# Tag Info

6

think about this with an example: the sine and cosine functions. They both average individually to zero over an interval. You can multiply those averages and still obtain zero. But if you multiply sin by itself and then average, you get a very distinct non-zero result. When the functions are arbitrary, the average of the product quantifies statistical ...

6

Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates. First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article: $$x=a\cosh\mu\cos\nu\cos\phi\\ y=a\cosh\mu\cos\nu\sin\phi\\ z=a\sinh\mu\sin\nu$$ where ...

5

The exact solution is $${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$ Clearly the field inside the conductor (that is, for $r<R$) vanishes. Here $Q$ is the total charge on the disk. The field, for large values of $r$, looks ...

5

As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a ...

4

So you were on the right track with integrating over r and over t. Here's how you could do it: The acceleration at any radius, r (if we assume Earth is a point mass) is: $$a=-{GM\over r^2}$$ The minus sign is because the acceleration is anti-radial. Then you can do the following: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ ...

4

Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}$$ For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...

4

I prefer to use Killing vectors and conservation laws to solve stuff like this, so let's analyze the problem using Killing vectors, and see if the results agree with your Euler-Lagrange equations. Notice that the metric is invariant under translations of $v$. The associated killing vector is $\partial_v$ which in turn gives the following conserved ...

4

Since radioactivity is a random process, you'd expect some fluctuation in the number of decays, i.e. if you wait for the half life, there's no guarantee that exactly half of the nuclei will have decayed! Based on your 3 estimates of the half life, you could just take the mean and go with that. And how to find the isotope? There's lists for that. I just ...

4

The velocity of the orbiting space junk is a vector, with both a radial and a tangential component. $$\vec{v}_f = \dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}$$ (my $r_f$ is your $r$) The equation for conservation of angular momentum involves only the tangential component of velocity, because it comes from the cross product of the radius vector and the ...

4

Here are real events relating to the last page of the pdf link you gave: Fig.1 This bubble chamber picture shows some electromagnetic events such as pair creation or materialization of high energy photon into an electron-positron pair (green tracks), the Compton effect (red tracks), the emission of electromagnetic radiation by accelerating charges ...

4

The simple answer is that you can't fully solve this problem--because as you note it is under-constrained--under the assumptions that are made when you first start doing statics (that objects are completely rigid). The introduction of finite strains bring in additional relationships.

3

First of all you should recall that Schroedinger equation is an Eigenvalue equation. If you are unfamiliar with eigenvalue equations you should consult any math book or course as soon as possible. Answer 1 (my apologies, I will use my own notation, as this is mainly copy-paste from my old notes): First define constants x_0 = ...

3

OP has already answered his own question with help from other answers, especially Peter Kravchuk's answer. Here we make some comments on how the fusion rule mentioned by Peter Kravchuk should be concretely realized. The first point is that the adjoint representation $Ad_{SU(N)}$ of $SU(N)$ is the real vector space of Hermitian traceless $N\times N$ ...

3

With the help of Peter Kravchuk and joshphysics, I have completed a proof of the trace identity. I will post it here as a reference. By the method of Kravchuk, we find $$\begin{split} \mathrm{tr}\big(t^a_Gt^b_Gt^c_Gt^d_G\big)&=2\big[ \mathrm{tr}\big(t^a_Nt^b_N\big)\mathrm{tr}\big(t^d_Nt^c_N\big)+ ... 3 Wow. I just had an amazing experience of discovering the following fact: It is known that for an element U of the group, in matrix sence:$$ Ad_Ux=UxU^{-1}.\,\,(1) $$Now, we note that the target space of the adjoint rep is spanned by N^2-1 traceless matrices t_a. So, if we add the unity matrix, we get a full basis in \mathrm{Mat}_N(\mathbb{C}). We ... 3 *The only problem I now have is that V1=V2=unknown which is making me think that he may have forgot to add that information to the question. * I belive that's the problem. You can't have V1=V2 in an adiabatic process (suposing that the number of particles is constant). Your tutor already pointed the right direction, combine ... 3 The differential dp(x) is the probability of finding the body in an interval of length dx centered at x. The quantity p you are looking for is the cumulative distribution function,$$P(x)=\int_{-\infty}^x \frac{dp}{dx}(x) dx,$$which is the probability that the particle will be to the left of the point x. Since the particle cannot be to the left of ... 3 On the Wikipedia page you link to there is a derivation of the commutation relation between \hat{a} and \hat{a}^{\dagger},$$ [\hat{a},\hat{a}^{\dagger}] = 1.$$This directly leads to (use the relation [AB,C]=[A,C]B+A[B,C])$$[\hat{a}^{\dagger}\hat{a},\hat{a}] = -\hat{a} , \qquad [\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] = +\hat{a}^{\dagger}.$$Up ... 3 Start with your \hat{H} = \hbar \omega \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right). I will omit hat notation from this point. The commutator then reads as \left[ H, a \right] = \hbar \omega \left[ \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) a - a \left( \hat{a}^\dagger\hat{a} + \frac{1}{2} \right) \right] = \hbar \omega ... 3 Just some pointers to get you started: The side length allows you to calculate the flux through a benzene ring given a magnetic field, which should then give you the induced current, if you claim that the electrons in the benzene ring are metallic (not a bad approximation due to their delocalized nature). Another thing you can look at is this article on ... 3 The key here is to notice that at any time t, one can expand the full time-dependent state \Psi(x,t) at that time in terms of a basis u_n;$$ |\Psi(t)\rangle = \sum_n c_n(t)|u_n\rangle $$The basic idea here is that the time-dependence of the state is being encoded in the expansion coefficients c_n(t). Having done this, notice now that$$ ...

3

There are many ways to go around this. You can start from the coherent states and apply the unitary $\hat{O}$ directly on them. That will not be that simple because you will get a term $\hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}$. Now, the typical approach would be to exchange the order of the operators to get something like ...

3

The commutators in the above expressions are sued to change the order of the Hamiltonian and annihilation or creation operators. I'll show you the first one in some detail, the second one should not give you problems afterwards. We start from $\hat{H}\hat{a}\psi_n$. Using the commutator $[\hat{H},\hat{a}] = \hat{H}\hat{a}-\hat{a}\hat{H} = ... 3 There are two perpendicular components of acceleration. 1)$a_t$along the direction of velocity,that increase the speed. so,$a_t=10 m/s^2$2)$a_c$centripetal acceleration,towards the center of rotation .$a_c=\dfrac{v^2}r=10 m/s^2$So, net$\vec a=\vec a_t +\vec a_c|a_c|=|a_t|$, so it's equally inclined(at$45^0$) to both components. Now you ... 3 Imagine your rod is made up from lots of little bricks stacked on each other. Then the total potential energy is the sum of the potential energy of each brick. The diagram shows the rod and one of the bricks of size$dx$and at a height$x$. If$\rho$is the mass per unit length, then the mass of the brick is$\rho \space dx$, and the potential energy is ... 3 Take its differential form : $$\mathrm{d} \rho = \frac{1}{4/3 \pi r^3} \mathrm{d}m - \frac{m}{4\pi r^4}\mathrm{d}r$$ The greatest variation in$\rho$will be achieved when all the terms add positively $$\delta \rho =\frac{1}{4/3 \pi r^3} \delta m + \frac{m}{4\pi r^4}\delta r$$ Factor of$-3$appears as matter of integration. Which you can recast into ... 3 The first resonant vibrational mode for a string clamped at both ends looks like: You should be able to deduce the wavelength from that diagram. The second mode looks like: Both of the images above are from http://www.clickandlearn.org/Physics/sph3u/Music/Music.htm and that site will spell it out in more detail for you. If your string length is ... 3 Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms. The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts The Power ... 3 When multiplying or dividing units, all you need to do is put the units in the numerator or denominator (wherever they appeared) of the answer. So: $$[e/M]={J\over kg}$$ $$[M/shc]={kg\over{J\over kg^oC}}={kg^2\,^{\circ}\rm C\over J}$$ But this is not the correct way of analyzing your units. You have$t = e / M / shc = e / (M * shc)$The units of this are: ... 3 I mean, can I replace this configuration by one capacitor with one resistor in series such that this resistor is equivalent to the other two? The answer is actually no. For a single resistor and capacitor in series, the real part of the impedance is independent of frequency, i.e., the real part acts like a resistor.$Z_s = R_s + \frac{1}{j \omega ...

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