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1 vote
1 answer
109 views

Spherical coordinate of a vector when divergence of the vector is zero

$\nabla \cdot \mathbf{\delta u_{perp}} = 0$ where $\mathbf{\delta u_{perp}}$ is a function of both x and y coordinates and perpendicular to z axis. Moreover, $\delta u_{perp}$ along z axis is $0$. I ...
1 vote
2 answers
286 views

What is the Laplacian of $k\hat{r}$ where $r=\sqrt{x^2+y^2}$ and $k$ is a constant? [closed]

Using 3D cylindrical coordinates, I get 0 as the answer. $$ \nabla^{2} (k \hat{r}) = \hat{r} (\frac{1}{r} \frac{\partial }{\partial r}\left(r \frac{\partial (k)}{\partial r}\right) + 0 + 0) + \hat{\...
1 vote
1 answer
34 views

Derivatives of the lagrangian of generalized coordinates [closed]

I know that $$U= \frac{1}{2} \sum_{j,k} A_{jk} q_j q_k \quad \quad T= \frac{1}{2} \sum_{j,k} m_{jk} \dot{q}_j \dot{q}_k $$ and the lagrangian is $$ \frac{\partial U}{\partial q_k} - \frac{d}{dt} \...
1 vote
1 answer
58 views

Energy change under point transformation

How do the energy and generalized momenta change under the following coordinate transformation $$q= f(Q,t).$$ The new momenta: $$P = \partial L / \partial \dot Q = \partial L / \partial \dot q\times ...
4 votes
1 answer
1k views

Explicit expression of gradient, laplacian, divergence and curl using covariant derivatives

For my course in General Relativity I am given the problem to find the expressions for the gradient, laplacian, divergence and curl in spherical coordinates using covariant derivatives. I have tried ...
0 votes
4 answers
5k views

Sign of acceleration from position-time graph

These three graphs are from my textbook. It states that the acceleration in 1) is positive, 2) is negative and 3) is zero and can be told by looking at the slope. What I understand from the graph is ...
0 votes
0 answers
108 views

What does $\nabla'$ mean? [duplicate]

In D.J Griffiths Electrodynamics (Page 173) it says, $\nabla' |\vec{x}| = \frac{\hat{x}}{x^2}$. However by my calculation $\nabla |\vec{x}| = -\frac{\hat{x}}{x^2}$ so what does the $\nabla'$ signify?
0 votes
1 answer
159 views

Relationship between derivatives of tensors in different Cartesian coordinate systems

I'm new to tensor calculus: I'm reading a little introductory book whose title is "Quick Introduction to Tensor Analysis", written by R.A Sharipov. I've reached the section called ...
0 votes
1 answer
31 views

Can someone please explain me how this came? [closed]

I am not getting how above equation is derived using cylindrical coordinates transformations. This is from page 36, Mathew Sadiku
0 votes
1 answer
53 views

What did I do wrong? I got $\nabla\cdot \vec A \neq div \vec A $ [closed]

We know, that in orthogonal Curvilinear coordinate system: $$ \nabla =\sum_{i=1} ^{3}{\hat{e_i} \over h_i}{\partial \over\partial u_i} $$ Let $$\vec A=\sum_{i=1} ^{3} A_i \hat e_i$$ Now $$ \nabla ...
1 vote
2 answers
635 views

Calculate the expression of divergence in spherical coordinates $r, \theta, \varphi$

Hi this is my first question in [Physics.SE] I saw a lot of posts and I liked them. I hope that my question will be answered too. While I'm solving a problem in vector calculus. I recognized that I ...
0 votes
2 answers
164 views

Derivation of $x \partial_y - y\partial_x = \partial_{\phi}$

On a $S^2$-sphere we can define the coordinates $$x = \sin(\theta)\cos(\phi)\\ y = \sin(\theta)\sin(\phi)\\z=\cos(\theta).$$ Then I want to show that $$x \partial_y - y\partial_x = \partial_{\phi}.$$ ...
1 vote
2 answers
111 views

Derivatives of polar coordiantes?

I'm a undergraduate and I was reading about the polar coordinate system specifically this paper. I don't understand the term: $$\frac{de_r}{d\theta} = e_\theta \text{, and } \frac{de_\theta}{d\theta}...
1 vote
1 answer
128 views

Generalized Coordinates Property for a System of Particles

I"m looking at "Principles of Dynamics: Second Edition" by Donald T Greenwood. I'm trying to figure out how he obtains Eq. (6-64) $$\frac{\partial\dot x_j}{\partial\dot q_i} = \frac{\partial x_j}{\...
2 votes
1 answer
838 views

Galilean transformation and differentiation

Given $x=x’-vt$ and $t=t’$, why is $\frac{\partial t}{\partial x’}=0$ instead of $1/v$? $t$ seems to depend on $x’$ because if $t$ changes, $x’$ changes. Also, in this problem, $dx=dx’$ as well, but I ...
1 vote
1 answer
104 views

Where is the error in this calculation of net curl for simple magnetic field?

I wasn't sure whether to post this on MSE, but PSE seems more appropriate. Let B be a static magnetic field in spherical coordinates, defined as $B=r\hat{\theta}$. Then, it's curl is $$\nabla \times ...
0 votes
2 answers
9k views

What is the derivative of an angle? [closed]

What is the derivative of an angle? I don't understand
0 votes
3 answers
3k views

Proof divergence of magnetic field is 0

I work in an R&D role that involves magnetism. I am refreshing my memory of electromagnetic and this stumps me. In polar coordinates, the magnetic field of a current loop for distances $R >>...
0 votes
3 answers
174 views

Question about differentiation of tensors

According to Arnab Rai Choudhuri, Astrophysics for physicists Page 363: $$\frac{\partial \overline A^i}{\partial \overline x^l}=\frac{\partial A^k}{\partial x^m}\frac{\partial x^m}{\partial \overline ...
0 votes
2 answers
164 views

Problem with derivatives for spherical coordinates [closed]

I got stuck with a derivative. I can't think of a solution for this, because I am taking the derivative of a function with respect to its integral. Theta and phi are generalized coordinates. I am ...
0 votes
1 answer
91 views

In central-force mechanics, how do we substitute $ξ=\frac{1}{r}$?

I have taken a look at central-force mechanics in the past, but I still cannot understand how $ξ=\frac{1}{r}$ is substituted to find $\frac{d^2r}{dt^2}$ in terms of ξ. So I know from $F=ma$ that: $$(...
0 votes
0 answers
28 views

Evaluating derivatives with respect to certain vector axis

So, I am trying to work in Spherical coordinates. I have a predefined fixed axis, $\hat{v}_0$, so that $\alpha=\vec{r}.\hat{v}_0$ Now, I am interested in the following: \begin{equation} f(r,\alpha)=\...
1 vote
1 answer
2k views

How to get the gradient potential in polar coordinate

In polar coordinate, $$\nabla U = \frac{\partial U}{\partial r}\hat{\mathbf{r}} + \frac{1}{r}\frac{\partial U}{\partial \theta}\hat{\mathbf{\theta}} .$$ Can anyone show me how to get this result?