– Jesse Sep 15 '13 at 21:53 • there's a messed up sign in there but i can't get mathjax to obey me, but do I have the right idea? – Jesse Sep 15 '13 at 21:54 • @Jesse not quite, I'll add the next step in my answer. You should have some terms with$\dot{\theta}$and others with$\dot{\phi}$. It's like a related rates problem from calculus I. – James S. Cook Sep 16 '13 at 1:28 • That was the issue -- I was getting an answer that made sense b/c I had a$\dot{\theta}$term but not a$\dot{\phi}$. But now I see that if I take the derivative w/r/t TIME the terms like$\cos \phi$should become$-\sin \phi \dot{\phi}$, correct? – Jesse Sep 16 '13 at 1:51 Start from$\mathbf{r}=r\hat{\mathbf{r}}$and take the derivative using the product rule $$\dot{\mathbf{r}}=\dot r\hat{\mathbf{r}}+r\dot{\hat{\mathbf{r}}}.$$ The first term is the first term of the answer you seek, so his focuses attention on$\dot{\hat{\mathbf{r}}}$. The derivative of a unit vector has no radial component, so can be expressed in terms of the other two unit vectors. At this point you can resort to explicit coordinates, but a graphical proof is more illuminating. Think about what happens when only$\dot{\theta}\neq 0$, and then when only$\dot{\phi}\neq 0$is changing. • @Jesse notice Austen is telling you to look at time-rates of change for$\theta$and$\phi$as they appear inside$\dot{\hat{r}}$which can be written$\frac{d}{dt}\hat{r}$. – James S. Cook Sep 15 '13 at 20:47 • Hm. When$\dot{\theta} \neq 0$that would mean r is changing direction, yes? but$\phi$isn't, so you should end up with$\frac{d \hat{r}}{d \theta}$but I have that already, no? If only$\phi$is changing then we have$\frac{d \hat{r}}{d \phi} = \bf \hat{x} \rm (-\sin\theta \sin\phi) + \bf \hat{y} \rm \cos \theta \cos \phi$– Jesse Sep 15 '13 at 20:48 • If$\dot{\theta}=0$then$\frac{d}{dt} \cos \theta = -\sin \theta \ \dot{ \theta} = 0$. – James S. Cook Sep 15 '13 at 20:52 • Whe I did this in 2 dimensional polar coordinates I did this:$d\hat{r} = d\theta \hat{\theta}$and$d\hat{\theta} = -d\theta\hat{r}$because when you go around the$\theta$axis you tilt the unit vectors. So I got$\frac{d\hat{r}}{d\theta} = \hat{\theta}$and$\frac{d\hat{\theta}}{d \theta} = - \hat{r}$. multiplying through by$\frac{d \theta}{dt}$I ended up with$\frac{d \hat{r}}{d \theta}=\hat{\theta} \frac{d\theta}{dt}\$. Now, would I do the same thing here but add another term to this? – Jesse Sep 15 '13 at 21:07
• Austen - thanks a lot-- I got it together with your help too... I didn't want you to think I liked your answer less! – Jesse Sep 16 '13 at 3:27