# Tag Info

3

Instead of scattering, think of it as diffuse reflection. The bidirectional reflectance distribution function (BRDF) describes optical surface properties. It's application is as well in computer graphics, as in-depth ray tracing simulations. It depends on angle of incident light $\vec \omega_i$ (2 dimensions) and angle of observation $\vec \omega_r$, also 2 ...

5

Yes. There is a standard way to generalize to field theory. Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$. To obtain the ...

1

As Emilio Pisanty asserts, the magnetic Lorentz force $\boldsymbol{F_m}= q \boldsymbol{v \times B }$ does not do work, since $\boldsymbol{F_m \cdot v} = 0$. I don't have access to Griffith, but I find Feynman's discussion (in v. II, ch 15) confusing, since he has magnetic forces doing work on the current while also asserting those same forces do no work. ...

6

Magnetic fields never do work directly. This is because the magnetic force on any charged particle, $$\mathbf{F}=q\mathbf{v}\times \mathbf{B,}$$ is always orthogonal to the velocity, and therefore the power transferred, $\mathbf{F}\cdot\mathbf{v}$, is zero. On the other hand, this seems to contradict much of our intuition about how magnets behave. If you ...

0

A dielectric is not a conductor, thus there are no electrons that are able to flow through it. However atoms or molecules within may be able to be polarised making an electric dipole, which can align to enhance or anti-align to reduce the applied field. This is bound charge. In a metal or in free space the electrons flow and are, in a sense, free. They are ...

-1

I think your question answers itself, indeed the area of what? Faraday's law is for a closed loop of wire, thus Faraday's law is inappropriate and we should look for an alternative, as you have done by considering the Lorentz force. If the metal were a closed loop of circumference $l$ then Faraday's law would be valid. The forces on the electrons from the ...

1

A quick answer, if I may. You need $\theta$ to be smooth since you want to derive it. So mathematics imposes you to choose $\theta$ smooth. Now the trick: choosing $\theta$ to be smooth means you can always impose $\mathbf{A}$ to be smooth, and use several patches related to each other by a gauge transform. Then you should always discuss smooth vector ...

1

Once you get $B$, then $$A(\vec{x}_1)-A(\vec{x}_0)=\int_{\vec{x}_0}^{\vec{x}_1}\vec{B}\cdot\mathrm{d}\vec{r}$$ This is usually done by setting some value to the potential at a particular point from the information provided. Be careful, even though a typical choice is to set $A=0$ at infinity, you can't do that in this case because your wire is infinite.

0

Recall I = n*q*(v_d)*A 1 Note that the electrical current depends only on the cross sectional area of the wire and not on its length... However if we manipulate eq.1 we find that I = (n*q*s*A)/t Since v_d = s/t if we continue further we see that I = (n*q*V)/t Where V is the volume of the Wire. I don't get your second question,please be more ...

2

OP's proposal $A_{\mu}=0$ are four$^1$ gauge-fixing conditions (in 4D), not just one. Usually in EM we only specify a single gauge-fixing condition $\chi=0$, such as, e.g. temporal gauge: $A_0=0$; axial gauge: $\vec{n} \cdot \vec{A}=0$, where $\vec{n}$ is some constant 3-vector; Coulomb gauge: $\vec{\nabla} \cdot \vec{A}=0$; Lorenz gauge: ...

4

Well $F_{\mu\nu}$ is gauge invariant, so you can't gauge it away. The other reason is to remember that the photon has two polarizations (this is really one and the same, as you can show two gauge redundancies), so you can't set more than two components of $A_\mu$ = 0. In short, if you start with the Lorentz gauge, $\partial_\mu A^\mu = 0$, you can further ...

1

As Ondřej points out, $\mathbf{S}$ is orthogonal to both $\mathbf{E}$ and $\mathbf{B}$ as it's defined as their cross product. However, $\mathbf{E}$ and $\mathbf{B}$ are in general not orthogonal to each other. This is obvious in the general case: you can make static fields using Helmholtz coils and electrodes, completely independent of each other, for ...

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