# Question about potential functions from conservative vector fields

1. Mar 30, 2013

### CAF123

1. The problem statement, all variables and given/known data
1) Show that $\underline{a} = \underline{r} f(r)$ is conservative and deduce a functional form for the potential if $f(r) = r^n$. For what value of n does the potential diverge at both $\underline{r_o} = 0$ and $\infty$?

3. The attempt at a solution

I have found the potential function but I appear to be out by a constant. f is conservative so take a straight line as the path over integration. So, $$\int_0^{\underline{r}} \underline{a}(\underline{r'}) \cdot d \underline{r'} = \int_0^1 \underline{a}(\lambda \underline{r}) \cdot d(\lambda \underline{r})$$

Inputting $f(r) = r^n$ gives the potential function as $\frac{1}{2}r^{n+2}$.

When I take the gradient of this, I get $f(r) \underline{r} (\frac{n}{2} +1)$. This factor of $(\frac{n}{2} + 1)$ is nonexistant in the expression for $\underline{a}$, so how should I interpret it?

I don't really understand the last part of the question. I would assume that a potential diverging is when it tends to $\infty$, although this was never defined. When $r_o = 0,$ I have that $$\int_0^1 r^{n+2} \lambda d \lambda = \frac{1}{2}r^{n+2}$$ while for $r_o = \infty$, I get$$r^{n+2} [ \frac{1}{2} - \operatorname{lim}_{t \rightarrow \infty} \frac{t^2}{2}]$$ The second term tends to ∞. I don't think this helps though.

Many thanks.

2. Mar 30, 2013

### voko

What does the underscore mean? That the variable is a vector? The expression for the force should have the unit-vector of displacement outside f(r), not just the displacement. To show that the force is conservative, you then integrate its work between some arbitrary $\vec{r}_i$ and $\vec{r}_f$, it should depend only the end positions, not the path.

Alternatively, you could integrate along a loop and apply the curl theorem.

3. Mar 30, 2013

### CAF123

Hi voko,
Yes, the initial position of the particle is at $\underline{r_o}$ and the final position is described by $\underline{r}$. It is the notation we are using in this class - I didn't use this in another class.

Yes, I showed that $\nabla \times \underline{a}= \underline{0}$, hence conservative. Then I integrated over a straight line between two points. I took the initial point to be at some $r_o = 0$ and the latter point some $r$. I then used a parametrisation r' = λr to obtain the potential. I am aware of other methods to find the potential.

It says in my notes that what I described above (in particular taking $r_o$ = 0) is only valid if the potential is finite or zero at r =0. In the case it was not, if the potential becomes finite or zero valued as r → ∞, then I set $r_o =∞$. This makes sense I think - an application would be the gravitational potential - bring a mass from infinity to some point r. The gravitational potential tends to infinity (I.e not single valued) at r=0, so we integrate from ∞ to r.

4. Mar 30, 2013

### voko

I do not understand how you get the 1/2 factor.

Is n restricted to be integer? Can it zero? Negative?

5. Mar 30, 2013

### CAF123

Let $f(r) = r^n$. Then $$\int_0^{\underline{r}} \underline{a}(r') \cdot d\underline{r'} = \int_0^1 \underline{a}(\lambda r) \cdot d \lambda \underline{r} = \int_0^1 \lambda \underline{r} r^n \cdot \underline{r} d\lambda = \int_0^1 r^{n+2} \lambda d\lambda = \frac{1}{2}r^{n+2}$$

It doesn't say on the sheet. I presume just an integer, then?

6. Mar 30, 2013

### voko

First of all, that should be $$\int_{\vec{r}}^{\vec{r_0}} \vec{a}(\vec{r'}) \cdot d\vec{r'}$$ where $\vec{r_0}$ is some arbitrary point where the potential is set to zero. You take that to be zero, which is OK but pay attention to the limits.

Secondly, your integrand should end up, ignoring consts, being $\lambda^{n + 1} d\lambda$, which should integrate to $\lambda^{n + 2} / (n + 2)$ - except the special case when n + 2 happens to be zero.

7. Mar 30, 2013

### CAF123

The parametrisation I used was, for a line, $\underline{r'} = \lambda \underline{r}$. At $\underline{r'} = 0 , \lambda = 0,$ while for $\underline{r'} = \underline{r} \Rightarrow \lambda = 1$. So my limits are 0 and 1. Did I make an error here or elsewhere?

Your result seems to make more sense with regard to answering the second part of the question, but I am still unsure where I made my error.

8. Mar 30, 2013

### voko

You limits are reversed. It has be from r to zero, not from zero to r.

I think you forgot to expand $f(\lambda r)$ properly.

9. Mar 30, 2013

### CAF123

Could you explain why this is the case? (In my notes, I have the limits from 0 to r)

Yes. r is scaled to λr.

10. Mar 30, 2013

### voko

Let's take a one-dimensional case for simplicity. $\frac {dU} {dr} = -F$. Then $$U(r) - U(r_0) = U(r) = - \int_{r_0}^r F dr = \int_{r}^{r_0} F dr$$.

11. Mar 30, 2013

### CAF123

This also makes sense. I think the issue could be that in physics, the convention is to write a vector field as the negative of the gradient potential (it makes the physics easier to visualize or conceptualize) while in mathematics, they omit the negative sign.

Incidentally, the next chapter of my notes then went onto discuss these different conventions. Do you think this is the reason for the limits being from 0 to r?

12. Mar 30, 2013

### voko

Well, this is a possibility. It does not change much in the second part - have you figured it out?

13. Mar 30, 2013

### CAF123

So I get the potential function to be $r^{n+2}$. (Although when I take the gradient of this, I get $(n+2) f(r) \underline{r} \neq f(r) \underline{r} = \underline{a}$ unless n = -1.) EDIT I see my error here - the potential should be $r^{n+2}/(n+2)$

For the last part of the question, do they just want me to say that if n = -2, then the potential tends to infinity (or diverges)?

EDIT I can add more reasoning: Let $r_o = \infty$. Then $$r^{n+2} \int_{\infty}^1 \lambda^{n+1} d \lambda = \frac{r^{n+2}}{n+2} - r^{n+2} \operatorname{lim}_{t \rightarrow \infty} \frac{t^{n+2}}{n+2}$$ The second term tends to $\infty$ for $n \geq -2$. When $r_o = 0$, the potetial diverged when n = -2 exactly. So the value of n I take is -2, right?

Last edited: Mar 30, 2013
14. Mar 30, 2013

### voko

Not quite. When $n = -2$, $\int \lambda^{n + 1} d\lambda = \int \lambda^{-1} d\lambda$ is not $\lambda^{n + 2}/(n + 2) + C$ - but what is it?

15. Mar 31, 2013

### CAF123

If n=-2, the integral evaluates to $\ln(\lambda) +C$

So, $r^{n+2}[\ln(1) - \ln(0)]$diverges when r_o =0 and $r^{n+2} [\ln(1) - '\ln(\infty)']$, when r_o = ∞, which also diverges. Is this the reasoning they wanted me to show?

Last edited: Mar 31, 2013
16. Mar 31, 2013

### micromass

Some remark on notation. You denote vectors by an underscore, so $\underline{r}$. This is the notation that people usually use on the blackboard, but it isn't standard notation to use in typed text. If you want to denote vectors in printed text, then people usually use bolded text such as $\mathbf{r}$.

17. Mar 31, 2013

### CAF123

I quite like denoting vectors with an underscore. I feel writing them like $\vec{a}$ can make the text get quite cluttered sometimes. I know of the bold face notation, but never used it. Given your advice, I'll try this out. Can I ask why the underline notation is not used in typed text?
Thanks,

18. Apr 1, 2013

### voko

Back before we had computer-aided typesetting, math/phys manuscripts sent for typesetting had all formulae written, well, manually. Naturally, that could not have anything in bold (unless the author was an artist in addition to the scientist), so the convention was that anything that was to be typeset as bold had to be underlined. The underscore was really a poor man's substitute for bold, and that was widely understood and used. These days I feel this is entirely outdated, even on the blackboard.

19. Apr 1, 2013

### CAF123

Is the above more explicit reasoning for why n=-2 causes the potential to diverge?

I have another quick question along the same lines: Find a potential function corresponding to $$\mathbf{a} = r \mathbf{c} + (\mathbf{c} \cdot \mathbf{r}) \mathbf{r}/r$$ By the same parametrisation as above, I have: $$\int_0^1 (\lambda r \mathbf{c} + (\mathbf{c} \cdot \lambda \mathbf{r}) \lambda \mathbf{r}/r ) \cdot \mathbf{r} d \lambda = 2 r (\mathbf{c} \cdot \mathbf{r})$$

However, when I take the gradient of this, it appears the 2 should not be there so as to recover the vector field $\mathbf{a}$

20. Apr 1, 2013

### voko

Your $/r$ term should really be $/(\lambda r)$.