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## Hot answers tagged physics

4

Suppose we have a parallel plate capacitor with area $A$ and thickness $d$. We will establish a coordinate system in which the plates are on the planes $x=0$ and $x=d$. Assume that the permittivity varies as a function of $x$ so that $\epsilon = \epsilon(x)$. Now, partition the interval $[0,d]$ into $N$ equal-sized subintervals of size $\Delta x=d/N$. ...

3

Let me try to explain what is happening with a simpler example. Consider the one dimensional problem $$u_t=u_{xx},\quad 0<x<1,\quad t>0,$$ with boundary conditions $u(0,t)=0$, $u(1,t)=1$, $t>0$ and initial condition $u(x,0)=0$, $0\le x\le1$. Here $\Omega=(0,1)$, $x=0$ would be the inner sphere and $x=1$ the outer sphere. To use separation of ...

3

The distinctions between the Riemann integral and the Lebesgue integral are entirely driven by pathological functions which are mathematically interesting, but have no relationship to the physical world. The most pathological that functions get in the physical world are ones where integrals need to be renormalized, but Lebesgue does not add anything to the ...

3

Is that what you are looking for? $$\begin{split} \text{Tr}(\partial^\mu(A G)\partial_\mu (AG)^{-1}) &= \text{Tr}(A\partial^\mu G\partial_\mu G^{-1}A^{-1})\\ &= \text{Tr}(\partial_\mu G^{-1}A^{-1}A\partial^\mu G) \\ &= \text{Tr}(\partial_\mu G^{-1}\partial^\mu G) \\ &= \text{Tr}(\partial^\mu G \partial_\mu G^{-1}) \end{split}$$

2

Ok, this is the physicists quick and dirty route. For a more formal analysis one could apply Laplace method or calculate the integral exactly in terms of Bessel function's (the solution is $\frac{m}{4\pi r} K_1(m r)$) and use their well known asymptotics. Due to the exponetial decay, the integral will be dominated by a small region around $s=m$. We put ...

2

I looked at your proof a bit and it seems to me you have the right ideas. But a little measure theory can make life easier. Well known result: If $f\in L^\infty(\mathbb {R}^3), g \in L^1(\mathbb {R}^3),$ then $$\tag 1 f\ast g (x) = \int_{\mathbb {R}^3} f(y)g(x-y)\,dy$$ is continuous on $\mathbb {R}^3.$ (And in fact $f\ast g (x)\to 0$ as $|x|\to \infty.$) ...

2

$$a_1 \cos (\omega t+ \phi_1)+a_2 \cos (\omega t+ \phi_2)=a \cos (\omega t+ \phi)$$ $$\Rightarrow a_1(\cos \omega t \cos\phi_1 - \sin\omega t \sin\phi_1)+a_2(\cos \omega t \cos\phi_2 - \sin\omega t \sin\phi_2)=a(\cos \omega t \cos\phi - \sin\omega t \sin\phi)$$ $$\Rightarrow (a_1 \cos\phi_1 + a_2 \cos\phi_2)\cos \omega t - (a_1 ... 2 The change R(r)=\sqrt{r}\,\phi(k\,r) transforms the equation into$$ 4\,k^2\,r^2\phi''(k\,r)+4\,k\,r\,\phi'(k\,r)+(4 k^2 r^2-4 l^2+4 l-1)\,\phi(k\,r)=0. $$Now let k\,r=\rho and divide by 4 to get Bessel's equation$$ \rho^2\phi''(\rho)+\rho\,\phi'(\rho)+\Bigl(\rho^2-\Bigl(l-\frac12\Bigr)^2\Bigr)\,\phi(\rho)=0. $$2 It's the Laplacian. I.e. \nabla^2 f=\sum \frac{\partial^2}{\partial x_i^2} f 1 Working backwards and assuming no rotational effects:$$d=\frac{1}{2}at^2F=maF_{total}=F_{right}-F_{left}F_{right}=F\cos\thetaF_{left}=F_{normal}\ \muF_{normal}=mg-F\sin\theta1 If you assume additionally that the friction is proportional to surface area and to velocity, then you can run the following logic to get a pretty simple ODE for v: \begin{align} r & =c_1t + c_2 \\ m & =c_3 r^3 \\ F & =c_4 m - c_5 r^2 v = c_6 r^3 - c_5 r^2 v \\ F & =m'v+mv' \\ & = 3c_3 r^2 r' v + c_3r^3 v' \\ & = 3c_3 c_1 r^2 v + ... 1 As A.G. stated, the (strict) local maxima of potential energy are unstable equilibria. E.g., a ball at rest at the top of a sphere, or a pendulum balanced at its highest position. These can't be sustained in reality, due to inevitable perturbations (air motion, etc). A non-strict local maximum may happen to be neutrally stable if the potential energy is ... 1 Use a right triangle to visualize this situation. You know certain facts 1) the speed of the particles can be represented on the two legs 2) the hypotenuse can be the distance between the two particles Thus, the function representing the distance will be based on the Pythagorean theorem. 1 The particles can be viewed as one moving along the negative x-axis away from the origin, the other moving towards the origin along the negative x-axis. The equations for position can be given as  r_1=(-25t,0) and r_2=(0, 20t-20) This distance between them is the hypoteneuse of the right angles triangle formed by the x coordinates of the first particle ... 1 There are two possibilities here. Which one is correct depends on the context: Add up the ratios as you've said, then divide by the count of numbers you've added together (from your statement, it sounds like you have more than 1 data item per row - you need to count data items summed, not rows). You add up all the \omega_f and add up all the \omega_i ... 1 Partial work. Let define the following function:g(\xi)=\int_{\mathbb{R}}\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x.$$Notice that:$$g'(\xi)=-i\int_{\mathbb{R}}x\exp\left(-\frac{x^2}{2a}\right)\exp(-i\xi x)\,\mathrm{d}x=-\frac{\xi}{a}g(\xi).$$I used derivation under the integral and then integration by parts. Moreover, using substitution, ... 1 The moment of inertia of the shell would be given by:$$\frac{2m}5 * \frac{R^5-a^5}{r^3-a^3} https://physics.ucsd.edu/neurophysics/courses/physics_2bl/p2bl_experiment_2_notes.pdf has the whole calculation under 0.0.1. The final answer is just a factorized form, if that's what's confusing you, look at the second last step and so on.

1

Second answer (no measure theory): Let $f$ be a nice bounded function on $D$ (like $k\rho$). For suitable $g,$ define $Tg(x) = \int_D f(y)g(x-y)\,dy, x \in \mathbb {R}^3.$ One suitable $g$ is any $g\in C_c$ (where $C_c$ is the set of continuous functions on $\mathbb {R}^3$ with compact support). Other examples are $g(y) = y_k/|y|^3.$ These are the only ...

1

Yes, it is correct. Indeed, let $\phi \colon \mathbb{R}^N \to \mathbb{R}$ be of class $C^2$, then \begin{align} \operatorname{div}(\nabla\phi) := \sum_{i = 1}^N\frac{\partial}{\partial x_i}(\nabla\phi)_i = \sum_{i = 1}^N\frac{\partial^2 \phi}{\partial x_i^2} =: \Delta\phi. \end{align}

1

We have $I_\text{rms}^2 = {1 \over \pi}{ \pi \over 3}(I^2+4 I^2 + I^2) = 2 I^2$. As an aside, with a pure sine wave, we have $I_\text{rms} = {1 \over \sqrt{2}} I_\max$, and here we have $I_\max = 2I$, which would correspond to $I_\text{rms} = \sqrt{2} I$ and a fundamental of $2I$, so these results show that the fundamental above is consistent with this. ...

1

To draw the problem in a coordinate system, choose a positive $x$ axis in the downward direction since the motion is vertical. Take the spigot at $x = 0$, the top of the tank at $x = 5$, the surface of the water at $x = 16$ and the floor at $x = 20$. The partition of water is therefore the closed interval $[16,20]$ on the $x$ axis. Take an element of volume ...

1

Your mistake comes from the assumption that each component of the force should be proportional to $\dfrac 1 {r^2}$. Clearly, this cannot be, since then how would you tell the difference between $F_x$, $F_y$ and $F_z$? Obviously, each of these must somehow contain a $x$, a $y$ and a $z$ respectively. This suggests $F_x \propto \dfrac x {r^3}$ etc. Can you ...

1

Yes, convex implies simply-connected. In fact, convex implies contractible: choose a point $x_0$ in the set. For any other point $x$, define $h(t,x) = (1 - t)x + tx_0$. $h$ is a contraction of the set to a point. Any closed curve in the set also contracts to a point under $h$, so the set is simply-connected. No, the converse is not necessarily true: ...

1

The infinitesimal surface element on the sphere can be thought of to be a rectangle with the sides $\sin\theta d\phi$ and $d\theta$. The first is parallel to the $xy$-plane, thus its projection is going to have the same length. Now the second side (i.e. $d\theta$) has a slope. This second side and the projection of it build a right angle (vertical) triangle, ...

1

To answer your questions: 1) What is your basis for assuming what the amplitude of a wave originating from the slit is? Your assumption is not clear to me and does not appear to be correct. 2) See 1). 3) What is it that you want to do? I will derive for you an expression for the field amplitude at a point on the screen. Let us start from first ...

1

Approximation around $\frac{\pi}{2}$ is only valid for a small (technically infinitesimally small) deviations, which implies that the result is only valid when $\frac{g}{R\omega^2}$ is very small. This means that $\cos\theta = \cos(\frac{\pi}{2}-\frac{g}{R\omega^2}) = \sin\frac{g}{R\omega^2} \approx \frac{g}{R\omega^2}$, pretty close to what conjecture 1 ...

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