Recent content by chingel

Why can't we use the Boltzmann distribution to calculate what fraction of particles are in the liquid and gas states? If in liquid state the potential energy is smaller by amount ##L## per mole lets say, the using the Boltzmann distribution (##r\propto \exp(-\Delta E/RT)##) the fraction of...

Thanks for the responses. I understand that the actual total potential energy of all the charges would be the total work done on the system, but we use the average field inside the dielectric, which removes some of the field (and so potential) that a charge of every dipole feels. Actually every...

I can calculate the potential energy of a charge on the capacitor by calculating the work done by the electric field when I take it to a plane which goes through the middle of the plates that is parallel to them (because on this plane the electric field is normal to the plane by symmetry and...

If I think in terms of the charge on the plate, then the energy is smaller with the dielectric inside $$E = \frac{Q^2}{2C} = \frac{Q^2}{2\epsilon C_\text{vac}}. $$ It seems to me that some of the energy of the charge on the dielectric surface cancels out the energy of the charge on the capacitor...

It is known that the energy required to charge a parallel plate capacitor (thickness ##x##, surface area ##S##) with a dielectric (with dielectric constant ##\epsilon##) is $$E=\frac{xQ^2}{2S\epsilon\epsilon_0}.$$ I would think that the work done goes only into the electrostatic energy between...

Let's assume for simplicity that there is no gravity. I understand that canthotaxis says that in the figure in the OP on the right ##\varphi## can be between 135 and 315 degrees and the contact line is pinned to the edge of the needle in the whole range. My question is what happens after that...

Let's say we are pushing water down a needle, the contact angle between the needle and the water is ##\mathrm{45°}##. For simplicity let's assume there is no gravity. As the water is initially moving down in the left image, the angle ##\theta=\mathrm{45°}## and the surface of the water is a...

So for anybody who is interested, the argument is quite simple in my opinion. First we know from angular momentum addition that by adding m copies of spin 1/2 (I mean by doing a tensor product of them) and symmetrizing, we get a m/2 representation of SU(2). So a (m/2,m/2) representation of the...

Thanks, I managed to take a quick look at the book and chapter 8 seems to cover exactly what I was asking about (and that I didn't find in my QFT books). On a quick skim apparently if we had for a four vector ##p_\mu\sigma^\mu## and transforming this under (1/2,1/2) gave the proper...

At least for the (1/2, 1/2) representation I figured it out by surfing on the internet. I can choose for A representation the ##\sigma/2## matrices and for B take ##-\sigma^*/2##, then the two index object ##\phi^{ij}## transforms as $$\phi \rightarrow \exp(i(\theta_k-i\beta_k)\sigma_k/2)\...

I've been trying to understand representations of the Lorentz group. So as far as I understand, when an object is in an (m,n) representation, then it has two indices (let's say the object is ##\phi^{ij}##), where one index ##i## transforms as ##\exp(i(\theta_k-i\beta_k)A_k)## and the other index...

Thank you for the link. There they calculate that the expectation value of the change of the spin or angular momentum is zero. While the expectation value does not change, they still seem to say that Zitterbewegung causes the spin to change in some sense. I still don't quite understand why or...

The Hamiltonian I am considering is $$H=\vec{\alpha} \cdot \vec{p} + \beta m,$$ where ##\alpha^i = \gamma^0\gamma^i##, ##\beta=\gamma^0##, ##p_k = i\partial_k##, ##\vec{p} = -i\vec{\partial}##, so that ##i\partial_t \psi=H\psi## gives just the Dirac equation ##(i\gamma^\mu\partial_\mu -m...

Here I am considering the one particle free Dirac equation. As is known the spin operator does not commute with the Hamiltonian. However, the solutions to the Dirac equation have a constant spinor term and only an overall phase factor which depends on time. So as the solution evolves in time...

The battery slightly changes voltage as it charges and discharges. A charged car battery might get close to 13V, while if it drops below 12V you might have trouble starting the car. You cannot quickly increase the battery voltage if you suddenly need it. You can only increase it during charging...