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5

$$\begin{align}F \color{orange}{-} (m_1a+\mu_xm_1g)-\mu_xm_2g &= m_2a\\ F \color{orange}{-} m_1a\color{orange}{-}\mu_xm_1g-\mu_xm_2g &= m_2a \\ F \color{red}{- m_1a}-\mu_xm_1g-\mu_xm_2g &= m_2a\\ F -\mu_xm_1g-\mu_xm_2g &= \color{red}{ m_1a} + m_2a \\ F -\mu_xm_1g-\mu_xm_2g &= \color{blue}{(m_1+m_2)}a\\ \frac{F ...


4

Once you choose a consistent system of units, and express the variables and parameters in those units, the units should cancel and leave you an equation that involves only pure numbers. Otherwise, your equation is wrong. For example, an equation $A + B = 0$ where $A$ is a length and B is a mass can't be correct. Look up "dimensional analysis" for more on ...


3

To the best of my knowledge there is no such book. David Hestenes has a web page "Selected Papers on Geometric Algebra in Quantum Mechanics": http://geocalc.clas.asu.edu/html/GAinQM.html


3

Everything you quoted (planetary theory, hydrostatics, mechanics, astronomy, optics, electricity & magnetism, elasticity) in some form or another are still compulsory subjects in maths departments in universities: for example, most academic universities in Russia (MSU, MPTI, NSU, SPbSU to name the top ones), well-reputed Grandes Ecoles in France ...


2

Only in the twentieth century did a substantial amount of mathematics come into being that was not connected with physics. mathematical logic / proof theory; (some) abstract algebra; computability; type theory; ... Thus it is now possible to have a mathematical career untouched by physics. There is at least one substantial body of such people. We call ...


2

It depends. As Robert points out in the comments, there are trivial cases where there are no such points, but in some cases (more than one point mass), there always must be. Mathematically speaking, what you are asking is the following: given the gravitational potential $V(x)$ due to some set of massive bodies, does there exist any point $x$ where the force ...


2

Here is a way to formalize your first idea, assuming $B$ is compact. Consider $B$ as a metric space where the distance between two points, $d(x,y)$, is given by the length of the shortest path within $B$ joining $x$ and $y$. (This is just the standard definition of distance on a compact Riemannian manifold.) Then a ball of radius $r$ around a point $x$ is ...


2

Although they are not dedicated to QM, the following two books devote a few chapters to the modeling of QM using Geometric Algebra : Doran & Lasenby, "Geometric algebra for physicists", CUP, 2003 (chapter 8 is about "Quantum theory and spinors", chapter 9 is relevant also) William Eric Baylis, "Clifford (geometric) algebras with applications to ...


2

The lecture note of Physical Applications of Geometric Algebra is a very good review of GA in physics, including some chapters on QM.


2

The key thing to remember is that the linking rods are rigid and must maintain their length throughout. Put the origin at O. The point P is instantaneously $b\cos\theta, b\sin\theta$ which automatically satisfies $|OP|=b$. If $Q=(z,0)$, one must have $|PQ|=c$. This gives you $$ (z-b\cos\theta)^2 + (b\sin\theta)^2 = c^2 \\ z^2 - 2bz \cos\theta + b^2-c^2 = ...


2

Consider a thin horizontal sheet of water, going from distance $y$ from the bottom of the tank to distance $y+dy$. The volume of this water is approximately $\pi x^2\,dy$. This water has to be lifted through distance $4-y$. The work done is approximately $$(62.5)(4-y)(\pi x^2)\,dy.$$ "Add up" (integrate) from $y=2$ to $y=3$. Note that $x^2=4y$, so we want ...


2

From the first line to the second, they added $m_1a$ to both sides and used the distributive law on the right. From the second to the third, they divided by $(m_1+m_2)$, then plugged in numbers from somewhere.


2

See the notion of fibration: A fibration is like a fiber bundle, except that the fibers need not be the same space, rather they are just homotopy equivalent.


2

The projection $\mathbb R^3\setminus \{0\}\to \mathbb R: (x,y,z)\mapsto x$ is a smooth submersion between connected smooth manifolds and all its fibers are connected. However the fibers are not diffeomorphic: the fiber of $0\in \mathbb R$ is difffeomorphic to $\mathbb R^2\setminus \{0\}$ whereas all other fibers are diffeomorphic to $\mathbb R^2$.


2

Projectile equation: $$y=x\tan\theta-\frac{gx^2}{2v^2}\sec^2\theta$$ When $y=0$ (i.e. at launch and at landing), $$\begin{align} x\left(\frac{gx}{2v^2}\sec^2\theta-\tan\theta\right)&=0\\ x&=0,\frac{2v^2}g \frac{tan\theta}{\sec^2\theta}\end{align}$$ Hence range $$R=\frac{v^2}g ...


1

You have your Vector Identity messed up, it should be: $\nabla \times \left( \nabla \times \mathbf{B} \right) = \nabla(\nabla \cdot \mathbf{B}) - \nabla^{2}\mathbf{B}$, where $\nabla^{2}=\Delta$ is the Vector Laplacian. And we know from Maxwell's equations that $\nabla \cdot \mathbf{B} = 0$, and so we are left with $$\nabla \times \left( \nabla \times ...


1

It states on page 40 that "$\omega$ ... gives the infinitesimal angles", and it states right after your equation on page 42 that it is an "infinitesimal version" of an equation which follows it. So basically yes, we would only expect them to match to first order in $\omega$.


1

The wave amplitude in the focal plane is the Fourier transform of the wave amplitude in the pupil plane of the lens $L$. Assuming we are dealing with a diffraction-limited lens, the relationship between object amplitude $A_o$ and image amplitude $A_i$ is thus a convolution: $$A_i(x,y) = \iint_{\mathbb{R}^2} dx' dy' \, L(x',y') A_o(x-x',y-y')$$ Note that ...


1

$E_{ij} (E^{-1})^{jk} = (A-Cz^2) \delta_i^k + (B+C)z_iz^k +(A+C) \varepsilon _{ijm}g^{jk} z_m$. According to the definition, $E_{ij} (E^{-1})_{jk} = \delta_i^k$, so we must have : $(B+C) = (A+C)=0, (A-Cz^2)=1$, that is the result. we use $ \varepsilon _{ijk} \varepsilon ^{ilm} = (\delta^l_j \delta^m_k - \delta^l_k \delta^m_j )$ in the computation.


1

To check your answer, consider an arbitrarily small "cell" of water inside the tank after it is filled. If the water in that cell has mass $\mathrm d m$ and is at height $y$ above the ground, the amount of energy you have to expend to raise it from the ground to its final position is $y\;\mathrm d m.$ For each such cell of water above the level of the ...


1

This is a valid manipulation of formal Laurent series $\mathbb{C}[[e^{ix},e^{-ix}]]$. Note that formal Laurent series don't form a ring because some pairs of elements can't be multiplied, but we indeed have that $\sum_{k=0}^{\infty}{e^{ikx}}$ is the multiplicative inverse of $1-e^{ix}$ and similarly for negative powers.


1

The derivative of your function $d$ w.r.t. $\theta$ is $$\frac{\partial d}{\partial\theta} = \frac 1 g\left(v\cos \theta\left( \frac{v^2\sin\theta\cos \theta}{\sqrt{2gh+v^2\sin^2\theta}}+v\cos\theta\right)-v\sin \theta\left(\sqrt{2gh+v^2\sin^2\theta}+v\sin \theta\right)\right)$$ Setting this to zero and solving for $\theta$ (in a specific range such as ...


1

Parametrize the curve $c(t)=(\cos t,\sin t), t\in [0,2\pi].$ Thus, given $u=(\alpha x,-\alpha y),$ we have $$\int_C (\alpha x, -\alpha y) \cdot dr =\int_0^{2\pi} (\alpha(\cos t,-\sin t)\cdot c'(t)) dt \\= \alpha\int_0^{2\pi} ((\cos t,-\sin t)\cdot (-\sin t,\cos t))dt \\=\alpha\int_0^{2\pi} (-\cos t\sin t-\sin t\cos t)dt \\=-\alpha \int_0^{2\pi}\sin(2t)dt=0. ...


1

$$\begin{align} e^{-r}\int^r e^\sigma\,\sigma^{-1/2}\,d\sigma&= e^{-r}\Bigl(e^\sigma\,\sigma^{-1/2}\Bigr|^r+\frac12\int^re^\sigma\,\sigma^{-3/2}\,d\sigma\Bigr)\\ &=r^{-1/2}+\frac{e^{-r}}{2}\int^re^\sigma\,\sigma^{-3/2}\,d\sigma\\ &=r^{-1/2}+O(r^{-3/2}), \end{align}$$ since $$ ...


1

Since it seems you are not incorporating a damping term, I'll proceed under that assumption. Let $k_i$ denote the spring constant for the spring above mass $m_i$, $i=1,2$. Then by Newton's Second Law $F=ma$ and Hooke's Law which says that the restorative force of the $y$ units beyond its equilibrium position, i.e. $F_\text{restorative}=-ky$, we get ...


1

$\ddot{r}=\dfrac{-Gm_e}{r^2}-r\dot{\theta^2}$, $\ddot{\theta}=-2\dfrac{\dot{r}\dot{\theta}}{r}$, It is now classical. To derive satellite orbits we use two equations of equilibrium in curvilinear coordinates. To get both components, twice differentiate the vector $ p=r e^{ i \theta} $ to get the radial and circumferential force components with respect to ...


1

I think (I can't actually remember at the moment) that the dragging factor is certainly proportional to the speed. So, the last equation you wrote yields two ordinary differential equations: \begin{align*} &\frac{d^2}{dt^2}r_x(t) + \frac{c}{m}\frac{d}{dt}r_x(t) = 0\\ &\frac{d^2}{dt^2}r_y(t) + \frac{c}{m}\frac{d}{dt}r_y(t) + g = 0 \end{align*} To ...


1

Variation of momentum means force: $F = \dot p$. In a very non-rigorous way, we have: $$ F = \frac{dp}{dt} \quad\Longrightarrow\quad Fdt = dp \quad\Longrightarrow\quad F = \frac{\Delta p}{\Delta t} $$ The force in this case, it is likely assumed to be the gravitational force: $F = mg\sin\theta$, where $\theta$ is the angle of the ladder and the horizontal ...



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