For an example in a harmonic oscillator energy eigenstate, position x and momentum p both have Gaussian probability distributions around zero. We have tiny but not zero probality to observe any large p as well as x.
For this exercise for the biginners I prefer observing what happens in finite well potential V system in the limit of ##V \rightarrow \infty## to discussing mathematical subject of ##V = \infty##. We may keep the interval ##[-\infty,+\infty]## and apply usual Fourier transform method to get...
Yes in the well but it is zero outside. That matters.
It is superposition of various mometum eigenstates, the coefficiets of which provide probability to observe specific momentum value by product with complex conjugate.
Yes inside but it includes no zero potential energy outside. That matters.
As a first try I take it easy as follows.
Say B is vacuum, gas in A expands free adiabatically. Temperature does not change from ##T_A## . ##P_A## drops to ##\frac{V_A}{V_A+V_B}P_A##.
Say A is vacuum, gas in B expands free adiabatically. Temperature does not change from ##T_B##. ##P_B##...
If
\Delta G=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kG\right),
for any G, then choose G as
G= \Delta^{n-1} F
I cannot find meaning of odd power of nabla applying scalar function. For example can you show where we meet...
If you have already proved the formula in OP of
\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)
I think you have a good reason of recurrence formula
\Delta^n F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert}...
As for 2D lattice we can make one of the lattice points is (0,0) without losing generality.
Say other points are ##(n_1,n_2),(m_1,m_2)##
n_1^2+n_2^2=A
m_1^2+m_2^2=A
(n_1-m_1)^2+(n_2-m_2)^2=A
where A is square of the side length. You will find contradiction in this set of formla.
Rather than "the nearest integer", I prefer floor function
\lfloor x \rfloor
, maximum integer that does not exceed x.
\lfloor 1.2 \rfloor=1,\ 1.2=1+0.2
\lfloor -1.4 \rfloor=-2,\ -1.4=-2+0.6
z=(1-2i)+(0.2+0.6i)
1
Say a wall in the container moves outward for volume v. The gas expands adiabatically
\frac{P_2}{P_1}=\frac{V^\gamma}{(V+v)^\gamma}<1
where
\gamma = \frac{C_p}{C_v}>1
Then we insert a new partion where there was the wall. The number of gas molecules in the container is
N_2=N_1\frac{V}{V+v}...
From your sketch, say S is area we want, it seems
2S=\frac{1}{2}*20^2 (4\alpha)-4*\frac{1}{2}*15\cos \alpha*5-2*\frac{1}{2}*5^2(\pi+2\alpha)
where
\sin\alpha =\frac{1}{3}
Is it same as your idea ?
I would make
0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)+U(u))~du
with U(u)<0 or
0 = m(t)v(t) + \int_{0}^{t} -\dot m(u) ( v(u)-U(u))~du
with U(u) >0 for normal thrust.
Thus
\frac{d}{dt} [m(t)v(t)]=\dot m(t) v(t)+F(t)
RHS first term : Minus. Loss of momentum which fuel to be burnt has hold in...
More comprehensive treatment. Say rocket is at rest at t=0. Momentum of the system is conserved zero.
0=m(t)v(t) + R \int_0^t ( v(u)+U )du
differentiating by t,
\frac{d}{dt}[m(t)v(t)]+R [ v(t)+U ]=\frac{d}{dt}[m(t)v(t)]+Rv(t)-F=0
where m(t) is mass of rocket body and fuel in tank...
m(t) is defined by man made convention of removing ejected fuel mass so d/dt [m(t)v(t)] ##\neq## F does not surprise me. For simplicity say F=0, fuel are just disposed with no thrust or kept in tank but we take off account for m(t). From my post #20
\frac{d}{dt}[m(t)v(t)]+Rv(t)=0
We are...
The system consists of Rocket body and fuel. All the fuel is contained in rocket at time t. Some fuel are ejected from Rocket during t and t+##\triangle## t. All the fuel ejected before time t, which continues inertial motion of constant mometum any way, does not matter here. If we choose...
At time t
P=m(t)v(t)
At time t + ##\triangle t##
P=m(t+\triangle t) v(t+\triangle t) + R\triangle t \{v(t)+U\}
where U is speed of ejected fuel relative to rocket. U < 0,
m(t)=m(0)-Rt
RHS first term is momentum rocket body and remaing fuel hold.
RHS second term is momentum that ejected fuel...
If we omit ##-B^2 y## in RHS, we can solve the simplified ODE,
y=B\int_0^t du \ e^{\frac{iA}{\omega}\cos \omega u}
We may be able to expect that in a short time the solution of the original ODE does not so much different from it. I would appreciate it If you could check the difference with...
[m3/m3] has no physical dimension but [molecules/cm3] has physical dimension of L^-3. I am afraid some more details of the exercise are required to help you.
Supplement to my post #2
Say three points ##(a_1,a_2),(b_1,b_2),(c_1,c_2)## is on the circle which is caratterized by (l,m, n)
\begin{pmatrix}
a_1& a_2 & 1 \\
b_1& b_2 & 1 \\
c_1& c_2 & 1 \\
\end{pmatrix}
\begin{pmatrix}
l \\
m \\
n \\
\end{pmatrix}
=-
\begin{pmatrix}
a_1^2+a_2^2 \\...
Contd. from my previous post
The equation of line which connects two commom points of the cirlces i and j, is
(l_i-l_j)x+(m_i-m_j)y+n_i-n_j=0
The system of such six lines meet at a point (x,y) are recuced to three equations.
\begin{pmatrix}
l_1-l_2 & m_1-m_2 & n_1-n_2 \\
l_2-l_3 & m_2-m_3 &...
Interesting exercise. I take Cartesian coordinates to look at as attached figure.
We can write the equation of circle touching three points as
x^2+y^2+lx+my+n=0
Giving three (x,y)s of touching points, we get a set (l,m,n) . We get four (l,m,n) sets for the four circles. The system of four...
I tried
=-2\pi V_0 A^2 \int_0^R (1- \cos 2k_1r)r^2dr =-\frac{\pi A^2}{4k_1^3} V_0 \int_0^{2k_1R} (1- \cos y)y^2dy
=-\frac{\pi A^2}{4k_1^3} V_0 \{ \frac{u^3}{3}-(u^2-2)\sin u -2u \cos u \}
where
u=2k_1R
Though I am careless in calculation, do you share it with me ? Does it have anything good to...
I do not go into the mathematics but I am convinced that velocity of fluid at particle surface has no radial component. In fact
\mathbf{\hat{r}} \cdot \textbf{v}_s(\hat{\textbf{r}}) = \sum {2\over{{n(n+1)}}} B_n (\mathbf{\hat{r}}\cdot (\hat{\textbf{e}}\cdot \hat{\textbf{r}}) \hat{\textbf{r}} -...
Polar form of complex number,
-1+i= \sqrt{2}\ e^{3/4\ \pi\ i}, is easy to handle power 1/3. Dividing phase by 3 and multiplying cubic roots of 1,
(-1+i)^{1/3}= \sqrt[6]{2}\ e^{1/4\ \pi i + 2/3\ n\pi i}
where n=-1,0,1.
Do you see continuing from post #4,
...=-\frac{\hbar^{2}}{2m} \int_{0}^{\infty} dr \psi^* \frac{\partial}{\partial r} (4\pi r^2 \frac{\partial \psi}{\partial r})=-\frac{\hbar^{2}}{2m} \int_{0}^{\infty} 4\pi r^2dr \frac{1}{r^2} \psi^*\frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial...
Last year I read the news article https://scitechdaily.com/physicists-prove-that-the-imaginary-part-of-quantum-mechanics-really-exists/ which says i is necessary for QM. You may be interested in it also.