# Tag Info

43

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 ...

21

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 ...

19

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 ...

15

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

I suppose you mean $k_e=\frac1{4\pi\epsilon_0}$. That comes from the fact that Coulomb's law can be stated as : $$F= \frac1{\epsilon_0}\frac1{4\pi r^2}q_1q_2$$ Now, $\epsilon_0$ is the electric constant, or the permittivity of free space, and it essentially scales the force. The $4\pi r^2$ comes from the surface ...

9

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 ...

9

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

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: $$\Phi_{inner} = \Phi \,q\, ... 9 Permittivity \varepsilon is what characterizes the amount of polarization \mathbf{P} which occurs when an external electric field \mathbf{E} is applied to a certain dielectric medium. The relation of the three quantities is given by$$\mathbf{P}=\varepsilon\mathbf{E},$$where permittivity can also be a (rank two) tensor: this would be the case in an ... 8 It's a good observation that the electric and gravitational fields both satisfy Poisson's equation$$ \nabla^2\Phi_G = 4\pi\rho_G, \qquad \nabla^2\Phi_E = -\frac{\rho_E}{\epsilon_0} $$where \Phi_G, \Phi_E are the gravitational and electric potentials and \rho_G,\rho_E are the mass and charge densities. It would seem from the perspective of Newtonian ... 7 The short answer is yes, and in fact you only need one single Maxwell equation, Gauss's law, together with the Lorentz force, to get Coulomb's law. More specifically, you need Gauss's law in its integral form, which is equivalent to the differential form for well-behaved fields because of Gauss's theorem. Thus, you use the law$$ ...

7

+1, Good question,. While I don't think your idea has much of a physical implications, it is a good analogy (in my opinion, at least). A fair approximation to General Relativity is Newtonian Gravity. A better one is Newtonian Gravity with some special relativistic corrections (I mean a modification to Newton's gravity where the masses $m$ are replaced ...

7

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 $$\vec{F} = K \frac{q_1 q_2}{r^\color{red}3} \color{red}{\hat{r}}$$

6

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

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

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.

5

The length scale $L$ has to be present in the denominator for dimensional reasons – only logarithms of dimensionless quantities are really "well-defined" unless one wants to introduce bizarre units such as the "logarithm of a meter". On the other hand, the dependence on $L$ is largely trivial and unphysical for most purposes. Replace $L$ by $K$ and you will ...

5

As the two charged bodies attract, they have unlike charges. So, Assuming your two charged bodies as conductors and charged equally, the system may be considered as a Capacitor. If you place a dielectric like glass of some Relative permittivity $\epsilon_r$ (3.7 to 10) which fills the empty space between the bodies, then the capacitance would be ...

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

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 $$\frac{ \text C^2}{ \text {N ... 5 The short answer is "observations". In the case of the gravitational law, the orbits of the planets around the sun, the moon around the earth fit mathematically a force with an inverse square law for the distance. An inverse law does not. In the case of electricity this article points out the observational history: Early investigators who suspected ... 5 Quoting from my copy of the 2nd edition of Jackson's book on Classical Electrodynamics, section 1.2: Assume that the force varies as 1/r^{2+\epsilon} and quote a value or limit for \epsilon. [...] The original experiment with concentric spheres by Cavendish in 1772 gave an upper limit on \epsilon of \left| \epsilon \right| \le 0.02. followed a ... 5 Gauss' law and Coulomb's law are equivalent - meaning that they are one and the same thing. Either one of them can be derived from the other. The rigorous derivations can be found in any of the electrodynamics textbooks, for eg., Jackson. For eg., consider a point charge q. As per Coulomb's law, the electric field produced by it is given by$$\vec{E} = ...

5

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 ...

5

Gauss's law would not be valid. You can imagine the electric field "flowing out" from positive charges and "draining" into negative charges. The "amount" of electric field decreases at the same rate as it spreads (since area of a surface increases by the square of its scale). This means that if no matter how we expand our Gaussian surface, if we don't cross ...

4

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 ...

4

With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge. Relativity has ...

4

It's more the other way around, I would say. Gauss's law, together with the fact that we live in a world with 3 spatial dimensions, requires that the force between charges falls off as 1/r^2. But there are perfectly consistent analogues of electrostatics in worlds with 2 or more spatial dimensions, which each have their own Coulomb's law" -- with a ...

4

Frenkel and Smit definitely make a mistake. Eq. (12.1.3) page 294 is: $$-\nabla^2 \phi(\mathbf{r}) = 4\pi \rho(\mathbf{r})$$ then immediately afterwards, Eq. (12.1.4) is "the solution of this equation" "for a single charge $z$ at the origin": $$\phi(\mathbf{r}) = \frac{z}{4\pi |\mathbf{r}|}$$ This is a mistake: Eq. (12.1.4) is definitely not "the solution" ...

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