Is the time derivative of the Hamiltonian integral over phase space (the gamma distribution) the equivalent of an action which conserves the probability phase/volume.

Probability Wave Dispersion

The function S_0(l) is called the action of the path, and to each path we define an action,

S_0(l) = \int_{t_l} E dt

It is weighted against Planck’s constant, which also has units of action per radian.

The presence of an electromagnetic field adds the interaction term, (superscript I),

{S_0}^I(l) = q \int_{t_1}^{t_2} A^\alpha g_{\alpha\beta} \partial x^{\beta}

which is the action of the electromagnetic Lorentz force. This is the quantum mechanical Bohm-Aharonov Effect, where q is the charge and A^\alpha is the electromagnetic gauge potential. In the absence of a magnetic field this becomes the prepared state of a quantum computation.

Derive the action that shifts the phase angle \theta_I for a static EM field...?

This is the integral.

{d \over dt} \int_{\Gamma} f(x,p) dxdp = \int_{\Gamma}({\partial f \over \partial x} {dx \over dt} \,+\, {\partial f \over \partial p } {dp \over dt})dxdp

\;\;\;\;\;\;\;\;\;\; = \int_{\Gamma}({\partial f \over \partial x} {\partial \hat H \over \partial p} \,+\, {\partial f \over \partial p } {\partial \hat H \over \partial x})dxdp

Christ Vkothii!..what was the bit after " Is the time?'

Well, probably it will get flamed to buggery once again.

The probability phase is just an abstract angle, an amplitude.
(as if that's any help)

The lil' piccy in my avatar is one of these phase angle things.

This has something to do with the conservation of probability and Lorentz invariance. Probability density actually, because the information content - which essentially is fundamental spin has to be conserved. I'm trying to see why the density changes in the Lorentz frame.

I am assuming q here is charge? S is spin yes?

I absolutely find it annoying, (not just on this occasion), but many occasions, that people come here, write down equations, and never explain the variables, or their specific functions.

I'm not daft, and niether are other people, but to simply have load of equations written and then taken for granted we should know, is absolutely ridiculous.

I am assuming q here is charge? S is spin yes?

I absolutely find it annoying, (not just on this occasion), but many occasions, that people come here, write down equations, and never explain the variables, or their specific functions.

I'm not daft, and niether are other people, but to simply have load of equations written and then taken for granted we should know, is absolutely ridiculous.

q is a generalized coordinate, and S is the action. E presumably is the lagrangian density? This is a bit confusing, but ok.

So what's the claim? Or even the question?

Ok.

My question is, or rather, i implore you to make a rule in this forum for the future. If equations are written down, the variables must be explained, instead of taken for granted we should know.

Ok.

My question is, or rather, i implore you to make a rule in this forum for the future. If equations are written down, the variables must be explained, instead of taken for granted we should know.

Why do you feel like you have to be involved in every thread? If you're a physicist, as you've claimed before, and you've never seen an action integral, one has to be suspicious of your qualifications.

Why do you feel like you have to be involved in every thread? If you're a physicist, as you've claimed before, and you've never seen an action integral, one has to be suspicious of your qualifications.


Where did that come from?

If you check my history, i've involved myself in about four or five discussions, five different threads. I did not deserve that you say i involve myself in every discussion.

Secondly, i know how to work integrals. The reason why i have a [[problem]] with people using variables i may be not so knowable of, is because of reasons like these:

\mu_{S}=q \frac{q}{2m}S

Do you see two symbols used here identical to the symbols used in the OP? See how some people could become confused?

Now, as i said, i don't think i deserved that.

And eh, after readin your post a second time, i know what an action integral is. Where did i say i didn't?

And eh, after readin your post a second time, i know what an action integral is. Where did i say i didn't?
To defend BenTheMan's statement, if you were in any way, shape or form familiar with action integrals, you'd be familiar with Hamiltonian mechanics and the notation within it.

Is it a new craze on this forum for people to pretend to know about things they don't? Do they not realize how transparent it is?

Don't patronize me, or my knowledge on physics. There is a library of equations out there, and there is also area's of interest. You cannot simply come to recognize an equation (willy-nilly) all of the time.

Besides, i thought it was an action integral. I just didn't know the symbols; or to be more precise, i did not know what those particular symbols where highlighting.

Are you a physicist?

And can i also say, its been ages since i have worked on Hamiltonians, or Langrangians, which are similar.

You unfair bunch of peeps.

Well what I've said is completely true. If you knew advanced things like path integrals, you'd know about Hamiltonian mechanics and so you'd know the notation. There's absolutely nothing wrong with not knowing about those things, but it does seem daft to make out you know more than you do.
Are you a physicist?
I guess I'm more of a mathematician. I have published results in mathematical physics journals, if that counts?

No... dear guest... what you have said is something i guessed to be little interest to your job career.

Let me tell you, in physics we deal with many equations; what you said is only half true, with very little logic behind the twist of it all.

I could qoute a great deal of equations: hundreds to be precise, and i would still forget a number of equations, because i have not dealt with them in months, in some cases, years.

Do you see?

My "job career"?

Look, the more you post the more it seems you have a vivid imagination. Are you telling us, that hand-on-heart, you have a working knowledge of (for instance) Hamiltonian mechanics? Be honest.

Do you want me to go through some?

Your lack of wisdom in such matters may be your downfall.

Well I wouldn't have gone that far! But as an example, would you be capable of constructing the action-angle coordinates for the Hamiltonian:

H(r,\theta,p_r,p_\theta) = \frac{p_\theta^2}{2r^2} + \frac{p_r^2}{2} - \frac{1}{r}

?

You wouldn't have gone that far?

You hyocrotical man/woman... you just blatently admitted i know nothing about them... and then when i said i would go through some of what i know, you said ''i wouldn't go that far,'' and then posted an equation, as if to prove something.

And here is the pivotal peak of your ignorance. Just the other day, Vkothii was posting threads on physics, he even admitted to copying from a book, only for you to blame him of going about and trying tro floute he was more intelligent by doing so.

And today, you have proved yourself a dirty hypocrite.

Is that a round about way of saying "no, I'm not capable"?

No. Don't speak for me.

I hope you can gather i am quite capable of that myself. Now shall i go through what i know about a Hamiltonian, or are you going to display more astounding ego....?

If you can construct the action-angle coordinates, why don't you? Hell, anyone can google "Hamiltonian mechanics" and reproduce some of the findings!

As I said, there's nothing wrong with not knowing about Hamiltonian dynamics. It's another thing to pretend you know about them!

You ignorant person.

I have no intention of googling anything. I was going to recite the classes i had when i did cover such a subject.

You know, you are too confident for your own boots. What's worse, is that YOU ARE NOT a physicist... but being a mathematician you do have an upper hand in being able to look at an equation and work it like a second language.

But i bet you don't know the first thing about freedom expressions like

T^{\alpha}=de^{\alpha}+w^{\alpha}_{\beta}.e^{\beta }

or perhaps how to understand a spatial composition

\frac{d^{2}x^{\mu}}{d \tau^{2}}= \frac{1}{2} \eta^{mu}^{\lambda}\partial_{ \lambda}h_{00} \frac{dt}{d \tau}^{2}

Or maybe not?

But here is where this pathetic arguement has gone, and i feel displacing from it very fast.

It's now completely apparent that you don't know anything about Hamiltonian dynamics (which is basic undergrad physics). Actually, I'd be willing to bet you don't understand any undergrad physics. I think it's usually the case that those who know least feel that they can get away with lying the most.

I see things are going well...

This is not some attempt to either demonstrate a woeful lack of math-ability, or give anyone their opening to show everyone else how big their head can get, rather an attempt to explore a certain experiment - the rotation of electron phase in a curl-free region of space - from some other angle.
Anyways, there's a mass action term derived in the Schwarzchild metric, or "the result of the gravitational field acting on the wave function of the free electron."

I just thought I would see if there was any interest. It's about something called "the propagator" - a path integral.
I don't know enough, about a lot of somewhat tricky mathematical ideas. I tell myself nonetheless, I don't need to learn about all of them, or even all about any of them.

Then again, I think I'm dealing with people who might be unable to abstract the idea that at some point, reality replaces math, you get information from reality, from real atoms, not a math formula.

It's now completely apparent that you don't know anything about Hamiltonian dynamics (which is basic undergrad physics). Actually, I'd be willing to bet you don't understand any undergrad physics. I think it's usually the case that those who know least feel that they can get away with lying the most.

Go suck a lemon.

I honestly don't care whether you believe me or not. It doesn't effect me, or the people i work with, whether you believe me or not. And i am 100% positive it doesn't affect my credibility with anyone here, if you believe i am not a physicist or not.

Deal?

Yes Vkothii..

..My original discontempt was that symbols are not explained. It's nothing against you.

But guest has turned this into a flame fest.

I actually take 50% Of the blame however, because if i had read your post instead of JUST the equations, i would have seen you where talking about integrals of action.

But as i said, that's not the point, and CERTAINLY not the one Guest is trying to ellaborate on.

And (plus sorry for making three posts in a row), that i offered to give up what i know on Hamiltonians, dispite admitting it has been a while working on them. But then Guest blamed me for going to google it up.

I am interested in exactly what he/she thought i was going to google up... wiki perhaps..

..hahaha... get a life.

I honestly don't care whether you believe me or not. It doesn't effect me, or the people i work with, whether you believe me or not. And i am 100% positive it doesn't affect my credibility with anyone here, if you believe i am not a physicist or not.
You actually claim you're a professional physicist?!? Wow.

And?

How are you funded, may I ask?

You've still not covered the equations i gave you, and you basically site there like some queen of the internet as if that single equation you gave me now holds any relevance.

How are you funded, may I ask?

The governement. And my own savings... about 20,000 pounds of it.

\exp^\beta doesn't make sense and the second equation is ill-defined since you have an index mis-match. These mistakes are indicative of the fact you're not experienced with mathematics.

Oh i do apologize. I never looked back on them; and i have had a few drinks. I will fix them right now.

There. Happy are we?

The governement. And my own savings... about 20,000 pounds of it.
The "governement"? What funding body is that? I guess you acquired the funding by producing a past record of high-level publications?

Stop presuming. You've done a lot of that, and made a great many mistakes already.

Now, now, if you can't be nice, you'll just have to play somewhere else.

So you got government funding to produce research given no prior record of doing so?

Research?

What did i tell you of presuming you twat.

So you're a professional physicist that doesn't do any research?

Now, now, if you can't be nice, you'll just have to play somewhere else.

He/She started it.

So you're a professional physicist that doesn't do any research?


I am a physicist. That is all you need to know, and all i am going to say.

So you're a government funded physicist who doesn't do any research and doesn't understand Hamiltonian mechanics. Do you also live in house made of cheese?

Are you the rat that eats it all?

Grow up.

It's just that your claims don't seem to add up.

No. You have made a profile on me, with what little evidence i have given you, and i have enjoyed every minute of you messing it up with your childish presumptions. What is even more amazing, is that you can't see your messy presumptions.

Better yet, i have not lied one bit.

Just saying "no" doesn't actually change the fact your story is completely laughable. Come on, I assume you're a grown man, why do you feel the need to make up stories about being a professional physicist? I know you're not, you know you're not, anyone with an ounce of sense knows you're not. What are you hoping to achieve by perpetuating such ludicrous stories?

You know nothing, other than the fabrications of your own mind.

You've helped to fabricate a great many things tonight. You've ellaborated my life story as if you should have been God to write it with His Great Hand.

You have nothing to be ashamed than other your fantastic mind.

I know you're not a professional physicist.

And? I know i am.

What do you want me to do about it?

So just to reiterate: you claim to be be a government funded physicist who doesn't do any research. You've also made it apparent you don't know about undergraduate level physics. You see no holes in the story?

Yes.

However, you have painted a picture of me which is totally false to the actual respresentation. I have not said anything wrong.

In a defence of not giving much about my life out, i gave you what you needed to know. Instead of taking what i said, which a minoritized report, you have blatently exhausted what was given to you.

Now... Undergraduate physics? I can give plenty of questions that would fly over your beautiful face.

But that last sentance had no effect before, so it will have no effect now. You have a profound value of weighing up your mathematical knowledge, but very little respect of completing anyone elses questions on something they might excell at.

So why not crawl back into the corner you came from, and give me some peace for the rest of the night.

Goodbye.

So if you know undergrad physics, why can't you construct the action-angle coordinates for the Hamiltonian system I've given you? I really don't understand why you'd want to put yourself in this kind of position by coming on internet forums and posing as a professional physicist when you don't even understand undergraduate level material.

Honesty is the best policy.

Stop this. You are starting to annoy me more than usual.

And you have based your presumption on a single equation in haven't recognized in about a year. Your claims are very very very laughable.

As a professional physicist who doesn't do any research, what exactly is it that you get paid for?

As a professional physicist who doesn't do any research, what exactly is it that you get paid for?

Education. I am still in education.

See..? Direct questions equal direct anwers. Put that into an equation.

Hold on, so you're not a professional physicist? You're a student?

When I finish my doctorate in Anthropology, I'm going to give you guys a mention in my thesis.

Actually guest, Ben would be very dissapointed with your statement.

Vkothii

Thanks ?? :)

"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

-\frac{\hbar^2}{2m}\nabla^2\phi = i\hbar\frac{\partial}{\partial t}\phi

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

\frac{E^2}{c^2} - p^2 = m^2c^2

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi

and the wave function \phi\ , is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of \partial_t \phi\ , and not just \phi\ ,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

\rho=\phi^*\phi\,

and its current by

J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)

and the conservation of probability density has a local form:

\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing \nabla by \,\partial_t:

\rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*)

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both \phi\ , and \,\partial_t\phi can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists

Direct questions equal direct anwers.
Hold on, so you're not a professional physicist? You're a student?
Come on - stick to your promises!

It wasn't direct. You mixed two questions in the one. Even if they did potentially yield the same answer.

Please, be direct.

"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

-\frac{\hbar^2}{2m}\nabla^2\phi = i\hbar\frac{\partial}{\partial t}\phi

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

\frac{E^2}{c^2} - p^2 = m^2c^2

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi

and the wave function \phi\ , is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of \partial_t \phi\ , and not just \phi\ ,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

\rho=\phi^*\phi\,

and its current by

J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)

and the conservation of probability density has a local form:

\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing \nabla by \,\partial_t:

\rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*)

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both \phi\ , and \,\partial_t\phi can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists


Lovely. I actually know these coordinates off-hand... what book are you working from?

Are you admitting to the following: you are not a professional physicist, but are in fact a student?

..what book are you working from?It's just wikipedia's entry for the Dirac electron. Now to derive that action integral for a non-relativistic one.

Oh ok. :)

Are you admitting to the following: you are not a professional physicist, but are in fact a student?

I claim no more than the man i just qouted for claiming he was a physicist.

Now, are you calling Ben liar simultaneously me?

To try and get this tread back on topic:

1.) Saxion---Regardless of your qualifications, if you don't understand the notation that Vkothii has used (which is, I might add, something that an advanced undergraduate/average grad student should be familiar with), it is unlikely that you'll be able to contribute in a meaningful way to the thread.

2.) Vkothii---You never answered my question in the third(?) post. What does "phase of a probability" mean? Probabilities are measurable quantities, so they should give real numbers---that is, there should be no phase associated with them. So what's the deal?

What does "phase of a probability" mean?What indeed?
Let the function \theta_I be the probability amplitude that the particle actually takes the path l between A and B. It is referred to as a phase factor. If the point is held fixed, then the propagator is only a function of the location , and can simply be written as the probability wave solution of the Dirac equation for a free electron.
I.E. show that both the gravitational field and electromagnetic field act upon the propagator in the same manner.
Both induce a phase shift which causes the wave function to disperse in the direction which minimizes the action, thereby decreasing the relative energy of the wave. The difference is that the gravitational field contribution is the result of the mass/energy, and the electromagnetic field contribution is a result of the electromagnetic energy. Both fields are representative of the local environment of a test particle.
The title is an incomplete phrase. Perhaps "phase of a probability amplitude"? (??)
And represent, that's a tricky one when it comes to information content of an electron's spin (phase).

"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

-\frac{\hbar^2}{2m}\nabla^2\phi = i\hbar\frac{\partial}{\partial t}\phi

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

\frac{E^2}{c^2} - p^2 = m^2c^2

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi

and the wave function \phi\ , is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of \partial_t \phi\ , and not just \phi\ ,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

\rho=\phi^*\phi\,

and its current by

J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)

and the conservation of probability density has a local form:

\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing \nabla by \,\partial_t:

\rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*)

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both \phi\ , and \,\partial_t\phi can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists

I hope you realise that none of the equations you have copied and pasted are the Dirac equation. That is \left( i \gamma^\mu \partial_\mu -m \right) \psi = 0. A solution to the Dirac equation is automatically a solution to the Klein Gordon equation, since the KG equation is simply an expression of the relativistic dispersion relation. However, solutions to the Klein Gordon equation are not necessarily solutions to the Dirac equation.

Also, and a sincerely hope that wolv1 is not reading this, what you state about particles going backwards in time is not a very good physical interpretation of the solutions of KG. A better thing to say is that the density that you define; \rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*) , is not a probability density but a charge density. You can work out that the \phi has a charge of +1 under the conserved quantity and \phi^* a charge of -1 (depending on how you define things) meaning that all quantum numbers are the same except those of the conserved quantity under charge conjugation, which is the definition of particles and antiparticles.

the Dirac equation. That is \left( i \gamma^\mu \partial_\mu -m \right) \psi = 0. A solution to the Dirac equation is automatically a solution to the Klein Gordon equation, since the KG equation is simply an expression of the relativistic dispersion relation. However, solutions to the Klein Gordon equation are not necessarily solutions to the Dirac equation.K.
How would you rewrite that to get it into an action integral formulation - there are at least two. One mass and one charge integral.

This "exercise" I thought has an interesting angle on the AB effect; you can do a gauge transformation that shows:
"... The gauge transformation is the gradient of a scalar function, and that gradient corresponds to the gradient in the probability of an interaction.

Thus by means of a variable index of refraction resulting from the absorption and emission of photons the probability density is transformed into a function of the local density of matter and energy, and so are the coordinates! ..."

Hint: there is a bit more to do than just derive the action for the EM field, but that's a detail of this I want to work through - see if it commutes with another view of the physical rep. (of information entropy).
Thing is, information has to be 'conveyed' in a classical sense which is necessarily irreversible.

Oh yeah, the actual Dirac form:
\psi_\pm(x,p)\, = \, \varsigma^{\script l}(p) \exp [{i \over \hbar} p^{\alpha} g_{\alpha\beta} {\chi^\beta}]

Why don't you see if you can find a hammer and a nail somewhere?

I love the way you logical types establish "facts".
You should have taken predicate calc as well, it looks like your logic buffer has overflowed.

Guest and Saxion---
please stay on topic.

There is no topic. We have one guy who insists on copying things down from wiki and MIT notes and pretending to understand them. We have another that alludes to being a government funded physicist who does no research and doesn't know about Hamiltonian mechanics.

I have no idea why these people make up the stories they do, but I imagine that pandering to their need to sprout rubbish only serves to perpetuate their habit.

And we have another guy, who displays a rather interesting interaction dynamic, in which their contribution to the entropy of information isn't measured in their local frame??.

Who spouts a lot of strange-looking "logic", and can't explain where it came from because it "didn't come from them". Interesting.

The topic of this thread is not: "act like a plonker", it's "derive the action for the electromagnetic Lorentz phase shift for Dirac electrons".

Also there's a vague link to: " gauge transformations".

Ed: some of you guys might not get it. There's a geometry and a topology. QM logic ties the gauge of the field together, see?
Maxwell's quaternions explain the large view - the far field effects where we apply the geometry of normal Einsteinian/Euclidian spacetime.
The quantum algebras that are accessible to Boole's network analysis formalism and on up to multi-dimensional algebras, explain the up-close view, where we can (maybe) prepare a resonance, and measure it.
But we have to do both as we do "out here", except up close it has do be done at the same time - you measure the output by "running the program".

And measurement at both scales speaks to our view of the whole show - if you consider it's a big information processor, see?
The big information-processing world, and the big curve in a field that makes particles with spin precess along a preferred axis - a geometry.
Then we get them to line up algebraically. They perform algorithmically, we drive the algorithm - what with?

With the local gradient of the field that is the equivalent of a spin-phase change. An applied field drives it.

There is no topic. We have one guy who insists on copying things down from wiki and MIT notes and pretending to understand them. We have another that alludes to being a government funded physicist who does no research and doesn't know about Hamiltonian mechanics.

I have no idea why these people make up the stories they do, but I imagine that pandering to their need to sprout rubbish only serves to perpetuate their habit.
Apologies BenTheMan, this post was uncalled for. Your forum, your rules, to which I should abide.

Vkothii has moved onto gauge theory and topology now, so I'm quite eager to see how the thread unfolds.

K.
How would you rewrite that to get it into an action integral formulation - there are at least two. One mass and one charge integral.

Why would you want to write the action as integrals over charge and mass? That would tell you how the solutions varied with charge and mass, but you aren't interested in that - we want to know how the Dirac particles behave as it moves in space time. The action is integrated over space time and it is

S = \int d^4 x \left(m \overline{\psi} \psi - \frac{i}{2} \overline{\psi} \gamma^\mu \left(\partial_\mu \psi \right)- \frac{i}{2} \left(\partial_\mu \overline{\psi}\right) \gamma^\mu \psi\right)


This "exercise" I thought has an interesting angle on the AB effect; you can do a gauge transformation that shows:
"... The gauge transformation is the gradient of a scalar function, and that gradient corresponds to the gradient in the probability of an interaction.

Thus by means of a variable index of refraction resulting from the absorption and emission of photons the probability density is transformed into a function of the local density of matter and energy, and so are the coordinates! ..."

Hint: there is a bit more to do than just derive the action for the EM field, but that's a detail of this I want to work through - see if it commutes with another view of the physical rep. (of information entropy).
Thing is, information has to be 'conveyed' in a classical sense which is necessarily irreversible.

Are you talking about the Aharonov-Bohm effect? I don't see what this has to do with quantum information theory. It's about showing the gauge field A_\mu is a more fundamental object that the electric and magnetic fields E and B.


Oh yeah, the actual Dirac form:
\psi_\pm(x,p)\, = \, \varsigma^{\script l}(p) \exp [{i \over \hbar} p^{\alpha} g_{\alpha\beta} {\chi^\beta}]

I presume you're using the greek letter chi for your space time coordinates here, despite the fact that psi is a function of x and p. This is quite an odd way of writing it, since chi is often used to write the Dirac spinor like this
\psi_s = \left(\begin{matrix}\phi \\ \chi \end{matrix}\right)
where phi and chi are 2 component Weyl spinors. I would write your solution like this:
\psi_\pm(x^\mu)\, = \, \left(\begin{matrix}\phi \\ \chi \end{matrix}\right) \exp [\pm i p_\mu x^\mu]

Final warning.

Either help Vkothii answer his question about Wilson lines, or go take your trolling to "About the Members".

psi is a function of x and p. This is quite an odd way of writing it, since chi is often used to write the Dirac spinor like this
\psi_s = \left(\begin{matrix}\phi \\ \chi \end{matrix}\right)
where phi and chi are 2 component Weyl spinors. I would write your solution like this:
\psi_\pm(x^\mu)\, = \, \left(\begin{matrix}\phi \\ \chi \end{matrix}\right) \exp [\pm i p_\mu x^\mu]


There's a way of writing it as a \psi_\pm(x,p) = ? You have to find a wave solution..? So you want the spinors in there - or the correct solutions for them..?

This is more a way to see the topology from another angle - the gauge transform. Is the Schwarzschild frame the only possibility here?
Do Dirac electrons as free particles, require that frame as a given, I mean are there other candidates (I haven't done any gauge stuff)?

Hang on just a minute! Since when are we talking about the Dirac equation in curved space? The machinery for quantum field theories in curved space is pretty limited compared to minkowski space QFT. The Dirac equation in Schwarzschild space is a highly non trivial problem - classically there are probably known solutions but quantum mechanically I doubt anything is in the literature. About the best we can do right now is QFT on a plane wave (http://en.wikipedia.org/wiki/Pp-wave_spacetime).

OK, I had some idea this guy's paper is going down a bit of a strange path.
I was more interested in the math, than the result itself.

I think the idea is to derive a relativistic eqn, do a gauge transform and assume a zero local curvature, to then derive a nonrelativistic form.
I don't want to get buried in it, just look at what's being done and why, hence the question about using a different metric for 'spacetime'.

The notion of a path integral is tied to the notion of 'communication' and irreversible changes, or events in spacetime. Right, it's a big subject and lots of ways to see the field/particle models.
I'm at the start of a QIS course, on my own bat I want to explore the basics, if I get a handle on the ideas then the math just seems to come more easily (I prefer to know why to derive something than how to just 'get a result').
I also realise some of this will bump into various ontological objections, etc.

P.S. This from the intro to the paper:

" ..the attempt is made where possible to translate the fluent geometrical language of General Relativity into the less rigorous, but more intuitive language of electrical engineering. A basic understanding of Quantum Mechanics, vectors and the index notation of General Relativity are required... The equivalent circuit for the transmission of probability waves through space-time is shown to be analogous to the transmission of waves through wave guides in quantum optics, or along transmission lines in electrodynamics. With this interpretation it is shown that the relationships between gravitation and electrodynamics compliment each other in the same fashion that the components of an electronic circuit compliment the sources of electrical energy.

What is presented here shows how the gravitational field can be modeled by variable component values in the equivalent circuit. In a linear circuit the component values do not depend on the strength of the sources."

So he assumes a completely linear "circuit" in each case (no corrections are assumed).

Looks like it could be an EE students attempt to explain the AB effect in GR terms. Maybe it makes some stretches that don't stretch that far?
Or it's an insight into gauge and spin, in terms of particle momentum - path integrals.

The question, ultimately, is WHAT is information? Well, it's what we say it is, essentially.
So how or why do we? (Maybe that's too simplistic for you guys)

Vkothii---

What are we discussing here? I am completely confused (which may be little surprise to some people here).

We started talking about Wilson Lines, and now we're doing Dirac equation in curved space-time?

What do you want to talk about? What is the question you want to answer with this threaD?

I didn't say enough in that last post?

Start at step #1, the derivation of Taylor's expansion of an exponential product with complex exponents.
Why do you get an order 3 approximation, and of what?

How come? The expansion 'oscillates', is how come. It drives the time-evolution of a 'discrete' precession angle. The algorithm is the spin wavefunction.
An applied potential isn't "just some number". It's the equivalent of what drives a digital computer's CPU - the system clock.
So gauge theories are the geometry/topology of this EM field which is QFT.

I didn't say enough in that last post?

Start at step #1, the derivation of Taylor's expansion of an exponential product with complex exponents.
Why do you get an order 3 approximation, and of what?

How come? The expansion 'oscillates', is how come. It drives the time-evolution of a 'discrete' precession angle. The algorithm is the spin wavefunction.
An applied potential isn't "just some number". It's the equivalent of what drives a digital computer's CPU - the system clock.
So gauge theories are the geometry/topology of this EM field which is QFT.

Now I'm more confused. You do know that the probability is the absolute square of the wavefunction, which kills the complex phase, right?

Yes, it's about (this thread is) looking at the engine - the field and what it "does" to fermionic spin.

Electrons aren't the only quanta in the field and we know how to play with a lot of 'quasiparticle' equivalents of wavefunction 'algorithms'. Measurement, or recovering a classical result is not trivial.
Understanding why/why not is another part of this QIS thing.

You guys have not ever done any algorithmics, huh? How about those topological gates, anyons in 2d?

What are you talking about? What field are you talking about?

Probabilities are real numbers, true or false?

Probabilities are real numbers, true or false?When we measure something, it's real, true or false?

I'm talking about the EM field - the one with electrons in it.

So what's the big problem with the phase 'collapsing' because it has to be the square integral of the wavefunction's amplitude?

When we measure something, it's real, true or false?

What do you mean "real"?

Well, is an event real, do we assign a real value to it?
Any observable is modeled as being selected, from a space; these 'spaces' are quite different classically and non-classically (i.e. in the quantum views of fields).
There are real probabilities assigned to classical variables, but not to quantum variables, observables evolve differently.
Or "we can't observe a wavefunction".

Semantically void symbol manipulation. You didn't answer Ben's question.

How about this question:
Are we talking about probability??

If we are, is an apple that's growing on an apple tree a probability?

Or does it "have" a probability, and what is "the real probability", since someone wants to know if a probability is a real number?

Did the question not go: "is a probability a real number?"

So, if it is, is an apple a real number? Or is something the apple "has", a real number?
Or, if it isn't, is an apple an imaginary number? Is whatever it "has" also imaginary, or WHAT??

Answer: apples are real, but they aren't "real numbers". Apples have a probability - actually several 'probabilities', which are events we associate with apples.
But the resident plonker would rather avoid that little mathematical problem, it bumps into way too much reality for them to handle, obviously.

I'm talking about the EM field - the one with electrons in it.

The EM field consists of photons, not electrons.

So electrons are in a different field, you mean? Not the EM field?
How about neutrons, what field are they in?

Don't you mean to say: "the quantum of the EM field is the photon".
The electron interacts with the bloody thing doesn't it? So electrons, protons, neutrons, any particle with electromagnetic charge, is "in the EM field" then.

Fuck this is absolutely fucking pathetic.

I might as well watch porn and masturbate like you fucks are.

Here's another obvious fuck-up that illustrates how opinionated you math-fucks are.

"The AB effect has nothing to do with quantum information, the same way QM has nothing to do with quantum information, or QFT, or QED, actually no quantum theory is remotely connected to information".

According to some expert here at this fucking pathetic forum.
I sure hope all that is sufficiently "semantically void manipulation" of the requisite symbology. Ben didn't fucking ask any FUCKING question.
What he did was bat my question away, because he is ignorant of what a quantum event is. Which means he's ignorant of what quantum information is.
Which means I am wasting my time asking any of you oh-so assured math types a bloody thing. Because you all have a fundamental lack of understanding of the goddam subject.

Most of them seem to think it isn't a subject, so much for that one, huh.

How do I keep missing getting in on the ground floor of these threads! Damn it! Wait.... I had better things to do on Friday and the weekend. Shazam!
When I finish my doctorate in Anthropology, I'm going to give you guys a mention in my thesis.I'm due to be mentioned in numerous people's rewriting of physics when they get their Theory of Everything published ;)
Also there's a vague link to: " gauge transformations".

Ed: some of you guys might not get it. There's a geometry and a topology. QM logic ties the gauge of the field together, see?
Maxwell's quaternions explain the large view - the far field effects where we apply the geometry of normal Einsteinian/Euclidian spacetime.
The quantum algebras that are accessible to Boole's network analysis formalism and on up to multi-dimensional algebras, explain the up-close view, where we can (maybe) prepare a resonance, and measure it.
But we have to do both as we do "out here", except up close it has do be done at the same time - you measure the output by "running the program".

And measurement at both scales speaks to our view of the whole show - if you consider it's a big information processor, see?
The big information-processing world, and the big curve in a field that makes particles with spin precess along a preferred axis - a geometry.
Then we get them to line up algebraically. They perform algorithmically, we drive the algorithm - what with?

With the local gradient of the field that is the equivalent of a spin-phase change. An applied field drives it.I think my spleen just ruptured....
This is more a way to see the topology from another angle - the gauge transform. Is the Schwarzschild frame the only possibility here?
Do Dirac electrons as free particles, require that frame as a given, I mean are there other candidates (I haven't done any gauge stuff)?Without knowledge of gauge transformations, you should not be asking ANYTHING about spinors in curved space-time. Spinors in curved space-time require the understanding of spin connections. Which are an extension of metric connections on space-time manifolds, as well as an understanding of how to describe the residual freedoms of a given spinor representation, which is then going to bring in connections on gauge bundles which are the generators of precisely that, gauge transforms.

Spinors in curved space-time are the kind of thing you get in the 10th chapter of a book on differential geometry, after chapter 2-9 covering G-bundles and vector bundles, their horizontal spaces, their vertical spaces, lifting paths in the base space into sections of the bundle and then relating the two. Vector bundles aren't too much trouble if you're familiar with GR on a working level. G bundles are the kind of thing you need for gauge transformations. And unfortunately, there's an isomorphic relation between the two, once you've defined the representation of your vector bundle. Obviously! :rolleyes: Who'd be stupid enough to forget the representation! :shrug:

Who'd be stupid enough to forget the representation! Someone who doesn't know what a probability is? Maybe they might forget that you represent it with a number.

Or they might forget that you can transform a measurement basis, because something as simple as a pair of sunglasses illustrates that information is what we say it is, because that is how we measure it.
It doesn't matter, except in terms of how much work you do mathematically analysing the process, if you consider the polaroid filter is the channel, or the photons that it polarizes are the channel - the content is the same.
But someone who doesn't really think about what measurement and probability are, say, probably wouldn't be able to see that. That it is pretty simple.

I'm finding it difficult to see your motivation for posting all this drivel. Everyone knows you don't know what you're talking about, so what's the end-game here?

I want to show everyone how to get a moron to keep repeating drivel.

You seem to be living up to expectation - you haven't said anything yet, just a lot of whining, and drivel.

I want to show everyone how to get a moron to keep repeating drivel.
Then you've succeeded in showing us. Well done.