Potential for a Central Force Field

1. Feb 6, 2014

Rococo

1. The problem statement, all variables and given/known data

I'm having trouble understanding these notes. Can anyone help me understand how Equation 1 is arrived at, and then how Equation 2 is arrived at?

A central force field $\underline{F}(\underline{r})$ is a field of forces, which are directed from a mass m inwards or outwards from a fixed point O, with magnitude depending only on $r= \mid{\underline{r}}\mid$ of m from O:

$\underline{F}(\underline{r}) = f(r) \hat{\underline{r}} = f(r) \frac{\underline{r}}{r}$

As a consequence, $\mid \underline{F}(\underline{r}) \mid = f(r)$ = constant on a sphere of radius $r = \mid \underline{r} \mid$.

Potential for a central force field:

$U(r) = -\int_{r_0}^{r} dr' f(r') + U(r_0)$ (Equation 1)

with $U(r_0)$ = constant. (Often chosen such that $U(r_0)=0$)

Let us calculate the force field:

$\underline{F}(\underline{r}) = -\underline{\nabla}_r U(r)$

$\underline{F}(\underline{r}) = \left.(\underline{\nabla}r)f(r')\right|_{r'=r} = f(r)\underline{\nabla}r$ (Equation 2)

2. Feb 6, 2014

BvU

Wiki article
Has something to do with the fact that a force that only depends on $|\vec r|$ exposes a certain symmetry. Symmetries lead to conservation laws (very basic in all, even very, very advanced, physics!)
Moving around on the sphere mentioned doesn't cost any work and the work going to another sphere with a different radius is recovered when returning to the original sphere, whatever the path.
You probably already are familiar with the earth gravitational field (potential energy = mgh, force = -mg), this is an extension. If you want to learn, work out ∇ x $\vec F$ for such an F(|$\vec r$|). Make it easy on yourself and use polar coordinates.

Equation 2: Check by writing out ∇ in polar coordinates.

Last edited: Feb 6, 2014
3. Feb 6, 2014

Rococo

In polar coordinates: $∇=\hat{r}\frac{∂}{∂r} +\frac{1}{r}\hat{θ}\frac{∂}{∂θ}$

In polar coordinates: $\vec F=F_r\hat{r} + F_θ\hat{θ}$

Hence: ∇ x $\vec F$ = $\frac{∂}{∂r}F_θ - \frac{1}{r}\frac{∂}{∂θ}F_r$

Now,

$F_θ=0$ (see text)

$\frac{∂}{∂θ}F_r=0$ (see text)

Therefore: ∇ x $\vec F$ = $0$

We have: $∇=\hat{r}\frac{∂}{∂r} +\frac{1}{r}\hat{θ}\frac{∂}{∂θ}$

Hence : $∇r=\hat{r}\frac{∂r}{∂r} +\frac{1}{r}\hat{θ}\frac{∂r}{∂θ}$

Now,

$\frac{∂r}{∂θ}=0$ (from the geometry)

$\frac{∂r}{∂r}=1$

Therefore: $∇r=\hat{r}$

Substituting this into the given equation: $\underline{F}(\underline{r}) = f(r) \hat{\underline{r}}$

We obtain: $\underline{F}(\underline{r}) = f(r)∇r$, which is Equation 2.

However I still do not understand how the text has arrived at Equation 1. That is, how this equation is obtained: $U(r) = -\int_{r_0}^{r} dr' f(r') + U(r_0)$

In addition, my lecturer arrived at Equation 2 directly after consideration of Equation 1. The following fact was used to help us understand:

$\frac{d}{dx}\int_{x_0}^{x}dx'h(x') = h(x)$

I understand this, however my difficulty is in seeing how it can be used to obtain Equation 2 from Equation 1.

4. Feb 6, 2014

BvU

Ok, I understand the ∇r now.

Remember how in work was $W=\int \vec F \cdot \vec {ds}$ and in 1 dimension it led us to potential energy in the gravity field ? This is comparable. The force is conservative, so it is convenient to introduce the potential, because it is conserved (consequence of the symmetry).

Once U is defined, the steps to eq 2) are just a check, actually. It says $\vec F (\vec r) = \vec F (\vec r)$.

5. Feb 6, 2014

Rococo

We have, making the comparison, an expression for the work done in moving to a distance $r$ from 0, from a point $r_0$ from O.

$W=\int_{r_0}^{r}F(r')dr'$

Using the equation $W=-ΔU$:

$-ΔU = \int_{r_0}^{r}F(r')dr'$

$-[U(r)-U(r_0)] = \int_{r_0}^{r}F(r')dr'$

$[U(r)-U(r_0)] = -\int_{r_0}^{r}F(r')dr'$

$U(r) = -\int_{r_0}^{r}F(r')dr' + U(r_0)$

However this is not the same as Equation 1, because Equation 1 contains $f(r')$ instead of $F(r')$. If we replace $F(r)$ with $f(r')\hat{r}$, we obtain:

$U(r) = - \int_{r_0}^{r}f(r')\hat{r}dr' + U(r_0)$.

This is different from Equation 1 because it contains $\hat{r}$. So, I still am having difficulty in seeing how Equation 1 was arrived at in the text.

6. Feb 6, 2014

BvU

In $W=\int_{r_0}^{r}F(r')dr'$ there is an inner product you forget (the only component of the force that does work is the component along the trajectory): $\vec F(\vec r) \cdot d\vec r = f(r) \hat r \cdot d\vec r = f(r) \hat r \cdot \hat r dr = f(r) dr$

7. Feb 6, 2014

Rococo

Of course. So Equation 1 is obtained as follows:

$W=\int_{r_0}^{r}\underline{F}(\underline{r'}) \cdot d\underline{r'}$

$W=\int_{r_0}^{r}f(r')\hat{r} \cdot dr'\hat{r}$

$W=\int_{r_0}^{r}f(r')dr'$

$-ΔU=\int_{r_0}^{r}f(r')dr'$

$ΔU=-\int_{r_0}^{r}f(r')dr'$

$U(r)-U(r_0)=-\int_{r_0}^{r}f(r')dr'$

$U(r)=-\int_{r_0}^{r}f(r')dr' + U(r_0)$

$U(r)=-\int_{r_0}^{r}dr'f(r') + U(r_0)$

which is Equation 1 as given in the text.

However my main difficulty is how Equation 2 is obtained. That is, my lecturer has directly used Equation 1 along with the fact that $\underline{F}(\underline{r}) = -\underline{\nabla}_r U(r)$ to arrive at the the equation:

$\underline{F}(\underline{r}) = \left.(\underline{\nabla}r)f(r')\right|_{r'=r} = f(r)\underline{\nabla}r$

I am having difficulty seeing where this has come from, although the following equation was used as an explanation:

$\frac{d}{dx}\int_{x_0}^{x}dx'h(x') = h(x)$

I understand this equation, but cannot see how it has been used.

Last edited: Feb 6, 2014
8. Feb 6, 2014

BvU

So $\vec F(r)$ is minus the gradient of $U(r)$ Consider the gradiënt in polar coordinates. Only derivative left is $\frac {d}{dr}$ and it points in the $\hat r$ direction. For $U(r)$ you have an expression: an integral with upper limit r. Your last line in post #7 tells you that $f(r)$ is that derivative!

9. Feb 6, 2014

Rococo

So, we have:

$\underline{F}(\underline{r}) = -\underline{\nabla}_{r}U(r)$

$\underline{F}(\underline{r}) = -\underline{\nabla}_{r}[-\int_{r_0}^{r}dr'f(r')]$

$\underline{F}(\underline{r}) = f(r)$

However this is not Equation 2, because Equation 2 is:

$\underline{F}(\underline{r}) = f(r)\underline{\nabla}r$

Where have I gone wrong in my working out above?

10. Feb 6, 2014

BvU

Consider the gradiënt in polar coordinates. Only derivative left is d/dr and it points in the $\hat r$ direction. That means that your third line misses an $\hat r$ . Must be: on the left there is a vector and on the right you have a scalar. and $\hat r = \vec\nabla r$

11. Feb 6, 2014

Rococo

I see, so we have:

$\underline{F}(\underline{r}) = -\underline{\nabla}_{r}U(r)$

$\underline{F}(\underline{r}) = -\underline{\nabla}_{r}[-\int_{r_0}^{r}dr'f(r')]$

$\underline{F}(\underline{r}) = f(r)\underline{\hat{r}}$

However this is still not Equation 2. At this point in the lecture, it had not yet been shown that $\underline{\nabla}r=\hat{r}$.

In fact, that was only proved after Equation 2 had been obtained. I have attached a picture of the lecturer's notes, showing how he first shows that $\underline{F}(\underline{r})=f(r)\underline{\nabla}r$, and then he proves that $\underline{\nabla}r=\hat{r}$, and he then goes on to finally show the equation that was stated at the very beginning, $\underline{F}(\underline{r}) = f(r) \hat{\underline{r}}$

I would like to follow his steps explicitly, but I am having difficultly in seeing how he obtained Equation 2 directly from Equation 1, as shown in the notes.

Last edited: Feb 6, 2014
12. Feb 7, 2014

BvU

Agree: F = minus the gradient of U.

yes, U(r0) is a constant. I understand that the next step is the one you don't like:

because it does something you didn't know about:

and equation 2 was:

Correct ?

In (B), with which you still agree, you see that $\vec F$ is the gradient of something that depends on $r = |\vec r|$ only.

You apparently don't want to run ahead and use the gradient in spherical coordinates expression, but you were already handed something for the derivative of an integral:

At this point, I don't know what to add unless you can specify a more precise question. From the (later) lecture notes in the picture I gather you are used to gradients in xyz coordinates. Working around a gradient from $xyz$ to $r\theta\phi$ is what the picture does, so no point in repeating.

In case your problem begins a little earlier, namely where $\vec F = -\vec\nabla U$ is written as $\vec F(\vec r) = -\vec\nabla_{\vec r} U(|\vec r|)$ maybe the following helps a little: We've seen that both F and U depend on |r| only. From F (a vector) we've constructed a U (a scalar) by integrating. Now we are going back by taking a gradient, which is differentiating, but in a number of dimensions corresponding to the number of coordinates.

Since there is no $\theta$ or $\phi$ dependence in U, it is clear that differentiating wrt those gives zeros: there is no way to get a $\theta$ or $\phi$ component of $\vec F$, whatever the multiplying factors (like $\sin\theta$ or r or whatever).

The only change in U occurs when r changes. Therefore the only direction F can point is in the r direction. (perpendicular to the spheres with constant r, if you want).

Perhaps the idea of the chain rule being applied here makes it acceptable for you ?
U is a function of $r = |\vec r|$ only. A gradient is kind of a derivative wrt a vector (artistic liberty, translated as physicists liberty to use shorthand for mathematical operations...)

You have $\vec F = -\vec\nabla U = \frac {dU}{dr} \ \vec\nabla {r} =\frac {dU}{dr} \ \frac {dr}{d\vec r}$

If I confuse you or talk nonsense, I'd like to be corrected.

Last edited: Feb 7, 2014
13. Feb 7, 2014

Rococo

Yes, I think my lecturer has used the chain rule without stating it explicitly, which is the source of my confusion.

Using the last line in post #12, $\vec F = -\vec\nabla U = \frac {dU}{dr} \ \vec\nabla {r} =\frac {dU}{dr} \ \frac {dr}{d\vec r}$

We have:

$\underline{F}(\underline{r})=-\underline{\nabla}_\underline{r}U(r)$

$\underline{F}(\underline{r})=-\frac{dU}{d\underline{r}}$

$\underline{F}(\underline{r})=-\frac{dr}{d\underline{r}}\frac{dU}{dr}$

$\underline{F}(\underline{r})=-(\underline{\nabla}r)(-\left.f(r')\right|_{r'=r})$

$\underline{F}(\underline{r}) = \left.(\underline{\nabla}r)f(r')\right|_{r'=r}$

(Note that this is the exact equation that appears directly before Equation 2, so I think this is probably how my lecturer has done it.)

$\underline{F}(\underline{r})=\underline{\nabla}rf(r)$

$\underline{F}(\underline{r})=f(r)\underline{\nabla}r$

Which is Equation 2.

However, I don't understand the equation in the last line of post #12. Maybe it is the notation or the gradient function itself that I am having difficulty understanding. Specifically, that equation implies:

$\underline{\nabla}r=\frac{dr}{d\underline{r}}=\frac{d}{d\underline{r}}r$

I don't understand the meaning of this, i.e. how it can be that the vector $\underline{r}$ is written on the bottom of the derivative, and the magnitude of the vector is written on the top, taking into consideration $\underline{r}=r\hat{r}$.

For example, if we had $f=2x$, then $\underline{\nabla}f=\frac{df}{dx}i=2i$. That is, the variable $x$ appears on the bottom of the derivative, and the answer is the derivative of the function with respect to x, multiplied by the unit vector $i$ in the x direction. That is my understanding of the gradient function and how it is denoted.

So, I don't understand the equation $\underline{\nabla}r=\frac{dr}{d\underline{r}}=\frac{d}{d\underline{r}}r$. The symbol $r$ is just a number and does not have components in the directions of any axes, unlike f. The symbol $\underline{r}$ is the position vector of a point, not a co-ordinate, like x. So, I'm unable to draw any parallels with how I have seen the gradient function used before.

Sorry if this all sounds a bit confusing!

Last edited: Feb 7, 2014
14. Feb 7, 2014

BvU

On top is easiest: U is a function of |r| so that's from the chain rule.
As denominator ${d\over d\vec r}$ is shorthand for $\vec\nabla$. Don't show this to your math teacher. Even the physics teacher might frown on it.

$\vec\nabla \cdot h$ is the gradiënt vector of the scalar function h. In cartesian 3d it stands for the vector with components $({dh\over dx}, {dh\over dy}, {dh\over dz})$. You can work it over to cylindrical or spherical coordinates.