Is there really no meaning in potential energy and potential?

Question

I have been told all my physics life that potential energy between two mass/charge has no meaning and only their difference has meaning. The same goes for electric potential, only the difference matter.

Perhaps I am not understanding it correctly, but before I talk about masses, let's talk about potential energy/potential associated with two charges.

I am not sure where had I seen it, but a long time ago I was presented with a problem like this. Let's say I have a +Q and a -Q. What is the potential energy between them?

The change in potential energy is the negative work done by the conservative force namely (vector sign and dot product got rid of, since the cosine 1)

$\Delta U = -\int_{a}^{b} k(Q)(-Q) \frac{1}{r^2} = -k(Q)(-Q) \left. \frac{-1}{r} \right |_{a}^{b} = -k(Q)(-Q) \left (\frac{1}{b} - \frac{1}{a} \right) $

Now I assume that I brought it from infinitely far, so that 1/a = 0

In that case I am left with $\delta U = \frac{kQQ}{b}$

Here are my questions

1) A long time ago, I saw a formula looking EXACTLY like what I just did there, but it doesn't concern with the change, it's just gives me the potential energy. Now the formula I remember was $U = -\frac{kQq}{r}$ where there is a negative. I thought this formula already takes care of the signs? Or am I wrong?

2) Kinda the same concept. If I tell you some charge (not telling you the sign) has a electric field and you have another charge (which is the test particle, not telling you the sign again even though it is conventional to use + charge ) somewhere in that field. I tell you that the electric potential at that point (not the difference) is K (where K is positive number). What can you conclude, if anything? What if it were negative?

Suppose I tell you suddenly that the charges are the same signs, and I give you a location in which the electric potential is positive (I think it has to be). What does it mean?

EDIT: Let me also just clarify a bit that I was taught that electric potential (not difference, I stress again) is the work that someone does to bring a charge from infinity to some point whereever. In my book however, it's defined as $V = \int_{R}^{\infty} \vec{E} \cdot \vec{ds}$

Answer 1

It is indeed correct that only the difference between two potential energies is physically meaningful. An in-depth explanation follows. For the rest of this answer, forget everything you know about potential energy.

I suppose you know that when you have a conservative force $\vec{F}$ acting on an object to move it from an initial point $\vec{x}_i$ to a final point $\vec{x}_f$, the integral $\int_{\vec{x}_i}^{\vec{x}_f}\vec{F}\cdot\mathrm{d}\vec{s}$ depends only on the endpoints $\vec{x}_i$ and $\vec{x}_f$, not on the path. So imagine doing this procedure:

1. Pick some particular starting point $\vec{x}_0$

2. Define a function $U(\vec{x})$ for any point $\vec{x}$ by the equation

   $$U(\vec{x}) \equiv -\int_{\vec{x}_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$

This function $U(\vec{x})$ is the definition of the potential energy - relative to $\vec{x}_0$. It's very important to remember that the potential energy function $U$ depends on that starting point $\vec{x}_0$.

Note that the potential energy function necessarily satisfies $U(\vec{x}_0) = 0$. So you can write

$$U(\vec{x}) - U(\vec{x}_0) = -\int_{\vec{x}_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$

Now why would you do that? Well, suppose you choose a different starting point, say $\vec{x}'_0$, and define a different potential energy function

$$U'(\vec{x}) \equiv -\int_{\vec{x}'_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$

(Here I'm using the prime to indicate the different choice of reference point.) Just like the original potential energy function, this one is equal to zero at the starting point, $U'(\vec{x}'_0) = 0$. So you can also write this one as a difference,

$$U'(\vec{x}) - U'(\vec{x}'_0) = -\int_{\vec{x}'_0}^{\vec{x}} \vec{F}\cdot\mathrm{d}\vec{s}$$

The neat thing about this definition is that even though the potential energy itself depends on the starting point,

$$U(\vec{x}) \neq U'(\vec{x})$$

the difference does not:

$$U(\vec{x}_1) - U(\vec{x}_2) = U'(\vec{x}_1) - U'(\vec{x}_2)$$

Check this yourself by plugging in the integrals. You'll notice that anything depending on the starting point cancels out; it's completely irrelevant.

This is good because the choice of the starting point is not physically meaningful. There's no particular reason to choose one point over another as the starting point, just as if you're on a hilly landscape, there's no particular reason to choose any one level to be zero height. And that's why potential energy itself is not physically meaningful; only the difference is.

Now, there is a convention in (very) common use in physics which says that when possible, unless specified otherwise, the starting point is at infinity. This allows you to get away without saying "difference of potential energy" and explicitly defining a starting point every time. So when you see some formula for potential energy, like

$$U(\vec{r}) = -\frac{k q_1 q_2}{r}$$

unless specified otherwise it is actually a difference in potential energy relative to infinity. That is, you should read it like this:

$$U(\vec{r}) - U(\infty) = -\frac{k q_1 q_2}{r}$$

Note that the function $\frac{k q_1 q_2}{r}$ goes to zero as $r\to\infty$. That's not a coincidence. It was chosen that way to ensure that $U(\infty) = 0$, so that you could insert it the same way I inserted $U(\vec{x}_0)$ in the calculations above. (This is just another way of saying it was chosen to make the $\frac{1}{a}$ term in the integral you did go away, so you don't have to write it.)

There are some situations in which you can't choose the reference point to be at infinity. For example, a point charge with an infinite charged wire has an electrical potential energy of

$$U = -2kq\lambda\ln\frac{r}{r_0}$$

where $r$ is the distance between the point charge and the wire. This potential energy function decreases without bound as you go to infinite distance ($r\to\infty$), it doesn't converge to zero, so you can't use infinity as your starting point. Instead you have to pick some point at a finite distance from the wire to be your starting point. The distance of that point from the wire goes into that formula in place of $r_0$.

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By the way, electrical potential (not potential energy) is something a little different: it's just the potential energy per unit charge of the test particle. For a given test particle, it's proportional to electrical potential energy. So everything I've said about applies equally well to electrical potential.

Answer 2

The potential energy between two charges is $$ E_{\rm pot} = \frac{Q_1Q_2}{4\pi\epsilon_0 r} $$ If the charges have the same sign, the energy is positive. If they have the opposite charge, it's negativing (the charges may be bound).

In general, additive shifts $$ E \to E+ \Delta E$$ may often be unobservable, and only energy differences are physically meaningful, except that in many situations, there exists a damn good definition what the level $E=0$ means.

In relativistic theories, $E=0$ is the unique choice that preserves the Lorentz symmetry together with the zero momentum; the absolute additive constant is determined from Lorentz symmetry. In the same way, $\rho=0$ gives flat space in general relativity; other energy densities curve the space.

In electromagnetism, there is a natural additive shift coming from the condition $$E_{r=\infty}=0$$ which means that the potential energy vanishes at infinite separation of the charges. This convention was used above. Concerning your last question, $$V = \int_R^\infty \vec E\cdot \vec{ds}$$ is the work obtained by moving a unit charge from $R$ to infinity, up to the sign, so the two things you are describing are the same and there's no contradiction. In particular, $\vec E$ is the electric field $$ \vec E = - \nabla \phi $$ where $\phi$ is the electrostatic potential and $-\nabla \phi\cdot \vec{ds}$ is nothing else than $-\Delta \phi$ by the definition of the derivatives (and the gradient) for an infinitesimal $\vec{ds}$. At any rate, there's no contradiction in anything you wrote.