Recent content by hgandh

Without assuming a universal speed that is constant in all inertial reference frames, is it a necessary consequence of Galilean symmetry that interactions are instantaneous? If this is the case how can we prove this?

prime denotes derivative with respect to time

The underlying interactions at the microscopic level that give rise to friction are derivable from potentials.

The equations of motions for a closed system consisting of ##N## particles are: $$m_i \vec x_i'' = \sum_{j \neq i}^N \vec F(\vec x_i, \vec x_i', \vec x_j, \vec x_j')$$ $$ i = 1,..., N$$ Now if we impose the requirement that this closed system be symmetric under Galilean transformations, do we...

I know that the entropy of a system is the same in different inertial frames. Is this still the case for non inertial frames? For example, is the entropy of a body as seen from a rotating reference frame the same as the entropy seen from a fixed frame?

If I restrict it to positive values only, would this become valid then?

If we have a function ##f(x+\Delta x)## where ##\Delta x << x##, is it valid to approximate this as: $$f(x + \Delta x) \approx f(x) + f'(x)\Delta x$$ even if ##\Delta x## is not necessarily small? If not, what is the valid expansion to first order?

The body and the object are two different things. The body is in an external medium and can exchange both work and heat. The body can only exchange work with the object. So you have 3 things here: the medium, body and the object.

I am reading Landau's Vol 5 on Statistical Physics and have trouble grasping some concepts in Section 20. If I understand this correctly, the body and the medium are in direct contact and can exchange work and heat while the object can only exchange work with the body. So the minimum...

No it is a general Lorentz transformation. I believe my mistake was the the time evolution operator itself changes in the new reference frame since the Hamiltonian as seen in this new frame is in general not the same as in the original. $$U(t) \to U(\Lambda)U(t)U^{-1} (\Lambda)$$

Symmetry transformations are changes in our point of view that preserve the possible outcomes of experiment: $$\Psi \rightarrow U(\Lambda) \Psi$$ In the Heisenberg picture, observables in a fixed reference frame evolve according to: $$P(t) = U^\dagger (t)PU(t)$$ while in the Schrödinger...

Ashcroft & Mermin Chapter 25, the Gruneisen Parameters are defined as: $$\gamma_{ks}=-\frac V {\omega_{ks}} \frac {\partial {\omega_{ks}}} {\partial V}$$ where the normal mode frequencies are defined by the eigenvalue equation: $$ M \omega^2 \epsilon = D(k) \epsilon $$ The volume of the crystal...

What are the precise conditions for thermodynamic equilibrium? I know that a system in thermodynamic equilibrium must have constant temperature and that there can be no net macroscopic flow of energy or matter. However, is it possible for there to be a system in equilibrium that has a spatially...

I should have marked it as "A". I am studying QFT currently with all of the assumed pre requisites and some group theory.

From Chapter 5.9 Weinberg's QFT Vol 1, massless fields are defined as: \psi_l(x)=(2\pi)^{-3/2}\int d^{3}p\sum_{\sigma}[k a(p,\sigma)u_l(p,\sigma)e^{ipx}+\lambda a^{c\dagger}(p,\sigma)v_l(p,\sigma)e^{-ipx}] With coefficients defined by the conditions: u_{\bar{l}}(p,\sigma) =\sqrt{|k|/p^0}...