# Conservation of energy and time-varying forces

1. Dec 31, 2014

### Stephen Tashi

Are there elementary mechanics problems that illustrate the phenomena of time varying force fields and energies associated with such fields?

Problems in elementary mechanics are often solved by using the principle of conservation of energy even though "the system" (of masses, pulleys, inclined planes etc.) does not name the earth itself, which supplying the force of gravity. I think the fact that the conservation of energy works in such cases is because the force field of gravity is assumed to be conservative. So, by including the gravitational field in calculating potential energy, the effect of the earth is included in the system.

Can we add a time varying force field to an elementary mechanics scenario and teach anything useful? Would the scenario have to account for the mechanism causing the time-varying force in detail?

2. Dec 31, 2014

### Staff: Mentor

You could do something simple like a simple harmonic oscillator in a time varying potential, or a simple pendulum in a varying gravitational field.

You could also look at Lagrangians involving friction or other dissipative forces.

3. Dec 31, 2014

### Stephen Tashi

Does it simplify matters to have a system where something is oscillating ?

How complex is the energy analysis of a falling body, of given mass $m$ initially at rest a given height $h$ in a gravitational field where $g = k + k cos(\omega t)$ for given $\omega$ and given $k > 0$. Can one compute a total energy in some straightforward way and use it compute the velocity of the mass when it has fallen distance $h$ ?

4. Dec 31, 2014

### Staff: Mentor

There is no conserved energy in such a system.

5. Dec 31, 2014

### Stephen Tashi

But if we include enough other stuff in the definition of "the system", we should be able define a system where total energy is conserved. Otherwise classical physics is in trouble, isn't it? If we describe a mechanism (or even a "black box") that creates the gravitational field, what properties would we have to be given to assign a total energy? Would it have to have some internal supply of potential energy that is expended to create the varying gravitational field?

6. Dec 31, 2014

### Staff: Mentor

In order to have a conserved energy you need to have a Lagrangian which does not depend on time. That is a result of Noether's theorem. Yes, you can include enough in "the system" to make it have such a Lagrangian, but then it is no longer time-dependent as you asked in the OP.

7. Dec 31, 2014

### Stephen Tashi

My goal is to find a simple example of a system that includes a time varying force field as a subsystem. I don't require that the total energy of the whole system be time dependent. In fact, I don't want it do be.

8. Dec 31, 2014

### ShayanJ

Let's analyse this more carefully. We have two kinds of time-dependence of the variables. Implicit and Explicit.(Maybe not the same as when we say time appears implicitly or explicitly in equations.)
Implicit time-dependence means that things change with time, only because as time passes, the system is changing configuration or state. So for example gravitational force between two masses changes only because their relative position changes.
Explicit time-dependence means that things change with time, because there is actually something with the laws themselves that change with time. It maybe given by a black-box or by a well known mechanism.
It seems to me that Stephen Tashi means the second kind of time-dependence. The second kind has two possibilities:
A) The very basic laws of physics depend on time. So e.g. we have $\vec F=m \vec a+\vec S(t)$ and $S(t)$ happens to have a very small value for our time and so we can't observe its effects. Or maybe $\vec \nabla \cdot \vec B=\delta(t)$ and again $\delta(t)$ is very small for our time. This kind of course we don't want to assume because, at least now, we don't assume it to be true.
B) The second possibility is that there is something that we consider as background and its changing. We may then consider the background+the system to be a bigger system and now we have the time-dependent part as a subsystem. If by this procedure, we cover all things that affect the background+older system, then we actually have reduced the Explicit time dependence to an implicit time dependence. But there is a possibility that there still remain some explicit time dependence. Then this can only mean that there was something we couldn't see in the first place but was actually affecting the background+older system. So if we include that too, we again eliminate all explicit time-dependencies and everything depends on time implicitly. So as long as we don't want to consider the possibility A, there is no system that fulfils the requirements given by Stephen.

P.S.
I have an example for the process I explained and I really want to talk about it. Consider a part of spacetime with some masses in it. Now that part of spacetime may change because the masses are moving and also some gravitational wave is coming from some direction. If we define our system to be only the masses, then we're considering the spacetime as background and so we have a changing background and things depend explicitly on time. But we may then consider the masses+that part of spacetime as the bigger system. Now some explicit time-dependency is converted to implicit time-dependency because we know that part of spacetime is changing partly because of the motion of the masses. But now we have gravitational waves coming from outside of this part of spacetime and changing the bigger system and so we again have some explicit time-dependency. Now if we go far enough to reach e.g. a binary system that is sending those gravitational waves and add it to our system, we've converted the remaining part of explicit time-dependency to implicit one.

9. Jan 1, 2015

### Stephen Tashi

One attempt:

Suppose that along the x-axis, there is a spring with one end fixed at point P. This will be a massless spring that obeys Hookes law both on extension and contraction. The the right of P, attached to the other end of the spring , is a large homogeneous spherical mass M of radius R. The mass (and spring) are in steady state harmonic motion. We pick a time at which the surface of M intersects the x-axis farthest from P and call this t =0 and the point of intersection is x = 0. At distance x =h there is a small point mass m. At t = 0, mass m is at rest and has zero velocity. As time progresses m "falls" toward toward the moving surface of M due to gravitational attraction. Does conservation of energy apply for this system? (I'd say yes. Whatever forces act to hold point P in place aren't producing any motion.)

Things like binary stars are said to cause gravitational waves "according to the theory of relativity". Do such configurations cause a varying gravitational field from the classical point of view? I'm looking for a simple classical mechanical example. .

10. Jan 1, 2015

### ShayanJ

This is addressed in my post. If you consider only m to be your system, then it'll seem to you that there is a background which is changing with time and energy is not conserved. But if you consider M and m to be your system, energy is conserved. In fact the Lagrangian for this system is(assuming x to be the coordinate of M and y to be the coordinate of m and h be the equilibrium position of M(when mass m is infinitely far away)w.r.t. the point where the spring is fixed):
$L=\frac 1 2 M\dot x ^2+\frac 1 2 m \dot y^2-\frac 1 2 k (x-h)^2+\frac{GMm}{y-x}$
This doesn't seem to me any different than the Lagrangians we usually see.

Yes. Just consider some masses in a region of space and some other bunch of masses oscillating far from the mentioned region. We still have gravitational potential energy and changes in gravitational field to play with. The only difference is there is no gravitational wave because all changes are felt instantaneously. In fact you gave an example of this.

Last edited: Jan 1, 2015
11. Jan 1, 2015

### Stephen Tashi

That's very good point. A function that varies in space and periodically varies in time need not be a wave.

In classical physics, what kind of force fields require us to consider energy being radiated in order to balance the energy books? Must we do that with any force field that is described by a propagating wave? Or is it other characteristics of fields that cause energy to be radiated?

12. Jan 1, 2015

### ShayanJ

I think all kinds of force fields. But it also depends on what you mean! I think when you define energy and say its conserved, then saying that energy propagates becomes inevitable.
In classical physics, we may have two kinds of force fields: (maybe not standard terminology)
1) Newtonian: Where effects are propagated instantaneously. Newtonian forces(inverse square forces) maybe modelled as action at a distance(AD) or using potential fields with action at an instant(AI).(Its just made up by myself but I think it makes sense). The difference is, in the AD case, we're assuming nothing in between, nothing that carries the force. We just say the particles exert force on each other. This way, the only kind of energy is kinetic energy and its not conserved and there is nothing called radiation of energy. But in the AI case, we define a potential field which carries the effects, but infinitely fast. Now we can define potential energy and the total energy is conserved.(Of course, we use both AD and AI in classical physics, depending on the situation).
2) Special Relativistic: This is one example of theories that imposes a maximum speed for the propagation of the effects.(Not that there are actually other theories, but there can be!). Here, the most appealing(or maybe the only) formulation is using fields. So again we have energy propagating but there is an important difference between here and the AI case in the Newtonian kind. Because energy propagates with finite velocity here, there can be moments of time that some part of energy is in neither of the particles of the system. So if we want to define energy and say its conserved, we have to accept that the field we defined can contain energy too and has, to an extent at least, physical reality.(This is the case with classical electrodynamics.)

13. Jan 1, 2015

### ShayanJ

And about this interesting observation. I remember a time when I was thinking that why people talk about field theories and then sometimes call equations like Dirac equations, wave equations. It led me to the question that what's the difference between a field and a wave. I realized that they're actually the same and so you can't have waves without having fields, its meaningless. Its because a wave is defined as a disturbance in some medium that propagates. Now what is a disturbance? Its a difference between the current value of the field at some point and the equilibrium value of the field at that point. So in fact a wave equation, is just the dynamical equation of a field. But this paragraph only applies when in the special relativistic case of my last post.
When we're doing things in the Newtonian case(the AI case), we have a field, but there is no wave. That's because instantaneously propagated effects can't exist at fields points where there is no particle. There is no moment of time when they're "moving" and "haven't reached" the other particle. So no disturbance moves in the field because the field reacts to any change instantaneously.

14. Jan 1, 2015

### Jano L.

The fact that the mass $M$ is moving makes its gravity force on the mass $m$ time-dependent. Still, if the point P remains at rest, this is a system of bodies that experiences only conservative forces - gravity and Hooke force, so it has constant sum of kinetic and potential energy.

It has to do more with the relativistic character of the theory than with energy conservation really.

In non-relativistic theory (Newtonian gravitational system), energy of isolated system is conserved and can be expressed as a function of positions and velocities of particles. Considering energy as integral of density (function of fields) and moving in space can be done, but is not necessary because force fields are not necessary.

In relativistic theory (electrodynamics) force fields are vital part of the theory that cannot easily be eliminated. The equations are such that they allow energy density and current density to be defined, but there may not be global energy conservation - energy may systematically leave to infinity or fluctuate in time. Global energy conservation is not necessary to have local energy density and energy current density defined.

There are special versions of electrodynamics like Fokker-Tetrode and Feynman-Wheeler theory where special boundary conditions are assumed and fields can be eliminated. Energy can still be thought of as distributed in space, but again global energy conservation is not necessary.

15. Jan 1, 2015

### Staff: Mentor

I would do a pair of coupled harmonic oscillators. As one gets much larger than the other then you can treat it as a time varying field on the other.

16. Jan 1, 2015

### Stephen Tashi

It's still not clear to me (in classical physics) what properties of a force field lead to the phenomenon of energy being radiated. Is the radiation of energy just postulated to match experimental data? - for example, not postulated in a gravitational field, but postulated in some EM fields? Or is radiation a mathematical consequence of some properties of the field.

A field (as far as I know) is defined in terms of a force it exerts on some imaginary "test mass" or "test charge". In the absence of any actual test body at a given point in space, the field does no actual work at that point. Even on a actual body, the field won't do work (in the mechanical sense of work=force times distance) unless the body is free to move.

If a field is realized in some medium that limits the speed of waves of force (or if we assume empty space has properties like a physical medium) then it becomes plausible that field can transmit energy because the attempt to change the field at a given point is being resisted by a type of "inertia" in the medium. The wave succeeds in overcoming this inertia and so its plausible that a force does work to overcome the"inertia" of the medium.

However, I don't know if one can back up this intuition with a mathematical proof that energy is radiated just from the fact that effects in the field are transmitted with a finite speed.

To mathematically prove energy is radiated, must we also provide a model for the medium ? - like small particles moving as a sound wave is transmitted through a medium.

17. Jan 1, 2015

### Jano L.

Radiation of energy (its transport by other means than by moving matter) is not a phenomenon that could be directly observed. What could be observed (I believe) is time delay between a radar emits EM pulse and subsequent detection of scattered EM pulse. Current change in the antenna of the radar is followed by similar changes in a far away object some time later that produce their own disturbance which finally reaches the radar.

We have special relativistic theory for this delayed action - it is based on Maxwell's equations for EM fields and equations of motion for charged particles. These by themselves do not contain any quantity similar to energy - the positions, velocities and the fields determine the state, the equations connect subsequent states so energy is not needed.

We introduce energy as additional useful concept based on the Maxwell equations and the equations of motion, similarly to how energy of a spring is defined based on the work-energy theorem and the equation of motion

$$m\ddot{x} = -kx.$$

However, because Maxwell's equations are partial differential equations as opposed to the above ordinary differential equation, functions of position (fields) are used instead of simple real variables. If energy is still to be defined as a real number dependent on the state of the system, it is quite acceptable to consider the possibility it can be expressed as a spatial integral of some position-dependent density.

This may not be possible in all theories using fields (energy may not be definable), but in case of electromagnetic theory, it is possible to do just that.

Yes, and we still think of a region in which no work is going on as having definite values of fields and is appropriated with definite amount of energy. In special relativistic theory, there is no way to test this experimentally, it is just a useful picture based on math and general idea of energy (work-energy theorem). In general relativistic theory I suppose there should be some influence on the geometry but that is beyond my current scope.

This is appropriate when the field describes material medium, such as field of deviations of elements of a piece of rubber. EM field in vacuo is not like that; it is a concept that quantifies potentiality, a force that would act on a particle if it was there. It is more of a mathematical concept than physical object.

Wave equation by itself implies finite speed and thus describes radiation. It is true that it allows us to define quantities that are conserved in some sense (that mathematicians call energy), but there is infinity of different them and without other ideas it is not possible to decide which one should be called energy.

Energy in physics is the quantity that is connected to work by the work-energy principle - work done on the particles in some region of space equals fields energy loss from in that space plus energy that came through the boundary:

$$dW/dt = -d/dt (field~ energy~ in~ region~ V) ~+~ energy~ coming~in~ through~ the~ boundary~ of~ V~ per~ unit~ time.$$

By manipulating the Maxwell equations and the equations of motion of charged particles, we can express all the energy rates as functions of fields and the coordinates of the region. We then look at the formulation of the work-energy theorem and define quantities that we call energy density and energy current density.

No, it works for charged particles in vacuum. In macroscopic theory of material medium, things are complex, there is no single material equation and no general macroscopic expression for energy. There is one easy and often taught case - linear uniform medium, where one arrives at Poynting expressions.

Last edited: Jan 1, 2015
18. Jan 2, 2015

### Stephen Tashi

I think I understand that. As a simple example, a time varying force field has the potential to do work on a mass at a point in space, so at a given point in space and at given time t, we can define the rate at which it would do work on a unit mass that is free to move. (By dealing with the rate-of-work instead of work, we avoid the complication of the unit mass actually moving to a different point in space as the field varies.) From the rate-of-work on a unit mass we can use integration to define various potential energies.

This raises the interesting question of whether potetial energy has a location, or area or volume. For example, in the typical elementary physics problem, we think of a mass M at had height h of "having" the potential energy Mgh, as if the potential energy has the spatial position of mass M and is a property of the mass. But the potential energy of the mass actually is an effect of the gravitation field exerting the force Mg over the entire distance h. So the potential energy isn't localized in M.

It appears that radiation of potential energy can be defined. However, does this account for the fact that a source that radiates potential energy must use up energy? A primitive try would be to say that radiated potential energy leaves the source. This gets into giving energy a location and also letting it move!

19. Jan 2, 2015

### Jano L.

In non-relativistic physics gravity force acts instantaneously so there is no propagation of potential force and it makes no difference for the prediction or retrodiction of motion whether we think of potential energy as stored in space with some density or we think of it as just a real function of mutual distances. Both are OK as theoretical concepts, but they are not real as stones or mutual configuration of particles.

Either energy in the region of source decreases, or energy is simultaneously being supplied to the source as in the case of antennas.

Distribution function for energy in space may be introduced both in non-relativistic and relativistic theory, but for given state of particles and the field, there is infinity of different ways to do it and they give different values. That's part of Stratton's, Feynman's and others' reasons to convey the view that Poynting expressions are fine and may be useful, but one should not think their usage proves that definite value of energy "really" is there at some point of space. It is just one convenient picture of the concept of energy that works nicely. In relativistic electromagnetic theory, its usefulness is greater than in nr mechanics because one cannot use potential energy function to formulate relativistic version of the work-energy theorem.

20. Jan 2, 2015

### Stephen Tashi

Yes, there are mechanisms by which energy is supplied to a source radiating energy. But my question is why (from the viewpoint of classical physics) must a radiation source need energy to power it? If the radiation is only the "potential to do work" , we don't know the radiated field is ever called upon to do actual work.

To eliminate considerations of internal resistance, friction, etc, we could imagine two charged masses connected by an ideal spring and set in oscillation in a frictionless environment. Such a system should create an time varying EM field and (I think) should radiate energy. Is that correct? Must I imagine that there are other charged masses outside the system that are moved and acquire kinetic energy?