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## 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, [itex]\phi[/itex] and [itex]\theta[/itex] and their unit vectors stay the same, then we can safely say that [itex]\frac{d \hat{\boldsymbol\phi}}{dr} = 0[/itex] and [itex]\frac{d \hat{\boldsymbol\theta}}{d r} = 0.[/itex] (someone please tell me if i am wrong).

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

When I take the derivative of [itex]\hat{r}[/itex] with respect to [itex]\vartheta[/itex], 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 [itex] \hat{\boldsymbol{\theta}}[/itex]

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

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.

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