# Tag Info

46

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.

30

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

27

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 ...

21

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

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 ...

21

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 $x$ ...

19

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 $n(x)$ ...

17

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 ...

12

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

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 $\hat ... 10 Maybe this graph helps to illustrate the point. Consider a constant force that is applied for a finite interval$\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$(an instantaneous impulse), it is necessary to get an infinite slope, hence an infinite force is required which is physically not ... 9 It seems to me that you are confonding a generic notion of total derivative and the so called Lagrangian derivative (also known as material derivative). Let us start from scratch. In Cartesian coordinates, a fluid or a generic continuous body is first of all described by a class of differentiable (smooth) maps from$\mathbb R^3$to$\mathbb R^3$: ... 9 Consider an$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

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 ...

8

$$A(\lambda+\epsilon)B(\lambda+\epsilon) = (A(\lambda) + \epsilon \dot{A} )(B(\lambda) +\epsilon \dot B ) = A(\lambda)B(\lambda) + \epsilon(\dot AB+A\dot B) + o(\epsilon^2)$$

8

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 ...

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

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 $$i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm ... 8 Regarding your first question, note that$$\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}$$Regarding your second question, it is stated that V = ... 7 The result is sometimes called Flanders' lemma. The remarkable point is that it does not need that f is analytic, but just that it is C^\infty. So it does not relies upon the Taylor series as it could seem at first glance, since that series may not converge. It works in any open star-shaped neighborhood of points in \mathbb R^n. A set A\subset ... 7 The issue here is that \dot{r} is not the magnitude of \dot{\vec{r}}, but rather the rate of change of r, the magnitude of \vec{r}. Think about a particle moving in a circle. Since r is constant, \dot{r}=0, but the velocity is certainly not zero! The formula \vec{r}\cdot\dot{\vec{r}} = r\dot{r} makes perfect sense, because \dot{r} is the ... 7 Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form |\Psi \rangle only depends on time perhaps it is more correct to use the last one, but it is matter of taste. In order to get your first equation (a wave equation), you must project on \langle x|:$$H(P=-i\hbar ...

6

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 ...

6

I) The use of $\delta$ in the derivative $$C~=~\frac{\delta Q}{dT}$$ is because in thermodynamics, heat $Q$ is not a state function. In particular, the differential $\delta Q$ is in-exact. II) In detail, the heat capacity $C$ is not obtained by differentiation of some ordinary function wrt. temperature $T$. Rather it should be viewed as a ratio ...

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

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. ${\rm d}T~=~ |\nabla T| |{\rm d}{\bf ... 6 What Goldstein means by$\nabla_iV_i$is $$\nabla_iV_i=\left(\frac{\partial}{\partial x_{1,i}}\hat{x}_{1,i}+\frac{\partial}{\partial x_{2,i}}\hat{x}_{2,i}+\frac{\partial}{\partial x_{3,i}}\hat{x}_{3,i}\right)V_i$$ which is indeed a vector. Here,$\mathbf r_i=(x_{1,i},\,x_{2,i},\,x_{3,i})$is the position of the$i\$th particle (with respect to the origin), ...

6

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: $$\frac{d}{dt} m r(t)^2 \dot{\phi}(t) = m\frac{d}{dr}(r(t)^2)\frac{dr}{dt}\dot{\phi}(t)+mr^2\ddot{\phi} = 2mr\dot{r}\dot{\phi}+ mr^2\ddot{\phi}$$

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