# Tag Info

49

Well it has nothing to do with the Higgs, but it is due to some deep facts in special relativity and quantum mechanics that are known about. Unfortunately I don't know how to make the explanation really simple apart from relating some more basic facts. Maybe this will help you, maybe not, but this is currently the most fundamental explanation known. It's ...

26

Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real. When the principles of special relativity are ...

22

Yes, absolutely. In fact, Gauss's law is generally considered to be the fundamental law, and Coulomb's law is simply a consequence of it (and of the Lorentz force law). You can actually simulate a 2D world by using a line charge instead of a point charge, and taking a cross section perpendicular to the line. In this case, you find that the force (or ...

20

Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), ...

14

This is not paradoxical and it is not necessary for any physical phenomenon to a priori have to obey any particular law. Some phenomena do have to obey inverse-square laws (such as, particularly, the light intensity from a point source) but they are relatively limited (more on them below). Even worse, gravity and electricity don't even follow this in ...

12

Permittivity $\varepsilon$ is what characterizes the amount of polarization $\mathbf{P}$ which occurs when an external electric field $\mathbf{E}$ is applied to a certain dielectric medium. The relation of the three quantities is given by $$\mathbf{P}=\varepsilon\mathbf{E},$$ where permittivity can also be a (rank-two) tensor: this is the case in an ...

11

You are correct when you concluded that two classical point electrons could never touch each other. It would take infinite energy.

11

Short Answer You've hit upon the quirk that the SI and CGS systems not only measure electric charge with different units, but also assign them different dimensionality. In SI, the Ampere is a base unit. Amperes are not made out of anything else - they are primitive, like meters, kilograms, and seconds. One Ampere is one Coulomb per second, so the unit of ...

11

The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law. More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law $$... 10 James Clerk Maxwell thought about this one and showed the following. Suppose we have two concentric conducting spheres and we charge one up to a potential \Phi relative to some grounding plane. Then the voltage of the inner sphere relative to the same ground is:$$\Phi_{inner} = \Phi \,q\, ...

10

Concerning the factor $\frac{1}{2}$: It seems that OP in his classical reasoning only accounted for the Coulomb potential energy $$\tag{1}\langle U\rangle ~=~-k_e e^2 \langle \frac{1}{r} \rangle ~=~-\frac{k_e e^2}{a_0} ~<~0.$$ Here $k_e$ is Coulomb's constant and $a_0$ is the Bohr radius.$^1$ However we should also take the kinetic energy $\langle ... 10 If you want to avoid factors of$\pi$in the more fundamental equations like$\nabla . E = \rho / \epsilon_0$, you have to accept them where they belong, for instance in:$E = \frac{1}{\epsilon_0} \frac{Q}{4 \pi r^2}$. As remarked by others, Newton failed to put a factor$4 \pi$into his gravitation equation (he stipulated$g = G \frac{M}{r^2}\$, instead of ...

9

Coulomb's law becomes invalid at distances of the order of the electron Compton wavelength and smaller, due to vacuum polarization. To first order in the fine structure constant, the electric potential due to a charge q at the origin is given by: $$V(r) = \frac{q u(r)}{r}$$ where $$u(r) = 1 +\frac{2\alpha}{3\pi}\int_1^{\infty}du ... 8 There is another 'infinity' (among others) lurking in classical electrodynamics which is evident when one calculates the electrostatic energy W of a uniform spherical charge distribution of radius a and total charge Q$$W = \frac{3}{5}\frac{Q^2}{4\pi \epsilon_0 a}$$Thus, by this result, a point (zero radius) particle of charge Q has 'infinite' ... 8 The mistake you made is in the way you stated Coloumb's law. It's either$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\vec{r}} $$OR$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}2} \color{red}{\hat{r}} $$but definitely NOT$$ \vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}} $$8 It's a good observation that the electric and gravitational fields both satisfy Poisson's equation$$ \nabla^2\Phi_G = 4\pi\rho_G, \qquad \nabla^2\Phi_E = -\frac{\rho_E}{\epsilon_0} $$where \Phi_G, \Phi_E are the gravitational and electric potentials and \rho_G,\rho_E are the mass and charge densities. It would seem from the perspective of Newtonian ... 8 Pure convention. There is no reason alternative conventions couldn't be used, apart from the need to avoid confusion. Newton introduced the constant to make the force law simple, whereas the electrostatic definition with the 4\pi is designed to make Poisson's equation (one of the equations for the electric field) look simple. You can write a Poisson ... 8 Not a direct answer to your question but still a surprising derivation of Maxwells equations: Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle. 7 Yes the dimension is different. In SI the current (A) a base unit independent from length (m), mass (kg) and time (s) because we choose to, but in CGS Gaussian unit this is not (1 unit of current = 1 g1/2 cm3/2 s-2), by setting \epsilon_{0,SI} = \frac1{4\pi}. This also leads to some perhaps unintuitive results, like the unit capacitance in CGS Gaussian is ... 6 The physical reason for the appearance of a 4\pi somewhere in the theory is the spherical symmetry of the problem and is discussed more in other answers . Here I want to quote an interesting argument from Arnold Sommerfeld's Lectures on Theoretical Physics Vol III, which has a section dedicated to this issue. If you remove the 4\pi from the force law ... 6 The two forces, A on B, and B on A, are equal and opposite. They are not unequal. This fact is codified in Coulomb's Law: the force is proportional to the product of the two charges, regardless of which force you are talking about. Absolutely the same. If one of them had a much larger mass than the other, then the heavy one would have a much smaller ... 6 I would say yes ! Actually some theories explaining quantum gravity use also this reasoning: gravity is a very weak interaction at a quantum level because it "leaks" into other dimensions, not observable at our scale, but that are present at this scale. The mathematical tools are different, but if you just think about gauss's law you can imagine one ... 6 A short answer, is that to estimate interaction energy (which says if same charges attract or repel), you use propagators. Propagators come from the expression of Lagrangians. Finally, the time derivative part for dynamical freedom degrees in the action must be positive, and this has a consequence on the sign of the Lagrangian. Choose a metrics ... 5 Quoting from my copy of the 2nd edition of Jackson's book on Classical Electrodynamics, section 1.2: Assume that the force varies as 1/r^{2+\epsilon} and quote a value or limit for \epsilon. [...] The original experiment with concentric spheres by Cavendish in 1772 gave an upper limit on \epsilon of \left| \epsilon \right| \le 0.02. followed a ... 5 First of all note that k is not dimensionless, it is k = \frac{1}{4 \pi \varepsilon_0}, and \varepsilon_0 has dimensions of \frac{ \text C^2}{ \text {N m}^2}. So you have already \frac{ \text{V C}^2 \text m^2}{ \text {N m m} ^2}. Also, volt can be expanded as  \text V = \frac{ \text {N m}}{ \text C}, so one gets$$ \frac{ \text C^2}{ \text {N ...

5

The short answer is "observations". In the case of the gravitational law, the orbits of the planets around the sun, the moon around the earth fit mathematically a force with an inverse square law for the distance. An inverse law does not. In the case of electricity this article points out the observational history: Early investigators who suspected ...

5

In Newtonian physics, there was no problem with action at a distance, and indeed Newton explicitly formulated his theory of gravitation in such terms. It may be that this was criticised from a philosophical standpoint (I don't know whether it was or not), but there were no fundamental mathematical difficulties with the idea. However, in relativity the ...

5

I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations. It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct ...

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