# Light waves are one half kinetic and one half potential energy

1. Sep 16, 2011

### liometopum

Water waves are one half kinetic energy and one half potential energy. The quote below comes from the Wikipedia wave power article.

I assume that it is the same with light waves. Do people here agree?

"....E is the mean wave energy density per unit horizontal area (J/m2), the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy,[3] both contributing half to the wave energy density E, as can be expected from the equipartition theorem."

2. Sep 17, 2011

### Bill_K

No indeed. Light waves and water waves have almost nothing in common. Light waves do not have kinetic energy or potential energy.

The energy density in a light wave is proportional to E2 + B2, so you could say that half of it is electric and half of it is magnetic. But note that E and B are in phase: they reach their maximum value simultaneously, and are zero simultaneously, so the energy density is not constant. To get a meaningful value one needs to average the energy density over a cycle.

3. Sep 17, 2011

### Naty1

Seems like water waves would be more related to sound waves: both are longtitudional, both involve matter(mass) ,etc,etc.

4. Sep 17, 2011

### liometopum

Good points.
Let's try it differently.... let's say an material object is moving at just below light speed, as close to light speed as imaginable.

Then, would kinetic energy and potential energy be almost the same?

5. Sep 17, 2011

### Philip Wood

Taking up the point about a 'material object'...

Potential energy is energy (ability to do work) measured with respect to a reference point (a point where, by convention, we say that the potential energy is zero).

This implies that whether PE and KE are equal or unequal is dependent on this convention, and cannot, in general have any fundamental significance. It is changes in potential energy which are of physical significance. An object travelling near the speed of light, or indeed at any speed, may or may not be changing its potential energy. Depends on whether or not it’s in a field.

Last edited: Sep 17, 2011
6. Sep 17, 2011

### JeffKoch

7. Sep 17, 2011

### liometopum

1. BillK: Light does have kinetic energy. Some calculate it from Planck's constant times the frequency.
2. BillK: Light waves and water waves have a lot in common. They are both waves, travelling disturbances carrying energy. Both show interference. They are both waves. Basic energy characteristics should be the same for all waves.
3. Naty1: What?? Sound waves are longitudinal (compressional) and water waves are transverse.
4. JeffKoch: But the sound wave is a good point. It is not gravitationally-effected like the water wave, so we can eliminate gravity. Rephrase the question I asked to sound waves. I am sure a sound wave is half kinetic and half potential as well. And I am positive that light is the same too. It must be a characteristic of all waves, regardless of medium of travel.
5. Phillip: If all waves have both PE and KE, then maybe there is no need for a frame of reference for PE, except the existence of the KE.
6. JeffKoch: Potential energy exists in many forms and is not restricted to gravitational fields.

Last edited: Sep 17, 2011
8. Sep 17, 2011

### Ken G

Perhaps sound waves are easiest to imagine, so we can contrast them with water waves. Sound waves are all kinetic energy-- no potential energy at all. So why doesn't the virial theorem apply there? It has to do with what the restoring force is. For a water wave, or a wave on a slinky, the restoring force is conservative. There's a "real" force there doing the restoring, and it soaks up kinetic energy by having work done on it, only to release that energy by doing the work back. But in a sound wave, the "restoring force" is not a conservative force, it's a kind of make-believe force that shows up when you treat the particles collective behavior as a fluid. It has no potential energy associated with it, and the assumptions that go into the virial theorem don't apply. But maybe there is some way to build an equivalent concept, I don't know. Still, without some work to build an equivalent concept, we see that the half-and-half idea is not a general property of waves.

9. Sep 17, 2011

### JeffKoch

Sound is compression of a medium that is capable of pushing back, like a spring - which also can have potential energy. There is no medium for light waves, unless you wish to believe in the ethereal theory that was disproved in the 19th century. And if you are positive, why do you ask, and in what sense do you think a light wave has potential energy associated with it?

10. Sep 17, 2011

### Redbelly98

Staff Emeritus
No, not necessarily. In fact, the potential energy could even be zero or negative.

That's not kinetic energy, that's electromagnetic energy.

11. Sep 17, 2011

### liometopum

Thanks KenG. And of course, thanks to everyone for your input!
Regarding the potential energy of sound waves:

From Wikipedia, Sound: (http://en.wikipedia.org/wiki/Sound)
"The energy carried by the sound wave converts back and forth between the potential energy of the extra compression (in case of longitudinal waves) or lateral displacement strain (in case of transverse waves) of the matter and the kinetic energy of the oscillations of the medium."

But I see that JeffKoch spotted this too. The back and forth aspect of any wave implies a transition between kinetic and potential energy. For example, the Wikipedia article on "Vibration" states "... Thus oscillation of the spring amounts to the transferring back and forth of the kinetic energy into potential energy."

All waves oscillate. It is part of being a wave.

JeffKoch: I ask in part because I need to bounce some ideas off people and this forum is the only option for me.

Redbelly: Maybe the interpretation of it being electromagnetic energy is analogous to kinetic energy turning to heat after a moving object hits something. The kinetic energy of light is transferred as electromagnetic energy upon interaction with the receiving object.

Redbelly: If PE+KE=a constant, and you go negative with the PE, that means you have to go above the initial kinetic energy. And now we are entering a strange area of talk and we might want to not go there.

All: Do you notice how much confusion exists over potential energy? Or might I say energy in general?

Last edited: Sep 17, 2011
12. Sep 17, 2011

### Ken G

It can be that, but the most common example, sound in air, isn't like that. For sound in air, there's no conservative force, and no potential energy anywhere.

13. Sep 17, 2011

### Ken G

I haven't read the entry, but on the surface, that's just nonsense. The vast majority of our experience with sound is in air, and that entry makes no sense at all for sound in air. Maybe they specify somewhere they are talking about sound in an elastic medium, for which the restoring force is a conservative force. Air isn't like that at all, the restoring force is just freely flowing momentum flux, it's not really much of force at all-- it just acts like one in a fluid description.

14. Sep 17, 2011

### JeffKoch

A sound wave is a compression wave, which results in a local increase in air pressure above ambient (the RMS is the sound pressure, which can be related to a number in decibels), so in a sense it is like a spring. I personally don't think of sound waves in terms of potential energy, but I suppose one could do so.

15. Sep 17, 2011

### Ken G

I can't really see how potential energy would make much sense for sound in air, but I agree with you that the mathematics can look kind of similar. When you compress air, there is the potential to re-expand, but no potential energy is involved.

16. Sep 18, 2011

### Phrak

So you think light has energy. Really? Light energy is usually given in terms of the Poynting vector, P. <P> makes people happy because it lets them get energy from one place to another in a continuous manner without raising any hard questions. But as I recall, application of the Poynting vector is an example of junk physics. In the derivation of the Poynting vector, it is initially assumed that charge density over a region in question is non-zero. Charge density then calcels out, and for some reason the Poynting vector is still faithfully applied where there is no charge density at all. Where charge density is zero, the equation yields 0=0. Properly applied, Mr. Poynting's formula does not include the vacuum.

Last edited: Sep 18, 2011
17. Sep 18, 2011

### Philip Wood

Sound Waves in air
Although Robert Boyle (of Boyle's Law fame) wrote about "the spring of the air" we now know that the mechanical properties of gases at 'ordinary' temperatures and pressures are well described by the kinetic model, in which the gas consists of molecules in rapid random motion, and for which the energy is wholly kinetic.

When sound is travelling through air, the energy associated with the oscillatory motion of the molecules might be thought, as in most oscillations, to be partly kinetic and partly potential. But here, the 'potential' part is due to "the spring of the air" and is, fundamentally, (random) kinetic. The 'kinetic' part is due to the organised superimposed motion of the molecules due to the passage of the sound wave.

Electromagnetic Waves
The energy of an e-m wave is electric field energy and magnetic field energy.
The energy per unit volume in an electric field of magnitude E is $\frac{1}{2} \epsilon_{0} E^2$.
The energy per unit volume in a magnetic field of magnitude B is $\frac{1}{2} \epsilon_{0}c^2 B^2$.
At any point in an e-m wave these are equal.

In circuit theory (and possibly in a wider context) there is quite a good analogy between B-field energy and kinetic energy, and between E-field energy and elastic potential energy. But it's only an analogy.

Last edited: Sep 18, 2011
18. Sep 18, 2011

### Ken G

I agree that's the bottom line here-- what the OP is noting is likely a fruitful type of analogy to look for in many situations, but it is also not something that should be taken too literally when more precise meanings of the words are in use. Language like potential energy can be used to mean something quite specific, or it can be generalized, we just have to be clear which one we are doing.

19. Sep 18, 2011

### Phrak

What does "indeed" mean. Could you supply a source and derivation of energy density = E2 + B2, please?

Could you supply a source and derivation for this?

In dimensional analysis, I get

$E = E[MD/QT^2]$

$B = B[M/TQ]$

$1/\mu = 1/\mu [T^2Q^2/M^2]$

$\epsilon = \epsilon [T^4Q^2/M^2D^2]$

therefore

$B^2/\mu$ is unitless.
$E^2 \epsilon$ is unitless.

Neither represents energy. Anyone have a problem with this?

In skew differential forms I know of only one natural combination that yields terms in E^2 and B^2. This is the Faraday tensor with lower indices, F multiplied by it's dual, G (outer product multiplication).

$G_{\mu \nu} = {\epsilon_{\mu\nu}}^{\alpha\beta}F_{\alpha \beta}$

$G_{\mu \nu}\wedge F_{\alpha \beta} = 0$

The product G^F is generally covariant when G, F and epsilon are tensor densities. I know of none that are expressed in generally covariant form and are nonzero in local coordinates.

Last edited: Sep 18, 2011
20. Sep 19, 2011

### Philip Wood

Here's an unsophisticated plausibility argument for electric field energy.

Consider the uniform electric field inside a parallel plate capacitor of plate area A and plate separation b, for which capacitance C = $\epsilon$0 A /b

We know that the energy U stored in the capacitor is $\frac{1}{2}C V^2$.

This can be written $\frac{1}{2} \frac{\epsilon_{0}A}{b} V^{2}$.

But the electric field strength, E, in the gap is given by E = V/b.

Expressing V in the energy formula in terms of E...
$$U = \frac{1}{2} \epsilon_{0}Ab E^{2}.$$
So the energy per unit volume of interplate gap is $u = \frac{1}{2} \epsilon_{0} E^{2}.$.

So far, so good. Nothing controversial. The leap is to attribute the capacitor's energy U to energy stored in the field, in which case the last formula gives the energy in the field per unit volume.

A similar argument can be constructed just as simply for magnetic energy by considering the energy stored in the uniform field in a long current-carrying solenoid.

Last edited: Sep 19, 2011