B Spin and polarizations in momentum space

[Azurite]

Can momentum space also able to handle spin and polarizations?

I'm understanding it that in QM, you have position, momentum, spin, polarization as observables. Position and momentum can be equivalent via Fourier transform. So if you use momentum space instead of position, how do you handle spin or polarization inside the momentum space since they are all observables?

 

[PeterDonis]

Can momentum space also able to handle spin and polarizations?

No. Spin/polarization degrees of freedom are separate from position/momentum degrees of freedom.

 

[Azurite]

No. Spin/polarization degrees of freedom are separate from position/momentum degrees of freedom.

I take it momentum space also means 4-momentum. But how come in particle physics they use 4-momentum to analyze particle tracks which could include spin and polarizations?

 

[PeterDonis]

I take it momentum space also means 4-momentum.

If you're doing quantum field theory, yes.

how come in particle physics they use 4-momentum to analyze particle tracks which could include spin and polarizations?

The 4-momentum does not include the spin/polarization degrees of freedom.

 

[Azurite]

If you're doing quantum field theory, yes.
The 4-momentum does not include the spin/polarization degrees of freedom.

How does quantum field theory handle spin/polarization without using the 4-momentum?

 

[PeterDonis]

How does quantum field theory handle spin/polarization without using the 4-momentum?

By recognizing, just as in non-relavistic QM, that the spin degrees of freedom are separate from the 4-momentum degrees of freedom, just as I said in previous posts.

 

[Mentz114]

How does quantum field theory handle spin/polarization without using the 4-momentum?

By using matrices. See this Wiki page on spin operators and Pauli spinors. If you have a wave function of a spinless particle then to make a WF for 2-valued spin one uses two wave functions in a vector.

 

[Azurite]

Thank you for helping.

How about the so called phase space.. can you do spin/polarization on phase space (or does this only handle position and momentum)?

 

[PeterDonis]

How about the so called phase space

Phase space is just another way of looking at position/momentum space. It doesn't include any other degrees of freedom.

It might be helpful if you gave some more context of what you are trying to figure out.

 

[Azurite]

Phase space is just another way of looking at position/momentum space. It doesn't include any other degrees of freedom.

It might be helpful if you gave some more context of what you are trying to figure out.

In momentum space, it's energy against momentum plotted in the axis while in spacetime it's space and time plotted in the axis. I just want to know in the case of spin and polarization.. what are the respective two axes?

 

[PeterDonis]

in the case of spin and polarization.. what are the respective two axes?

There aren't two axes. Spin/polarization (these are not complementary observables like position and momentum, btw) are discrete observables, not continuous ones. They don't work the same.

 

[Azurite]

There aren't two axes. Spin/polarization (these are not complementary observables like position and momentum, btw) are discrete observables, not continuous ones. They don't work the same.

Position and momentum are continuous.
Spin/polarizations are discrete.

What are the terms that distinguished them? Why can't spin/polarizations become continuous and position and momentum discrete? What symmetry or rules possibly forbids them?

 

[Azurite]

Also the wave function of position is very related to momentum (fourier of position) so can't we say generally that momentum is the wave function of position?

 

[Mentz114]

Position and momentum are continuous.
Spin/polarizations are discrete.

What are the terms that distinguished them? Why can't spin/polarizations become continuous and position and momentum discrete? What symmetry or rules possibly forbids them?

As PeterDonis has said, spin and polarization are different things. Quantum spin is a property of Fermions and may be aligned wrt a spatial direction. Look up the Stern-Gerlag experiment to see how it may be measured.

Polarization is a property of light and is measured with a polarizer.

 

[PeterDonis]

Why can't spin/polarizations become continuous and position and momentum discrete?

Because that's not how their Hilbert spaces work. You can't just wave your hands and "make" an observable be continuous or discrete. You have to look at the physics of each observable and figure out whether it's continuous or discrete. Being continuous or discrete is simply "built in" to the physics of the observable. Your question is like asking why the number 4 has to be even instead of odd.

the wave function of position is very related to momentum (fourier of position) so can't we say generally that momentum is the wave function of position?

No. This doesn't even make sense.

What background do you have in QM? What textbooks have you studied? Your questions indicate that you would probably benefit from taking some time to learn the basics of QM and how it models physical systems and observables.

 

[Azurite]

No. This doesn't even make sense.

Let me explained. Are you familiar with crystallography? See http://phy-page-imac.princeton.edu/~page/phy312/lecture_notes_s11/fourier_optics2.pdf
The diffractions patterns in the screen of the double slit experiment are simple Fourier transform of the slits occurring in momentum space.

What background do you have in QM? What textbooks have you studied? Your questions indicate that you would probably benefit from taking some time to learn the basics of QM and how it models physical systems and observables.

I study textbooks in the beginning stage. I came across the above concept that the interference patterns of the double slits screen is the Fourier transform of the slits. I was asking that since the slits are the source of the intereference patterns as Fourier transform of it. What if the particle just gets guided to it? In this context.. the momentum is the Fourier transform of position. We need the Schrodinger Equation to solve for the wave function because we need to compute the diffraction patterns.. but if the diffraction patterns already exist in a real momentum space as proposed by Lee Smolin. Then why does the particle needs to have probability waves.. it can just navigate the existing interference patterns already there in momentum space. Can you refute this interpretation? I know all interpretations use the same math. But the math can produce this interpretation I'm describing if Lee Smolin is right the momentum space is objective. Is it not?

 

[PeterDonis]

The diffractions patterns in the screen of the double slit experiment are simple Fourier transform of the slits occurring in momentum space.

What specific equation in the reference you gave do you think says this? And what do you think it means?

 

[Azurite]

What specific equation in the reference you gave do you think says this? And what do you think it means?

It's equation 18.. I can't repeat the equation here but he description says:
"In words, we would state this as: the diffraction pattern for a single frequency is the square of the Fourier transform of the aperture distribution. In these domains, spatial distributions transform or map into angles.
This is a very powerful concept. It allows you to "see" the diffraction pattern of almost any aperture. Likewise, given a diffraction pattern, you can work in reverse and figure out what makes it. This is the basis behind electron diffraction, crystallography and many scattering experiments."

Do you not deny that the interference patterns seen in any double slit experiment is Fourier transform of the slits? Then if the momentum space as proposed by Lee Smolin is objective.. see the following reference:
https://www.newscientist.com/article/mg21128241-700-beyond-space-time-welcome-to-phase-space/

Then couldn't it be that the particle simply navigate the momentum space and not waving (probabilistically, Many worldly, etc.)? This interpretation may sound silly but I just want you to state clearly why it is silly so I can forget about it.

 

[PeterDonis]

Do you not deny that the interference patterns seen in any double slit experiment is Fourier transform of the slits?

Not the way you mean it. The reference you gave is saying that, as it summarizes at the end of the paragraph following equation 18, "spatial distributions transform or map into angles". Angles are not the same as momentum; the closest you could get would be to say they are limits on the directions of various possible momenta of particles when they leave the slits, but we're not talking about particles, we're talking about waves. So this reference, while it is certainly valid and interesting, does not mean what you think it means and does not support the claims you are making about quantum mechanics.

see the following reference

This is a pop science article. Furthermore, it is describing an extremely speculative hypothesis, not standard QM. So again it does not support the claims you are making.

couldn't it be that the particle simply navigate the momentum space and not waving (probabilistically, Many worldly, etc.)?

I can't tell if this is a reasonable description of Smolin's speculative hypothesis or not. But it is certainly not a reasonable description of standard QM. And standard QM is what we can discuss in this forum.

 

[Azurite]

Not the way you mean it. The reference you gave is saying that, as it summarizes at the end of the paragraph following equation 18, "spatial distributions transform or map into angles". Angles are not the same as momentum; the closest you could get would be to say they are limits on the directions of various possible momenta of particles when they leave the slits, but we're not talking about particles, we're talking about waves. So this reference, while it is certainly valid and interesting, does not mean what you think it means and does not support the claims you are making about quantum mechanics.

But is it not momentum is the Fourier transform of position? Optical transform is about getting the Fourier transform of position or vice versa. How do you understand optical transform?

This is a pop science article. Furthermore, it is describing an extremely speculative hypothesis, not standard QM. So again it does not support the claims you are making.
I can't tell if this is a reasonable description of Smolin's speculative hypothesis or not. But it is certainly not a reasonable description of standard QM. And standard QM is what we can discuss in this forum.

Ok let's not talk about phase space but just plain optical transform..

 

[PeterDonis]

is it not momentum is the Fourier transform of position?

No. A wave function in momentum space is the Fourier transform of a wave function in position space.

Optical transform

What do you mean by "optical transform"? And how does it relate to quantum wave functions, which is what you're making claims about?

 

[Azurite]

No. A wave function in momentum space is the Fourier transform of a wave function in position space.
What do you mean by "optical transform"? And how does it relate to quantum wave functions, which is what you're making claims about?

Consider the following optical transforms of the image on the left side. The image on the right is the optical transforms actually measured and computed. I found this on the net.

In the double slit experiment.. would the slit separation, sizes, etc actually produced the interference patterns that can be computed using optical transform? Or not? But in the above, the slits are actual and the right image are the diffraction and interference patterns of the left. So if typical double slit experiment slits and sizes were used.. I think it should produce the same interference patterns. Would it not?

The bottomline is. The interference patterns in the double slit experiment is the Fourier transform of the slits right?

You said "A wave function in momentum space is the Fourier transform of a wave function in position space."

In the double slit experiment depicted above. The position is not of the particle, but the position of the slits.. so the interference patterns are the Fourier transform of the slits. Therefore doesn't it hold that the position space of the slits (again not the particle) can be Fourier transformed producing the interference patterns in momentum space?

 

[PeterDonis]

I found this on the net.

Where? Please give a specific reference.

In the double slit experiment.. would the slit separation, sizes, etc actually produced the interference patterns that can be computed using optical transform?

You haven't said what an "optical transform" is, and you gave no reference for the picture you posted. So I have no information on which to base an answer.

The bottomline is. The interference patterns in the double slit experiment is the Fourier transform of the slits right?

What does "the Fourier transform of the slits" mean? If you gave a reference for the pictures you posted, with the math that was used to compute them, it would be a lot easier to respond to all these questions.

In the double slit experiment depicted above. The position is not of the particle, but the position of the slits..

That appears to be the case, yes. But you gave no reference so I don't know for sure.

so the interference patterns are the Fourier transform of the slits

I have seen no math so I don't know what this means.

Therefore doesn't it hold that the position space of the slits (again not the particle) can be Fourier transformed producing the interference patterns in momentum space?

The slits aren't moving so they have no "momentum space", and they aren't treated as quantum objects in the standard treatment so they have no "position space" either. The standard quantum treatment of the double slit experiment does not assign a quantum wave function to the slits, only to the particles (electrons, photons, whatever) that are passing through the slits. So it makes no sense in the standard QM treatment to talk about the "position space" or "momentum space" of the slits.

Earlier, you said:

I study textbooks in the beginning stage. I came across the above concept that the interference patterns of the double slits screen is the Fourier transform of the slits.

In what textbook did you come across this? So far, you appear to have misinterpreted the only mathematical reference you have given. So I really don't understand what concepts you think you are working with.

 

[Azurite]

Where? Please give a specific reference.
You haven't said what an "optical transform" is, and you gave no reference for the picture you posted. So I have no information on which to base an answer.
What does "the Fourier transform of the slits" mean? If you gave a reference for the pictures you posted, with the math that was used to compute them, it would be a lot easier to respond to all these questions.
That appears to be the case, yes. But you gave no reference so I don't know for sure.
I have seen no math so I don't know what this means.
The slits aren't moving so they have no "momentum space", and they aren't treated as quantum objects in the standard treatment so they have no "position space" either. The standard quantum treatment of the double slit experiment does not assign a quantum wave function to the slits, only to the particles (electrons, photons, whatever) that are passing through the slits. So it makes no sense in the standard QM treatment to talk about the "position space" or "momentum space" of the slits.

Earlier, you said:
In what textbook did you come across this? So far, you appear to have misinterpreted the only mathematical reference you have given. So I really don't understand what concepts you think you are working with.

Yes I think I have misunderstood it.

I mixed the concepts of Fourier transforms in optics and that of wave function. After pondering it more.. I realized how I got confused. According to http://cns-alumni.bu.edu/~slehar/fourier/fourier.html
"Fourier theory states that any signal, in our case visual images, can be expressed as a sum of a series of sinusoids. In the case of imagery, these are sinusoidal variations in brightness across the image."

in the quantum case, "A wave function in momentum space is the Fourier transform of a wave function in position space." as you described.

I got the distinctions now.. thanks for your patience. :)

 

[Azurite]

As PeterDonis has said, spin and polarization are different things. Quantum spin is a property of Fermions and may be aligned wrt a spatial direction. Look up the Stern-Gerlag experiment to see how it may be measured.

Polarization is a property of light and is measured with a polarizer.

of course I know the difference between spin and polarization. I included them in "spin/polarization" to denote they are discrete not like momentum that is continuous.

 

[cosmik debris]

Yes I think I have misunderstood it.

I mixed the concepts of Fourier transforms in optics and that of wave function. After pondering it more.. I realized how I got confused. According to http://cns-alumni.bu.edu/~slehar/fourier/fourier.html
"Fourier theory states that any signal, in our case visual images, can be expressed as a sum of a series of sinusoids. In the case of imagery, these are sinusoidal variations in brightness across the image."

in the quantum case, "A wave function in momentum space is the Fourier transform of a wave function in position space." as you described.

I got the distinctions now.. thanks for your patience. :)

An optical transform is just a 2D Fourier transform isn't it?

Cheers

 

[Azurite]

An optical transform is just a 2D Fourier transform isn't it?

Cheers

It's just the statement that the diffraction pattern off an obstacle, at large distances, is the Fourier transform of the obstacle.
Also note reciprocal space mean the same thing as momentum space and k-space.

Mathematically summarizing all. https://en.wikipedia.org/wiki/Reciprocal_lattice

"In physics, the reciprocal lattice represents the Fourier transform of another lattice (usually a Bravais lattice). In normal usage, this first lattice (whose transform is represented by the reciprocal lattice) is usually a periodic spatial function in real-space and is also known as the direct lattice. While the direct lattice exists in real-space and is what one would commonly understand as a physical lattice, the reciprocal lattice exists in reciprocal space (also known as momentum space or less commonly as K-space, due to the relationship between the Pontryagin duals momentum and position.) The reciprocal lattice of a reciprocal lattice, then, is the original direct lattice again, since the two lattices are Fourier transforms of each other.

The reciprocal lattice plays a fundamental role in most analytic studies of periodic structures, particularly in the theory of diffraction. In neutron and X-ray diffraction, due to the Laue conditions, the momentum difference between incoming and diffracted X-rays of a crystal is a reciprocal lattice vector. The diffraction pattern of a crystal can be used to determine the reciprocal vectors of the lattice. Using this process, one can infer the atomic arrangement of a crystal."

Now about position and momentum space.
https://en.wikipedia.org/wiki/Position_and_momentum_space

"Mathematically, the duality between position and momentum is an example of Pontryagin duality. In particular, if a function is given in position space, f(r), then its Fourier transform obtains the function in momentum space, φ(p). Conversely, the inverse transform of a momentum space function is a position space function."

This means even in classical physics.. a function in momentum space is the Fourier transform of a function in position space.
This can be applied to quantum case too.. where a wave function in momentum space is the Fourier transform of a wave function in position space.

My question then should be. If the momentum space is objective (or has ontology). Can the deBroglie pilot wave be located in momentum space?

 

[PeterDonis]

If the momentum space is objective (or has ontology)

What does this mean?

Can the deBroglie pilot wave be located in momentum space?

What does this mean?

You appear to be taking the term "space" way too literally. Momentum "space" is an abstract thing, just like position "space" is. Particularly when you are talking about quantum systems, where the wave function of a system with multiple particles has no intuitive representation in ordinary 3-dimensional "space" at all (for example, the position "space" of a two-particle quantum system has six dimensions, not three).

Furthermore, if you take seriously the fact that the momentum space wave function of a quantum system is the Fourier transform of the position space wave function of the same system, then both "spaces" are equally "real" (or equally abstract)--they're just different ways of representing the same thing. They aren't really separate "spaces" or separate degrees of freedom; they're just different ways of describing the same "space", the same degrees of freedom.

 

[Azurite]

What does this mean?
What does this mean?

You appear to be taking the term "space" way too literally. Momentum "space" is an abstract thing, just like position "space" is. Particularly when you are talking about quantum systems, where the wave function of a system with multiple particles has no intuitive representation in ordinary 3-dimensional "space" at all (for example, the position "space" of a two-particle quantum system has six dimensions, not three).

I got it that Quantum mechanics was about wave function evolving in a 3N dimensional space. But I was wondering about versions where quantum mechanics is about N particles evolving in three-dimensional space and I came across the following timely article.
https://www.jstor.org/stable/10.1086/518633

logging in inside I read the following:

"2. The Wave Function is Represented by a Property. On my picture of what quantum mechanics is about, quantum mechanics is about particles, and system of particles, all evolving in three-dimensional space. Quantum mechanics tells us a bit about the properties of these particles, but to really get a full story, one needs an interpretation of quantum mechanics. A modal interpretation, for example, will specify the circumstance under which these particles have definite properties. The many-worlds interpretation would specify that there are actually different worlds, and the particles in the different words have different properties. For whatever interpretation you pick, the interpretation gives a fuller account of what properties these particles have.

Where does the wave function fit into all this? The wave function is a representation of the quantum state of the closed system. As long as this quantum state is pure - and, since I'm restricting the discussion to closed systems, it's reasonable to assume that it is pure - then it's the eigenstate of some observable. I endorse the eigenstate-to-eigenstate half-link- if not for every quantum system, then at least for the system of all the particles in the universe (or if the universe is too big for you, then whatever closed system you are interested in). The eigenstate-to-eigenvalue half-link holds that if the system is in an eigenstate of some observable, then the system actually has the property represented of some observable, then the system actually has the property represented by the eigenvalue associated with that eigenstate. So, since the quantum state of all the particles in the universe (or whatever closed system you're interested in) is an eigenstate of some observable, then by the eigenstate-to-eigenvalue half-link, the system actually has the property represented by the eigenvalue. That's how the wave function fits into my picture: the wave function doesn't exist on its own, but it corresponds to a property possessed by the system of all the particles in the universe (or whatever closed system you're interested in"

May I know what particular interpretation doesn't use the 3N configuration space but strictly N particles evolving in 3 dimensional space only?

Furthermore, if you take seriously the fact that the momentum space wave function of a quantum system is the Fourier transform of the position space wave function of the same system, then both "spaces" are equally "real" (or equally abstract)--they're just different ways of representing the same thing. They aren't really separate "spaces" or separate degrees of freedom; they're just different ways of describing the same "space", the same degrees of freedom.

 

[PeterDonis]

was wondering about versions where quantum mechanics is about N particles evolving in three-dimensional space

There aren't any. I can't read the paper you linked to (perhaps others here on PF might have the access to do so), but it's published in a philosophy journal, not a physics journal, so I would not rely on it as an accurate description of the actual physics.

 

[Azurite]

There aren't any. I can't read the paper you linked to (perhaps others here on PF might have the access to do so), but it's published in a philosophy journal, not a physics journal, so I would not rely on it as an accurate description of the actual physics.

There is a "Read Online" button.. just press it.. enter any email address and anyone can read the whole article. But if the wave function is just mathematical tool.. it can occur in any 3N dimensional.. but it doesn't mean it correlates to any real N dimensional space. Our reality is three-dimensional only so perhaps the wave function is more of mathematical tool than actually there.. isn't it.

 

[PeterDonis]

Our reality is three-dimensional only

Really? How do you know this?

 

[PeterDonis]

There is a "Read Online" button.. just press it.. enter any email address and anyone can read the whole article

No, set up a free JSTOR account and anyone can read the whole article. Too much friction, particularly since, as I said, the article is philosophy, not physics.

 

[Azurite]

Really? How do you know this?

I mean.. even if our reality is 8 dimensional or 20 dimensional.. it doesn't have to correspond to the 3N dimensions of the configuration space (for example 500 dimensional) of Quantum mechanics, isn't it?

 

[PeterDonis]

even if our reality is 8 dimensional or 20 dimensional

What does it mean for "reality" to have a particular "dimension"?

 

[Azurite]

What does it mean for "reality" to have a particular "dimension"?

In the sense of the extra large dimensions and compactified small dimensions of for example string theory. Is the 3N configuration space of quantum mechanics saying that the larger dimensions is equal to the 3N configuration space of QM?

 

[PeterDonis]

In the sense of the extra large dimensions and compactified small dimensions of string theory.

Ok, then those are spacetime dimensions, so they're not the same as the dimensions of QM configuration space. But we don't have any way of knowing which, if either, is the "dimensions" of "reality". We don't even know if that makes sense.

 

[PeterDonis]

the 3N configuration space of quantum mechanics

It's also worth nothing that this "3N configuration space" is for a non-relativistic quantum system consisting of N particles. But that's only an approximation anyway. Our best fundamental quantum theory is quantum field theory, which, to the extent the idea of "number of dimensions of configuration space" makes sense at all, has a continuous infinity of dimensions of its "configuration space".

This is an example of why I find philosophers' commentaries on physics to be not worth reading. Philosophers of QM are still talking about a version of QM that has been outdated since the late 1920s, as though it were the latest, most exact version of QM, the one that physicists think best describes "fundamental reality". But no physicist thinks that, or has since, as I said, the late 1920s.

 

[Azurite]

Ok, then those are spacetime dimensions, so they're not the same as the dimensions of QM configuration space. But we don't have any way of knowing which, if either, is the "dimensions" of "reality". We don't even know if that makes sense.

If we don't know how to tell.. you mean the dimensions of QM configuration space can become the spacetime dimensions? Does this make sense?

 

[PeterDonis]

you mean the dimensions of QM configuration space can become the spacetime dimensions?

No. I mean that all of these "dimensions" are properties of abstract models. We don't know which, if any, of these "dimensions" are the "dimensions of reality". We don't even know if "reality", whatever it is, can even be usefully described as having "dimensions".

 

[Azurite]

No. I mean that all of these "dimensions" are properties of abstract models. We don't know which, if any, of these "dimensions" are the "dimensions of reality". We don't even know if "reality", whatever it is, can even be usefully described as having "dimensions".

spacetime dimensions = lorentzian spacetime manifold
QM configuration dimensions = non-lorentzian spacetime manifold

Is there a way to relate configuration dimensions to spacetime manifold?

 

[PeterDonis]

spacetime dimensions = lorentzian spacetime manifold

Yes.

QM configuration dimensions = non-lorentzian spacetime manifold

No. Configuration space is not spacetime, not even a non-lorentzian spacetime.

Is there a way to relate configuration dimensions to spacetime manifold?

It depends. For the non-relativistic case of $N$ particles, if space (not spacetime, since we're non-relativistic) has $D$ dimensions, then the configuration space of the quantum system has $N * D$ dimensions.

But in the relativistic case, where we have to use quantum field theory, there is a continuous infinity of "dimensions" (degrees of freedom would be a better term) in "configuration space" (which isn't a good term here because it isn't the space of configurations of anything), and there's no way to tell from this how many dimensions spacetime itself has.

 

[Azurite]

Yes.
No. Configuration space is not spacetime, not even a non-lorentzian spacetime.

I meant QM configuration dimensions = non-"lorentzian spacetime manifold".

It depends. For the non-relativistic case of $N$ particles, if space (not spacetime, since we're non-relativistic) has $D$ dimensions, then the configuration space of the quantum system has $N * D$ dimensions.

Since relativity is true and it should be spacetime.. then it's not space (which I understood as Newton space)... meaning "space" is incomplete formalism. But then is it not we can still go back to absolute time by making the length contraction and time dilation occurring in the object? So Newton space is still valid if the relativistic mechanics is really occurring in the atomic processes? Is this a right distinction?

But in the relativistic case, where we have to use quantum field theory, there is a continuous infinity of "dimensions" (degrees of freedom would be a better term) in "configuration space" (which isn't a good term here because it isn't the space of configurations of anything), and there's no way to tell from this how many dimensions spacetime itself has.

 

[PeterDonis]

is it not we can still go back to absolute time by making the length contraction and time dilation occurring in the object?

No. Length contraction and time dilation are perspective effects, like the way the apparent size of an object changes when you view it from different angles. They are not actual things that happen to the object.

So Newton space is still valid if the relativistic mechanics is really occurring in the atomic processes?

No, because "relativistic mechanics is really occurring in the atomic processes" is wrong. See above.

 

[Azurite]

No. Length contraction and time dilation are perspective effects, like the way the apparent size of an object changes when you view it from different angles. They are not actual things that happen to the object.

I know the above is vintage original Einstein Special Relativity. But is it not in Lorentz Ether Theory.. it occurs in the objects themselves?
https://en.wikipedia.org/wiki/Lorentz_ether_theory

"Lorentz proposed three possible explanations for this relative contraction:[B 3]

The body contracts in the line of motion and preserves its dimension perpendicularly to it.

• The dimension of the body remains the same in the line of motion, but it expands perpendicularly to it.
• The body contracts in the line of motion, and expands at the same time perpendicularly to it."

So absolute time and length contraction occurring in the object is still valid!

No, because "relativistic mechanics is really occurring in the atomic processes" is wrong. See above.

 

[PeterDonis]

is it not in Lorentz Ether Theory

Please read the PF rules; see the section on "non-mainstream theories" here:

https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

 

[Azurite]

Please read the PF rules; see the section on "non-mainstream theories" here:

https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

Ah you meant "Attempts to promote or resuscitate theories that have been discredited or superseded (e.g. Lorentz ether theory); this does not exclude discussion of those theories in a purely historical context"

Can you at least point to reference where this Lorentz ether theory is discredited? I thought since LET occurs in all frames.. there is no way to tell.. so it's not actually discredited but just unprovable (there is a difference).. so it's just unprovable and not discredited.. correct? If it's not discredited.. then please let Greg change the wording in the forum rules above.

 

[Azurite]

Yes.
No. Configuration space is not spacetime, not even a non-lorentzian spacetime.
It depends. For the non-relativistic case of $N$ particles, if space (not spacetime, since we're non-relativistic) has $D$ dimensions, then the configuration space of the quantum system has $N * D$ dimensions.

If space is an incomplete formalism. Why should we even mention "configuration space" where it still uses the outdated concept of space I wonder??

But in the relativistic case, where we have to use quantum field theory, there is a continuous infinity of "dimensions" (degrees of freedom would be a better term) in "configuration space" (which isn't a good term here because it isn't the space of configurations of anything), and there's no way to tell from this how many dimensions spacetime itself has.

 

[Nugatory]

If space is an incomplete formalism. Why should we even mention "configuration space" where it still uses the outdated concept of space I wonder??

In the phrase "configuration space", the word "space" is being used in the mathematical sense (a set with some added mathematical structure) and there is nothing outmoded about it.

 

[PeterDonis]

Can you at least point to reference where this Lorentz ether theory is discredited?

No. The phrase you quoted from the rules says "discredited or superseded". LET is superseded (by the standard intepretation of SR, which simply uses Occam's Razor to say that, since the "absolute rest frame" is unobservable, you can just ignore it).