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13

Your intuition about the meaning of the divergence operator is wrong. In physics it's easiest to think intuitively about divergence by using the divergence theorem which states $$\int_V dV \ \nabla \cdot \mathbf{B} = \int_{\partial V} \mathbf{B} \cdot d\mathbf{S}$$ where $\partial V$ is the surface area surrounding the volume $V$. The magnetic field has ...


8

Although you might not like to hear it, the answer really DOES lie in the definition of $\mu_0$ (and $c$). $\mu_0$ is defined to be exactly $4\pi *10^{-7}\ \text{H m}^{-1}$. Similarly, $c$ is defined as exactly $299792458\ \text{ms}^{-1}$. It immediately follows from the relation $$\epsilon_0=\frac{1}{\mu_0 c^2}$$ that $\epsilon_0$ also has no uncertainty. ...


6

The first equation is only valid for massive particles. If you see the formula of the Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ If $v=c$ (the case of massless particles), it is undefined. You can also see as $v \rightarrow c$, $\gamma \rightarrow \infty$, which "compensates" for $m_0=0$. The second doesn't imply zero energy, because ...


6

Divergence means the field is either converging to a point/source or diverging from it. Divergence of magnetic field is zero everywhere because if it is not it would mean that a monopole is there since field can converge to or diverge from monopole. But magnetic monopole doesn't exist in space. So its divergence is zero everywhere. Mathematically, we get ...


5

Covariant notation is a simple way to say how something transforms under Lorentz. An object with an index, e.g. $z^\mu=(z^0,\vec z$), transforms under Lorentz as, $$ {z^{\prime}}^{\nu} = {\Lambda^\nu}_\mu z^\mu $$ where $\Lambda^\mu_\nu$ is the matrix you have written down in your question. $z^\mu$ is called a four-vector. This transformation property is ...


3

Are there experiments that could show that light waves resemble more say square waves than sine waves? Are there experiments that could show that light waves resemble more say square waves than sine waves? Temporarily looking at your example of a square wave, a square wave of spatial wavenumber $k$ can be represented in a Fourier expansion as a ...


3

It's because magnetic and electric fields transform into one another depending on your intertial frame, and therefore field lines aren't an invariant intrinsic property of space. In one frame, you could have a region where there's only a uniform magnetic field $\vec B$, so that any stationary charge remains classically at rest. Yet viewed from a frame ...


3

Just to add to Danu's Answer, which I believe to be right. The relative scalings of the "electro" and "magnetism" parts of the unified electromagnetism whole are somewhat arbitrary; we're only required to ensure that $c=\frac{1}{\sqrt{\mu_0\,\epsilon_0}}$ to achieve a valid set of Maxwell equations. As we change these relative scalings, we change the ...


2

Simply consider Maxwell equation : $$\vec{\nabla}\wedge\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$ If you interger this on a given closed surface $\Sigma$, it follows : $$\oint_\Sigma \left(\vec{\nabla}\wedge\vec{E}\right) \cdot d\vec{S} =-\frac{\partial}{\partial t}\oint_\Sigma \vec{B}\cdot d\vec{S}$$ where $d\vec{S}=dS\,.\vec{n}$ with $dS$ the ...


2

The rate of formation is much higher in the presence of dust. There needs to be a mechanism for the energy of formation of the hydrogen molecule to be dissipated. Dissipating energy via a photon involves a forbidden transition. Instead, the energy can be transferred to the vibrational lattice of a dust particle. See The Interstellar Abundance of the ...


2

The metal is detuning both the tag's antenna and depending on how close the phone is, the phone's RFID antenna too. When a piece of metal is placed in the near field area of an antenna it becomes coupled to the antenna and it's resonance frequency drops, the impedance decreases (causing a large signal loss) and the bandwidth widens (Q decreases). In an ...


2

You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator $(b^2-c^2)^2$: $$ \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ...


2

If you want to include "all real world effects" in your analysis, you need to make sure you include all effects. At the very least, include parasitics. And include the fact that your "real world voltage source" has finite impedance, output capacitance, inductance in the leads, ... So when you state Say it starts of at a voltage V when you connect it to ...


1

The scalar potential and magnetic vector potential are combined into a four-vector, $A_{\mu}=(\phi,\vec{A})$ which is a gauge field, and in the language of differential geometry, a 1-form. The Lagrangian of the field theory (i.e. Maxwell theory) is, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ where $F_{\mu\nu} = \partial_{[\mu}A_{\nu]}$ is the ...


1

One has to define in what framework one is talking of electric and magnetic fields. In the classical framework the field is defined, for simplicity lets take a point charge, as proportional to 1/r^2 and exists up to infinity. Thus classically there is no transmission for a static charge, it just is. When a charge is moving, i.e. changing its (x,y,z,t) ...


1

So, my best understanding: The basic solution to the wave equation is $$\Psi(x,t)=Ae^{ikx-i\omega t}$$ Where the signs are arbitrary. If you combine this with the good old Euler Formula this expands to $$\Psi(x,t)=A\cos(kx-\omega t)+B\sin(kx-\omega t)$$ Where the imaginary part is absorbed into that B


1

So firstly, it is not the strong nuclear force that keeps electrons in fixed orbits around the nucleus. The strong nuclear force, that is, the Quantum Chromodyanmic interactions, hold the protons and neutrons together in the nucleus, as well as holding the quarks and/ or antiquarks together in other hadrons and mesons. These interactions do not come into ...


1

As you have written, $$ \epsilon = {\Delta \phi \over \Delta t}$$ This equation says that $ \epsilon $ is greatest when the change in flux with respect to time is greatest, not when the flux itself is greatest. In order to find when the change in flux is the greatest, you need to come up with an equation for the flux, then take the derivative with respect to ...


1

But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ...


1

Maybe you have seen the experiment where a permanent magnet is dropped through a pipe made of conducting metal. The magnetic flux through a cross-section of the pipe will be changing, so a current is induced. The induced magnetic field is such as to oppose the change in magnetic flux, so it will slow down the falling magnet. Eventually an equilibrium is ...


1

The conceptual problem here is that of EMF, $\mathcal{E}$ vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ...


1

The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ...


1

Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals. If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then ...


1

$F_{\mu\nu}$ is a Lorentz tensor, easy to see by $\partial_\mu A_\nu - \partial_\nu A_\mu$, which is a 2-form. Contractions of Lorentz tensors are Lorentz tensors. $\tilde{F} = \star F$ is the Hodge dual of $F$, which is also a 2-form, hence a Lorentz tensor, therefore the same applies about its contractions. By these definitions, they are also tensors in ...


1

The term "Rosenberg-Coleman effect" originates from the article Heliographic latitude dependence of the dominant polarity of the interplanetary magnetic field. It is also referred to as the "dominant polarity effect". As the Earth orbits the Sun, the Earth travels above and below the equator of the Sun. According to Rosenberg and Coleman, the polarity ...



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