Integrating EM Energy Density: Is E=hf?

[tim_lou]

when you integrate the energy density (from electromagnetic field) times the differiential volume of the whole 3D space for a photon...would you get the energy of it? E=hf ?

one more question... if there is a positive charge in 3d space... when i integrate the electromagnetic energy density times the differiential volume over the whole 3d space... i get infinity (assuming $$E=\frac{1}{4 \pi \epsilon_0{r^2}}$$)! so is there something wrong?

edit: (come on, this is not homework... i havn't even started school yet... i just thought about it when i was taking showers couple days ago)

 

[quasar987]

when you integrate the energy density (from electromagnetic field) times the differiential volume of the whole 3D space for a photon...would you get the energy of it? E=hf ?

I've always wondered myself what is considered to be 1 photon. We always see things like "When two of these particle collide, they produce this and that particle, and ONE photon." The hell does that look like?!

one more question... if there is a positive charge in 3d space... when i integrate the electromagnetic energy density times the differiential volume over the whole 3d space... i get infinity (assuming $$E=\frac{1}{4\pi\eps_0{r^2}}$$)! so is there something wrong?

Haha. Other will probably be able to extrapolate more on this intersting subject, but I have an explanation of this for you. If the book that teaches you E-M is anything like mine, it derived the equation of the energy in the electric field

$$W=\frac{\epsilon_0}{2}\int \vec{E}\cdot \vec{E} dV \ \ \ \mbox{(1)}$$

by mathematical manipulations of the equation

$$W=\frac{1}{2}\int_{body} V(\vec{r}')\rho(\vec{r}')dV' \ \ \ \mbox{(2)}$$

which in turn is a "continuous version" of the discrete one

$$W=\frac{1}{2}\sum_{i=1}^N q_iV(\vec{r}_i) \ \ \ \mbox{(3)}$$

where $V(\vec{r}_i)$ means "the potential at the position of $q_i$ due to all the OTHER charges". But in (2), under a regular rho (i.e. one NOT representing a point charge), V is well defined even where rho is not zero so it's ok. If rho represents a point charge ($\rho(\vec{r}')=q\delta^3(\vec{r}')$), V "explodes" and the integrals (2) and (1) too.

(3) represents the energy required to bring the N charges from infinity to a certain finite distance of one another. Similarily, (2) represents the energy required to construct the object of density rho by bringing charges from infinity. In analogy, we could say that (2) explodes for a point charge because it represents the energy necessary to CREATE a point charge out of smaller parts. Since doing so would mean bringing charges from infinity to exactly the same point in space, the energy required would be infinite, given the 1/r² repulsion to be overcome.

Finally, I recall Feynman saying in one of his "lectures on physics" that he considers this infinite energy of the point charge to be a real problem in classical E-M. Though I'm not sure I see in what sense he meant that, because it seems to me that if we are to use (2) (or (1)) for regular charge densities and (3) for point charges, then there is no problem. It is just that the integral machinery is in this case not an adequate mathematical tool to do what we want it to do. The REAL equation for the energy is (3) anyway, since there are only point charges in the universe.

However, I cannot imagine that Feynman has not seen this, so I must be wrong in at least one place, so I keep an open mind about this.

Was the explanation satisfactory?

 

[pseudovector]

The behaviour of the classic electric field in the vicinity of a point charge reveals that the classical theory has an inherent error, and therefore cannot fully describe reality. Very close to charged particles, this theory breaks down and aught to be replaced with a more precise theory, like qed.

 

[tim_lou]

thx for the clarification. you explained it very clearly that it takes an infinite amount of energy to bring charges into a "infinitely small point", and i see why the energy density can be considered "infinite". well, i guess the problem lies inherently in the concept of point-like particles...

well, this "infinite energy" connot be true since charged particles can be created from photons... the classic E-M says this requires infinite amount of energy while it shouldn't...

 

[tim_lou]

I've always wondered myself what is considered to be 1 photon. We always see things like "When two of these particle collide, they produce this and that particle, and ONE photon." The hell does that look like?!

same here... i always wonder why there is no 1.5 particles, pi particles or e particles... it just doesn't make sense to me how there has to be 1 and EXACTLY 1, 2, 3 particles and so on...

anyway, can i calculate the energy of "1" photon using this integration method?