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You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

3

The answer to that is because the moment of inertia is not the same for the solid cylinder than for the hollow one. As you write the formula for the moment of inertia, it depends on the distribution of the mass. The further away the mass is from the rotation axis, the more contributes to the moment of inertia (as in distance squared $r^2$). So, since the ...

3

Your question actually is one of the most important questions in analytic mechanics. This is because, when you explicitly write the Eulero-Lagrange equations for any constrained system with $n$ degrees of freedom and Lagrangian of the form: $$L(t, {\bf q},\dot{\bf q}) = T(t, {\bf q},\dot{\bf q}) - U(t, {\bf q},\dot{\bf q})$$ where $T$ is quadratic in ...

2

For first question, you must simly take $l = 0$(in your notations, $l$ is curved). It's because angular momentum is conserved in radial-symmetry field problem, and you can simply take $L^2 = \hbar^2 l(l+1)$. So, for $l = 0$ states you simply ommit $\frac{L^2}{2mr^2}$ term. Potential is given in problem. It behaves like constant for some $r$ values, and ...

2

I really don’t like that diagram. No, REALLY. I think it conveys a bad intuition that may confuse you. I don’t like what it does with the connexion coefficients (Christoffel symbols). Here’s why. In a general manifold, tangent spaces of course are not comparable at different points. This is in contrast with the situation in Euclidean space, which can be ...

2

Here's a close analogy which (besides having has its own practical use) can shed some light, if you're looking at the problem from the probabilistic-mathematical point of view. Imagine we play the following game ($A$) : we throw $N$ balls into $K$ boxes, with uniform probability. Let call the result (or configuration) $X=(x_1,x_2 ...x_k)$, where $x_i$ is ...

2

Inside vs outside, there is a sign change inside the square root, so that changes the nature of the "phase" $\phi(r)$. Normally, when you match wave functions you require that $\psi_\mathrm{left}(x) = \psi_\mathrm{right}(x)$ (continuity) and that the derivative changes according to what you get when you integrate the Schrodinger equation: ...

2

In principle, yes, you can specify the stress tensor and solve the resulting equations, but in practice, this is hard to do because the field equations are non-linear PDEs...darn. The simplest possible example is the case in which the stress tensor vanishes; $T_{\mu\nu} = 0$ namely the vacuum equations. The field equations with vanishing cosmological ...

2

Your calculation is nice, and the answer seems correct. Another way (less nice) to calculate it is to imagine the charge in the center as a small charged ball (to avoid infinities) and calculate by how much is the field energy lower when it is in the center than when it is far away. Due to symmetry and the Gauss law, the electric field in the former case is ...

2

The matrix you quote has the following determinant $$\det\left( \begin{array}{c c c} -1+2/8& 2/8& 2/8\\ 2/8 & -1+(2/8) & 2/8\\ 2/8 & 2/8 & -1+(2/8) \end{array} \right) = -1/4$$ which is not unitary required by quantum mechanics. It implies that your matrix is not possible to be constructed using any standard quantum gates which ...

2

The basic idea for this is to use the momentum space version of the Schroedinger equation: $$\hat{p}\to p,\quad\hat{x}\to i\hbar\frac{\partial}{\partial p}$$ and then solve the system1, $$\left[\frac{p^2}{2m}+img\hbar\frac{d}{dp}\right]\phi=E\phi$$ which should be solvable (e.g., complex exponentials). You can then Fourier transform to physical space to ...

2

The string contacts the point on two infinitesimally close points with different slopes. Imagine a small pulley end the two points are the entry and exit point of the string. If the string is between points A on the left and point B on the right (with B lower) then we call the angles of the string from horizontal $\theta_A$ and $\theta_B$. If the mass is ...

1

Knowing the reason why you think it is $2A$ rather than $4A$ could help with a more focussed answer, but here it goes anyway. First, note that in one period the block starts from $x=A$ and ends at the same position. Second, the block does not first turn around at $x=0$, but rather at $x=-A$. Now try breaking the motion up into two half-periods. That is, ...

1

You are right in thinking that the car's acceleration is what keeps it in place, but it is important to remember that an object moving at a constant speed in a circle is accelerating (despite not speeding up). The reason for this is that acceleration is defined as a "change in velocity," and velocity is a vector quantity (i.e. it has magnitude and ...

1

As I had written in the comments, it is the second term that you have gotten incorrect. Focusing exclusively on this term (leaving aside the $1/8$ factor), we have $$|\psi_2\rangle \langle \psi_2| = \frac{1}{3}(|00\rangle \langle 00|+ |10\rangle \langle 10|+ |11\rangle \langle 11| - |00\rangle \langle 10|-|10\rangle \langle 00| + \ ...),$$ where I have ...

1

You didn't specify in what direction the force of hand is applied, so for simplicity I assume that you are applying the force perpendicular to the desk. Now there are four forces on the book: 1) Gravity ($mg$) is trying to take the book down; it has a component $mg\cos\theta$ that is perpendicular to the desk and a component $mg\sin\theta$ that is parallel ...

1

I find this much better motivated in Weinberg's text "Quantum Theory of Fields", which starts from the idea of particles, and what we measure, rather than from a Lagrangian formulation. Essentially, what we are after is a Poincar\'e invariant and unitary S-matrix. One crucial thing that we require is a Hamiltonian density, transforming as a scalar, and ...

1

I'll try to explain parallel transport first: This image shows how a vector generally does not remain parallel to itself under parallel transport. The vector here is moved along the path A-N-B and is constrained to remain in the tangential plane to the spherical surface at all points of the path because the geometry does not know of the "third" dimension ...

1

Well, you are not yet done! You should also fix $A,B,C,D$ by imposing that the wavefunction is $C^1$ crossing the discontinuities of the potential. Following that way you obtain that only an overall arbitrary constant remains. You have thus a function like this $k\psi_E$, where $k\neq 0$ is any complex number, can be fixed arbitrarily (without imposing ...

1

They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

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