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In Berkeley Physics Course volume 1 on Mechanics, on page 144 it says "The law of conservation of energy states that for a system of particles with interactions not explicitly$^1$ dependent on time, the total energy of the system is constant.

$^1$ Consider the system with particles permanently frozen in place: then a force that depends on time is said to depend explicitly on time."

What does it mean in simple freshman level terms for something to be explicitly dependent on time, dependent on time but not explicitly, and not dependent on time at all? The footnote makes no sense to me as an example of anything.

Does this have anything to do with distinguishing conservative from non-conservative forces or is it just a general statement about the validity of energy conservation.

Here are two examples that seem to be answering the question in non freshman level terms. link link

Qmechanic
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trying_to_learn
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2 Answers2

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Think about the energy for a particle of mass $m$ in a uniform gravitational field, $$E=\frac{1}{2}mv^{2}+mgz.$$ This energy is conserved, because it does not depend explicitly on $t$. That essentially just means that $t$ does not appear in the mathematical expression. However, if the mass were changing (if, say, it was being eroded by wind), making the mass a function of $t$—something like $m(t)=m_{0}e^{-t/T}$, for some constants $m_{0}$ and $T$, then the energy $$E(t)=\frac{1}{2}m_{0}e^{t/T}v^{2}+m_{0}e^{t/T}gz$$ is clearly not going to be constant in time. The energy will decay along with the mass of the object.

This should all be fairly straightforward. The confusing bit of terminology appears because, the energy $E$ is defined in terms of other dynamical quantities (in this case, the velocity $v$ and the vertical position $z$). And both of these are going to be functions of time in the usual way: $$v(t)=v_{0}-gt\\ z(t)=z_{0}+v_{0}t-\frac{1}{2}gt^{2}.$$ So you could still consider $E$ as a function of $t$, through its dependence on $v$ and $z$, as in $$E(t)=\frac{1}{2}m\left[v(t)\right]^{2}+mgz(t).$$ What the statement of the book is trying to convey is that when the time dependence of the energy $E$ is only of this form—only implicit, through the time dependences of the position and velocity—then the total energy is conserved. In fact, if you calculate the derivative $dE/dt$ from the above equation, using the multivariable chain rule, you will get an expression that is zero if you plug in the $v(t)$ and $z(t)$ that obey the equations of motion.

Having a usual kinetic term $\frac{1}{2}mv^{2}$ and a potential energy $V(x)$ is one way to get a time-independent (and thus conserved) $E$. However, the statement that if $E$ has no explicit time dependence, then it is conserved, is more somewhat more general.

Buzz
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It’s a bit odd. I’ve never heard it put that way. The author might just be a bit enamored with some application(s) where energy is not conserved and explains the violation in terms of some explicit time dependence. So for starters I wouldn’t worry too much about it. They will probably give you a case that makes it clear.

What I thought of was a dashpot or viscous damping. If we push on a spring, the work is $\int F ~dx$. The rate of going over $dx$ is irrelevant. If it resisted not just displacement but also velocity, $F\neq -kx$ but $F= -kx -vc$ then it would be less work going fast versus going slow. Energy would not be conserved. But that’s really just saying we have friction in my opinion. I’d bet when you discover what they’re talking about, it is some cases of friction energy loss. And it is time dependent. But you’ll see.

That said, conservation of energy means forces are conserved. If we push something one direction, or accelerate or decelerate something, it shouldn’t matter when we do it, or how fast. If things were ever expressed with $t$’s in them, something would be wrong. Nothing about $F=ma$, or $E=\tfrac{1}{2}mv^2$, or $F= k \frac{q_1 q2}{r^2}$, or gravity, or heat flow $dE=dq=r dT$ etc etc, has a “t” term. If the work done by an expanding gas wasn’t $\Delta E= P ~ dA$ but instead depended on how long it took $\Delta E= Pt^2~ \Delta A$, then of course energy wouldn’t be conserved because we could go fast compressing it and slow expanding it. Maybe the “explicit” is because you do see a $t$ in $v=\tfrac{dx}{dt}$, and in $a$, but that doesn’t mess up conservation of energy.

I promise that’s a good enough view for now.

Al Brown
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