Point-like particles, Lorentz invariance and QM/QFT

[PhilDSP]

Can you show that derivation?

It will take a while to write up properly and is in some ways involved enough to warrant a paper in itself. That probably makes it off-topic in this forum much as I would appreciate feedback.

 

[TrickyDicky]

It will take a while to write up properly and is in some ways involved enough to warrant a paper in itself. That probably makes it off-topic in this forum much as I would appreciate feedback.

Ah,ok.

 

[Dickfore]

As I have written in the text, Eq. (3) is valid for ARBITRARY coordinate transformations, not only for rotations. If you were familiar with topics such as general relativity or differential geometry, you would probably know it.

And if you knew relativity, you would know that the electric field is not a 4-vector, but, together with the magnetic field constitutes a rank-2 antisymmetric 4-tensor.

When you classify something as a 3-vector (as you did in your "derivation"), it has to do with rotations in 3D-space only! This is not the same as Galilean transformations, which involve a relative translation of one frame relative to another.

I suggest you have a look at the document linked in post #25.

 

[TrickyDicky]

Maybe the conditions under which the Gauss law is Galilean invariant should have been better specified, certainly Galilean invariance holds for a static electric field or for spherically symmetric situations. It won't hold for situations where the curl of the magnetic field doesn't vanish though. In such cases either the Lorentz group or a more general group of transformations is required.

So as long as one assumes isotropy it is possible to formulate electrodynamics consistently without relativity considerations, after all Maxwell did it, didn't he?

 

[Dickfore]

It won't hold for situations where the curl of the magnetic field doesn't vanish though.

Exactly! Notice that invariance (covariance) of an equation means that if it holds in one reference frame, then it holds in ALL reference frames.

In such cases either the Lorentz group or a more general group of transformations is required.

So, that implies relativity. As for "other" transformations, what else would you suggest?

So as long as one assumes isotropy it is possible to formulate electrodynamics consistently without relativity considerations, after all Maxwell did it, didn't he?

Isotropy of what? Maxwell sure did it, but then people thought of something called aether. I wonder why?

 

[TrickyDicky]

Isotropy of what? Maxwell sure did it, but then people thought of something called aether. I wonder why?

Spatial isotropy, a spherically symmetric charge distribution would allow Galilean invariance for the Gauss' law. Spatial isotropy is a reasonable assumption in our universe.

 

[ytuab]

Would the need for considering those effects be by-passed if we assume the form of the Dirac equation is correct and can subsequently demonstrate that the energy equation E^2 = p^2c^2 + m^2c^4 can be rigorously derived from purely non-relativistic considerations?

As far as I know, Dirac equation is equal to "special relativity".
Substituting relativistic x and t into usual accelaration equaion (of Newtorian mechanics).
Ant if we use the force F of v=0, we can get your relativistic momentum and energy.
(And the solution of Dirac equation uses four vector momemtum, energy, time, position variables, which are based on SR.)

Special relativity = photon particle + speed limit = c. (It denies ether.)
Of course, relativistic QED is based on photon particle.
But QED uses Maxwell theory, too. (I don't understant the reason why.).
Maxwell theory is not completely equal to the relativity.
When we use the "Lorentz gauge condition" intentionally, Maxwell theory is equal to Special relativity.

All particle need to satisfy the special relativity according to QFT.

 

[Dickfore]

Spatial isotropy, a spherically symmetric charge distribution would allow Galilean invariance for the Gauss' law. Spatial isotropy is a reasonable assumption in our universe.

But, a spherically symmetric charge distribution means there is a well defined center. Where is the center of the Universe?

 

[Demystifier]

When you classify something as a 3-vector (as you did in your "derivation"), it has to do with rotations in 3D-space only!

This is simply wrong. Just as vector may have to do with arbitrary coordinate transformations (not only rotations) in 4 dimensions, it may have to do with arbitrary coordinate transformations in any number of dimensions, including 3.

It is a common misconception among physicists that general coordinate transformations are only relevant in general theory of relativity. But this is wrong. In 3 dimensions, they are relevant even in non-relativistic physics.

 

[Demystifier]

I suggest you have a look at the document linked in post #25.

In the conclusion they say:

"The following 3 assumptions are inconsistent:
1. Galilean invariance
2. Continuity equation.
3. Magnetic forces between electric currents."

Fine, but this means that if we do not require one of these items, then we still have a consistent theory. In particular, for the consistency of the Gauss law (which is what we are talking about) you don't need magnetic forces so you don't need to require 3. And then 1. and 2. are consistent.

To remove confusion, let me remind you about the history of our discussion:
- In #19 you said that one cannot formulate consistent non-relativistic electrodynamics.
- In #20 I asked why not.
- In #21 you said because Gauss's Law is not covariant w.r.t. to Galilean transformations.
I think you made a mistake in #21. I agree now that one cannot formulate consistent non-relativistic electrodynamics, but I don't agree that Gauss's Law is not covariant w.r.t. to Galilean transformations. However, since we agree on the central statement that one cannot formulate consistent non-relativistic electrodynamics, I think we should not longer argue on the auxiliary statement that Gauss's Law is not covariant w.r.t. to Galilean transformations.

 

[TrickyDicky]

But, a spherically symmetric charge distribution means there is a well defined center. Where is the center of the Universe?

If we add the usual assumption of homogeneity I don't think you need to have a center.

 

[Dickfore]

This is simply wrong. Just as vector may have to do with arbitrary coordinate transformations (not only rotations) in 4 dimensions, it may have to do with arbitrary coordinate transformations in any number of dimensions, including 3.

It is a common misconception among physicists that general coordinate transformations are only relevant in general theory of relativity. But this is wrong. In 3 dimensions, they are relevant even in non-relativistic physics.

Acceleration remains invariant under Galilean transformations:
<br /> \mathbf{a}&#039; = \mathbf{a}<br />
but velocity transforms as:
<br /> \mathbf{v}&#039; = \mathbf{v} - \mathbf{V}<br />
where \mathbf{V} is the relative velocity of the reference frame K' w.r.t. K.

Since these two do not transform in the same way under Galilean transformation and both a 3-vectors, it means that your "rule" is wrong.

 

[Dickfore]

If we add the usual assumption of homogeneity I don't think you need to have a center.

But the center of the charge distribution is a privileged point, so homogeneity is broken.

 

[Dickfore]

In the conclusion they say:

"The following 3 assumptions are inconsistent:
1. Galilean invariance
2. Continuity equation.
3. Magnetic forces between electric currents."

Fine, but this means that if we do not require one of these items, then we still have a consistent theory. In particular, for the consistency of the Gauss law (which is what we are talking about) you don't need magnetic forces so you don't need to require 3. And then 1. and 2. are consistent.

To remove confusion, let me remind you about the history of our discussion:
- In #19 you said that one cannot formulate consistent non-relativistic electrodynamics.
- In #20 I asked why not.
- In #21 you said because Gauss's Law is not covariant w.r.t. to Galilean transformations.
I think you made a mistake in #21. I agree now that one cannot formulate consistent non-relativistic electrodynamics, but I don't agree that Gauss's Law is not covariant w.r.t. to Galilean transformations. However, since we agree on the central statement that one cannot formulate consistent non-relativistic electrodynamics, I think we should not longer argue on the auxiliary statement that Gauss's Law is not covariant w.r.t. to Galilean transformations.

I wanted you to show what assumptions you used to satisfy Gauss's Law. Then, there is at least one Maxwell equation or force law that is not Galilean invariant. This means that electrodynamics as we know it, and is verified by a large class of experiments, is inconsistent.

But, you didn't even manage to grasp transformation properties of fields under Galilean transformations, so I think my attempt remains futile.

 

[granpa]

I'd very much like to know which law isn't invariant.

 

[Dickfore]

I'd very much like to know which law isn't invariant.

Why don't you have a look at the paper cited in post #25 then?

 

[TrickyDicky]

But the center of the charge distribution is a privileged point, so homogeneity is broken.

This is fun: No, in an expanding universe is not broken, no privileged points.

 

[Demystifier]

Acceleration remains invariant under Galilean transformations:
<br /> \mathbf{a}&#039; = \mathbf{a}<br />
but velocity transforms as:
<br /> \mathbf{v}&#039; = \mathbf{v} - \mathbf{V}<br />
where \mathbf{V} is the relative velocity of the reference frame K' w.r.t. K.

Since these two do not transform in the same way under Galilean transformation and both a 3-vectors, it means that your "rule" is wrong.

That's a good objection. More precisely, that means the following. The 3-velocity transforms as a vector under arbitrary space transformations, but it does not transform as a vector under arbitrary space-time transformations. Galilei transformation is a space-time transformation (because the transformation x'=x-vt mixes space and time), so the 3-velocity does not transform as a 3-vector under Galilei transformation.

Now the question is: How the electric field transforms under Galilei transformation? If not as a 3-vector, then can you write down the transformation law explicitly? But do it for nonrelativistic physics, in which magnetic field does not exist (because it is an effect of the order v/c, which vanishes in the nonrelativistic limit).

Anyway, I claim that in nonrelativistic limit E'=E. Indeed, this is consistent with your result that a'=a, because acceleration of a charged particle is proportional to the external electric field.

 

[Dickfore]

But do it for nonrelativistic physics, in which magnetic field does not exist (because it is an effect of the order v/c, which vanishes in the nonrelativistic limit).

How about a bar magnet? Velocity is zero, so you are surely in a non-relativistic regime. But, there is a non-zero magnetic field around a bar magnet.

 

[Dickfore]

This is fun: No, in an expanding universe is not broken, no privileged points.

What are you talking about? You keep invoking some extra condition that have nothing to do with the problem at hand.

 

[TrickyDicky]

What are you talking about? You keep invoking some extra condition that have nothing to do with the problem at hand.

Well those extra conditions happen to be fulfilled by the current model of universe.
And in fact to come back to the OP the homogeneity condition is required for the point-like particles QFT assumption.

 

[Dickfore]

I'm out of this thread.

 

[TrickyDicky]

I'm out of this thread.

A not so polite way of conceding. But thanks for your contribution.

 

[Demystifier]

How about a bar magnet? Velocity is zero, so you are surely in a non-relativistic regime. But, there is a non-zero magnetic field around a bar magnet.

The velocity of magnet is zero, but it does not mean that the velocity of microscopic particles (electrons in atoms) which create this magnetic field is zero.

 

[PhilDSP]

I have no idea what you mean, but spin orbit coupling is not especially a quantum mechanical effect but could in principle also be observed in classical systems.
See e.g. here:
https://www.physicsforums.com/showthread.php?t=161632&page=2

Thanks for the link. Jackson has a right-to-the-point treatment of Thomas precession starting at page 548 in "Classical Electrodynamics" and identifies it as a correction to the 'classical' (i.e. naive) equation of motion for electron angular momentum. His characterization of the situation is

In 1926 Uhlenbeck and Goudsmit introduced the idea of electron spin and showed that, if the electron had a g factor of 2, the anomalous Zeeman effect could be explained, as well as the existence of multiplet splittings. There was a difficulty, however, in that the observed fine structure intervals were only half the theoretically expected values. If a g factor of unity were chosen, the fine structure intervals were given correctly, but the Zeeman effect was then the normal one. The complete explanation of spin, including correctly the g factor and the proper fine structure interaction, came only with the relativistic electron theory of Dirac.

I believe Jackson is referring to Dirac's 1928 paper and his electron equation or the various different forms of it. The point being that the need for the Thomas precession correction disappears in the Dirac treatment.

 

[PhilDSP]

As far as I know, Dirac equation is equal to "special relativity".
Substituting relativistic x and t into usual accelaration equaion (of Newtorian mechanics).
Ant if we use the force F of v=0, we can get your relativistic momentum and energy.
(And the solution of Dirac equation uses four vector momemtum, energy, time, position variables, which are based on SR.)

In addition to the fact that the Dirac equation is highly billed as being relativistic, Dirac shows in his 1928 paper that it is invariant under the Lorentz transformation. But it may be worth investigating exactly where and why SR is invoked and where it is not required.

The gamma matrices contain fixed numeric values don't they? They don't vary with regard to a Lorentz transformation. A very large part of the motivation of the development of the Dirac equation revolves around the split of the momentum into components for each spatial dimension. Symmetry is invoked for the manipulation of the spatial components and consideration of spin, not from a relativistic viewpoint but from a geometric viewpoint.

 

[ytuab]

In addition to the fact that the Dirac equation is highly billed as being relativistic, Dirac shows in his 1928 paper that it is invariant under the Lorentz transformation. But it may be worth investigating exactly where and why SR is invoked and where it is not required.

The gamma matrices contain fixed numeric values don't they? They don't vary with regard to a Lorentz transformation. A very large part of the motivation of the development of the Dirac equation revolves around the split of the momentum into components for each spatial dimension. Symmetry is invoked for the manipulation of the spatial components and consideration of spin, not from a relativistic viewpoint but from a geometric viewpoint.

As far as I know, Dirac respected Einstein's special relativity.
Dirac knew about Maxwell's electromagnetic theory after knowing the special relativity according to this book.
(The Strangest Man by Graham Farmelo. )

The gamma matrix and Pauli patrix are fixed values. But their eigenfunctions (= spinor) vary with regard to Lorentz transformation.
For example, when the spin points to z direction at first, its direction may change under Lorentz transformation.
Sigma matirix (gamma matrix) is fixed, so their spinors (= eigenfunctions) need to change under Lorentz trandformation.

As you know, Klein-Gordon equation (= Lorentz invariant) is completely equal to the special relativity.
And Dirac equation also satisfies Klein-Gordon equation.
According to the QFT, the "forms" of Dirac equation need to be the same under Lorentz trandformation, because it is based on the Lorentz invariant equation of the special relativity.

Under the Lorentz transformation,

(-i\hbar\gamma^{\mu} \partial_{\mu} + mc) \psi (x) \quad \to \quad (-i\hbar\gamma^{\mu} \partial_{\mu}&#039; + mc) \psi&#039; (x&#039;)

To satisfy this relation, the spinor (=psi) needs to satisfy the next relation of

\psi&#039; (x&#039;) = S \psi (x) \quad S = \exp (-\frac{i}{2} \omega_{\mu\nu} S^{\mu\nu})

(Of course, "kx" = momentum or energy x space or time of psi also need to satisfy SR.)

By the way, if you don't use the special relativity, how can you get the next relation (= origin of K-D and Dirac Eq.) ??

\vec{p}^2 - \frac{E^2}{c^2} + m^2 c^2 = 0

So Dirac equation is dependent on the special relativity.
(The spin relation can be gotten also from Schrodinger equation, I think. The difference is that Dirac's is 4 x 4 matrices. )

 

[PhilDSP]

By the way, if you don't use the special relativity, how can you get the next relation (= origin of K-D and Dirac Eq.) ??

\vec{p}^2 - \frac{E^2}{c^2} + m^2 c^2 = 0

Yes, that's exactly what we were discussing a little earlier in this thread. (We're working on that)

\vec{p}^2 - \frac{E^2}{c^2} + m^2 c^2 = 0 \ \ \ -&gt; \ \ \ \vec{p}^2c^2 + m^2 c^4 = E^2

 

[juanrga]

As we know nonrelativistic quantum mechanics doesn't have the Lorentz invariance property and yet it makes a number of powerful predictions and gives rise to all the fundamental quantum properties (HUP, tunnelling effec, harmonic oscillator, superposition, wave-particle duality etc).
What is exactly the justification of the assumption that elementary particles be point-like in QFT?

I fail to understand your question. First, in non-relativistic quantum mechanics elementary particles are point-like as well. An electron in the Schrödinger equation does not have size. Therein the position eigenfunctions in non-relativistic quantum mechanics are Dirac deltas. The consideration of point-like particles is based in experimental data.

Second, QFT theory would continue to work for today typical applications even if in the next decade elementary particles are found to be non point-like.

 

[TrickyDicky]

I fail to understand your question. First, in non-relativistic quantum mechanics elementary particles are point-like as well. An electron in the Schrödinger equation does not have size. Therein the position eigenfunctions in non-relativistic quantum mechanics are Dirac deltas.

You are right in the usual interpretation of QM but dirac deltas are simplified models, theoretical idealizations of point charges that can be mathematically represented by the dirac delta, and this is not directly related to Schrodinger's equation.

The consideration of point-like particles is based in experimental data.

Experimental data is compatible with non point-like particles too.

Second, QFT theory would continue to work for today typical applications even if in the next decade elementary particles are found to be non point-like.

Well, it depends on how you define typical applications. The internal consistency of the Lorentz symmetry of QFT theory demands point-like interacting particles if a field is represented as a canonical system with an infinite number of degrees of freedom and particles are quantized excitation of propagating fields.

 

[juanrga]

You are right in the usual interpretation of QM but dirac deltas are simplified models, theoretical idealizations of point charges that can be mathematically represented by the dirac delta, and this is not directly related to Schrodinger's equation.

Any physical model is simplified. Moreover, extended models of electrons have been studied since Poincaré times and abandoned due to experimental difficulties and/or internal inconsistencies.

Dirac deltas are directly related to the Schrödinger equation if you expand the ψ in the position basis.

Experimental data is compatible with non point-like particles too.

Yes in the sense that current data constraints our physical exploration of electrons up to 10^-22 meters. No in the sense that no experiment up to date suggests that electrons have internal structure and finite shape. As a consequence our better models of the electron define it as a structureless point-like particle. And the same for the other elementary particles.

Well, it depends on how you define typical applications. The internal consistency of the Lorentz symmetry of QFT theory demands point-like interacting particles if a field is represented as a canonical system with an infinite number of degrees of freedom and particles are quantized excitation of propagating fields.

For example current application in particle accelerators. If tomorrow a theory of non-point-like particles was needed, this theory would reduce to QFT in some well-defined limit compatible with the scales accessible to current experiments.

 

[TrickyDicky]

For example current application in particle accelerators. If tomorrow a theory of non-point-like particles was needed, this theory would reduce to QFT in some well-defined limit compatible with the scales accessible to current experiments.

Sure, but it would be a non-Lorentz invariant theory, that would be Lorentz invariant(or as you say it would reduce to QFT) only at the infinitesimal limit that for particles is perfectly compatible with the accuracy limits of current experiments(except the Opera experiment in case it is confirmed).

 

[TrickyDicky]

Any physical model is simplified. Moreover, extended models of electrons have been studied since Poincaré times and abandoned due to experimental difficulties and/or internal inconsistencies.

Dirac deltas are directly related to the Schrödinger equation if you expand the ψ in the position basis.

But this simplification is well known, and the position eigenket is known to be just a fair approximation not strictly correct: from the HUP we know the lower limit to how well localized a particle can be is not zero as it would be the case if it was a point-like particle, this limit is actually about the particle's compton wavelengh's order.

 

[juanrga]

Sure, but it would be a non-Lorentz invariant theory, that would be Lorentz invariant(or as you say it would reduce to QFT) only at the infinitesimal limit that for particles is perfectly compatible with the accuracy limits of current experiments(except the Opera experiment in case it is confirmed).

There exist Lorentz invariant models of non-point-like objects. E.g., String theory models [*].

[*] I am not endorsing string theory by any bit of imagination.

 

[TrickyDicky]

There exist Lorentz invariant models of non-point-like objects. E.g., String theory models [*].

[*] I am not endorsing string theory by any bit of imagination.

Right, but they need a bunch of non-observed dimensions.

 

[juanrga]

But this simplification is well known, and the position eigenket is known to be just a fair approximation not strictly correct: from the HUP we know the lower limit to how well localized a particle can be is not zero as it would be the case if it was a point-like particle, this limit is actually about the particle's compton wavelengh's order.

HUP works in QM, where position eigenstates are well-defined.

The problem that you are referring to is specific of the Dirac (and Klein-Gordon) early (and inconsistent) formulations of RQM. A problem which was somehow inherited by QFT (where position is downgraded to unobservable parameter).

The origin of the problem is other. For the Dirac theory the problem is associated to the fact that in his inconsistent formulation the position operator is not hermitian [*], with the non-Hermitian part giving a non-localisation term of the order of the compton wavelengh's order.

This problem is not present in modern approaches as the Stuckelberg-Horwitz-Piron theory where relativistic position eigenfunctions in a generalized 4N-dimensional Hilbert structure are well-defined.

[*] A consequence of Dirac incorrect linearisation procedure.