## Conservation of Energy

a continuation of conservation of momentum

note: minor edits to fix sign errors and whatnot 2/23/09

Newton’s second law:

$F = m\frac{dv}{dt}$.

$F - m\frac{dv}{dt} = 0$.

I would like to imagine a new rule about force, which is that as far as any one particle is concerned, force is a function of position alone (not time, except insofar as position changes with time, and not any of the time derivatives of position, and for good measure, not anything else your sick little mind can conceive).

$F(x) - m\frac{dv}{dt} = 0$

I would like to integrate over time to obtain a conservation law, but that would only tell me about the change in momentum. I can make things more interesting if I first multiply everything by velocity.

$F(x)v - m v \frac{dv}{dt} = F(x)v - \frac{1}{2}m \frac{d(v^2)}{dt} = 0$.

Integrating, and selectively choosing to turn one of those $v$‘s into a $dx/dt$, we get

$\int_{t_a}^{t_b}dt \left(F(x)\frac{dx}{dt} - \frac{1}{2}m\frac{d(v^2)}{dt}\right) = 0$.

And so

$-\frac{1}{2}m\left(v(t_b)^2 - v(t_a)^2\right) + \int_{x_a}^{x_b}dx F(x) = 0$

for arbitrary times $t_a, t_b$, with $x_a, x_b$ the locations of the particle at times $t_a, t_b$. The integral of force is over the path followed by the particle between the start and end times.

Defining a function

$U(x) = -\int_{x_0}^{x}dx F(x)$

called the “potential”, we have a new conserved quantity:

$-\frac{1}{2}mv^2 - U(x)$.

We generally multiply this by $-1$ to say instead that

$\frac{1}{2}mv^2 + U(x)$

is conserved. (They’re both conserved, of course. It’s pure convention.)

However, the value of $U(x)$ will vary depending on where we choose $x_0$. This will change the value of the conserved quantity, but will not change the fact that it’s conserved. The new conserved quantity is called “energy”.

Like momentum, it need not be conserved. Again, faulty measurement, problems with Newton’s laws themselves, or variable mass could all be to blame. The existence of extra particles cannot be to blame, because our derivation this time didn’t care about extra particles one way or the other. What it did need was for force to be a function of position alone.

If there are other particles around, they may exert some force on the test particle we’re interested in. That’s fine, but if the other particles start moving, the force they exert will probably change, even if the test particle itself is stationary. That’s anathema. In that case, energy will not be conserved (not for this one particle! It may or may not be conserved for all the particles put together, but that’s something else entirely. We haven’t even defined energy for multiple particles.)

Also, energy may not be conserved if force is a function of time, or of velocity. Those break the assumption that $F = F(x)$. Imagine if the gravitational constant lessened with time. Then the potential energy of the brick you’re holding above your head (you are doing that, right?) is decreasing while the velocity remains zero, and the total energy decreases. On the other hand, a friction force is velocity dependent (it senses the direction of velocity, but not its magnitude. That still counts as velocity dependence.) and does not conserve energy.

An interesting case is again the electromagnetic force. In the derivation of the energy conservation law, we integrated the term $F(x)v$ over time. A velocity-dependent force may be written as whatever force would be there for zero velocity, plus a delta to bring the force up to whatever it truly is when the velocity is present.

$F(x,v) = F_s(x) + F_v(x,v)$.

$F_s$ indicates “stationary force” and is defined by

$F_s(x) = F(x,0)$.

$F_v$ indicates “velocity-dependent force” and is defined by

$F_v(x) = F(x,v) - F(x,0)$.

Then

$Fv = F_sv + F_vv$

As it turns out, in electromagnetism, although the Lorentz force is velocity-dependent, its dot product with the velocity is zero, so

$F_vv = 0$

and

$Fv = F_s(x)v$

so the proof still holds. Energy is still conserved when considering the Lorentz force. Nonetheless, a charged particle that accelerates does not conserve energy! An accelerating particle creates an electromagnetic field that is acceleration-dependent. The particle then interacts with the field it created. So in effect, the force is acceleration-dependent. The assumption $F = F(x)$ fails, and energy is not conserved for the particle. (You might try to define energy of the field, and by doing so with sufficient cleverity retain an energy conservation law, but that’s beyond the scope of the current blog post.)

Such considerations of energy and radiation have historical importance. Before people knew about quantum mechanics, they realized that electrons orbiting a nucleus were accelerating charged particles and should not conserve energy. Instead, they should lose all their energy and collapse into the nucleus of the atom on very short time scales. The electrons gave no sign of doing that, and it was weird. End of history lesson.

We’ve seen the perils of force laws that are not purely position-dependent. Even those force laws that are solely position dependent could fail us. While it would always be true that

$\frac{1}{2}mv^2 + \int_{x_a}^{x_b}dx F(x)$

would be conserved, for many functions $F(x)$ it would be impossible to write down a sensible function

$U(x) = -\int_{x_0}^{x}dx' F(x')$.

The reason is that the value of $U(x)$ might depend on just which path we choose to go from $x_a$ to $x_b$. Maybe there is a highway and some back roads that both go between the two places. A particle takes the highway, integrates the force as it goes along, and gets the value for $U(x)$. The next day it takes the back roads, integrates the force, and gets a different value for $U(x)$, meaning that the function is not well defined.

Notice that if the function is well-defined, then the highway and back roads must have the same integral of the force. So if you go from A to B by the highway, and then go backwards from B to A on the back roads, the two integrals must cancel each other out, leaving you with zero total integral for the entire trip. This means that the function $U(x)$ will be well-defined whenever all closed path integrals of $F(x)$ come to zero. When we make those path integrals really tiny we write this as

$\nabla\times F(x) = 0$.

next time: energy of multiple particles