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6

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...


4

Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


2

I don't want to be too much precise, but the Schrödinger equation ($i \dot{\psi}= H\psi$ to avoid confusion) has at most an unique solution under very general assumptions on the Hamiltonian operator $H$, even if you see it as a liner equation in the more general setting of Banach spaces. In particular, it is not a priori necessary that $H$ is self-adjoint ...


1

The E field inside the conducting wire is 0, so what is it really doing? The potential difference between two points is related to the electric field along the path between them: $$V_{ba}=\int_a^b \vec{E}\cdot{}\vec{dl}$$ So the fact that the high-conductivity material forces a (near-)zero electric field is exactly why the two ends of the conductor ...


1

Think of a gas. We will ignore gravitational potential, and we'll consider the situation to be at steady state. Compressing the gas takes energy, so we conclude that the entire container of the gas will be at a constant level of compression - in this case, constant density as well. You may intuitively understand that this is true no matter how oblong the ...


1

There is a small misprint in the third line (the 3rd term). If you try to differentiate the second part, you will get exactly part of the force: $\frac{d}{dt}\frac{\partial}{\partial \dot{r}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2 r}\right)=\frac{d}{dt}\left(\frac{2\dot{r}^2}{c^2 r}\right)=\frac{2 \ddot{r}}{c^2 r}-\frac{2 \dot{r}^2}{c^2r^2}$ And also your ...


1

You should look at an electric circuit a like a river or a water slide attraction in an amusement park (see my little artist impression below). The resistors are the steep parts: that's where the potential energy is lost. The wires are the horizontal parts, so there no potential energy is lost. But as the water is already moving, it doesn't stop moving in ...


1

Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla = \langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, \frac{\partial}{\partial{z}}\rangle$$ ...


1

The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...



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