Skip to main content

All Questions

Filter by
Sorted by
Tagged with
0 votes
1 answer
225 views

The time derivative of a vector not defined in terms of the time variable $t$

Recently I got a question where I needed to determine the time derivative of a position vector. However, the vector didn’t have the variable $t$ but instead had $x$, $y$, and $z$ as its terms, so I ...
Andrew Norfield's user avatar
0 votes
3 answers
785 views

Derivative of cross product equation

Recently I got a problem that equated the time derivative of a cross product d/dt (P x Q) with a function of time (like t + t^2). Ex. d/dt (P x Q) = 5t - 6t^2 My question is, how can you have an ...
Andrew Norfield's user avatar
2 votes
3 answers
193 views

Is $ d \mathbf v · d \mathbf v = d \mathit v^2 $?

My teacher has proved the following: $$ \mathit v^2 = \mathbf v·\mathbf v = \frac{d\mathbf r}{dt}·\frac{d\mathbf r}{dt} = \left(\frac {ds}{dt}\right)^2 \Rightarrow \mathit v = \frac{ds}{dt} $$ Because ...
Pascu22's user avatar
  • 23
4 votes
1 answer
167 views

What motivates defining vectors as first order differential operators?

I have read some introductions to geometrical ideas and tensors and physics and what some of them do (see, for example, Frankel's Geometry of Physics) is define a vector as a first order differential ...
JDThinking's user avatar
1 vote
1 answer
303 views

How to ("geometrically") differentiate unit vectors of spherical coordinates?

I have been trying to derive the expressions of partial derivatives of unit vectors with respect to each other in the spherical coordinate system. I was able to get all of them except $\frac{\partial \...
Mohit Kumar's user avatar
1 vote
6 answers
713 views

Why does differentiating a scalar give a vector? [closed]

I was wondering why $F=-\frac{dU}{dr}$ would give me a vector quantity when a scalar quantity is differentiated. There are similar pre-existing queries but I think this issue has yet to be properly ...
NG E EN CASPER HCI's user avatar
4 votes
4 answers
413 views

Suppose there is a vector $\vec v$ which is a function of time, then will $\dfrac{d}{dt}|\vec v|$ be a vector quantity or a scalar quantity?

Suppose there is a vector $\vec v$ which is a function of time, then will $\dfrac{d}{dt}|\vec v|$ be a vector quantity or a scalar quantity? I think it should be scalar because, let's assume $\vec v=...
Akshaj Bansal's user avatar
0 votes
0 answers
152 views

Product rule for 4-vectors and derivation of 4-force form

In deriving the form for the 4-force in special relativity, we begin with $$\frac{d}{d\tau}p^{\alpha}=m\frac{d}{d\tau}u^{\alpha}=f^\alpha$$ where $\tau$ is the proper time, m is rest mass. Since $...
lilolalorphism's user avatar
0 votes
1 answer
87 views

What mean this momentum-derivative?

I'm working with quantum gravity. I have to make a Taylor-series. I got some help for this, but I have problem with understanding the formalism. So, I have the operator $A((P-p)^2)$, which needs to ...
Atka's user avatar
  • 3
0 votes
1 answer
2k views

Derivative of a metric tensor

I would like to ask you a question - maybe simple - but bothering me. We have two four-position vectors product in curvilinear coordinates given by $(1) \quad X^{\alpha}g_{\alpha \beta}X^{\beta} = \...
pog's user avatar
  • 9
1 vote
0 answers
66 views

Differentiating the four-velocity contracted with itself

Let us denote $u^\mu$ as the contravarient component of a four velocity at a point in some coordinate system for a pseudo-Riemannian manifold. I want to examine the following equation. $$\partial_\nu(...
Nugi's user avatar
  • 551
0 votes
2 answers
85 views

How to Differentiate $\vec{r}$ in Polar Form?

Question First off, I do realize that: $$\vec{r} = r \hat{r}$$ $$\dot{\vec{r}} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$$ $$\ddot{\vec{r}} = \ddot{r} \hat{r} + \dot{r} \hat{r} + \dot{r} \dot{\...
Athenian's user avatar
0 votes
1 answer
197 views

4-velocity lowering index question

The 4-velocity in contravariant form is given by $$V^\mu=\frac{dx^\mu}{d\tau}$$ for some general co-ordinates $x^\mu$ and proper time $\tau$. Is the 4-velocity in covariant form given by $$V_\nu=V^\...
John Eastmond's user avatar
1 vote
1 answer
50 views

Path Coordinates: direction problem (doubt) in derivative of tangential vector

Why is the direction of derivative of tangential vector perpendicular to the direction of the tangential vector?
Prakul Virdi's user avatar
0 votes
1 answer
509 views

Divergence of a vector multiplied by dot product [closed]

If I am correct, then $\operatorname{div} [(\vec A\cdot \vec B)\vec C] = (\vec A \cdot \vec B) \operatorname{div} \vec C + \vec C \cdot \nabla (\vec A\cdot\vec B)= (\vec A \cdot \vec B) \operatorname{...
Kubrick's user avatar
  • 113
1 vote
1 answer
554 views

Meaning of normal acceleration?

acceleration means the rate of change in velocity (vector quantity) and the differentiation means to divide a certain quantity into small elements (i.e $dx$) as we do to find the acceleration at any ...
Kareem Ahmed's user avatar
0 votes
1 answer
78 views

Why do we neglect $\Delta t^2(\frac{d\vec{r}}{dt}\frac{d\vec{\hat{r}}}{dt})$ at Taylor Expansion?

I'm just started to Ankara University Physics Department two weeks ago. I have missed my 2 hours of PHY105 course that is the last week Wednesdey. The subject that i missed was Derivatives of Vectors. ...
M. Çağlar TUFAN's user avatar
0 votes
1 answer
248 views

Non-uniform circular motion with constant radius of curvature [closed]

$\let\oldhat\hat \renewcommand{\vec}[1]{\mathbf{#1}} \renewcommand{\hat}[1]{\oldhat{\mathbf{#1}}}$ Suppose we have a car moving on a circular track of radius $b$ and speed $v=ct$, where $t$ is time ...
coreyman317's user avatar
-1 votes
2 answers
158 views

What does it mean for velocity to be a derivative of position, if position a point, not a function? [closed]

So in mass-spring simulation I encountered that one simulates particles by using positions and velocities of particles etc. People may say that velocity is the derivative of position. But isn't "...
mavavilj's user avatar
  • 459
6 votes
2 answers
1k views

Terminology for time derivative of speed (not velocity)

Is there any standard terminology for the derivative of the magnitude of velocity with respect to time (suitable for use in first-year Calculus)? The word ‘acceleration’, in its technical sense, is ...
Toby Bartels's user avatar
3 votes
1 answer
225 views

Derivative with respect to vector

How in Lagrangian and Hamiltonian mechanics we take derivatives with respect to velocity and momentum respectively if they are vectors? Can we take derivative with respect to a vector?
Antonios Sarikas's user avatar
2 votes
1 answer
105 views

Work-Kinetic energy theorem derivation

So I came across this derivation in the book Classical Mechanics by Herbert Goldstein. I don't follow from the second step onwards. I understand that there's a dot product, but how do you compute it? ...
Pugs's user avatar
  • 87
1 vote
1 answer
415 views

Vector calculus notation, maybe?

I just got a new book on turbomachinery that uses some notation I'm not familiar with. $$ \nabla \lor \vec{W} = -2\vec{\Omega} $$ The del-(something)-vector format makes me think its vector calculus....
ericksonla's user avatar
0 votes
1 answer
204 views

Can you apply product rule to arg of a bra-ket?

I found the following expression in a paper: $$ \frac{d}{dt}\arg\langle\phi_+|\dot{\phi_-}\rangle $$ where the $\arg$ term is the argument of the complex number given by inner product between two ...
TribalChief's user avatar
1 vote
0 answers
74 views

Does the proper four-acceleration $A^{\mu} = (0,0)?$

Let the proper four-position vector $x^{\mu}(\tau) = (0, \tau)$. Differentiating this successively wrt $\tau$ I get the four-velocity $u^{\mu}(\tau) = (0, 1)$ and then the four-acceleration $A^{\mu}(\...
Physiks lover's user avatar
13 votes
7 answers
3k views

Can we divide a vector by another vector? How about this: $a = vdv/dx?$

My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$ It says acceleration vector equals velocity (as ...
4d_'s user avatar
  • 876
1 vote
1 answer
2k views

Is $\nabla=\nabla'$? Nabla operator acting on $r^n$

I have been taught that $$\nabla r^n =\text{gradient of }r^n =n r^{n-1}\ \hat{\boldsymbol r}$$ but in introduction to electrodynamics by Griffith (4th edition) on page 173, $\nabla' r^n$ is given by $-...
A lotta doubts's user avatar
2 votes
1 answer
278 views

Dot product in cylidrical coordinates

I'm given the vector: $$\vec{V}{(r,θ,z)}=\frac{1}{r}\hat{e_r} + (r\cosθ)\hat{e_θ}+\frac{z^2}{r^2}\hat{e_z}$$ I want the scalar product ${\vec{\nabla}}\cdot{\vec{V}}$ We know that in cylindrical ...
Jim Β's user avatar
  • 111
0 votes
3 answers
1k views

How to determine the direction of instantaneous acceleration in a 2D motion? [duplicate]

How do we determine the direction of instantaneous acceleration when the body is moving in a plane (or a 3D space)? This question has been truly bothering me for nearly two weeks. I looked it up, ...
4d_'s user avatar
  • 876
0 votes
2 answers
84 views

Acceleration in a non-inertial reference frome - derevation

The general velocity equation for a point B in on body rotating and translating about point A with respect to the inertial reference frame say 'xyzo' can be expressed as, $\vec{r_{B/o}} = \vec{r_{A/o}...
Raptor's user avatar
  • 17
0 votes
3 answers
84 views

How is velocity defined in circular motion in central force field?

In my view the velocity is change of displacement in the increasing direction of displacement. Now in circular motion in central force field the particle is changing its direction then how is the ...
Nobody recognizeable's user avatar
-1 votes
1 answer
149 views

Basic vector calculus: Show that $\nabla \vec{r} = \vec{1}$ [closed]

Show that $\nabla \vec{r} = \vec{1}$ My instructor in my E & M class put the $r$ and $1$ in bold. I am not sure what a bold one means. From my work I get $1ii + 1jj + 1zz$.
WAS's user avatar
  • 119
-1 votes
2 answers
273 views

How can I show that the acceleration vector for uniform circular motion undergoes uniform rotation?

Does it suffice to show that the dot product between the acceleration vector and the derivative of the acceleration vector = 0?
slothropp's user avatar
4 votes
2 answers
261 views

The acceleration of circulation motion

We know that in circular motion the position vector is $r\hat{r}$. Then the velocity is the time derivative of it. So it gives $$dv/dt = r d\hat{r}/dt + \frac{dr}{dt} .\hat{r}.$$ now I can't ...
Nobody recognizeable's user avatar
0 votes
1 answer
156 views

Which relation is correct for resultant instantaneous velocity in 2d?

Please forgive me if the following question sounds silly and I can't exactly pin point where exactly the problem is but there is some problem with my understanding of vectors. In Cartesian ...
horaceZettai's user avatar
1 vote
1 answer
2k views

Time derivative of vector in rotating frame with angular velocity about a rotating axis

In general, I know that if you have a vector $\vec{F}$ in a rotating frame, and the frame has an angular velocity $\vec{\Omega}$ that the time derivative of $\vec{F}$ in a fixed frame would be $$\frac{...
WnGatRC456's user avatar
0 votes
3 answers
1k views

Feynman's Four Gradient in Special Relativity

I am having trouble understanding Feynman's explanation of four gradient. In section 25-3 of Vol. 2 of the Feynman lectures, he explains why the four gradient is not $(\frac{\partial}{\partial t},\...
A. Remorov's user avatar
1 vote
1 answer
130 views

On the derivative of a vector function

In "An Introduction to Mechanics" by Kleppner and Kolenkow, in the section on the time derivative of a vector: Given $A(t)$ is a vector valued function, then, $$\Delta A = A(t + \Delta t) - A(t)$$ ...
trynalearn's user avatar
4 votes
2 answers
5k views

How is dot or cross product possible using the del operator?

Yesterday in class my teacher told me that the del operator has a direction but no value of its own (as its an operator). So it can't be called exactly a vector. But in vector calculus we see that div ...
Theoretical's user avatar
  • 1,432
-1 votes
1 answer
651 views

What is difference between $d\vec{l}$ and $\vec{dl}$? [closed]

What is difference between $d\vec{l}$ and $\vec{dl}$? $d$ means differential as usual.
Magneto's user avatar
  • 15
0 votes
1 answer
482 views

Partial Derivative of a scalar (absolute distance) with respect to its position vector

Imagine we want to take the partial derivative of a quantity, we will call it $\rho_i = f(F(r_{ij}))$ with respect to a particle's position vector, $\vec{r}_k$. In mathematical terms, this would be ...
cwm5412's user avatar
  • 15
5 votes
2 answers
2k views

Why the unit vector is represented as a partial derivative in GR?

Can someone give a good intuitive explanation why we represent the unit vector as a partial derivative in GR and what does it mean?
Noam Chai's user avatar
  • 595
1 vote
5 answers
7k views

Direction of velocity vector in 3D space

According to a well-known textbook (Halliday & Resnick), the direction of a velocity vector, $\vec v$, at any instant is the direction of the tangent to a particle's path at that instant, as is ...
Mihail's user avatar
  • 113
4 votes
4 answers
1k views

Does $\nabla \cdot \vec{E}$ have the same meaning of dot product $ \vec{a}\cdot \vec{b}=|a| |b| \cos(θ)$?

When the del operator is involved, does the meaning of dot and cross product of vectors change? That is, $\nabla \cdot \vec h$ is defined as below: $$\nabla \cdot \vec h = \frac{\partial h_x}{\...
Satheesh Paul Antonysamy's user avatar
0 votes
1 answer
259 views

Confusion in adding constant to magnetic vector potential

The $x$-component of $B$ is: $B_x=\dfrac{\partial {A_z}}{\partial y}-\dfrac{\partial {A_y}}{\partial z} =\dfrac{\partial {(A_z+C_1)}}{\partial y}-\dfrac{\partial {(A_y+C_2)}}{\partial z}$ where $...
Joe's user avatar
  • 137
0 votes
1 answer
274 views

Kinetic energy derivation: Why is $\frac{d \mathbf v}{dt} \cdot \mathbf v= \frac 12 \frac{d}{dt}(v^2)~?$

In Goldstein's Classical Mechanics 3rd edition, page 3, the Kinetic energy is derived by considering the work done on a particle by an external force $\mathbf F$ from point $1$ to point $2$ $$W_{12}=\...
user avatar
2 votes
4 answers
318 views

Why is $\vec{v}\cdot d \vec{v} = v dv$? [closed]

Can someone help me understand why is this true: $$\vec{v} \cdot d \vec{v} = v \cdot dv$$ where $v$ is speed? I found somewhere that $\vec{v} \cdot d \vec{v}=|\vec{v}||d \vec{v}| \cos \phi$. And I ...
Tine Mlač's user avatar
3 votes
2 answers
134 views

Generalization of $F=mv\frac{dv}{dx}=\frac{m}{2}\frac{d}{dx}(v^2)$ to 3-dimensions in a compact notation

Starting from $F=ma=m\frac{dv}{dt}$, in 1-dimension, it is easy to show that $$F=mv\frac{dv}{dx}=\frac{m}{2}\frac{d}{dx}(v^2).\tag{1}$$ Can we generalize this formula in 3-dimensions? In 3D, $$\textbf{...
Solidification's user avatar
4 votes
2 answers
861 views

Integration of tangential acceleration with respect to time

Here, by tangential acceleration, I mean the component of acceleration along the velocity vector. What do you get when you integrate tangential acceleration with respect to time? What does the '$v$' ...
xasthor's user avatar
  • 1,106
0 votes
2 answers
471 views

Why exactly in general relativity are tangent vectors defined as maps from functions to $\mathbb{R}$?

I am basing this on the lectures from the hereaus international winter school on gravity and light. If $M$ is the manifold of physical spacetime, then at any point $p \in M$, we have a tangent space ...
user56834's user avatar
  • 1,880