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A problem I am trying to work out is as follows:

A particle moves in a force field given by $\vec F =\phi(r) \vec r$. Prove that the angular momentum of the particle about the origin is constant.

I set it up as follows:

$\vec F = m {d^2\vec r \over dt^2}$

$\vec v = \int {\frac {\vec F}{m} }\ dt = \int {\frac {\phi(r) \vec r}{m} }\ dt$

which is equal to :

${\frac {\phi(r) t \vec r}{m} } + c$

(I am not sure what I am doing at this point. Is my integrated expression correct?)

Assuming it is, we get:

Angular Momentum $L = m (\vec r \times \vec v) = \vec r \times (\phi(r) t \vec r + c)$

Now I don't know what to do with the constant term, but I do know that

$\vec r \times k\vec r = 0$

However, the problem states that we have to prove the result is a constant, so I think I'm wrong. Specific places where someone could help me out are:

(1) Is my integration correct? If not, how does one integrate a force (given in terms of position vector notation) w.r.t. time?

(2) What happens to the constant? Cross-product of a vector and a scalar doesn't make any sense.

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When you want to prove that some quantity $XYZ$ doesn't change with time, the most intuitive way to start is to try to prove that $XYZ(t=t_1) = XYZ(t=t_2)$ for any two times $t_1$ and $t_2$. But that is almost never the easiest way to proceed. A wiser approach is to try to prove that the time derivative of XYZ is zero. – Steve B Jun 8 '12 at 13:09
up vote 13 down vote accepted

If you want to prove that $\vec{L}=\vec{r}\times \vec{p}$ is constant with respect to time for a particle in a central force field $\vec F = \phi(r) \vec r$, just show that the angular momentum doesn't change with time, i.e. $\frac{d}{dt}\vec{L}=0$.

Using the product rule we get two terms:

$\frac{d}{dt}\vec{L}=\frac{d}{dt}(\vec{r}\times \vec{p}) = \frac{d\vec r}{dt} \times \vec p + \vec r \times \frac{d \vec p}{dt}$.

Since $\vec p = m \frac{d \vec r}{dt}$ and $\frac{d \vec r}{dt}$ are obviously parallel, the first term vanishes. In the special case of a central force $\vec F = \phi(r) \vec r$ the second term vanishes too: We have $\frac{d\vec p}{dt} = \vec F \propto \vec r$, so the two vectors in the second term are parallel, causing the cross product to become zero.

Therefore $\frac{d}{dt}\vec{L}=0$ and $\vec{L}$ is a constant with respect to time.

To answer your questions:

(1) No, you can't integrate like that. The position of the particle $\vec r$ changes with the time, so you can't treat it as a constant in your integration. If you want to solve this integral, solve the equations of motion $\frac{d \vec p}{dt} = \vec F$ first.

(2) If your integration would have been correct (for instance if the particle position were constant), the integration constant would have been a vector too. Then the cross product would make sense again.

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Since $ F = \phi(r) \vec r $, you can find the torque around the origin.

Torque $ \tau = \vec F \times \vec r = \phi (r) \vec r \times \vec r$

But $\vec r \times \vec r$ is zero, so the torque around the origin is also zero.

Since torque is just rate of change of angular momentum $\frac{ d\vec L}{dt}$, the angular momentum doesn't change, which is what you wanted to prove.

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+1 for an elegant solution – Bryson S. Nov 19 '14 at 21:26

You can't do that integral because we expect $\vec{r}$ to change with time. If the potential is central, then it only depends on the norm of the radius vector, but this of course can change. However, in your integral you're mixing numbers and vectors, $\vec{c}$ in any case is a vector.

Edit: I misunderstood the question, the above answer is good

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