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2

You were pretty close already. There is a handy table on Wikipedia for a variety of coordinate systems. But for the polar system: $$ \vec{\nabla} \cdot \vec{U} = \frac{\partial U_r}{\partial r} + \frac{1}{r} \frac{\partial U_\theta}{\partial \theta} $$ and you can look up the curl in the same table. These can be derived from the Cartesian definitions by ...


1

In electrostatics, there is always the scalar potential (voltage) such that $\vec\nabla\phi = -\vec E$. This does not hold true in general electromagnetism, which is described by the electromagnetic four potential $A$ consisting of the electrostatic potential $\phi$ and the magnetic vector potential $\vec A$ such that $$ \vec E = -\vec\nabla\phi - ...


2

I'll do all calculations assuming the lagrangian $\mathcal{L}$ acts on a 1-dimensional manifold $M$. I believe you'll find the generalization absolutely trivial, and this will spare me of writing tons of sums. Let \begin{equation} \mathcal{L}: \mathbb{R} \times T M \rightarrow \mathbb{R} \end{equation} be a lagrangian over $T M$, with time in ...


2

You are writing (apart from a minus sign) the tidal tensor, which is the traceless part of the Hessian of gravitational potential, for a point mass in the origin. This tensor contains information about the piece of gravitational force which can't be removed by choosing a free falling reference frame. I'm not sure to understand your problem, however the tidal ...


3

An electric charge has an electrostatic potential associated with it. This potential is a scalar, so it has no direction associated with it, and its value at any distance $r$ from the charge is simply: $$ V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r} $$ If we graph the potential created by the charge as a 2D plot then we get something like this: (picture ...


0

You are free to define the vector $\bf{F_{EM}}$, but I don't believe this vector would have any value. It wouldn't obey any simple laws, and it would not be found to have any practical use in the lab.


1

I will find the Maxwell word in the Oersted Medal Lecture 2002: Reforming the Mathematical Language of Physics by Hestenes, on pages 25/26 That formidable text presents a better math formalism for physics, imo. Starting with $ F(x,t) = E(x,t) + i B(x,t) $ ... The 4 equations of Maxwell (64..67) that describe two viewpoints ( E and B ) of a single ...


12

Let me try this more clearly than the other answers, which aren't wrong. You ask: So, can someone please elaborate what this EM field is with respect to $\vec E$ and $\vec B$ in the context of Helmholtz decomposition? There is no "EM field in the context of Helmholtz decomposition". Helmholtz just says that every vector field $\vec V$ is decomposable ...


4

If the field is not stationary, curl of $\vec{E}$ does not vanish. So generally you cannot identify electromagnetic field with the curl-free part of the decomposition. However, you can indeed introduce a complex vector combination of electric and magnetic field, in a certain system of units it is $\vec{E}+i\vec{H}$. This is the so-called Riemann-Silberstein ...


5

If fits very well so we can write that electromagnetic field is equal $\bf{F_{EM}}=\bf{E+B}=-\nabla\phi+\nabla\times\bf{A}$ or can we? No! For the love of god, no! Do not just add those fields together... it's not a useful quantity. In the SI system $E$ and $B$ have different units. Another good indicator that you don't want to just add ...


0

Actually the electromagnetic field can be seen as a tensor. The combination Wikipedia talks about is this, $E$ and $B$ are organised in an antisymmetric matrix $F_{\mu\nu}$ with $\mu,\nu = 0,\ldots, 4$ so the number of independent components is $6$.


1

You're close: we can write the "operator" $p.\nabla$ as the "scalar" $$p.\nabla = p_x{\partial\over\partial x}+p_y{\partial\over\partial y}+p_z{\partial\over\partial z}$$ and $$E = E_x i + E_y j + E_z k = (E_x,E_y,E_z)^T$$ Then the result will be the vector $$\left(p_x{\partial E_x\over\partial x}+p_y{\partial E_x\over\partial y}+p_z{\partial ...


2

The definition of divergence is $$\textrm{div}\,\vec{F} = \lim_{V \to p}\iint_{S(V)} \frac{\vec{F}\cdot\vec{n}}{|V|}dS, \qquad [1]$$ where $\vec{F}$ is the vector field, $V$ is the volume surrounding the point $p$ where the divergence is calculated, $\vec{n}$ is a unit-length normal vector of the surface, $S(V)$, of the volume, and $|V|$ is the total volume. ...


1

Vectors can be defined in multiple different ways. Here I will show the four ways and explain how they are equivalent$^1$. An equivalence class of curves. Given a curve parameter $t$, curves are considered equivalent if they have equal zeroth- and first-order derivatives at $t=0$. In other words, a vector at a point $p$ is an equivalence class $[\gamma]$ ...


1

When physicists say that a vector is an n-tuple that transforms according to... they expect you to guess a lot that isn't said. What they mean is that for each basis you are given an n-tuple of numbers. And when you take the matrix that gives you the change of bases for any two bases and you apply the formula in their "definition" the first n-tuple goes to ...


0

From a strictly mathematical point of view, the reason why the notion of: $$\text{“$n$ globally defined scalar fields $f_i\colon M \to \mathbb R $”}$$ is different from the one of: $$\text{“a vector field $X\colon M \to TM$”}$$ is indicated in Qmechanic's answer. The notion of vector field is intrinsic, that is, it doesn't depend from the choice of a ...


2

OP wrote (v3): Now, given a set of component functions $X^1,\dots, X^n$ I can't see how they fail to make up a good vector field. If the functions are differentiable, continuous, and if they obey the property that $\pi \circ X= \operatorname{id}_{M}$ then we are good to go. It is (implicitly) implied by OP's notation that the component functions ...


0

There's been some good discussions linked and 0celo7 wrote the answer with great brevity, but I'll take a crack at adding some exposition. Let's consider situations of rotation: We can consider clockwise rotation (for our cartesian basis $\{e_i\}$) for some vector $V$, $ \hat x' = \hat x\cos(\theta)+\hat y\sin(\theta) $ $ \hat y' = -\hat x\sin(\theta)+\hat ...


0

I think that the physicists and mathematicians in this situation take two different ways to obtain the same thing. As Borges would have said, everyone is either platonist or aristotelian; in this case (and probably always) the mathematicians being the former and the physicists the latter. From the mathematical point of view, you the general structure ...


0

Take Cartesian coordinates for the real plane and transform them into polar. Does the set of coordinates $(x,y)$ transform as a vector? If you work out this example you'll see that this transformation, unlike linear ones, doesn't involve the Jacobian of the transformation. In the linear case this is just an accident.


1

The Lie derivative has a geometrical meaning: it measures the change of a tensor field (including scalar function, vector field and one-form), along the flow of another vector field. For example, the Lie derivative of the metric tensor along a Killing vector is zero (this defines the Killing vector equation). It means that a tensor (for example the metric) ...



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