Determining the potential energy

[Andreas C]

I have found that even though knowing the potential energy is vital for classical mechanics, most of the times what you know is actually the force, so you have to determine the potential energy based on that. So, here's the issue:

The relationship between the force and the potential energy is this:

$F=-\nabla V(q_{1},q_{2},...,q_{i})$

If I want to know the potential energy, what do I do? I had an idea, but I don't know if it is correct:

If $F=-\nabla V(x,y,z)$, then V(x) is just minus the integral of F over x by treating y and z as a constant, and so on and so forth, and then if you want to know what V(x,y,z) is, you just add everything together. But then you'd get 3 different constants of integration that I don't know what they are.

Any help would be appreciated.

 

[Jonathan Scott]

In classical mechanics, the potential energy is relative (all that matters is differences) and you can choose any reference level to treat as zero. When there are specific sources, the zero potential is conventionally assumed to be at infinite distance (i.e. what it would be in the absence of all the sources being considered).

 

[Dale]

But then you'd get 3 different constants of integration that I don't know what they are.

The constants of integration don't matter. You can set them to any convenient constant, usually 0.

 

[Andreas C]

The constants of integration don't matter. You can set them to any convenient constant, usually 0.

Sure, but is that method right?

 

[Dale]

Sure, but is that method right?

Not in general, but it does work in the case of spherical symmetry or other situations where you have only 1 coordinate to integrate over. Otherwise integration isn't the inverse of partial differentiation.

 

[vanhees71]

The potential, if it exists, is given by
$$V(\vec{x})=-\int_{C(\vec{x}_0,\vec{x})} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$
Here $C(\vec{x}_0,\vec{x})$ denotes an arbitrary path connecting some arbitrary fixed point $\vec{x}_0$ with the variable point $\vec{x}$.

The potential exists, if $\vec{\nabla} \times \vec{F}=0$ for simply connected regions of space. Then the integral is independent of the choice of the path connecting $\vec{x}_0$ and $\vec{x}$ within this simply connected region.

 

[Andreas C]

The potential, if it exists, is given by
$$V(\vec{x})=-\int_{C(\vec{x}_0,\vec{x})} \mathrm{d} \vec{x}' \cdot \vec{F}(\vec{x}').$$
Here $C(\vec{x}_0,\vec{x})$ denotes an arbitrary path connecting some arbitrary fixed point $\vec{x}_0$ with the variable point $\vec{x}$.

The potential exists, if $\vec{\nabla} \times \vec{F}=0$ for simply connected regions of space. Then the integral is independent of the choice of the path connecting $\vec{x}_0$ and $\vec{x}$ within this simply connected region.

Hmm, could you offer an example? Say Fx=3x, Fy=4y, the body starts from point (X,Y), and ends up in (0,0). What's the potential energy?

 

[vanhees71]

That doesn't make sense. The potential is given by the formula I've given not from what you suggest. So your force (in 2D obviously) is
$$\vec{F}=a x \vec{e}_x + b y \vec{e}_y.$$
Obviously that force is defined in the entire plane and analytic there. You have also obviously $\vec{\nabla} \times \vec{F}=0$. So it's a conservative force, and we can use my formula (although here it's much simpler to solve the partial differential equations defining the potential, but let's do my formula).

I can choose any path I like and any fixed starting point of the path I like. So let's take $\vec{x}_0=(0,0)$ and as a path the straight line to $\vec{x}=(x,y)$. To evaluate the line integrals you need a parametrization of the path, which is very simple here
$$\vec{x}'(\lambda)=\lambda \vec{x}, \quad \lambda \in [0,1].$$
The integral thus reads (because of $\mathrm{d} \vec{x}'=\mathrm{d} \lambda \vec{x}$)
$$V(\vec{x})=-\int_0^1 \mathrm{d} \lambda \vec{x} \cdot \vec{F}(\vec{x}') = - \int_0^1 \mathrm{d} \lambda \lambda (a x^2+b y^2)=-\frac{1}{2} (a x^2+b y^2).$$
Now let's check, whether this is correct by evaluating the gradient. Indeed, we have
$$-\vec{\nabla} V=-\vec{e}_x \partial_x V-\vec{e}_y \partial_y V=a x \vec{e}_x+b y \vec{e}_y.$$
Obviously you have an anistropic harmonic oscillator in the plane (already in normal coordinates).

 

[Andreas C]

That doesn't make sense. The potential is given by the formula I've given not from what you suggest. So your force (in 2D obviously) is
$$\vec{F}=a x \vec{e}_x + b y \vec{e}_y.$$
Obviously that force is defined in the entire plane and analytic there. You have also obviously $\vec{\nabla} \times \vec{F}=0$. So it's a conservative force, and we can use my formula (although here it's much simpler to solve the partial differential equations defining the potential, but let's do my formula).

I can choose any path I like and any fixed starting point of the path I like. So let's take $\vec{x}_0=(0,0)$ and as a path the straight line to $\vec{x}=(x,y)$. To evaluate the line integrals you need a parametrization of the path, which is very simple here
$$\vec{x}'(\lambda)=\lambda \vec{x}, \quad \lambda \in [0,1].$$
The integral thus reads (because of $\mathrm{d} \vec{x}'=\mathrm{d} \lambda \vec{x}$)
$$V(\vec{x})=-\int_0^1 \mathrm{d} \lambda \vec{x} \cdot \vec{F}(\vec{x}') = - \int_0^1 \mathrm{d} \lambda \lambda (a x^2+b y^2)=-\frac{1}{2} (a x^2+b y^2).$$
Now let's check, whether this is correct by evaluating the gradient. Indeed, we have
$$-\vec{\nabla} V=-\vec{e}_x \partial_x V-\vec{e}_y \partial_y V=a x \vec{e}_x+b y \vec{e}_y.$$
Obviously you have an anistropic harmonic oscillator in the plane (already in normal coordinates).

Oh no, I didn't claim the potential is given by a different formula, I just asked for an example with numbers. So for that force (Fx=3x, Fy=4y) and for that path, the potential is -(3x^2+4y^2)?

 

[vanhees71]

Although your force doesn't make sense, because it has the wrong dimensions, of course you can plug in numbers for $a$ and $b$. I guess what you wanted to write is $a=3 \text{N}{\text{m}}$, $b=4 \text{N}/\text{m}$. Then, as I've shown in my previous posting $V=-(3x^2+4y^2)/2 \text{N}/\text{m}$. Then of course, you have an reversed harmonic oscillator, i.e., a repelling force.

 

[Andreas C]

Although your force doesn't make sense, because it has the wrong dimensions, of course you can plug in numbers for $a$ and $b$. I guess what you wanted to write is $a=3 \text{N}{\text{m}}$, $b=4 \text{N}/\text{m}$. Then, as I've shown in my previous posting $V=-(3x^2+4y^2)/2 \text{N}/\text{m}$. Then of course, you have an reversed harmonic oscillator, i.e., a repelling force.

Ooooh that's what you meant! Ok, yeah, I know, the dimensions were a bit wrong, but what I meant was that the magnitude of the x component of the force is proportional to 3 times the x distance from the origin. I just ignored the units.