How can you conclude that the angular momentum of a particle moving in a central field of force will be constant? When determining the magnitude of the angular momentum vector $|\vec L |$, for a particle in a central field, my textbook concludes the following:  
$L=|\vec{r}\times m\vec{v}| = mr^2\dot{\theta}$ = constant
"for any particle moving in a central field of force." I understand how the magnitude of angular momentum was derived, but I don't understand why you can conclude that it will always be a constant in a central field. Wouldn't it be possible for $r$ and $\dot{\theta}$ to be non-constant quantities as functions of time? For example, a particle moving in an elliptical orbit would have a non-constant $r$.
I can see that $\vec L$ must be a constant when you consider that $\frac{d\vec{L}}{dt}=0$, but how can you conclude that from the line of reasoning presented above?
 A: The derivative of $\mathbf r$ is $\mathbf v$, while that of $\mathbf v$ is the acceleration $\mathbf a$. Now use the fact that the force is central and the skew-symmetry of the cross product to get to $\dot{\mathbf L}=0$.
It is possible for the quantities you have mentioned to depend on time (in most cases this will be the case), but such dependence is such that their combination as in that formula won't change over time.
The angular momentum $\mathbf L$ is not the only constant vector of the motion in this scenario, when the potential is that of Kepler's problem, but there is another one which, in a sense, comes from a hidden symmetry.
A: A "central force field" is a force field that admits the rotation group as a generator of symmetries. Recall Newton's second law
$$\vec F=-\nabla V$$
If $V$ has rotational symmetry, then $V=V(r)$. The radial coordinate is defined by
$$r\equiv\sqrt{\sum_i(x^i)^2}$$
Taking the derivative, we have $r\,dr=\sum_ix^i\,dx^i$ or
$$\frac{\partial r}{\partial x^i}=\frac{x^i}{r}$$
So with $V=V(r)$, we have (prime denoting $d/d r$)
$$F^i=-\frac{\partial V}{\partial x^i}=-\frac{d V}{d r}\frac{\partial r}{\partial x^i}=-\frac{x^i}{r}V'$$
Let $f(r)\equiv-V'(r)/r$. Then we obtain angular momentum conservation immediately. To see this, multiply Newton's equation
$$m\ddot{x}^i=f(r)x^i$$
by $x^j$ so that $m\ddot{x}^i x^j=f(r) x^i x^j$. Subtract from this the equation with $i\leftrightarrow j$. Regardless of $f(r)$, we find
$$x^j\ddot x^i-x^i\ddot x^j=0$$
But this is the same as
$$\frac{d}{dt}(x^j \dot x^i- x^i \dot x^j)=0$$
The quantity $l^{ij}=x^j \dot x^i- x^i \dot x^j$ is the angular momentum per unit mass. The previous equation says that it is constant, i.e. conserved.
A: You can prove that it must be so by considering a central force acting on a body such that it is only a function of radius only and has no angular dependence. $\underline{F}=F\left(\underline{r}\right)\hat{\underline{r}}$.
So the torque on the body is defined as $\underline{\Gamma}=\underline{r}\times \underline{F}=\frac{\partial\underline{L}}{\partial t}$ 
So from the form of the force above you can see that the torque is $\underline{r}\times F\left(\underline{r}\right)\underline{\hat{r}}=0$ as $\underline{r}\times \underline{\hat{r}}=0$
So if there is no torque on the body the angular momentum must be conserved. This constrains $\underline{r}$ and $\underline{p}$ for a given initial angular momentum. You are indeed right that $\underline{r}$ is not constant for an elliptical orbit. If you consider the definition of the angular momentum vector $\underline{L}=\underline{r}\times \underline{p}$ then for smaller $\underline{r}$ the momentum of the body must increase for the angular momentum to remain constant, which is indeed what happens.
