7
$V=0$ at the center, but how does $V$ change as you move away from that point? $\nabla V$ doesn't care about the value of $V$, it cares about the rate of change of $V$.
More directly, the appropriate thing to do here would be to calculate $V(x,y)$ for arbitrary $x$ and $y$. From there you can take your derivatives, and only at the end would you set $x$ and ...
5
Here are the key points, which I'll express for an arbitrary number of spacetime dimensions because that makes the pattern more clear:
Inside $\sqrt{g}$, the thing denoted $g$ is the magnitude of the determinant of the metric tensor $g_{ab}$.
$g^{ab}$ are the components of the inverse of the metric tensor, defined by the condition $\sum_b g^{ab}g_{bc}=\...
5
If you expand $U$ to linear order in $t$, your density matrix will also only have trace one to linear order $t$, so $\mathrm{tr}(\rho(t))=1+O(t^2)$. As long as you get this, you did everything fine. Of course, your results will only be correct as long as the terms of order $t^2$ and higher will be small compared to the rest.
4
I cannot imagine what discrepancy you are talking about. The EOM yields
$$ \partial_ r \left (\frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2}\right )=-\dot p_r, $$
which amounts to
$$
\frac{p_\theta^2}{mr^3} = \dot p_r.
$$
It is quite misleading to think of it as "the" radial force, as it willfully skips the centripetal acceleration.
The radial ...
4
Since the question talks about mass (and not light) maybe the expected answer regards excape velocity, which for a mass big enough in a radius small enough would exeed the speed of light. (So relativity is involved, but with a quite notorius fact).
Now excape velocity is a High School topic I think...
3
I'll give a more general approach. Recall the basic setup of Noether's theorem. If $\delta\phi_i$ is any variation of the fields of your theory you have $$\delta{\cal L}=\dfrac{\partial\cal L}{\partial \phi_i}\delta\phi_i+\dfrac{\partial\cal L}{\partial(\partial_\mu\phi_i)}\partial_\mu\delta\phi_i+\cdots+\dfrac{\partial \cal L}{\partial (\partial_{\mu_1}\...
3
Since all calculations did by hft actually is right, we only proposed picture for this question how coil looks like from three different point.
3
To compute electric field from potential you have to have a continuous potential function $ V(x,y,z)$ first. Only then you can take the gradient and put $ x=a,y=b,z=c$ to calculate $\vec{E}$ at any particular point $(a,b,c)$.
3
Here is the proof. Consider two events $p$ and $q$ connected by a timelike segment, i.e., a timelike geodesic of Minkowski spacetime:
$$\gamma(\tau) = p + \tau{\bf n}\:, \quad {\bf n}:= \frac{\vec{pq}}{\sqrt{- g(\vec{pq}, \vec{pq})}}\:.$$
Since the constant tangent unit vector ${\bf n}$ to the segment is future directed and timelike, we can fix a Minkowski ...
3
As Jakob commented, to prove identities of that kind it is often good to go back to the definition of the adjoint operator as arising from an inner product. Given an inner product $(\cdot,\cdot)$ and an operator $\hat{A}$, one defines the adjoint operator $\hat{A}^\dagger$ to be the operator that satisfies
$$(v,\hat{A}w) = (\hat{A}^\dagger v,w)$$
for all ...
3
The "-50 J" work done by the resistive force will decrease the kinetic energy of the body. This comes from the work-energy theorem. Assuming the body started from rest at the given acceleration, when the resistive force starts operating, it'll decrease the net acceleration in a way that depends on the force( whether its constant, or variable). ...
2
Your relation dictates $(\hat{n}\cdot \vec{\sigma})^2=I$, hence, for integer m,
$$
(-\hat{n}\cdot \vec{\sigma})^{2m}=I, \qquad (-\hat{n}\cdot \vec{\sigma})^{2m+1}= -\hat{n}\cdot \vec{\sigma}.
$$
So, what is
$$
\exp \Bigl ( i\phi (-\hat{n}\cdot \vec{\sigma}) \Bigr )= \cos \Bigl ( i\phi (-\hat{n}\cdot \vec{\sigma}) \Bigr ) +i\sin \Bigl ( i\phi (-\hat{n}\cdot \...
2
If you are pushing a box at constant force $F$ and the box is moving at constant speed (acceleration $a=0$) this means that the toal force $F_T$ acting on the box is $F_T=0$ (because $F_T=ma=0$). So there must be another force $F_f$ due to friction acting on the box that cancels out your force
$$F_f=-F$$
(the same applies for a car, with $F$ the "...
2
In the 1D world-line, the square root
$$\sqrt{g_{tt}}~=~e~>~0$$
of the metric $g_{tt}$ is an einbein. And the inverse metric is
$$g^{tt}~=~\frac{1}{e^2}.$$
Therefore Witten's Lagrangian (1) is
$$L~=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}.\tag{1'}$$
Variation wrt. the einbein $e$ leads to
$$\frac{\dot{x}^2}{e^2}+ m^2~\approx~0,\tag{2'}$$
which is Witten's ...
2
The minima is found by differentiating and setting the derivative to zero.
$$
\frac{dU}{dr}= -4\epsilon \left[12 \left(\frac{\sigma}{r}\right)^{12}\frac{1}{r}- 6 \left(\frac{\sigma}{r}\right)^{6}\frac{1}{r}\right]
$$
Setting the term in the square bracket to zero yields the correct expression for the position of the minima:
$$
\left(\frac{\sigma}{r}\right)^{...
2
In the referenced paper we read
It is straightforward to establish (for example by
Fourier transforming with respect to $z$) that
\begin{equation}
\boxed{\:\:
\lim_{\upsilon\boldsymbol{\rightarrow} 1}\dfrac{1\boldsymbol{-}\upsilon^2}{\left[\left(z\boldsymbol{-}\upsilon\,t\right)^2\boldsymbol{+}\left(1\boldsymbol{-}\upsilon^2\right)r^2_{\boldsymbol{\perp}}\...
2
The question is ambiguous, as the answer depends on how $m_2$ is placed on $m_1$. If it is moving with $m_1$, then you're adding kinetic energy to the system:
$$ \frac 1 2 m_1 v^2 \rightarrow \frac 1 2 (m_1+m_2) v^2$$
and if it's stationary, then you are taking energy out (but conserving momentum):
$$ \frac{p^2}{2m_1} \rightarrow \frac{p^2}{2(m_1+m_2)}$$
...
2
The first method is not correct by assuming a shperical symmetric geometry. For an element charge locates on the cylinder $\vec{r}' = (r', \phi', z')$. It contribution to the field point $\vec{r} = (r, \phi=0, z=0)$, $d\vec{E}$ will be along the direction $\hat{n}$ unit vector along the $\vec{r} - \vec{r}'$.
The square distance
$$
|\vec{r} - \vec{r}'|^2 = ...
2
You don't need to think of the whole loop all at once. Each of the four legs of the "loop" can be considered one at a time.
Also, the problem is in a sense two-dimensional, since the way it is specified the loop is fixed such that only torque about the x-axis matters. So if it is easier, you can just draw two-dimensional diagrams only showing y and ...
2
You are correct. The reason is that endurance $S_e$ is tested in a uniaxial test machine with alternating stress (zero midrange stress). This defines points B and B' below. Also in static uniaxial stress test point A is defined with median stress = actual stress. Or with zero stress amplitude. So a constant load defines point A and the purely alternating ...
2
The equation $$x = x_0 + v_0t + \frac{1}{2} at^2$$ will find the distance your object travels. Set $x_0$ to 0 and set $v_0$ = 0. the acceleration is 2 m/$sec^2$ best we can tell is that acceleration is constant. The expression for acceleration vs. time must be give to find the distance traveled for a non-constant acceleration.
2
The formula holds, because there is a sum over all ${\bf p}$. After that sum is performed, ${\bf v}\cdot{\bf v}'$ is essentially the only vector structure that is consistent with rotation symmetry.
To prove the formula, first rewrite it so that the ${\bf v}$ and ${\bf v}$' are outside the sum:
$$\sum_{{\bf p}}({\bf p}\cdot{\bf v})({\bf p}\cdot{\bf v}')F(p^{...
2
Taking Hermitian Conjugate is simply taking Complex Conjugate and then Transpose, both operations are linear so their composition is also linear:
$$(\hat{A}+\hat{B} )^{\dagger}$$
$$=((\hat{A}+\hat{B} )^{T})^{*}$$
$$=(\hat{A}^{T}+\hat{B}^{T} )^{*}$$
$$=(\hat{A}^{T})^{*}+(\hat{B}^{T} )^{*}$$
$$=\hat{A}^{\dagger}+\hat{B}^{\dagger}$$
$$=\hat{A}+\hat{B} $$
In ...
2
If you want an exact solution, you can't use perturbation theory.
Try a change of variables $x\rightarrow y$ so that:
$$ \frac 1 2 m \omega x^2 + qE x = \frac 1 2 m\omega (y-a)^2 - b^2 $$
where $a$ is an offset to the (classical) ground state, and $-b^2$ is a global energy shift.
Classically, that energy shift is stored in the spring, and quantum ...
2
This isn't intended as a perturbation theory problem. (It actually can be solved, to all orders in perturbation theory, but that would be unthinkably arduous.) The actual point is the notice that the potential, including the linear potential due to the electric field, is still a quadratic function of the position,
$$V(x)=\frac{1}{2}m\omega^{2}x^{2}+qEx=\...
2
HINT: When proving conservation laws in the Lagrangian formalism, it's usually easier to avoid explicitly introducing second derivatives. Instead, leave the left-hand side of the Euler-Lagrange equations in the form
$$
\frac{d}{dt} \left[ 2 Q_{ij} \dot{q}_j \right],
$$
and note that
$$
v_i \frac{d}{dt} \left[ 2 Q_{ij} \dot{q}_j \right] = 2 \frac{d}{dt} \...
2
The measurement is a projection so does not preserve the norm. Thus, if
$M_m$ takes $\vert \psi_i\rangle $ to the unnormalized state
$\vert\phi\rangle := M_m\vert \psi_i\rangle$, its normalized version is
\begin{align}
\vert\psi_i^m\rangle &= \frac{\vert \phi\rangle}{\sqrt{\langle \phi \vert \phi\rangle}}=\frac{M_m\vert \psi_i\rangle}{\sqrt{\langle \...
1
If you denote the external points by $x,y$ and the internal points by $1,2,3$ than the diagrams of the last line in your picture are $ x1, 11,12,23,23,23,3y$ and $x1,12,12,12,23,33,3y$ with integration over the internal points and what is between brackets is contracted. Yes, they are the same.
1
Regarding your results: Both coincide, since $\langle 0|1\rangle \in \mathbb{C}$.
Edit: In fact, the result is zero, because both states are orthogonal, which is also used in the calculation performed in the textbook.
I hope this helped.
1
After asking my son to ask the question to his physics teacher, her response was that the question does not make much sense with what they learnt so far and that the only thing they could try is to check the order of magnitude of the force and realize that it is "large".
When I compare what they learn to what I had in high school (same profile, ...
Only top voted, non community-wiki answers of a minimum length are eligible
