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I am looking at my lecture notes for mechanics and it says that for a central force $\vec F=f(r)\hat r$,

$$\vec F=-\nabla V $$ and so $$dV=-\vec F\cdot d\vec r$$

Then $$V=\int \vec F \cdot d\vec r=\int f(r)\frac{\vec r}{r}\cdot d\vec r $$

I'm fine up until this point. However, next it says that $\vec r \cdot d\vec r=r dr$ and so $$V=\int f(r) \ dr $$

I don't understand why $\vec r \cdot d\vec r=rdr.$

Could someone explain this please?

MHW
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2 Answers2

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Note that the differential displacement, $d\vec r$, can be written in spherical coordinates $(r,\theta,\phi)$ with orthogonal unit vectors $(\hat r,\hat \theta,\hat \phi)$ as

$$d\vec r=\hat r dr+\hat \theta rd\theta+\hat \phi r\sin(\theta) d\phi$$

Then, taking the inner product with the position vector, $\vec r=\hat rr$, expressed in spherical coordinates, we find

$$\vec r \cdot d\vec r=\hat rr \cdot (\hat r dr+\hat \theta rd\theta+\hat \phi r\sin(\theta) d\phi)=rdr $$

since $\hat r\cdot \hat \theta=\hat r \cdot \hat \phi=0$.

Mark Viola
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  • It's highly unconventional to retract a best vote. And how is the answer posted a day later more illuminating? It is a much less direct approach. – Mark Viola Mar 18 '16 at 02:20
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$\displaystyle r\, dr =\frac{1}{2} dr^{2} =\frac{1}{2}d(\mathbf{r} \cdot \mathbf{r}) =\frac{1}{2}(d\mathbf{r} \cdot \mathbf{r}+\mathbf{r}\cdot d\mathbf{r}) =\mathbf{r} \cdot d\mathbf{r}$

Ng Chung Tak
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