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8

As you said, $|d{\bf v}/dt|$ is the magnitude of acceleration. $d|{\bf v}|/dt$ is the rate of change of speed. In general they are not the same. For example, in uniform circular motion $$\bigg|\frac{d{\bf v}}{dt}\bigg|=\frac{|{\bf v}|^2}{R}$$ while $$\frac{d|{\bf v}|}{dt}=0$$


4

It seems you have some misprints in your formula. Start by calculating the Cristoffell symbols for the new metric $$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + g^{cd}(g_{db}\nabla_a \ln\Omega + g_{ad}\nabla_b \ln\Omega- g_{ab}\nabla_d \ln\Omega)$$ or $$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + (\delta_b^c\nabla_a \ln\Omega + \delta_a^c\nabla_b \ln\Omega- ...


3

Actually, your heart is in the right place; you just have the wrong inner product in mind. Instead of taking the inner product of $f$ and $g$ to be$$\int_0^\infty f(t)g(t)\,dt,$$you should take it to be the average value of $f(t)g(-t)$, in an appropriate sense; this will be proportional to$$\int_{-\infty}^\infty f(t)g(-t)\,dt,$$and thus you can use this in ...


3

The condition that $k$ is a gradient implies that it is curl-free; i.e. $\nabla_i k_j = \nabla_j k_i$. Let $\gamma$ be an integral curve of $k^i$; i.e. $\dot \gamma^i(t) = k^i(\gamma(t))$. Then replacing $\dot \gamma$ with $k$ in the covariant acceleration and applying the curl-free condition we get $$\ddot \gamma^i = \dot \gamma^j \nabla_j \dot \gamma^i = ...


2

If we are dealing with motion in one dimension, then the velocity function is a scalar (i.e. $v: \mathbb{R} \to \mathbb{R}$), and it makes sense to talk about "positive velocity" and "negative velocity" (and likewise for acceleration). I'm going to assume that in what follows. First, it may be helpful to know that $\frac{d}{dt} |t| = sgn(t)$, where ...


2

No, they do not essentially have the same meaning. As stated in the Wikipedia lemma, the 'mathematical' notion of configuration space is closely related to the state space in physics. On the other hand, they don't 'just happen to share the same name', because the notions behind the constructions are similar in some way. To be very brief, the physical notion ...


2

Let be $$\mathcal A_{k l}^m = \int_{-1}^1 P_k^m \left({x}\right) P_l^m \left({x}\right) \, \mathrm d x $$ where the associated Legendre functions are $$P^m_l \left({x}\right) = \left({1 - x^2}\right)^{m/2} \dfrac {\mathrm d^m P_l \left({x}\right)} {\mathrm d x^m}={\left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - ...


2

Start with the fact that the Einstein tensor (and thus $T$) is divergence-free: $$\nabla_a T^{ab} = \nabla_a(\rho u^a)u^b + \rho u^a \nabla_a u^b = 0. \tag{1}$$ The product rule expansion I've chosen is quite suggestive: in the first term we see the divergence term that appears in the continuity equation, while in the second we see the covariant ...


2

What you are looking for is the derivative of the distance between points with respect to time, that is their relative radial velocity. If $r=\sqrt{(x_B-x_A)^2}$ then $$ v_{rel}={dr\over dt}={(x_B-x_A)\cdot(v_B-v_A)\over r}. $$ Notice that this also works if $x_A$ and $x_B$ are vectors: in that case you have a dot product in the numerator.


2

(My understanding is) This is not a purely mathematical problem, but a mixture of physics and math. And the key lies in the physical picture. The heat bath is a trillion times larger than the system. The energy fluctuation of the system with respect to the bath is like a small ripple on the Pacific ocean (let's say there are no waves or currents). So first ...


2

I'm not sure exactly what problem you are trying to solve (or even if you are still trying to solve it...) but I have used the Crank-Nicolson method (i.e., the average of the forward and backward Euler methods) to solve something potentially similar. Rather than using a ground-state potential explicitly you might be able to approximate it using a soft ...


1

Suppose the Cartesian equation of the curve $\gamma$ is given by $y=y(x)$ and $A$ be a fixed point on it. Let $P(x, y)$ be a given point on $\gamma$ such that $\mathrm{arc}(AP)=s$. We know that $$\frac{\mathrm d y}{\mathrm d x}=\tan \psi\tag 1$$ where $\psi$ is the angle made by the tangent to the curve $\gamma$ at $P$ ...


1

Method 1: $a=\cos x - b\sin x$ $\cos x = a+b\sin x$ $\cos^2x = a^2+2ab\sin x+b^2\sin^2x$ $1-\sin^2x = a^2+2ab\sin x+b^2\sin^2x$ $(a^2-1)+2ab\sin x+(b^2+1)\sin^2x=0$ Then quadratic. Method 2: $$\sin(\alpha-x)=\sin\alpha\cos x-\cos\alpha\sin x$$ $$\frac{\sin(\alpha-x)}{\sin\alpha}=\cos x-\frac1{\tan\alpha}\sin x$$ Let $\frac1{\tan\alpha}=b$, then ...


1

The Associated Legendre Polynomials (or Functions) occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz. These functions are of great importance in quantum physics, for example, because they appear in the solutions of the Schrodinger equation in spherical polar coordinates. The ...


1

You are correct that for $\Omega(x)>0$ one can always write $\Omega(x)=\exp(\ln(\Omega(x)))=\exp(\ln(\Omega(x_0+(x-x_0))))$ and then Taylor expand the logarithm. The Taylor expansion of the logarithm is $\ln(\Omega(x_0))+\frac{x-x_0}{\Omega(x_0)}$, so we are left with $\Omega(x_0)\exp \left ( \frac{x-x_0}{\Omega(x_0)} \right )$ when we perform this ...


1

One possibility is to have $$h(x)=[f(x)+g(x)]/2+\cos(x) [f(x)-g(x)]/2 $$ You can replace the $\cos(x)$ with any trigonometric function that oscillates between -1 and 1. The first term is the average between the curves, the second contains the distance from the average to each of the curves


1

The reason we are told that $40~\text{N}$ of force is required to compress the spring $3~\text{cm}$ is that this allows us to solve for $k$. We need to consider the units. Since a Newton ($\text{N}$) is a unit of force and $F = ma$, where $m$ is the mass and $a$ is the acceleration, $$\text{N} = \text{kg} \cdot \frac{\text{m}}{\text{s}^2}$$ where ...


1

The velocity of $A$ relative to $B$ is $$\underline{v}_A-\underline{v}_B$$ Note that the velocities must be expressed in vector form


1

So with the help of @Winther, the solution is as follows: $$ U = \frac{1}{r^3}\left(\vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2}\right) \equiv \frac{1}{r^3}P_iT_j^iQ^j\\ \vec{P}\cdot\vec{Q} - \dfrac{3(\vec{P}\cdot\vec{r})(\vec{Q}\cdot\vec{r})}{r^2} = P_iT_j^iQ^j\\ $$ $$ P^ke_kQ_le^l- ...


1

Even though the question is suited for Physics SE, I'll answer it. We use the term instantaneous velocity to show that the velocity is something at a particular instant (or the smallest amount of time that can be measured). Here, the function $v(t) = 3 - \frac{t}{2}$ shows how the velocity changes with time. If we put $t=6$, we get that $v(6) = 0$. What ...


1

You mean "numerically solve", right? In this case I would suggest you to give FiPy an earnest shot. It's reasonably simple and fairly well documented. It may be the answer to your problem. PS: FiPy is based on the Finite Volume Method (FVM).


1

EDIT: This answer is too wrong to correct. Striking it out. If $v:\mathbb R\to \mathbb R$, then numerically, they are the same where they both exist. That's a consequence of the fact that $(-v)' = -(v')$. But where $v$ changes sign, $|v|$ may not be differentiable. This is because the absolute value function $x\mapsto |x|$ isn't differentiable at $x=0$.


1

In an algebraic expression, all terms which are added or subtracted must have the same dimensions. This implies that each term on the left-hand side of an equation must have the same dimensions as each term on the right-hand side. So you have that $$ [F]=[x]=[x^3]=[M][L][T]^{-2} $$ It's obvious thatt if $x$ and $x^3$ have the same dimension, there exist at ...


1

We have: $$4\left(y-\frac{v^2}{4g}\right)^2=4\left(\frac{v^2\sin^2(\alpha)}{2g}-\frac{v^2}{4g}\right)^2=\frac{v^4\sin^4(\alpha)}{g^2}-\frac{v^4\sin^2(\alpha)}{g^2}+\frac{v^4}{4g^2}$$ $$=\left(\frac{v^4\sin^2\alpha}{g^2}\right)(\sin^2\alpha-1)+\frac{v^4}{4g^2}$$ $$=-\left(\frac{v^4\sin^2\alpha}{g^2}\right)(1-\sin^2\alpha)+\frac{v^4}{4g^2}$$ ...


1

Anytime you see a finite domain, standard heat equation, think separation of variables. Try to use $u(x,y,t) = T(t)X(x)Y(y),$ and you'll get 3 ODE's to solve, something along the lines of \begin{align} T'(t) + c^2(\lambda_x + \lambda_y)T(t) &= 0\\ X'' - \lambda_x X &=0\\ Y'' - \lambda_y Y &=0. \end{align} This should be the motivated approach ...


1

From the context, it seems that the indices $a$, $b$, $c$ can only take the values 1, 2 and 3, and then the formula means that $\{L_1,A_1\}=0$, $\{L_1,A_2\}=+A_3$, $\{L_1,A_3\}=-A_2$, and so on. (The positive sign if $abc$ come in the right cyclic order, negative sign if the wrong order, or zero if there is repeated index.) In detail: summation over $c$ is ...



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