Recent content by Higgsono

Thanks for your answers! Maybe some of the confusion lies in my assumption that we are dealing with classical particles. In that case, intuition tells me that there should be a continnum of directions regardless of the energy of the particle.

Doesn't answer my question. What do you mean? Being in a moment state means it has a certain momentum.

The multiplicity of states for a particle in a box is proportional to the product of the volume of the box and the surface area of momentum space. $$ \Omega = V_{volume}V_{momentum}$$ The surface area in momentum space is given by the equation: $$p^{2}_{x}+ {p}^{2}_{y}+{p}^{2}_{z} =...

huh? Q and V are not constants. I must be able to double the charge and the relation should be the same right?

The capacitance C of a conductor is given to be a constant relationship between charge Q and potential V of the conductor given by Q = CV. But how can C be a constant? Because the potential of the conductor will not be a linear relationship of the charge that I add. THe more charge there is on...

Because microscopically all processes have a time reversal symmetry.

Both the heat equation and the diffusion equation describe processes which are irreversible, because the equations have an odd time derivative. But how can these equations describe the real world when we know that all processes in nature are reversible, information is always conserved? But these...

So if it has the same form, the transformation would be $$\delta f= f(x') - f(x)$$ intead of $$\delta f= f'(x') - f(x)$$ I'm always confused why they put a prime on the function when they consider coordinate transformations.

f is a scalar field, and I am considering variations of this field in the Lagrangian. But if $$ f'(x') = f(x)$$ by definition. What is the point of varying the fields?

I don't know. But they always write a prime on the function as well.

Given a coordinate transformation $$x' = g(x)$$

Given a scalar function, we consider the following transformation: $$\delta f(x) = f'(x') - f(x) $$ Given a coordinate transformation $$x' = g(x)$$ But since ##f(x)## is a scalar isn't it true that ##f'(x') = f(x) ## Then the variation is always zero? What am I missing?

oh wait, is it because ##\epsilon^{\mu \nu}## for fixed indices define an independent parameter so that we can wary each individual parameter by itself and equality should still hold? Like what you do when you equate the coordinates between two vectors if you know that A=B. But in this case, we...

I don't understand. It is simply not true that ##\epsilon^{\mu \nu} = +-1## in general. Why do you make that assumption? ##\epsilon^{\mu \nu}## have 6 independent parameters.

It's from "A first course in string theory by Barton Zwieback"