All Questions
23 questions
1
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1
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109
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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 ...
- 31
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.
$$ \nabla^{2} (k \hat{r}) = \hat{r} (\frac{1}{r} \frac{\partial }{\partial r}\left(r \frac{\partial (k)}{\partial r}\right) + 0 + 0) + \hat{\...
- 19
1
vote
1
answer
34
views
Derivatives of the lagrangian of generalized coordinates [closed]
I know that
$$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} \...
- 11
1
vote
1
answer
58
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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: $$P = \partial L / \partial \dot Q = \partial L / \partial \dot q\times ...
- 990
0
votes
0
answers
108
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What does $\nabla'$ mean? [duplicate]
In D.J Griffiths Electrodynamics (Page 173) it says, $\nabla' |\vec{x}| = \frac{\hat{x}}{x^2}$. However by my calculation $\nabla |\vec{x}| = -\frac{\hat{x}}{x^2}$ so what does the $\nabla'$ signify?
0
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1
answer
159
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Relationship between derivatives of tensors in different Cartesian coordinate systems
I'm new to tensor calculus: I'm reading a little introductory book whose title is "Quick Introduction to Tensor Analysis", written by R.A Sharipov. I've reached the section called ...
- 540
4
votes
1
answer
1k
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Explicit expression of gradient, laplacian, divergence and curl using covariant derivatives
For my course in General Relativity I am given the problem to find the expressions for the gradient, laplacian, divergence and curl in spherical coordinates using covariant derivatives.
I have tried ...
- 53
0
votes
1
answer
31
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Can someone please explain me how this came? [closed]
I am not getting how above equation is derived using cylindrical coordinates transformations.
This is from page 36, Mathew Sadiku
- 27
0
votes
1
answer
53
views
What did I do wrong? I got $\nabla\cdot \vec A \neq div \vec A $ [closed]
We know, that in orthogonal Curvilinear coordinate system:
$$ \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 ...
- 123
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 ...
user262095
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 $$x \partial_y - y\partial_x = \partial_{\phi}.$$
...
- 147
1
vote
2
answers
111
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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: $$\frac{de_r}{d\theta} = e_\theta \text{, and } \frac{de_\theta}{d\theta}...
- 13
1
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1
answer
128
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Generalized Coordinates Property for a System of Particles
I"m looking at "Principles of Dynamics: Second Edition" by Donald T Greenwood. I'm trying to figure out how he obtains Eq. (6-64)
$$\frac{\partial\dot x_j}{\partial\dot q_i} = \frac{\partial x_j}{\...
- 149
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 ...
- 341
1
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1
answer
104
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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 $$\nabla \times ...
0
votes
4
answers
5k
views
Sign of acceleration from position-time graph
These three graphs are from my textbook. It states that the acceleration in 1) is positive, 2) is negative and 3) is zero and can be told by looking at the slope.
What I understand from the graph is ...
- 493
0
votes
2
answers
9k
views
What is the derivative of an angle? [closed]
What is the derivative of an angle? I don't understand
0
votes
3
answers
3k
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Proof divergence of magnetic field is 0
I work in an R&D role that involves magnetism. I am refreshing my memory of electromagnetic and this stumps me. In polar coordinates, the magnetic field of a current loop for distances $R >>...
- 23
0
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3
answers
174
views
Question about differentiation of tensors
According to Arnab Rai Choudhuri, Astrophysics for physicists Page 363:
$$\frac{\partial \overline A^i}{\partial \overline x^l}=\frac{\partial A^k}{\partial x^m}\frac{\partial x^m}{\partial \overline ...
- 859
0
votes
2
answers
164
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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 ...
- 143
0
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1
answer
91
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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:
$$(...
- 21
0
votes
0
answers
28
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Evaluating derivatives with respect to certain vector axis
So, I am trying to work in Spherical coordinates. I have a predefined fixed axis, $\hat{v}_0$, so that $\alpha=\vec{r}.\hat{v}_0$ Now, I am interested in the following:
\begin{equation}
f(r,\alpha)=\...
- 490
1
vote
1
answer
2k
views
How to get the gradient potential in polar coordinate
In polar coordinate,
$$\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?
- 573