Proof of Conservation of Vector Calculus: Force on a Mass with Position Vector r

[supermanii]

Homework Statement

The Force on a mass with position vector r satisfies:

m\frac{d^{2}\textbf{r}}{dt^{2}}=F=f(\textbf{r})\textbf{r}

where f(r) is scalar function of r. Show that L:

L=\textbf{r}\times\frac{d\textbf{r}}{dt}

is conserved.

Homework Equations

The Attempt at a Solution

So for L to be conserved I have to prove that r x dr/dt is a constant however I am not sure where to start.

 

[justPAB]

I'm sorry to interrupt your post, what math is this? Calculus 3?

 

[supermanii]

A 2nd year undergraduate physics module called mathematical methods.

 

[LCKurtz]

If you want to show L is a constant vector, show its derivative is the zero vector. Differentiate it and use what you are given. Do you know if you have a product rule for derivatives of cross products?

 

[supermanii]

Yes I know the product rules for cross products. So:

\frac{dL}{dt}=\frac{dr}{dt}\times\frac{dr}{dt}+\frac{d^{2}r}{dt^{2}}\times r

All the derivatives of and r itself are in the same direction so all go to 0 thus proving L is conserved :). Thanks for the help.

 

[LCKurtz]

Yes I know the product rules for cross products. So:

\frac{dL}{dt}=\frac{dr}{dt}\times\frac{dr}{dt}+\frac{d^{2}r}{dt^{2}}\times r

All the derivatives of and r itself are in the same direction so all go to 0 thus proving L is conserved :). Thanks for the help.

Not quite. r and its derivatives are not generally in the same direction. For example, acceleration need not be same direction as velocity. You have a bit left to do to complete your argument.

 

[supermanii]

Ok I was wrong, but dr/dt cross itself goes to zero so:

\frac{dL}{dt}=\frac{d^{2}r}{dt^{2}}\times r?

I have to prove that \frac{d^{2}r}{dt^{2}} and r are in the same direction. Am I right in thinking that the very first equation shows exactly this? As both m and f(r) are scalars so do not affect the direction.

 

[LCKurtz]

Ok I was wrong, but dr/dt cross itself goes to zero so:

\frac{dL}{dt}=\frac{d^{2}r}{dt^{2}}\times r?

I have to prove that \frac{d^{2}r}{dt^{2}} and r are in the same direction. Am I right in thinking that the very first equation shows exactly this? As both m and f(r) are scalars so do not affect the direction.

Yes. You have to do an additional step and use the given info about the second derivative.

You would expect angular momentum to be preserved given that the acceleration is in the radial direction, eh?