Point-like particles, Lorentz invariance and QM/QFT

[Demystifier]

please give a reference for this conclusion.

I don't have one, because I have derived it by myself. Please find the error (if any) in my derivation.

 

[DrDu]

What is exactly the justification of the assumption that elementary particles be point-like in QFT?

In interacting quantum field theories, point like particles don't exist. Also the classification of a particle as "elementary" is not fundamental but only refers to the way how it winds up in perturbation theory from some non-interacting elementary particles whose choice is more due to convenience than tp principle.

 

[Dickfore]

I don't have one, because I have derived it by myself. Please find the error (if any) in my derivation.

Please give an exposition of your derivation.

 

[Demystifier]

Please give an exposition of your derivation.

Please see the attachment.

 

[TrickyDicky]

But it says that the non-relativistic treatment gives a factor that is twice as big for the spin-orbit coupling term.

https://www.physicsforums.com/showpost.php?p=3540286&postcount=34

 

[Dickfore]

Please see the attachment.

No one asked you how the E-field transforms under rotations, but under Galilean transformations. So, Eq.
(3) is not valid.

 

[Dickfore]

https://www.physicsforums.com/showpost.php?p=3540286&postcount=34

I haven't gone through the argument in your quote, but I noticed this:

As yet, no other authors on the arXiv cite Jabs' paper.

Also, I was asking a different question. Does the non-relativistic treatment give the correct factor for the spin-orbit interaction?

 

[TrickyDicky]

Also, I was asking a different question. Does the non-relativistic treatment give the correct factor for the spin-orbit interaction?

Yes.

 

[DrDu]

Yes.

Are you sure? I remember the derivation of the SO interaction to involve calculating the magnetic field in the momentary rest frame of the spin. As the spin is accelerated, there is an additional Thomas precession term which is of purely relativistic origin.

 

[TrickyDicky]

Are you sure? I remember the derivation of the SO interaction to involve calculating the magnetic field in the momentary rest frame of the spin. As the spin is accelerated, there is an additional Thomas precession term which is of purely relativistic origin.

Maybe I should have added that I was just basing my opinion on this heuristic from the Wikipedia entry on spin-orbit coupling: "Using some semiclassical electrodynamics and non-relativistic quantum mechanics, in this section we present a relatively simple and quantitative description of the spin-orbit interaction for an electron bound to an atom, up to first order in perturbation theory. This gives results that agree reasonably well with observations."
I'm sure more exact results would demand the full QFT/QED treatment (that involves not only relativistic considerations but regularization/renormalization)

 

[DrDu]

I found an excellent source for the relativistic contribution to SO-coupling on my shelf, although it is in German:
W. R. Theiss, Grundzuege der Quantentheorie, Teubner, Stuttgart, 1985

The magnetic field seen by the spin is due to the Lorentz transformed Coulomb field of the nucleus. The Thomas precession makes up ( at least in lowest order of v/c) for a contribution of different sign which compensates half of the effect size. So it is not a small higher order correction.

Addendum: The Thomas precession would even lead to SO splitting if the spin where bound to the nucleus not by a Coulombic force but by some other force which does not interact with the magnetic moment of the spin, e.g. gravitationally.

 

[Demystifier]

No one asked you how the E-field transforms under rotations, but under Galilean transformations. So, Eq.
(3) is not valid.

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.

 

[PhilDSP]

I found an excellent source for the relativistic contribution to SO-coupling on my shelf, although it is in German:
W. R. Theiss, Grundzuege der Quantentheorie, Teubner, Stuttgart, 1985

The magnetic field seen by the spin is due to the Lorentz transformed Coulomb field of the nucleus. The Thomas precession makes up ( at least in lowest order of v/c) for a contribution of different sign which compensates half of the effect size. So it is not a small higher order correction.

Addendum: The Thomas precession would even lead to SO splitting if the spin where bound to the nucleus not by a Coulombic force but by some other force which does not interact with the magnetic moment of the spin, e.g. gravitationally.

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?

 

[TrickyDicky]

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?

Can you show that derivation?

 

[DrDu]

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?

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

 

[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:
$$\mathbf{a}' = \mathbf{a}$$
but velocity transforms as:
$$\mathbf{v}' = \mathbf{v} - \mathbf{V}$$
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:
$$\mathbf{a}' = \mathbf{a}$$
but velocity transforms as:
$$\mathbf{v}' = \mathbf{v} - \mathbf{V}$$
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.