Hot answers tagged

47

It means don't be a jerk. The third derivative of position (i.e. the change in acceleration) is called "jerk", though it's a little used quantity. It's called jerk because a changing acceleration is felt as a "jerk" in that direction.


31

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


31

The simplest way to explain the Christoffel symbol is to look at them in flat space. Normally, the laplacian of a scalar in three flat dimensions is: $$\nabla^{a}\nabla_{a}\phi = \frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$$ But, that isn't the case if I switch from the $(x,y,z)$ ...


31

Typically: $\rm d$ denotes the total derivative (sometimes called the exact differential):$$\frac{{\rm d}}{{\rm d}t}f(x,t)=\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{{\rm d}x}{{\rm d}t}$$This is also sometimes denoted via $$\frac{Df}{Dt},\,D_tf$$ $\partial$ represents the partial derivative (derivative of $f(x,y)$ with respect to $x$ ...


30

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 ...


30

In physics, sometimes the third derivative of position with respect to time is called jerk.


25

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 ...


22

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 $n(x)$ ...


21

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 ...


20

Killing fields are one of the most important concepts in general relativity both in its classical as well as quantum versions. Classically, one thing we are always interested in is the world-line/trajectory of a free-falling observer in curved space-times. These world-lines are described as geodesics and satisfy the equation $$ \frac{d^2 x^\mu}{d\tau^2} + ...


12

Regarding your first question, note that $$\frac{d}{dx}\ln c = \frac{1}{c}\frac{dc}{dx}$$ Thus $$\frac{1}{a}\frac{da}{dx} + \frac{1}{b}\frac{db}{dx} = \frac{1}{c}\frac{dc}{dx}$$ Then, as physicists often do, 'cancel' the $dx$ on both sides to yield $$\frac{da}{a} + \frac{db}{b} = \frac{dc}{c}$$ Regarding your second question, it is stated that $V = ...


11

what if distance and time both become zero as at origin in the graph is It appears that you're trying to say that the velocity is equal to the position of the particle divided by the clock time at that position: $$v= \frac{x(t)}{t}\; ? $$ But this isn't correct. Average velocity $\bar v$ is defined as displacement $\Delta x = x(t_f) - x(t_i)$ ...


11

On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...


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 ...


11

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 ...


10

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) ...


10

It seems to me that you are confonding a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from $\mathbb R^3$ to $\mathbb R^3$: ...


10

Maybe this graph helps to illustrate the point. Consider a constant force that is applied for a finite interval $\Delta t$. The slope in the $p$ graph at any given time is $\frac{dp}{dt}$. Now, in order to have a step in $p$ (an instantaneous impulse), it is necessary to get an infinite slope, hence an infinite force is required which is physically not ...


9

This is an axiom, not a result. When defining the covariant derivative, we choose for it to obey a number of properties. Carroll's Spacetime and Geometry summarizes these well -- look at the preprint for Chapter 3 here. We of course want $\nabla$ to act linearly on its argument and to obey the product rule. We also demand that it commute with contractions ...


9

The reason you see the $da$, $db$, or $dc$ is precisely because it isn't specified what the derivative is taken with respect to. What happens is the following: First, they write the differential element of the logarithmic function: $$d\,\ln f=\frac{df}{f}$$ This is always true no matter what function $f$ is. When we take a derivative with respect to some ...


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 ., ...


9

This is true - in fact you could define $\nabla^\sigma = g^{\sigma\rho} \nabla_\rho$. I assume this meant to say $$ g^{\sigma\rho} \nabla_\nu \nabla_\sigma = \nabla_\nu \nabla^\rho. $$ Again, this is true, but for a slightly less trivial reason than (1). To employ (1) to prove this, you need to be able to switch $g^{\sigma\rho}$ with $\nabla_\nu$, which you ...


9

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 ...


9

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 ...


9

Curl can be equated with the closed line integral in the limit that the encircled area $\Delta S$ goes to zero. However, we would have to do this in three components because curl is a vector. $$ (\nabla \times \vec{v})_x = \lim_{\Delta S \rightarrow 0} \frac{1}{\Delta S} \oint \vec{v}\cdot d\vec{l} $$ in the $yz$ plane and so on. But what does it mean? ...


8

Despite what some of the other answers are mentioning, the following equation you have is correct $$ \vec{r} \cdot d \vec{r} = r dr $$ You can check this by noting $$ \vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k} $$ Then $$ \vec{r} \cdot d \vec{r} = x dx + y dy + z dz $$ Further note $$ r = ...


8

Imagine water flowing steadily in a stream, steadily enough that the surface of the water never changes shape. The water forms a three-dimensional manifold $M$. Watching how the water moves over a period of $t$ seconds gives a diffeomorphism $\phi^t \colon M \to M$. If the current is carrying a diatom down the stream, and you see it at the point $p$, you ...


8

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 ...


8

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 ...



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