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Energy density of electric- and magnetic fields


hobz
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So what you're saying is that because coulomb is defined they way it is, and that causes the introduction of [math]\epsilon_0[/math] (in Coulombs law), and similar for the definition of a second in the Biot-Savart law which requires [math]\mu_0[/math] to appear, and these in turn defines [math]c = \frac{1}{\sqrt{\epsilon_0 \, \mu_0}} \neq 1[/math], then the relation has to be like that, but the energy in the fields are equal?

I most physics book I am shown perpendicular E and B "waves" along a direction of propagation of the same magnitude? I suppose this is wrong if [math]E=c\, B[/math]?

Edited by hobz
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I most physics book I am shown perpendicular E and B "waves" along a direction of propagation of the same magnitude? I suppose this is wrong if [math]E=c\, B[/math]?

 

Do the diagrams actually show the scaling? Just because the curves have equal height does not mean they have the same magnitude; the value and units are different.

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How can it be that the electric field is [math]c[/math] times greater than the magnetic field while their energy densities are the same?

 

As Swansont noted the units for the E and B fields are different. Therefore you cannot say that the E field is greater than the B field.

 

That would be like saying that 1 kilogram is greater than 1/2 meter, and by the same logic then less than 50 centimeters -- an immediate contradiction.

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So what you're saying is that because coulomb is defined they way it is, ...

That is kind of what I was saying; I was a bit in a rush when I wrote it. Let me give a more elaborated version:

 

What you are asking is how 20 tomatoes can have the same mass as two pumpkins when at the same time tomatoes are larger than pumpkins by a factor of 10 tomatoes/pumpkin. This "larger by" is comparing apples with oranges - or tomatoes with pumpkins.

 

There is a bit more to it in this case. In a relativistic treatment it does make a lot of sense to measure time and distance in the same units, which yields [math]c=1[/math]. With these units, a lot of seemingly-different things turn out to be effectively the same, the most prominent example being the center-of-mass energy of an object and its mass (E=m). To some extend that holds true for electromagnetic fields. If you have a vacuum with no electrical charges around, a possible electric field is [math]\vec E(\vec x) = \vec E_0 \sin(\vec k \vec x - |\vec k| t)[/math] with [math] \vec k \vec E_0 = 0[/math] (in whatever units). In this case, the Maxwell equations demand that [math]\vec B(\vec x) = \vec B_0 \cos(\vec k \vec x - |\vec k| t)[/math] with [math]\vec B_0 \vec k= \vec B_0 \vec E_0 = 0[/math] and [math]|\vec B_0| = |\vec E_0|[/math] (you might want to check for yourself that this is true, btw). So the magnetic amplitude of this electromagnetic wave in "B-direction" and the wave's amplitude in "E-direction" are indeed the same (when c=1).

 

I'm a little short on time and I notice I've possibly run off into the wrong direction. A few comments on what I just said:

- This is a very special example of a plane wave in vacuum. For the electric field of a point charge, and no magnetic field around, no such scenario occurs.

- Even in this special case, [math]|\vec B (\vec x, t)| = |\vec E (\vec x, t)|[/math] usually does not hold true. The value [math] \vec B \vec B + \vec E \vec E[/math] is a constant, though, and is in fact related to the energy.

- The direction I probably should have run into: Under relativistic coordinate transformations, electrical fields and magnetic fields do mix. In this sense, they are indeed very related to some extent.

 

Bottom line: there is indeed some relation between electric fields and magnetic fields that becomes more apparent with c=1. Plainly saying E=cB without giving a context is just wrong. Keep in mind that a lot of these equations that people throw around as if they knew what they mean are only valid within the context a special scenario and/or with a special meaning of the symbols (c.f. "what is the equation for time dilatation with gravitational field and velocity?")

 

hope above makes kind of sense; it's a bit rushed.

 

EDIT: Actually, I think you could add a constant E- or B- field to the solutions I presented above and still fulfill the Maxwell equations. Makes my point about their relation even weaker, I guess. But perhaps to still justify why the example might be relevant: such electromagnetic waves are the standard basis for those ominous "photons" that everyone seems to talk about.

Edited by timo
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