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6

It's not possible to derive the orbital angular momentum $L = r \times p$ from the $\mathfrak{so}(3)$ commutation relations alone, since the spin operator $S$ also fulfills the same commutation relations, but certainly is different from $r \times p$.


6

I have read somewhere that commutation relations of the form \begin{equation} [a_i,b_j]=\epsilon_{ijk} c_k \end{equation} admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$ ...


2

The most straightforward answer is no, you cannot consider a component to be a vector. A vector is something which has an associated direction, but a component has no direction. It's just a number. For example, if $\vec{F}$ is a vector, the component of $\vec{F}$ in the $\hat{x}$ direction is $\vec{F}\cdot\hat{x}$, and hopefully you know that the value of a ...


2

This particular case is easy, because we're working in the $xz$ plane so the sums are straightforward. The general case for two arbitrary vectors would be a lot harder. Anyhow, let's take the particular example you describe: The angle (1 0 a) makes with (1 0 0) is just tan$^{-1}$(a) and the angle (1 0 b) makes is tan$^{-1}$(b). You want the $a$ angle to ...


2

Electric flux is given by: Here, S (vector area is same) while the strength of the electric field E is more in A (length of arrows). $\theta=0$ in both cases. Hence flux is not same. A > B


2

Basically, vectors are called contravariant because their components transform oppositely to the basis vectors: if our change of coordinates is such that $$ \frac{\partial}{\partial x^i} = \frac{\partial y^j}{\partial x^i} \frac{\partial}{\partial y^j}$$ then if we have a vector $\mathbf{V}$, its components $V^i_x$ in the $x$ coordinates are related to its ...


1

This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.


1

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...


1

The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...


1

Short answer, you can get the result in electrostatics, for simple (path) connected surfaces, if the electric field isn't singular on the surface. How? For simple surfaces were can adjust the whole potential in one surface by a uniform amount and it doesn't change the electric field in that region. So pick a point A on the interface surface and why not ...


1

What is wrong with my reasoning? Opposite charges attract because one of the charges has a negative sign. The force on the negatively charged particle is thus $$\vec F_- = \frac{kQ(-Q)}{r^2}\hat r = -Q\,\frac{kQ}{r^2}\hat r = -Q\,\vec E_+ $$ The force on the negatively charged particle is opposite the direction of the field from the positively ...


1

In general the vector equation $$\mathbf{A}+\mathbf{B}+\mathbf{C}=0 $$ means $$A_x+B_x+C_x=0,\quad A_y+B_y+C_y=0 $$ or in terms of your angles and magnitudes of the vectors $$A \cos\alpha+B \cos\beta=0,\quad A\sin\alpha+B\sin\beta-C=0 $$ since $C$ is along a negative $y$. You know $\alpha,\beta, C$ so you can solve for $A$ and $B$.



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