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4 votes
2 answers
2k views

Confusion with partial derivatives as basis vectors

So I have seen that the directional derivative can be written as $$ \frac{df}{d\lambda} = \frac{dx^i}{d\lambda}\frac{df}{dx^i} $$ And we can identify $ \frac{d}{dx^i} $ as basis vectors and $ \...
Hermitian_hermit's user avatar
7 votes
4 answers
16k views

Is Del (or Nabla) an operator or a vector?

Is Del (or Nabla, $\nabla$) an operator or a vector ? \begin{equation*} \nabla\equiv\frac{\partial}{\partial x}\vec{i}+\frac{\partial}{\partial y}\vec{j}+\frac{\partial}{\partial z}\vec{k} \end{...
Sofiane's user avatar
  • 129
2 votes
1 answer
125 views

Vector Derivative: General Case

From "An Introduction to Mechanics" by Kleppner & Kolenkow, SIE-2007, Chapter 1 (Vectors and Kinematics), Section 1.8 - "More about the derivative of a vector". In this section, towards the end, ...
seavoyage's user avatar
  • 203
1 vote
2 answers
108 views

Two different formulas

My problem is simple : given a particle of mass $m$, charge $q$ and velocity $\bf{v}$. If $\bf{A}$ denotes the magnetic potential satisfying $\bf{B}= \nabla \times \bf{A}$. I want to etablish the ...
M.LTA's user avatar
  • 123
0 votes
1 answer
100 views

Why trajectories approach to origin tangent to the slower direction?

I am reading non-linear dynamics from Strogartz. Suppose, I have two solutions of a non linear system: $x(t) = x_0e^{at}$ and $y(t) = y_0e^{-t}$, where $a\in \mathbb{R}$. Now it is clear that,for $a&...
sm10's user avatar
  • 27
6 votes
4 answers
751 views

Do $\vec r$ and $d \vec r$ have the same direction?

One question is bugging me for a long time but I couldn't make out anything nor could my friends. Here it goes: We know, $\vec r$ is regarded as the position vector. So we can say, $$\vec r \cdot\vec ...
SchrodingersCat's user avatar
4 votes
1 answer
3k views

Differentiation of a vector with respect to a vector

Does differentiation of a vector with respect to a vector make any sense? Even if it makes sense, how does it make any physical meaning? I mean what is the physical interpretation?
SchrodingersCat's user avatar
15 votes
5 answers
2k views

What does it mean for a physical quantity if its mixed second partial derivatives are not equal?

This goes for every problem (either in electromagnetism or fluid dynamics) that has to do with vector fields. Say we have a fluid flowing in a closed circular pipe (or an electromagnetic field, the ...
TheQuantumMan's user avatar
2 votes
3 answers
179 views

Difference between $|d{\bf r}|$ and $d|{\bf r}|$

What is the difference between $|d{\bf r}|$ and $d|{\bf r}|$ and why are both of them not always equal to each other? My question might seem stupid to some and will probably get downvoted but I have ...
Karan Singh's user avatar
2 votes
4 answers
20k views

How to find tangential/radial/angular velocity for motion in any curve? [closed]

Is the radial velocity responsible only for changing distance between objects and the component perpendicular to it only for change in direction? If so why? Please try to give a different explanation ...
Robin Hood's user avatar
-1 votes
3 answers
69 views

Vector question, differentials, Electromagnetism

I was reading this demonstration of electric potential in my book: Let $q$ be a point charge at point $P$ The Electric field created at point $M$ by $q$ is : $$\vec{E}(M) = \...
mwa1's user avatar
  • 685
0 votes
1 answer
268 views

Gradient of two-particle system

I'm working on problem 5.1a from Griffiths Intro to QM and given that: $$\mathbf R \equiv \frac{m_1\mathbf{r_1} + m_2 \bf r_2}{m_1+m_2}$$ and $\bf r \equiv \bf r_1 - \bf r_2$ I need to show that, $$\...
Logan's user avatar
  • 207
0 votes
1 answer
68 views

The vector r points from $P'(x',y',z')$ to $P(x,y,z)$ [closed]

For some reason this question is giving me a hard time :( The vector $r$ points from $P'(x',y',z')$ to $P(x,y,z)$. (a) Show that if $P$ is fixed and $P'$ is allowed to move, then $\nabla'(\frac{1}{r}...
sci-guy's user avatar
  • 817
1 vote
2 answers
4k views

Is there a difference in handwritten nabla $\vec{\nabla}$ with an overset arrow and typeset nabla $\nabla$?

According to some physicist at KIT it is usual to write the following when using pen and paper: whereas in typeset texts you write $\nabla$. Is that true? Are there sources for this convention?
Martin Thoma's user avatar
1 vote
4 answers
9k views

Dot product of vector and its derivative with respect to time? How does $L \cdot\frac{dL}{dt} = \frac{1}{2}\frac{d(L^2)}{dt}$? [closed]

How does: $$L \cdot\frac{dL}{dt} = \frac{1}{2}\frac{d(L^2)}{dt}$$ where L is a vector (I dunno how to make it bold in the equation). How do they reach to this right hand side equation? And what is ...
user135688's user avatar
1 vote
1 answer
253 views

Curl of a vector field with two different systems of coordinates

Let $$\mathbf{H} = H_x \mathbf{u}_x + H_y \mathbf{u}_y + H_z \mathbf{u}_z$$ be a vector field whose components are defined with respect to the unit vectors $\mathbf{u}_x$, $\mathbf{u}_y$ and $\...
BowPark's user avatar
  • 777
3 votes
1 answer
133 views

Contradiction of a scalar product

Can anyone resolve this contradiction: $$\vec{r}\cdot\dot{\vec{r}}=\frac{1}{2}\frac{d}{dt}\left(\vec{r}^2\right)=\frac{1}{2}\frac{d}{dt}\left(\left|\vec{r}\right|^2\right)\equiv\frac{1}{2}\frac{d}{dt}...
Andy's user avatar
  • 393
0 votes
2 answers
5k views

How is the direction of the instantaneous acceleration determined?

I know from the text book that the direction of velocity at any point on the 2D path of an object is tangential to the path at that point and is in the direction of motion. But how would one determine ...
KawaiKx's user avatar
  • 941
15 votes
3 answers
44k views

Derive vector gradient in spherical coordinates from first principles

Trying to understand where the $\frac{1}{r sin(\theta)}$ and $1/r$ bits come in the definition of gradient. I've derived the spherical unit vectors but now I don't understand how to transform ...
Lucidnonsense's user avatar
7 votes
6 answers
8k views

How is gradient the maximum rate of change of a function?

Recently I read a book which described about gradient. It says $${\rm d}T~=~ \nabla T \cdot {\rm d}{\bf r},$$ and suddenly they concluded that $\nabla T$ is the maximum rate of change of $f(T)$ ...
Inquisitive's user avatar

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