# Tag Info

## New answers tagged vector-fields

3

The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole. If you want to see gravitational waves, you need to allow the metric to evolve dynamically, ...

15

The fact is that, in the general case $$\vec{E} = -\vec{\nabla}V - \frac{\partial\vec{A}}{\partial t};$$ (signs depend on conventions used) where $\vec{A}$ is called vector potential. You can consult for example Wikipedia. Let us consider homogeneous Maxwell equations: $$\begin{cases} \vec{\nabla}\cdot\vec{B} = 0,\\ \vec{\nabla}\times\vec{E} + ... 2 For dynamic electric and magnetic fields, there is a piece of the electric field that depends on the vector potential:$$ \vec{E} = - \vec{\nabla} V - \frac{\partial \vec{A}}{\partial t}, \qquad \vec{B} = \vec{\nabla} \times \vec{A}. $$Taking the curl of the first equation yields Faraday's Law (with the V-dependent term dropping out as you note); taking ... 6 When there is a time-varying magnetic field, the electric field is non-conservative and therefore cannot be written in the form \mathbf{E}=-\nabla V. 1 Assuming that \theta is the polar angle (angle between \vec r and \hat z) and \phi the azimuthal angle then the following relationships can be used. x = r \;\sin \theta \;\cos \phi y = r \;\sin \theta\;\sin \phi z= r\; \cos \theta 2 You said this is the r-component, then you've missed the \hat{\bf r}. Use$$r=\sqrt{x^2+y^2+z^2}\cos\theta=z/r=\frac{z}{\sqrt{x^2+y^2+z^2}}\hat{\bf r}=\frac{x\hat{\bf x}+y\hat{\bf y}+z\hat{\bf z}}{r}$$0 This diagram shows part of a surface cut up into small elements (shaded pink) of area dA. The normal to the area is the vector \vec {dA} and is of magnitude dA. What is required is the projection of the magnetic field \vec B onto the vector \vec{dA} which is B_\bot = B \cos \phi in the diagram. To do this the dot product is taken \vec B ... 0$$ \int \mathbf{B} \cdot \,d\mathbf{A} $$just means$$ \int B\,dA\,\cos\theta. $$d\mathbf{A} points in the direction perpendicular to the surface, so if this is in the xy plane, it'll be in \mathbf{\hat{z}}. I.e. your usual integration over the area ("integrand") weighted by a factor of \cos\theta. \theta is the angle with the z axis. Keeping ... 0 The integration variable is the surface vector: flux is an oriented thing. It is the "normal" integration of the scalar product of the field and the surface vector:$$\Phi=\iint_S \vec{B}\cdot\vec{\mathrm dS}$$Or, if \vec{n} is the normal to the surface at the point of integration:$$\Phi=\iint_S \vec{B}\cdot\vec{n}\ \mathrm dS$$0 Assuming your calculations are accurate, you could apply Gauss Theorem to get: \int d^{3}x \vec{\nabla}\cdot \vec{A} = \int_{S} \vec{A}\cdot \hat{n} \thinspace dS Since your integration occurs at the entire space, the surface S will be "at infinity" where your fields would enevntually go to zero, so the Right Hand Side would be null, proving your ... 2 I) In this answer we will consider the microscopic description of classical E&M only. The Lorentz force reads$$ \tag{1} {\bf F}~:=~q({\bf E}+{\bf v}\times {\bf B})~=~\frac{\mathrm d}{\mathrm dt}\frac{\partial U}{\partial {\bf v}}- \frac{\partial U}{\partial {\bf r}}~=~-q\frac{\mathrm d{\bf A}}{\mathrm dt} - \frac{\partial U}{\partial {\bf r}},  ...

3

There is no "physical aspect of this fact". The physical variables are the electric and the magnetic field, not the potentials. Introducing the potential is aesthetically and technically pleasing, but it is not necessary. A gauge symmetry is not a physical symmetry. The reason you can have a non-unique potential is that every divergence-free field such as ...

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