@RewCie I think you will find determinism (not to be confused with fatalism) interesting
Hey all! I was wondering if any GR expert could help me?
If we assume isotropy and homogeneity then we get the FLRW metric upto a constant (cosmological constant). Is there any counter argument why this constant cannot be a property of the stress energy tensor?
@MoreAnonymous The CC can be treated either as a curvature term, in which case we write it on the left side, or as a vacuum energy, in which case we write it on the right side and add it to the SE tensor.
The two are mathematically equivalent. Traditionally it tends to be treated as a curvature, but now that we have discovered dark energy I guess that would be added to the SE tensor.
@DIRAC1930 You may find the Bloch sphere helpful. "The points on the surface of the sphere correspond to the pure states of the system, whereas the interior points correspond to the mixed states".
@MoreAnonymous The SE tensor is the sum of the SE tensors from all the components present. So for example with a mixture of hydrogen and helium you can split the SE tensor into a contribution due to the hydrogen and a contribution due to the helium.
If we assume isotropy and homogeneity then we get the FLRW metric up to a constant (cosmological constant). Let's say I have $2$ isotropic and homogenous universes one with hydrogen and another with helium. Will they have the same cosmological constant?
The CC is separate from the matter density. Hydrogen, helium or any other matter will never result in a SE tensor that contains a cosmological constant.
To get a CC contribution you need a form of energy with an unusual equation of state, i.e. you need dark energy or something like it.
A component that acts like a CC has to have an energy density that is independent of the scale factor i.e. as the universe expands the energy density remains constant. This is equivalent to having a negative pressure.
I find it hard to see how any scattering process is going to satisfy this requirement.
You need the energy density to be independent of the particle density. Suppose you have N particles per cubic metre and take the pairwise sums to get the interaction energy. Call this V.
Since entropy is a state function that means that regardless how we reach from state A to state B (for a given system) entropy will be the same S_B - S_A
But if the process to reach state B from A is reversible then the entropy change is zero
and if it is irrever. then its bigger then zero
so this is a contradiction
So what's the correct thing to say or think about?
from wikipedia: "The thermodynamic variables of the grand canonical ensemble are chemical potential (symbol: µ) and absolute temperature (symbol: T). The ensemble is also dependent on mechanical variables such as volume (symbol: V) which influence the nature of the system's internal states. This ensemble is therefore sometimes called the µVT ensemble, as each of these three quantities are constants of the ensemble."
so fixed chemical potential, absolute temperature, and volume
in particular this means that the total number of particles and the system energy are not constrained and can fluctuate
(by contrast, the canonical ensemble assumes a fixed number of particles and the microcanonical ensemble further assumes a fixed energy.)
(Hence why wikipedia also denotes the microcanonical/canonical/grand canonical ensembles as the NVE, NVT, and µVT ensembles respectively.)
that's like saying that any ensemble (grand canonical/microcanonical/canonical) fixes the average pressure b/c one can compute that as a thermodynamic derivative
depends on which ensemble you're talking about. if you have a canonical ensemble, temeprature isn't understood as being an average
if the temperature can fluctuate, you're not in the canonical ensemble. (you could imagine a scenario where the temperature changes, but it'd have to do so slowly enough to remain in equilibrium)
and in the canonical ensemble, there's no fluctations in particle number
so if you've got a system where heat can flow in/out of a reservoir but particles can't, then there's no need to talk about average particle number
the quantum version of the canonical ensemble isn't too bad, happily. there's not really an easy version of the microcanonical ensemble, by contrast, owing to energy quantization. (energy in the microcanonical ensemble is supposed to be a free variable, and quantization obstructs that.)
the quantum grand canonical ensemble isn't too bad either
@M.ÇağlarTUFAN as ever: if you've got a question, ask, and if people are interested they'll respond
Testing: I think I typed so long I can't post my message.
I started 3rd grade this semester and I take quantum physics course. I have finished 4 weeks today. First 2 weeks of the course was like some chat about before and after of quantum physics.
My instructor talked about the experiments that can be explained with classical physics and other experiments that can not be explained with classical physics.
Then he talked about some new ideas like quantum structure of light, photons etc. And other experiments like blackbody radiation, photoelectric effect and so on.
He derived the Planck Radiation Formula in class, I don't know why he did this especially. I couldn't grab the ideas. Later on, last 2 weeks was really difficult for me.
In 3rd week, my instructor introduced us wavefunctions, probablity densities, average positions, average squares and square averages, some about Fourier transforms.
Then 4th week he introduced operators, position and momentum operators in position space and momentum space. And lastly energy operator and a little bit about Shrodinger equations.
Finally I posted it! Yay!
Anyways, like I was saying 4th week has just finished and I'm having so much trouble understanding things. My instructor has his notes on his website I took and photocopied them. I have been trying to work on the notes this whole weekend and I couldn't got any far.
I especially asked my instructor about book but he just said "search my name on google and go to the website, then download pdf notes" but he has recommended 2 books on his website
Let me find them quickly
1) Concepts of Modern Physics, Arthur Beiser, 6th ed, McGraw Hill, 2003 2) Quantum Physics, Berkeley Physics Course, Vol. 3, 1965
I wonder if you could take a look at my instructor's notes but they are in Turkish. Maybe you can just follow some names and understand them.
Also, first book on the recommendations seem to follow the same order of subjects my instructor explained in lectures (except 1st chapter). But the book has less calculations, mostly verbal explanations of things.
the intention, i imagine, is to start with classical mechanics (particle moving in 1D potential) and give the QM version of that (Schrodinger equation)
which is a very wave-mechanics way of approaching it
the trouble is that it involves a good deal of technical machinery to set up
and stuff you just sorta have to take on faith that it works
my own preference nowadays is routes which start from the Stern-Gerlach experiment and show you how the linear algebra machinery emerges in that
and it's the same idea with $x^2$ as an integral: If you have any $f(x)$, then the expected value is $\langle f(x)\rangle =\int_{-\infty}^\infty f(x)|\psi(x)|^2\,dx$
sorta yes sorta no. the problem is that, in the integer example, theres's a finite probability of getting each outcome
but the probability of picking 0 from a continuous distribution is strictly 0
so in that sense $|\psi|^2$ isn't the probability of getting any particular $x$, because that probability is zero
then you can approximate the left-hand side as $\int_a^{a+\Delta x}|\psi|^2\,dx\approx |\psi(a)|^2\Delta x$
so $\text{Pr}(a\leq x\leq a+\Delta x)\approx |\psi(a)|^2\Delta x$
it's more like: if you compute the probability to find the particle in a very small interval $\Delta x$, then this probability is approximately some multiple of $\Delta x$ itself
and that multiple is the probability density
in other words, while the probability gets smaller and smaller as you shrink the interval, the rate at which this probability vanishes occurs will not be zero
so it's more like "probability over a range" / "range" -> probability density
if this rings any bells, it's probably because this is exactly the situation you have for the Planck radiation law
@DIRAC1930 b/c all grand canonical ensemble cares about is if your particle number and energy operators exist. (that's the QM version of "if particle number and energy label your microstates".)
I really feel exactly like when I'm listening to my instructor in class. I could understand the $\langle x \rangle$ and $\langle x^2 \rangle$ and about dimensions of these integrals, but almost nothing other than these two :P
@DIRAC1930 in QM, the grand partition function is the density matrix $\hat{\rho}=\exp\left[(\Omega +\mu \hat{N}-\hat{H})/kT\right]$ where $\Omega$ is the grand potential determined by demanding that $\rho$ have trace 1
and that expression doesn't care about whether $N,H$ commute
"Note that for the grand ensemble, the basis states of the operators H, N are states with multiple particles in Fock space, and the density matrix is defined on the same basis. Since the energy and particle numbers are all separately conserved, these operators are mutually commuting."
If $\int{\rho dV}$ has unit of mass because we are integrating somrthing with units of mass/volume over volume, then with this respect if $\int_{-\infty}^{\infty}{|\psi|^2dx}=1$ and is unitless, then $|\psi|^2$ shouls have units of something/length here something is dimension of probablity density
