Deriving equations of motion in spherical coordinates

Homework Statement

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:

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vela
Staff Emeritus
Homework Helper

Homework Statement

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 $$\bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\ \hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\ \hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi \sin \phi \\$$

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

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

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

CAF123
Gold Member
Now, if I take it that position $$\bf \vec{r} \rm = r \bf \hat{r}$$ and say $$\bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\ \hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\ \hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi \sin \phi \\$$
The unit vector for $\hat{\phi}$ is incorrect, check this.

Now, if I look at $\bf \vec{r} \rm = r \bf \hat{r}$ and take the derivative w/r/t time, I should get $\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm$
Yes, that's correct.
I notice that this happens (and some of this is just seeing the notation):
$$\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm = \dot{r} \bf \hat{r} \rm + r \hat{\boldsymbol \theta} \rm$$
You have that $\dot{\hat{r}} = \hat{\theta}$ which is incorrect. You forgot the chain rule.

pasmith
Homework Helper

Homework Statement

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

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).
Those derivatives should be partial.

If we do the same thing with changing θ and $\varphi$ though, the result is different. When 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.
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.

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

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

vela
Staff Emeritus
Homework Helper
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 $d\hat{r} = d\theta \hat{\theta} + d \phi \hat{\phi}$
That doesn't work units-wise, does it?

CAF123
Gold Member
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 $\frac{d\hat{r}}{d\theta} = \hat{\theta}\frac{d \theta}{dt}$
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.