Quantum field theory, spacetime, and coordinates

[Demystifier]

[Moderator's note: This thread is spun off from another thread since it was dealing with a more technical point that is out of scope for the previous thread. The quote that starts this post is from the previous thread.]

In GR it's not conceivable to me, how to make sense of a relativistic mass at all. You can make some sense of this idea in SR, but why should one?

I feel the same about transformations of Dirac matrices and Dirac field under space-time coordinate transformations. Most QFT textbooks use pedagogy according to which field transforms and matrices don't, while in GR only the reverse pedagogy makes sense. So I propose to use the reverse pedagogy in SR too: https://arxiv.org/abs/1309.7070

 

[vanhees71]

I feel the same about transformations of Dirac matrices and Dirac field under space-time coordinate transformations. Most QFT textbooks use pedagogy according to which field transforms and matrices don't, while in GR only the reverse pedagogy makes sense. So I propose to use the reverse pedagogy in SR too: https://arxiv.org/abs/1309.7070

Interesting idea. In GR, in the usual picture one needs to use tetrades to make sense of spinors. As you formally treat the Dirac field as a scalar, is there a way not to use tetrades and does the theory and lead to something else sense the Einstein-Cartan spacetime of spin-1/2 fields? On the other hand your vector formalism seems to be equivalent to the usual spinor approach. So one also should get the same GR generalization of Dirac fields in your realization as in the usual one.

 

[vanhees71]

Well, it helps to motivate why energy is the source of gravity in GR.

This follows from the Einstein-Hilbert action without much ado. See Landau-Lifshitz vol. II.

 

[vanhees71]

In post 82, I said that it help to motivate why energy is the source of gravity in GR.

Here are examples:

Blau, Lecture notes on gravity http://www.blau.itp.unibe.ch/newlecturesGR.pdf, gives heuristic motivation for relativistic gravity (p20) with statements like: "We already know (from Special Relativity) that ρ is not a scalar but rather the 00-component of a tensor, the energy-momentum tensor".

Schutz, Gravity from the Ground up http://www.gravityfromthegroundup.org/ also makes use of the notion of relativistic mass.
p190 "As an object moves faster, its of an object increases with its speed. We noted above that no force, inertial mass increases, so it is harder to accelerate it. This enforces the speed of light as a limiting speed: as the object gets closer to the speed of light, its mass increases without bound"

On why rest mass is not the correct generalization for the source of gravity in GR:
p242 "What would happen to a gravitational field created by rest-mass when rest-mass is turned into energy by nuclear reactions? Would gravity disappear? This seems unreasonable. Rest mass is a dead end."

p242 "the active gravitational mass generates the curvature of time, which is the most important part of the geometry of gravity. Its density is defined as the density of ordinary mass-energy, plus three times the average pressure divided by c2."

Of course, I prefer Blau's heuristics much over Schutz's since Blau's is much more intuitive for me, after I got rid of all the old-fashioned ballast treating relativity not in a manifestly covariant way. Of course, if it comes to applications (e.g., in classical electrodynamics), switching to the (1+3) formalism may be necessary, but it's just not manifestly covariant, but derivable from the manifestly covariant formalism.

However, as I said, I have no idea, how to make any sense of non-covariant quantities in GR, where covariance means even general GL(4) covariance and also the "gauge freedom" of GR, having a massless spin-2 field rather than a massless spin-1 field as in E&M, is somewhat more involved, it's hard to conceive how to interpret such non-covariant quantities physically at all. So using non-covariant quantities is an unnecessary burden in SR but not even interpretable in GR, or how does Schutz define a physically interpretable "inertial mass"? I think it's better to teach already in SR that the inertia is not related to mass (i.e., invariant mass) but to energy, and that energy is the time component of a four-vector should be standard in teaching relativity in the 21st century anyway!

 

[Demystifier]

Interesting idea. In GR, in the usual picture one needs to use tetrades to make sense of spinors. As you formally treat the Dirac field as a scalar, is there a way not to use tetrades and does the theory and lead to something else sense the Einstein-Cartan spacetime of spin-1/2 fields? On the other hand your vector formalism seems to be equivalent to the usual spinor approach. So one also should get the same GR generalization of Dirac fields in your realization as in the usual one.

My vector formalism is equivalent to the usual spinor approach in Minkowski spacetime. In Minkowski spacetime it is not necessary to use tetrads. In more general spacetimes one must use tetrads and replace ordinary derivative with the appropriate covariant derivative, which is where the additional physical effects come from.

 

[vanhees71]

Indeed, it's clear that your formalism is equivalent to the usual one in flat Minkowski space. My question was, whether you get something different when extending it to GR. In the standard formalism you necessarily have to extend the Lorentzian manifold of GR to an Einstein-Cartan manifold with torsion. Now that your Dirac fields are scalars, is this not the case anymore? On the other hand you still need the Dirac matrices, and this might also work only with the tetrad formalism and in this way also makes the extension from Lorentzian to Einstein-Cartan manifolds necessary.

 

[rubi]

My vector formalism is equivalent to the usual spinor approach in Minkowski spacetime. In Minkowski spacetime it is not necessary to use tetrads. In more general spacetimes one must use tetrads and replace ordinary derivative with the appropriate covariant derivative, which is where the additional physical effects come from.

It fails not only in GR. Also in Minkowski spacetime, you get non-trivial spin connection coefficients if you switch to curvilinear coordinate systems. Treating the Dirac spinor as a scalar will always get you in trouble.

 

[vanhees71]

Well, in the Eur. J. paper Demystifier simply moves the Dirac-representation matrices over to the $\gamma$ matrices and declares $\Psi=S^{-1} \psi$ a scalar field, leading effectively to the description to transform the $\gamma^{\mu}$ as matrix-valued components of a four-vector. On the level of a classical field theory in Minkowski space the two ways to realize Lorentz transformations should thus be equivalent.

If it comes to QFT, I'm a bit sceptical too since there you want a Hilbert space on which (at least proper orthocrhonous) Poincare transformations are realized via a unitary transformation (the unitary ray transformations are all liftable to a proper unitary transformations in the case of the Poincare group), and since the corresponding generators are usually built with field operators, which include creation and annihilation operators for Fock states, one should use the usual formalism with $\psi$ being a bi-spinor.

It's also clear that Minkowski spacetime stays a flat affine space, no matter in which coordinates you calculate. You do not get a non-flat spacetime nor a space with torsion. Of course, the Christoffel symbols in general curvilinear coordinates occur in the covariant derivatives but they are not indicating any curvature or torsion of spacetime, which still stays the good old flat Minkowski spacetime.

 

[rubi]

Well, in the Eur. J. paper Demystifier simply moves the Dirac-representation matrices over to the $\gamma$ matrices and declares $\Psi=S^{-1} \psi$ a scalar field, leading effectively to the description to transform the $\gamma^{\mu}$ as matrix-valued components of a four-vector. On the level of a classical field theory in Minkowski space the two ways to realize Lorentz transformations should thus be equivalent.

They are only equivalent if you restrict to cartesian coordinates, but the only way to have a formulation that is invariant under all coordinate transformations (even curvilinear ones) is to treat $\Psi$ as a (bi-)spinor.

It's also clear that Minkowski spacetime stays a flat affine space, no matter in which coordinates you calculate. You do not get a non-flat spacetime nor a space with torsion. Of course, the Christoffel symbols in general curvilinear coordinates occur in the covariant derivatives but they are not indicating any curvature or torsion of spacetime, which still stays the good old flat Minkowski spacetime.

Yes, of course there is no curvature or torsion in curvilinear coordinates on Minkowski space, because these are coordinate invariant concepts. The Dirac field is also a coordinate invariant object and so is the Dirac equation. But Demystifiers formulation will spoil this fact in curvilinear coordinate systems, because it doesn't account for the transformation of the spin connection coefficients. You will only get these coefficients if you treat $\Psi$ as a (bi-)spinor. (It's easy to see: The covariant derivative of a scalar has no connection coefficients in any coordinate system. Demystifyer just got lucky, because the spin connection coefficients vanish in cartesian coordinates.)

 

[vanhees71]

Ok, then Demystifyer's didactics is misleading, because it's not covariant!

 

[Demystifier]

It fails not only in GR. Also in Minkowski spacetime, you get non-trivial spin connection coefficients if you switch to curvilinear coordinate systems. Treating the Dirac spinor as a scalar will always get you in trouble.

Maybe it's just a matter of terminology, but Minkowski spacetime, by definition, is flat spacetime in non-curvilinear coordinates. Rindler spacetime, for instance, is not Minkowski.

 

[Demystifier]

one should use the usual formalism with $\psi$ being a bi-spinor.

I have no idea what do you mean by saying that $\psi$ is a bi-spinor.

 

[Demystifier]

Ok, then Demystifyer's didactics is misleading, because it's not covariant!

It's covariant with respect to Lorentz transformations in MInkowski spacetime, and it can be generalized to QFT without substantial changes. (If you disagree, please present your argument in terms of equations.)

 

[rubi]

Maybe it's just a matter of terminology, but Minkowski spacetime, by definition, is flat spacetime in non-curvilinear coordinates. Rindler spacetime, for instance, is not Minkowski.

No, that's not right. A spacetime is always a coordinate-independent object. It's a manifold with a (pseudo-Riemannian) metric. Minkowski spacetime is $\mathbb R^4$ with the Minkowski metric $\eta$. And that's not my definition, it's standard GR as taught in every standard textbook such as Wald, MTW, Schutz, Carroll and so on. Rindler spacetime is not Minkowski, because it is just a wedge of Minkiwski, not because of the coordinates. Minkowski spacetime is an extension of Rindler spacetime.

 

[Demystifier]

Yes, of course there is no curvature or torsion in curvilinear coordinates on Minkowski space, because these are coordinate invariant concepts. The Dirac field is also a coordinate invariant object and so is the Dirac equation. But Demystifiers formulation will spoil this fact in curvilinear coordinate systems, because it doesn't account for the transformation of the spin connection coefficients. You will only get these coefficients if you treat $\Psi$ as a (bi-)spinor. (It's easy to see: The covariant derivative of a scalar has no connection coefficients in any coordinate system. Demystifyer just got lucky, because the spin connection coefficients vanish in cartesian coordinates.)

I agree with all this (except that I don't understand why would $\Psi$ be a bi-spinor). But my intention in the paper was not to describe physics in curvilinear coordinates. Just the usual flat Lorentz coordinates in Minkowski spacetime.

 

[Demystifier]

No, that's not right. A spacetime is always a coordinate-independent object. It's a manifold with a (pseudo-Riemannian) metric. Minkowski spacetime is $\mathbb R^4$ with the Minkowski metric $\eta$. And that's not my definition, it's standard GR as taught in every standard textbook such as Wald, MTW, Schutz, Carroll and so on. Rindler spacetime is not Minkowski, because it is just a wedge of Minkiwski, not because of the coordinates. Minkowski spacetime is an extension of Rindler spacetime.

I appreciate such a definition, but, for instance, Birrell and Davies in "Quantum Fields in Curved Space" use a different definition. So I think we can agree that definitions are not uniform in the literature.

 

[rubi]

I agree with all this (except that I don't understand why would $\Psi$ be a bi-spinor).

It's a bispinor, because it's the direct sum of two spinors. Mathematically speaking, it's an object in the $(\frac 1 2,0)\oplus(0,\frac 1 2)$ representation of the Lorentz group, so you have one left- and one right-handed spinor.

But my intention in the paper was not to describe physics in curvilinear coordinates. Just the usual flat Lorentz coordinates in Minkowski spacetime.

If a poor student takes your formulation and tries to calculate the energy levels of the relativistic hydrogen atom by going to spherical coordinates, they will obtain invalid results, because they will apply the wrong transformation laws. There's a good reason for using coordinate-invariant formulations in physics, so I don't see the point of your formulation. Especially for didactics.

I appreciate such a definition, but, for instance, Birrell and Davies in "Quantum Fields in Curved Space" use a different definition. So I think we can agree that definitions are not uniform in the literature.

Birrell and Davies write (section 3.1): "We assume spacetime to be a $C^\infty$ $n$-dimensional, globally hyperbolic pseudo-Riemannian manifold." They don't mention a preferred coordinate chart either. I don't think there exist relativists who think that a spacetime includes a preferred coordinate chart.

 

[Demystifier]

It's a bispinor, because it's the direct sum of two spinors. Mathematically speaking, it's an object in the $(\frac 1 2,0)\oplus(0,\frac 1 2)$ representation of the Lorentz group, so you have one left- and one right-handed spinor.

OK, it's clear now.

If a poor student takes your formulation and tries to calculate the energy levels of the relativistic hydrogen atom by going to spherical coordinates, they will obtain invalid results, because they will apply the wrong transformation laws.

Nonsense! The relativistic hydrogen atom is solved e.g. in Bjorken Drell 1, in spherical coordinates without an explicit study of Dirac equation in arbitrary coordinates. In my approach, the hydrogen atom is solved in the same way as in Bjorken Drell.

Birrell and Davies write (section 3.1): "We assume spacetime to be a $C^\infty$ $n$-dimensional, globally hyperbolic pseudo-Riemannian manifold." They don't mention a preferred coordinate chart either. I don't think there exist relativists who think that a spacetime includes a preferred coordinate chart.

See Birrell Davies Eq. (2.1) and note that the Box operator is defined with ordinary (not covariant) derivatives. But I really don't want to argue over various definitions and terminologies.

 

[rubi]

Nonsense! The relativistic hydrogen atom is solved e.g. in Bjorken Drell 1, in spherical coordinates without an explicit study of Dirac equation in arbitrary coordinates. In my approach, the hydrogen atom is solved in the same way as in Bjorken Drell.

No, it's far from being nonsense. You don't need to study the Dirac equation in arbitrary coordinates for that. You just need to be able to apply the correct coordinate transformation law in order to switch to spherical coordinates, which Bjorken&Drell do. If you consider $\Psi$ to be a scalar, you will miss the connection coefficients, because the transformation of a scalar just does not produce connection coefficients. (And one doesn't need to know what connection coefficients are in order to obtain them. The terms will arise naturally from the coordinate transformation.) If you think, you can do it in your formulation, then please show your work.

See Birrell Davies Eq. (2.1) and note that the Box operator is defined with ordinary (not covariant) derivatives. But I really don't want to discuss various definitions and terminologies.

Well, they have written down the KG equation in cartesian coordinates. Nothing wrong with that, but they don't claim that "Minkowski" refers to cartesian coordinates. On the contrary, the quote I have given shows that they use the standard terminology, i.e. a spacetime is a coordinate-independent object.

 

[Demystifier]

No, it's far from being nonsense. You don't need to study the Dirac equation in arbitrary coordinates for that. You just need to be able to apply the correct coordinate transformation law in order to switch to spherical coordinates, which Bjorken&Drell do. If you consider $\Psi$ to be a scalar, you will miss the connection coefficients, because the transformation of a scalar just does not produce connection coefficients. (And one doesn't need to know what connection coefficients are in order to obtain them. The terms will arise naturally from the coordinate transformation.) If you think, you can do it in your formulation, then please show your work.

Where exactly do Bjorken and Drell study connection coefficients resulting from the fact that their $\psi$ is not a scalar? As far as I can see, in the solution of the hydrogen atom they only take care of coordinate transformations of the derivatives, not of the spinors.

 

[Demystifier]

Well, they have written down the KG equation in cartesian coordinates. Nothing wrong with that,

If there is nothing wrong with writing an equation in Cartesian coordinates, then why is it so wrong when I do that for the Dirac equation?

 

[Demystifier]

Let me also quote from S. Weinberg, "Gravitation and Cosmology", page 367 (my bolding):
"A coordinate scalar or coordinate tensor transforms as a scalar or a
tensor under changes in the coordinate system. A Lorentz scalar or Lorentz tensor
or Lorentz spinor transforms according to a rule like (12.5.12), with D(A) the
identity or a tensor representation or a spin or representation of the infinitesimal
Lorentz group, under changes in the choice of the locally inertial coordinate frame.
For instance, a field such as (12.5.4) is a coordinate scalar and a Lorentz vector,
the Dirac field of the electron is a coordinate scalar and a Lorentz spinor."

In other words, according to Weinberg, if I change global coordinates (and not change locally inertial coordinate frame), then the Dirac field transforms as a scalar.

 

[rubi]

Where exactly do Bjorken and Drell study connection coefficients resulting from the fact that their $\psi$ is not a scalar? As far as I can see, in the solution of the hydrogen atom they only take care of coordinate transformations of the derivatives, not of the spinors.

The connection coefficients arise in the calculation that leads to (4.13).

If there is nothing wrong with writing an equation in Cartesian coordinates, then why is it so wrong when I do that for the Dirac equation?

Because there's a difference between writing down a coordinate-independent equation in a coordinate chart and writing down a coordinate-dependent equation. A transformation of the KG equation will produce the correct equation in a different coordinate chart, while a transformation of your equation will not be satisfied by a transformed solution of the Dirac equation, because your equation transforms according to different transformation law than the solution of the Dirac equation, i.e. the two transformations won't cancel. Try it.

Let me also quote from S. Weinberg, "Gravitation and Cosmology", page 367 (my bolding):
A coordinate scalar or coordinate tensor transforms as a scalar or a
tensor under changes in the coordinate system. A Lorentz scalar or Lorentz tensor
or Lorentz spinor transforms according to a rule like (12.5.12), with D(A) the
identity or a tensor representation or a spin or representation of the infinitesimal
Lorentz group, under changes in the choice of the locally inertial coordinate frame.
For instance, a field such as (12.5.4) is a coordinate scalar and a Lorentz vector,
the Dirac field of the electron is a coordinate scalar and a Lorentz spinor.

This is fine, but in a coordinate transformation, you also transform the connection coefficients, not only the scalars, tensors and spinors. And since your equation contains a derivative acting on a scalar, you don't get any transformed quantities from that, while the Dirac equation has a derivative acting on a spinor, which acquires tranformed connection coefficients under a coordinate transformation.

 

[Demystifier]

The connection coefficients arise in the calculation that leads to (4.13).

As far as I can see, the implicit connection coefficients in (4.13) arise only from transforming derivatives to spherical coordinates, not from transforming spinors to spherical coordinates. This in fact must be so, because the transformation of spinors is not even defined for general coordinate transformations.

 

[rubi]

As far as I can see, the implicit connection coefficients in (4.13) arise only from transforming derivatives to spherical coordinates, not from transforming spinors to spherical coordinates. This in fact must be so, because the transformation of spinors is not even defined for general coordinate transformations.

Connection coefficients always arise from transforming (covariant) derivatives. They are part of the covariant derivative, so to speak. And the point is that the terms that come from the transformation of the connection coefficients cancel exactly the (representation of) the Jacobian matrix that you have to put in front of the tensor/spinor. A transformation of spinors is well-defined for isometries, i.e. transformations with $SO(3,1)$ Jacobians $J$. For a Dirac spinor, you take a corresponding $SL(2,\mathbb C)$ matrix $\hat J$ and the $(\frac 1 2,0)\oplus (0,\frac 1 2)$ representation of the Lorentz group (let's call it $S$) and compute $S(\hat J(x))$.

 

[rubi]

Let me explain it with formulas:
The standard Dirac equation is $(\mathrm i \gamma^\mu\partial_\mu - m)\Psi=0$. If we use a linear coordinate transformation $x\rightarrow\Lambda x + a$, then we have a Jacobian $J=\Lambda$ and the transformations $\Psi\rightarrow S(\hat\Lambda)\Psi$ and $\partial_\mu\rightarrow\Lambda^\nu_{\phantom{\nu}\mu}\partial_\nu$, so we get:
$(\mathrm i\gamma^\mu\partial_\mu^\prime-m)\Psi^\prime=(\mathrm i\gamma^\mu\Lambda^\nu_{\phantom{\nu}\mu}\partial_\nu-m)S(\hat\Lambda)\Psi=\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1}\partial_\nu S(\hat\Lambda)\Psi-mS(\hat\Lambda)\Psi$
And now comes the crucial point: We need to pull $S(\hat\Lambda)\Psi$ through the derivative $\partial_\nu$:
$\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1}\partial_\nu S(\hat\Lambda)\Psi-mS(\hat\Lambda)\Psi=\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1} S(\hat\Lambda)\partial_\nu\Psi-mS(\hat\Lambda)\Psi = S(\hat\Lambda)(\mathrm i\gamma^\nu\partial_\nu-m)\Psi=0$
So we see that the transformed $\Psi$ satisfies the Dirac equation again. However, if the coordinate transformation isn't linear anymore, we get an $x$-dependent Jacobian $J(x)$ and then obviously, we can't pull $S(\hat J(x))$ through the derivative $\partial_\nu$ anymore. In order to fix this, we need to introduce a covariant derivative acting on spinors, with corresponding spin connection coefficients, such that the equation $\nabla_\nu S(\hat J(x))\Psi(x) = S(\hat J(x))\nabla_\nu\Psi(x)$ holds. Hence, we see that we only get the correctly transformed Dirac equation, if we use the correct transformation law for the spin connection coefficients, which also means that $\Psi$ can't be a scalar, because in that case, the derivatives don't acquire connection coefficients.

Edit: Fixed a typo.

 

[PeterDonis]

Minkowski spacetime, by definition, is flat spacetime in non-curvilinear coordinates.

That depends on the definition. Some definitions define "Minkowski spacetime" in terms of being flat, not in terms of choosing any particular coordinates.

 

[Demystifier]

Let me explain it with formulas:
The standard Dirac equation is $(\mathrm i \gamma^\mu\partial_\mu - m)\Psi=0$. If we use a linear coordinate transformation $x\rightarrow\Lambda x + a$, then we have a Jacobian $J=\Lambda$ and the transformations $\Psi\rightarrow S(\hat\Lambda)\Psi$ and $\partial_\mu\rightarrow\Lambda^\nu_{\phantom{\nu}\mu}\partial_\nu$, so we get:
$(\mathrm i\gamma^\mu\partial_\mu^\prime-m)\Psi^\prime=(\mathrm i\gamma^\mu\Lambda^\nu_{\phantom{\nu}\mu}\partial_\nu-m)S(\hat\Lambda)\Psi=\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1}\partial_\nu S(\hat\Lambda)\Psi-mS(\hat\Lambda)\Psi$
And now comes the crucial point: We need to pull $S(\hat\Lambda)\Psi$ through the derivatice $\partial_\nu$:
$\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1}\partial_\nu S(\hat\Lambda)\Psi-mS(\hat\Lambda)\Psi=\mathrm iS(\hat\Lambda)\gamma^\nu S(\hat\Lambda)^{-1} S(\hat\Lambda)\partial_\nu\Psi-mS(\hat\Lambda)\Psi = S(\hat\Lambda)(\mathrm i\gamma^\nu\partial_\nu\Psi-m)\Psi=0$
So we see that the transformed $\Psi$ satisfies the Dirac equation again. However, if the coordinate transformation isn't linear anymore, we get an $x$-dependent Jacobian $J(x)$ and then obviously, we can't pull $S(\hat J(x))$ through the derivative $\partial_\nu$ anymore. In order to fix this, we need to introduce a covariant derivative acting on spinors, with corresponding spin connection coefficients, such that the equation $\nabla_\nu S(\hat J(x))\Psi(x) = S(\hat J(x))\nabla_\nu\Psi(x)$ holds. Hence, we see that we only get the correctly transformed Dirac equation, if we use the correct transformation law for the spin connection coefficients, which also means that $\Psi$ can't be a scalar, because in that case, the derivatives don't acquire connection coefficients.

That's all right and very deep, but I claim that this is not what Bjorken and Drell used when they solved Dirac equation in spherical coordinates. Instead, they used a dirty but smart trick that bypasses all this deep math. Let me explain it in detail.

Being deep:

In general curved coordinates the Dirac equation has the form
$$[i\gamma^{\mu}(D_{\mu}-ieA_{\mu})+m]\psi=0 \;\;\;\;\;\; (1)$$
where $A_{\mu}$ is the electromagnetic potential and $D_{\mu}$ is the covariant derivative that carries the information about curved coordinates and local tetrads. The equation above can be written as
$$[i\gamma^{\mu}D_{\mu}+e\gamma^{\mu}A_{\mu}+m]\psi=0 \;\;\;\;\;\; (2)$$
The quantity $D_{\mu}$ is quite complicated to compute in practice, which is the price for being deep. Fortunately, there is a trick that in some cases can be used to bypass the computation of $D_{\mu}$. That trick is used in Bjorken and Drell, which is what I discuss next.

Being smart:

Let us start from the laboratory Lorentz frame with coordinates $x^{\mu}$ in which Eq. (2) takes the form
$$[i\gamma^{\mu}_{\rm fix}\partial_{\mu}+e\gamma^{\mu}_{\rm fix}A_{\mu}+m]\psi_{\rm lab}=0 \;\;\;\;\;\; (3)$$
where $\gamma^{\mu}_{\rm fix}$ are the standard fixed Dirac matrices. This equation is not general covariant, but is valid in one smartly chosen system of coordinates. Now let us introduce some new curvilinear coordinates $x'^{\nu}$. Instead of properly transforming everything to new coordinates (which would be quite complicated in practice), the smart dirty trick is just to use the identity
$$\partial_{\mu}=\frac{\partial x'^{\nu}}{\partial x^{\mu}} \partial'_{\nu} \;\;\;\;\;\; (4)$$
which implies that (3) can be written as
$$[i\gamma^{\mu}_{\rm fix}\frac{\partial x'^{\nu}}{\partial x^{\mu}} \partial'_{\nu}+e\gamma^{\mu}_{\rm fix}A_{\mu}+m]\psi_{\rm lab}=0 \;\;\;\;\;\; (5)$$
In this way we have written the Dirac equation in terms of curvilinear coordinates without calculating the complicated quantity $D_{\mu}$. The price paid is that (5) is very very non-covariant. Yet it is correct in the given circumstances. And it is much simpler to deal with than would be the fully covariant equation. Basically, Bjorken and Drell solved Eq. (5) in the special case $x'^{\nu}=(t,r,\theta,\varphi)$, $A^{\mu}=(A^0(r),0,0,0)$.

Now let us see how my formalism works in this case. My formalism (just as the formalism used by Bjorken and Drell) is covariant only with respect to Lorentz transformations, not with respect to general coordinate transformations. Nevertheless, it can be used to solve Dirac equation in curvilinear coordinates by using the same smart dirty trick above. In my formalism, the Dirac equation in the laboratory Lorentz frame is
$$[i\Gamma^{\mu}_{\rm lab}\partial_{\mu}+e\Gamma^{\mu}_{\rm lab}A_{\mu}+m]\Psi=0 \;\;\;\;\;\; (6)$$
Having in mind that my formalism is related to the standard Bjorken and Drell formalism via
$$\Psi=\psi_{\rm lab}, \;\;\; \Gamma^{\mu}_{\rm lab}=\gamma^{\mu}_{\rm fix} \;\;\;\;\;\; (7)$$
we see that my Eq. (6) is equivalent to the standard Eq. (3). Therefore one can proceed with (4) and (5) as above.

The moral: Being deep is for mathematical physicists. Practical physicists must be smart.

 

[rubi]

That depends on the definition. Some definitions define "Minkowski spacetime" in terms of being flat, not in terms of choosing any particular coordinates.

"Some" is quite an understatement. I don't think there are any relativists who use a different definition.

That's all right and very deep, but I claim that this is not what Bjorken and Drell used when they solved Dirac equation in spherical coordinates. Instead, they used a dirty but smart trick that bypasses all this deep math. Let me explain it in detail.

I didn't claim that they performed the transformation by computing the connection coefficients directly. Of course, in this case, everyone would just apply the chain rule, but you will end up with a a transformed equation where you can read off the connection coefficients if you arrange it properly. However, in your formalism, if you perform the transformation by applying the chain rule, the resulting equation will not be consistent with your formalism anymore, because your formalism isn't covariant. You will get different results in your formalism if you transform your quantities using the transformation rules for scalars and covariant derivatives acting on scalars, which shows that your objects just cannot be scalars and covariant derivatives acting on scalars. A scalar is an object that transforms as a scalar and your $\Psi$ does not transform as a scalar. You can't fix invalid math using a trick.

Now let us introduce some new curvilinear coordinates $x'^{\nu}$. Instead of properly transforming everything to new coordinates (which would be quite complicated in practice), the smart dirty trick is just to use the identity
$$\partial_{\mu}=\frac{\partial x'^{\nu}}{\partial x^{\mu}} \partial'_{\nu} \;\;\;\;\;\; (4)$$

Your smart trick is just the chain rule and this is exactly the problem with your formalism. The transformation rules of scalars/vectors/tensors/spinors/connection coefficients is designed to always give the results that can be obtained by application of the chain rule. If an application of the chain rule gives you a different result than the application of the corresponding transformation rules, it proves that the objects in your formalism don't obey said transformation rules.

Now let us see how my formalism works in this case. My formalism (just as the formalism used by Bjorken and Drell) is covariant only with respect to Lorentz transformations, not with respect to general coordinate transformations. Nevertheless, it can be used to solve Dirac equation in curvilinear coordinates by using the same smart dirty trick above. In my formalism, the Dirac equation in the laboratory Lorentz frame is
$$[i\Gamma^{\mu}_{\rm lab}\partial_{\mu}+e\Gamma^{\mu}_{\rm lab}A_{\mu}+m]\Psi=0 \;\;\;\;\;\; (6)$$
Having in mind that my formalism is related to the standard Bjorken and Drell formalism via
$$\Psi=\psi_{\rm lab}, \;\;\; \Gamma^{\mu}_{\rm lab}=\gamma^{\mu}_{\rm fix} \;\;\;\;\;\; (7)$$
we see that my Eq. (6) is equivalent to the standard Eq. (3).

However, as a matter of fact, if you apply the transformatio rule of scalars to your objects, instead of applying the chain rule, you will get a different result, which proves that your objects are not scalars/covariant derivatives acting on scalars. In other words: If the application of the chain rule doesn't recover the transformation rules of scalars/..., then your objects don't satisfy these transformation rules of scalars/... and thus aren't scalars/...

 

[Demystifier]

because your formalism isn't covariant.

You keep saying that, without specifying with respect to which transformations. So let me repeat, what I already said a dozen of times.
1. The formalism in Bjorken Drell is covariant with respect to Lorentz transformations. So is mine.
2. The formalism in Bjorken Drell is not covariant with respect to general transformations. Neither is mine.
Can we agree on those statements?

 

[rubi]

You keep saying that, without specifying with respect to which transformations. So let me repeat, what I already said a dozen of times.

That's not usual terminology. A formalism being covariant always refers transformation rules under arbitrary coordinate changes.

2. The formalism in Bjorken Drell is not covariant with respect to general transformations. Neither is mine.
Can we agree on those statements?

No, we can't. The "formalism of Bjorken Drell" is covariant with respect to general transformations. Your's is not. In fact, Bjorken Drell don't use a different formalism, because for Bjorken Drell it doesn't matter whether they apply the chain rule or the transformation rules, since the objects in their equation (the ordinary Dirac equation) are designed to transform consistently with the chain rule. In your case, you will get different results depending on whether you apply the chain rule or apply the transformation rules.

 

[Demystifier]

The "formalism of Bjorken Drell" is covariant with respect to general transformations.

OK, now we know what is the source of all disagreement between you and me. I propose to not mention my formalism any more (because we agree on it), and concentrate on the issue whether the Bjorken Drell formalism is general covariant or not. I still claim that it is not and don't understand why do you think that it is.

 

[Demystifier]

That's not usual terminology. A formalism being covariant always refers transformation rules under arbitrary coordinate changes.

Except in the 99% of literature on particle physics.
But let's not argue over different terminologies in different subfields again.

 

[rubi]

I propose to not mention my formalism any more (because we agree on it)

I'm not sure whether we fully agree. Do you agree that your formalism will produce invalid results if I try to compute the hydrogen spectrum with it by switching to spherical coordinates using the transformation rules (of scalars and covariant derivatives acting on scalars in your case) rather than the chain rule? If we can agree on that, I'm fine, because that's my issue with your formalism.

and concentrate on the issue whether the Bjorken Drell formalism is general covariant or not.

Okay, so do you think that an application of the spinor transformation rules and the rule for transforming the spin connection coefficients will give different results than the application of the chain rule in the case of the ordinary Dirac equation? Because as I said, the covariant formalism is designed to reproduce the results of the chain rule exactly.
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Except in the 99% of literature on particle physics.
But let's not argue over different terminologies in different subfields again.

Well, the usual Maxwell equations presented using $\nabla, \vec E, \vec B$ are also invariant under Lorentz transformations, but when we speak of the covariant Maxwell equations, we usually refer to the $F_{\mu\nu}$ version. But ok, let's ignore the terminology discussion.

 

[Demystifier]

I'm not sure whether we fully agree. Do you agree that your formalism will produce invalid results if I try to compute the hydrogen spectrum with it by switching to spherical coordinates using the transformation rules (of scalars and covariant derivatives acting on scalars in your case) rather than the chain rule? If we can agree on that, I'm fine, because that's my issue with your formalism.
​

Well, it's again a terminological issue, i.e. depends on what one means by "scalar". If I stipulate that my Psi transforms as a scalar under global coordinate transformations (diffeomorphisms) and as a spinor under local Lorentz transformations of tetrads, then the covariant derivative acts non-trivially on my Psi, in which case my formalism (with $\partial_{\mu}\rightarrow D_{\mu}$) will produce valid results. Otherwise, it will not.

Okay, so do you think that an application of the spinor transformation rules and the rule for transforming the spin connection coefficients will give different results than the application of the chain rule in the case of the ordinary Dirac equation? Because as I said, the covariant formalism is designed to reproduce the results of the chain rule exactly.

It reproduces the chain rule for covariant derivatives exactly. But my point is that Bjorken and Drell do not even define the covariant derivatives, they work only with ordinary derivatives. Nevertheless, once the ordinary derivatives are replaced with covariant ones (which Bjorken and Drell didn't do), then the two procedures will give the same results, as will my formalism with the stipulation above.​