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23

You are essentially asking about the material derivative when discussing a total derivative with respect to time. Let's say you are looking at the velocity of the air in your room. There is a different velocity everywhere, and it changes with time, so $$v = v(x,y,z,t)$$ When you take a derivative like $$\frac{\partial v}{\partial t}$$ you are saying "I ...


18

The Laplacian measures what you could call the « curvature » or stress of the field. It tells you how much the value of the field differs from its average value taken over the surrounding points. This is because it is the divergence of the gradient..it tells you how much the rate of changes of the field differ from the kind of steady variation you expect ...


18

I usually explain it this way: $$\rho = \rho(t,x(t),p(t))$$ $$\frac{\partial\rho}{\partial t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t),p(t))-\rho(t,x(t),p(t))}{\Delta t}$$ $$\frac{d\rho}{d t} = \lim_{\Delta t \to 0} \frac{\rho(t+\Delta t,x(t+\Delta t),p(t+\Delta t))-\rho(t,x(t),p(t))}{\Delta t}$$


18

The symbol $\Delta$ refers to a finite variation or change of a quantity – by finite, I mean one that is not infinitely small. The symbols $d,\delta$ refer to infinitesimal variations or numerators and denominators of derivatives. The difference between $d$ and $\delta$ is that $dX$ is only used if $X$ without the $d$ is an actual quantity that may be ...


17

joseph f. johnson already gave a motivation for the one-dimensional case. I think the physical example diffusion equation $$\frac{\partial}{\partial t}n(x)=D \frac{\partial^2}{\partial x^2}n(x)$$ is the best way to illustrate it: If the function looks like $x^2$ (positive curvature), like left and right, then in the next time step the function $n(x)$ ...


11

It does follow from calculus. Here's the standard way this is treated (I'm not going to be explicit about mathematical details such as smoothness assumptions here). Definition of $\delta q$. Given a parametrized path $q:t\mapsto q(t)$, we consider a deformation of the path which we call $\hat q:(t, \epsilon)\mapsto \hat q(t,\epsilon)$ satisfying $\hat ...


9

Consider an $n$-dimensional space (two dimension in the picture), and let $f(\vec x)$ be a non-constant scalar function, like a temperature distribution in your case. Let $\vec y(t)$ be any curve in the space such that the function $f(\vec y(t))=c$ is constant along that trajectory (the colored lines). Now compute the scalar product $\left\langle ., ...


8

Nothing really new to add to the two great answers given already - just a particular example that helped me. When you examine the finite-difference version of Laplace's equation in 2 dimensions, you find that the discretized $\phi$ satisfies Laplace's equation if, in this picture of part of the grid, $\phi_{i,j}$ at the centre is the average of the ...


7

If you want a simple intuitive explanation, you can get a lot from vehicles. In a car traveling at a constant speed, suppose there is a white dot painted on the top of the steering wheel. If that dot is in the center, you are traveling in a straight line. If you turn it some angle to the left, say 90 degrees, then the car is traveling in a circular arc at a ...


7

If you have just given the voltage signal with $$ \def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V $$ then the current at $t=2\rm s$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that $$ i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ...


7

The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that $f$ is analytic, but just that it is $C^\infty$. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in $\mathbb R^n$. A set $A\subset ...


6

The equation ${\rm d}T~=~ \nabla T \cdot {\rm d}{\bf r}$, says that the change in T, namely ${\rm d}T$, is the scalar product of 2 vectors, $\nabla T$ and ${\rm d}{\bf r}$, which can also be written as the magnitude of the 1st vector times the magnitude of the 2nd vector times cosine the angle between them. ${\rm d}T~=~ |\nabla T| |{\rm d}{\bf ...


6

I) Here we discuss the problem of defining a connection on a conformal manifold $M$. We start with a conformal class $[g_{\mu\nu}]$ of globally$^{1}$ defined metrics $$\tag{1} g^{\prime}_{\mu\nu}~=~\Omega^2 g_{\mu\nu}$$ given by Weyl transformations/rescalings. Under mild assumption about the manifold $M$ (para-compactness), we may assume that there ...


6

In the classical mechanics picture, where an object's position is a real-valued function of the real parameter time, then yes, you are correct. If the object is at rest over any extended interval, then it's position as a function of time over that interval is simply $x(t) \equiv 0$, which of course means $x^{(k)}(t) = 0$ for any $k$-th derivative at any ...


6

It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) ...


5

Maybe an intuitive answer is best given in terms of classical physics. Suppose you are looking at the movement of a classical particle. The relevant variables here are position and momentum. If you solve the motion of your system, you are presented with functions $x(t)$ and $p(t)$. Now, there are a lot of derived quantities you can build from these ...


5

You are right that the result you see is due to the chain rule. The author uses either spherical or cylindrical coordinates, so \begin{equation} r = \sqrt{x^2 + y^2 + z^2} \end{equation} or \begin{equation} r = \sqrt{x^2 + y^2} \end{equation} which you can differentiate to obtain \begin{equation} \frac{\partial{r}}{\partial{x}} = \frac{x}{r} ...


5

Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form $|\Psi \rangle$ only depends on time perhaps it is more correct to use the last one, but it is matter of taste. In order to get your first equation (a wave equation), you must project on $\langle x|$: $$H(P=-i\hbar ...


5

Yes, it's just the second derivative of some function, it doesn't matter that this function is organized as a component of a tensor, $h_{\mu\nu}$. The identity above – assuming the function is differentiable and smooth etc. (add some "niceness" conditions on the function) – follows from the rules of calculus and is formally proven by the ...


5

The ratio is meant to denote $$ p = \lim_{\Delta A\to 0} \frac{\Delta F}{\Delta A}$$ where $\Delta A$ is the area of a particular piece of the surface whose magnitude we send to zero; and $\Delta F$ is the correspondingly small force that acts on this small area. In practice, it's enough to choose $\Delta A$ small enough so that the pressure is constant (the ...


5

You're almost good, you just needed to use the chain rule, which you did perhaps without knowing it. It is clearer if you write it this way perhaps: $$ \frac{d}{dt} m r(t)^2 \dot{\phi}(t) = m\frac{d}{dr}(r(t)^2)\frac{dr}{dt}\dot{\phi}(t)+mr^2\ddot{\phi} = 2mr\dot{r}\dot{\phi}+ mr^2\ddot{\phi}$$


4

$\partial_t\equiv\frac\partial{\partial t}$ and $\partial^\mu\equiv g^{\mu\nu}\frac\partial{\partial x^\nu}=\left(\sum_{\nu=0}^3g^{\mu\nu}\frac\partial{\partial x^\nu}\right)_{\mu=0}^3$ are differential operators. $\partial^\mu$ is formally contravariant (upper index) and obeys the corresponding transformation laws. $\partial_t$ has a lower index and is (up ...


4

The physicist's derivative notation denotes the components of a Frechet derivative in the direction of the delta-function supported at $y$. This is one of those places where the habit of denoting the function $f$ by its value $f(x)$ gets confusing. It's somewhat clearer if you write $\delta_y$ for the delta function at $y$, and $$ \frac{\delta F}{\delta ...


4

1.) The differentiation operator acting will give rise to Kronecker-Deltas since $\frac{\partial x_a}{\partial x_b}=\delta_{ab}$ This will kill one summation. More specifially: $\frac{\partial U}{\partial x_a}=-1/2 \sum_{ij}b_{ij}(\delta_{ai}x_j+\delta_{aj}x_i)=-1/2( \sum_{j}b_{aj}x_j+\sum_{i}b_{ia}x_i)=-\sum_{j}b_{aj}x_j$. Rename j to be i and you're ...


4

This is more like a maths question to me. This is just an identity, which is true and facilitates the calculation and it is valid for any vector field. The proof, using Einstein summation convention would be something like: $$ (\nabla \times \vec u )\times \vec u = \epsilon_{ijk}(\nabla \times u)_j u_k = \\ \epsilon_{ijk}\epsilon_{jlm}\partial_l (u_m) u_k ...


4

It is actually a result from maxima and minima in basic calculus. ]Notice that at the peak of this function, the derivative goes to 0. This is because if a function is increasing at first, and then decreases, it means it has reached a maximum. For example, if a train is moving slowly, then starts to pick up speed[, it reaches its maximum, aafter which its ...


4

We take: $$x=r\cos\theta\cos\phi$$ $$y=r\cos\theta\sin\phi$$ $$z=r\cos\theta$$ Now, you know the definition of the gradient in spherical coordinates: $\vec{\nabla}=\frac{\partial}{\partial x}\hat{x}+\frac{\partial}{\partial y}\hat{y}+\frac{\partial}{\partial z}\hat{z}$ Now, we use the chain rule or each component. For instance, $$\frac{\partial}{\partial ...


4

I) The point of Ref. 1 is similar to why the generalized positions $q^j$ and the generalized velocities $\dot{q}^j$ in the Lagrangian $L(q,\dot{q},t)$ are independent variables, see e.g. this Phys.SE post. A less confusing notation would probably be to denote the generalized velocities $v^j$ instead of $\dot{q}^j$. Ref. 1 is referring to the non-commutative ...


3

Here we will only consider the added last subquestion (v4): $$\tag{1} \frac{d}{d\lambda}e^{\hat{A}} ~=~ \int_0^1\!ds~e^{(1-s)\hat{A}}\frac{d\hat{A}}{d\lambda}e^{s\hat{A}} .$$ The identity (1) follows by setting $t=1$ in the following identity $$\tag{2} e^{-t\hat{A}} \frac{d}{d\lambda}e^{t\hat{A}} ~=~ ...



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