Recent content by Stealth95

Hello, I am a Physics undergraduate, currently beginning my third (out of four) year of studies. I have already taken courses about Real and Complex Analysis, Linear Algebra, ODEs and PDEs (for Maths), Classical Mechanics (Lagrangian and Hamiltonian formalism), Electromagnetism, Special...

Saying Fourier's trick, I mean the usual method one can use to find the coefficients (here ##A_m## and ##C_m##) in series with ##\cos m\varphi ## and ##\sin m\varphi ## (Fourier Series). For example if ##U(r,\varphi )## is given by the last equation you wrote in your first post, then it is...

As ZetaOfThree wrote, you can use the boundary conditions on the cylinder's surface. From the figure in your first post (if you assume that the distance between the quarters is negligible) you have ##U(a,\varphi )## (the potential at ##r=a##) for every ##\varphi \in [0,2\pi)##. It is: ##...

You can get this by using the current density vector \displaystyle{\vec{J}}. This is the current that flows through a unit area of the wire crossection (or for uniform distributions the ratio current/area). For this case the current flows only in a specific direction, so it's not necessary to...

The Amperian loops are always closed curves (that's because the line integral at LHS is closed), so they are always the circumference of a surface. In our case this is the surface of a circle with radius \displaystyle{s}. The current at RHS is the total current that goes through this surface...

I think you already have the equations. Conservation of Momentum: \displaystyle{p_\gamma =p_\Lambda \Rightarrow E_\gamma =p_\Lambda c\Rightarrow E_\gamma ^2=E_\Lambda ^2-m_\Lambda ^2c^4} If you combine this with the conservation of energy you have two equations and two uknowns...

Yes, the system is nonlinear. But based on the fact that the problem asks to find an annihilation operator I wasn't very strict with Maths. So I assumed that: \displaystyle{B=\frac{i}{\sqrt{2mk\hbar }}} and then I used one of the equations to find \displaystyle{A}. Note that the value I chose...

If you use \displaystyle{[a_b,a_{b}^{\dagger}]=1} you get an equation for \displaystyle{A,B} (and their conjugates). Then you use \displaystyle{a_{b}^{\dagger}a_b=\frac{H}{\hbar k}-\frac{1}{2}}. But now it's not so trivial to find the equations for \displaystyle{A,B}, because there are also...

I recently started self-studying Quantum Mechanics so I am not really sure for my answer, but I can find an operator that works for some cases. I assumed that \displaystyle{a_b=AX+BP} for some constants \displaystyle{A,B \in \mathbb{C}^*}. Then \displaystyle{a_{b}^{\dagger}=A^{*}X+B^{*}P}...

I think that ehild proposed the easiest approach for this problem. But note that using Coulomb's law you can find the force only for the case in which the dipole is in this specific point of space. If you move the dipole in another place then you must apply Coulomb's law again to find the new...

Hello. It is possible to solve this equation without using Legendre polynomials. Just remember that: \displaystyle{\frac{\partial}{\partial \rho }\left(\rho ^2\frac{\partial V}{\partial \rho } \right)=0\Rightarrow \rho ^2\frac{dV}{d\rho }=C} which is easy to solve. You have to find two constants...

Firstly, here I think you must write \displaystyle{dA=rdrd\varphi } (\displaystyle{\varphi } is the azimuth). Note that if you have two charges at the same distance from the origin but in different angles they don't create the same field at the point along the z-axis. The two fields have the...

I am not sure if you can integrate this equation, because you set \displaystyle{N=0}, so it's valid only for a specific moment. But you can solve the exercise easier without integrating. Just take the Newton's 2nd law for the normal components of the forces. Imagine that every moment you take...

I hope I understood your problem correct! It's correct to use Gauss law to find the field of the line, but I think that you made a mistake because your answer is independent of \displaystyle{\lambda _1}. Also your answer is not general, because point P is not a random point, it is rod's...

I think that your equation must be: \displaystyle{P_1 \left(\frac{n_2 R T_1}{P_1}\right)^{\gamma} = P_0 \left(\frac{n_2 R T_2}{P_0}\right)^{\gamma}} Although we have \displaystyle{n_1} moles, only \displaystyle{n_2} expand adiabaticaly. The rest \displaystyle{n_1 -n_2 } moles leave the chamber...