Quantum field theory, spacetime, and coordinates

[rubi]

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

No, that's not a terminology issue, but a math issue. How can you stipulate that? If your $\Psi$ transforms as a spinor under local Lorentz transformations, then it also transforms as a spinor under global Lorentz transformations (a global transformation is also a local transformation), where the Jacobian is just $\Lambda$, i.e. $\Psi\rightarrow S(\hat\Lambda)\Psi$ and this is in conflict with the definition of your $\Psi$, which is supposed to be constant in that special case. If you decide to give up this property, then all you have is just the ordinary Dirac equation and not a new formalism.

It reproduces the chain rule for covariant derivatives exactly.

No, the whole point is that you are allowed to treat all derivatives as ordinary partial derivatives and when you carry out the chain rule completely, then in the end you end up with the properly transformed equation, because the definition of tensors and so on is exactly made in order to match the results of coordinate transformations by the chain rule. The transformation rules in the covariant formalism are derived from the chain rule, so it's no surprise that they match.

But my point is that Bjorken and Drell do not even define the covariant derivatives

And they needn't to that, because they are smart physicists who know that they can get the same results by just applying the chain rule.

they work only with ordinary derivatives. Nevertheless, once the ordinary derivatives are replaced with covariant ones (which Bjorken and Drell didn't do)

They are working with covariant derivatives, too. It's just that the Dirac equation in cartesian coordinates doesn't have any spin connection coefficients, because these coefficients happen to vanish in cartesian coordinates.

then the two procedures will give the same results, as will my formalism with the stipulation above.

As I said, your stipulation is inconsistent, while in the case of the Dirac equation, it's perfectly fine.​

 

[Demystifier]

If your $\Psi$ transforms as a spinor under local Lorentz transformations, then it also transforms as a spinor under global Lorentz transformations (a global transformation is also a local transformation),

Oh no, not again! Please read what Weinberg said in post #22. I just rephrase what Weinberg says, but for some reason you accept the Weinberg's words and don't accept mine.

Next time I will trick you. I will exactly copy the Weinberg's words, but will pretend that those are my words, to see if you will agree.

 

[rubi]

Oh no, not again! Please read what Weinberg said in post #22. I just rephrase what Weinberg says, but for some reason you accept the Weinberg's words and don't accept mine.

Weinberg says the same as I do. You didn't rephrase it, you made an inconsistent statement. He discusses coordinate transformations in the tetrad formalism. A coordinate transformation consists of two things: Relabeling the coordinates ($x\rightarrow x(x^\prime)$) and a change of the reference frame ($e_x,e_y,e_z\rightarrow e_r,e_\theta,e_\phi$) or in the case of spinors, the reference spin frame. What he means when he says that something is a coordinate scalar is that in order to perform the relabeling the coordinates, you just need to make the replacement $x\rightarrow x(x^\prime)$. If you leave it at that, you are still working with two coordinate systems. In order to complete the transformation, you also have to express the objects in the new reference frame and here get the transformation matrices, which in case of spinors are $S(\hat J(x))$. Please note, that Weinberg also says that vectors ("12.5.4") are coordinate scalars. Nevertheless, they transform with the Jacobian. Spinors are no different than vectors in that respect. In fact, the tensor product of two spinors is a vector, so if spinors didn't transform with the Jacobian, then neither would vectors and every undergrad knows that this is wrong.

As an example: The electromagnetic vector potential $A^\mu(x)$ is a coordinate scalar and a Lorentz vector in Weinberg's terminology. Do you disagree that a coordinate transformation induces a multiplication by a Jacobian?

Next time I will trick you. I will exactly copy the Weinberg's words, but will pretend that those are my words, to see if you will agree.

Well, you can try it, but it really has nothing to do with you or Weinberg. It's still true that your formalism is inconsistent and Weinberg surely would agree.​

 

[vanhees71]

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

You can also call it Dirac spinor. The Dirac representation is built by the two irreducible representations (1/2,0) and (0,1/2) of the orthochronous proper Poincare group, which allows to extend it to admit space reflections and parity-invariant models like QED and QCD.

 

[vanhees71]

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.

As Euclidean space can be described by curvilinear coordinates (to be more precise parts of it by maps and then connecting multiple maps to an atlas covering the entire Euclidean space), you can describe also Minkowski spacetime by curvilinear coordinates. Rindler coordinates cover only a wedge-shaped part of Minkowski space but still do not lead to a spacetime with curvature or/and torsion.

 

[Demystifier]

The electromagnetic vector potential $A^\mu(x)$ is a coordinate scalar and a Lorentz vector in Weinberg's terminology.

No it isn't. In his terminology $A^\mu(x)$ is a coordinate vector. It is $^*\!A^\alpha(x)$ that is a coordinate scalar in his terminology.

 

[rubi]

No it isn't. In his terminology $A^\mu(x)$ is a coordinate vector. It is $*A^\alpha(x)$ that is a coordinate scalar in his terminology.

This is only for the sake of that particular chapter in order to be able to separate the relabeling of coordinates and the change of the reference frame. You can only have the stared objects if you have fixed a background reference frame and a background spin reference frame. But this is not what a coordinate transformation does. A coordinate transformation always induces a relabeling of the coordinates as well as change of the reference frame, e.g. $e_x,e_y,e_z\rightarrow e_r,e_\theta,e_\phi$ in the case of spherical coordinates. That's also what Bjorken Drell do (without saying it explicitely) and it's also what we do if we transform to spherical coordinates in the standard non-relativistic hydrogen atom Schrödinger equation. And you also can't use it to fix your formalism, because there is an error in your calculation that you only recognize if you allow curvilinear coordinates. The error is that you cannot anymore exchange $\partial_\mu$ and $S(\hat J(x))$. This was a crucial step in the proof of the equivalence in case of Lorentz transformation. Without it, you don't get equivalence.

 

[Demystifier]

A coordinate transformation consists of two things: Relabeling the coordinates ($x\rightarrow x(x^\prime)$) and a change of the reference frame ($e_x,e_y,e_z\rightarrow e_r,e_\theta,e_\phi$)

I think I see now what is the source of our disagreement. You think that it is not allowed to do one of those two things without doing the other. I think that it is allowed. In other words, it is allowed to use a non-coordinate basis. See e.g. Appendix J in Carroll.

 

[rubi]

I think I see now what is the source of our disagreement. You think that it is not allowed to do one of those two things without doing the other. I think that it is allowed. In other words, it is allowed to use a non-coordinate basis. See e.g. Appendix J in Carroll.

It is allowed to use a non-coordinate basis, but this is not what you do in your paper and also not what you proposed in post #35. And it's also not what Bjorken Drell do and not what is done in the standard quantum mechanics of the hydrogen atom.

 

[stevendaryl]

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.

That approach works fine if you're working in flat spacetime.

If you go to curved spacetime, though, I assume that in a small region of spacetime, you can locally do the same trick, but then it's not clear how you're supposed to match up the local solutions of the Dirac equation to get a global solution.

 

[Demystifier]

That approach works fine if you're working in flat spacetime.

If you go to curved spacetime, though, I assume that in a small region of spacetime, you can locally do the same trick, but then it's not clear how you're supposed to match up the local solutions of the Dirac equation to get a global solution.

Yes, I agree with that.

 

[rubi]

That approach works fine if you're working in flat spacetime.

If you go to curved spacetime, though, I assume that in a small region of spacetime, you can locally do the same trick, but then it's not clear how you're supposed to match up the local solutions of the Dirac equation to get a global solution.

No, it doesn't even work in flat spacetime if one uses curvelinear coordinates. Demystifiers $\Psi$ just isn't a scalar. If he postulates it to be a scalar, he will get inconsistent results with the Dirac equation, because his equivalence proof uses a crucial step that only works for Lorentz transformations and not for more general coordinate transformations. In fact, from the mathematical point of view, there isn't much difference between a curved spacetime and a flat spacetime, when working in coordinates. The formulas work exactly the same way. It's just that in the curved case, some curvature invariants become non-zero, but the calculation doesn't care about that.

 

[Peter Morgan]

You can also call it Dirac spinor. The Dirac representation is built by the two irreducible representations (1/2,0) and (0,1/2) of the orthochronous proper Poincare group, which allows to extend it to admit space reflections and parity-invariant models like QED and QCD.

Or you can call it an element of a left module of the Dirac algebra, the latter being a Clifford algebra over the complex field using a (1,3) metric, which is isomorphic to a matrix algebra M₄(ℂ).
The left module structure is not reducible under the whole Lorentz group, which is, I venture to say, why mathematical physics more uses (or, rather, seems in my experience and in my preference more to use) Dirac spinors than Weyl spinors.
The Dirac spinor also supports a representation of the connected component of the conformal group in 1+3- or 3+1-dimensions, because (going a little too fast here, and it's surely not the only route) the Clifford algebras over ℝ using either a (2,3) or a (4,1) metric are both isomorphic to a matrix algebra M₄(ℂ). [I'm slightly correcting vanhees71 mentioning the Poincaré group instead of the Lorentz group, which I think he intended here; time now for someone to slightly correct me, I expect.]

 

[Demystifier]

It is allowed to use a non-coordinate basis, but this is not what you do in your paper and also not what you proposed in post #35.

In a sense, this is exactly what I propose, even though I have not expressed it explicitly in this form. First I do a coordinate transformation which induces also a change of tetrads. With respect to this transformation $\psi\rightarrow \psi'$ transforms as a spinor under local Lorentz transformations. So far you agree. But then I do an additional local Lorentz transformation that is not accompanied by a corresponding change of coordinates. According to Carroll, I have a freedom to do such an additional transformation as well. Hence the full transformation is $\psi\rightarrow \psi'\rightarrow \Psi$. I choose the second transformation such that it cancels the first local Lorentz transformation. In this way the full transformation $\psi\rightarrow \Psi$ looks like a transformation of a scalar under both coordinate and local Lorentz transformations. In essence, this is what Eq. (53) in my paper does.

Of course, in general, the second transformation may not exist everywhere, so the procedure may be impossible in general. However, when the initial coordinate transformation is the global Lorentz transformation in Minkowski spacetime, then the procedure above is well defined.

 

[rubi]

In a sense, this is exactly what I propose, even though I have not expressed it explicitly in this form. First I do a coordinate transformation which induces also a change of tetrads. With respect to this transformation $\psi\rightarrow \psi'$ transforms as a spinor under local Lorentz transformations. So far you agree. But then I do an additional local Lorentz transformation that is not accompanied by a corresponding change of coordinates. According to Carroll, I have a freedom to do such an additional transformation as well. Hence the full transformation is $\psi\rightarrow \psi'\rightarrow \Psi$. I choose the second transformation such that it cancels the first local Lorentz transformation. In this way the full transformation $\psi\rightarrow \Psi$ looks like a transformation of a scalar under both coordinate and local Lorentz transformations. In essence, this is what Eq. (53) in my paper does.

You may do all these transformations, but it won't lead to an equivalent formulation, because the equivalence proof requires that $S(J(x))$ be pulled through the derivative, but this does not work in general coordinates, hence the two formulations will be non-equivalent.

In order to see that, just try it: Use your equation (53) with an $x$-dependent Jacobian, so you have $\Psi(x) = S^{-1}(J(x))\psi(x)$ and carry out all the steps of your equivalence proof. You will find that one step produces extra terms due to the product rule, which can't be made to go away.

 

[Demystifier]

You may do all these transformations, but it won't lead to an equivalent formulation, because the equivalence proof requires that $S(J(x))$ be pulled through the derivative, but this does not work in general coordinates, hence the two formulations will be non-equivalent.

And I fully agree with you that it doesn't work in general coordinates. That's why I treat the hydrogen atom in spherical coordinates differently, by using the "dirty smart trick" I described before. And in genuinely curved spacetime, even that kind of trick will not work. I agree with all that. In general, I cannot avoid the explicit use of tetrads and covariant derivatives, which makes the computations rather complicated. But in some cases I can make things simpler, so this is what I do in my paper - I make things simpler in cases in which it is possible.

 

[vanhees71]

Or you can call it an element of a left module of the Dirac algebra, the latter being a Clifford algebra over the complex field using a (1,3) metric, which is isomorphic to a matrix algebra M₄(ℂ).
The left module structure is not reducible under the whole Lorentz group, which is, I venture to say, why mathematical physics more uses (or, rather, seems in my experience and in my preference more to use) Dirac spinors than Weyl spinors.
The Dirac spinor also supports a representation of the connected component of the conformal group in 1+3- or 3+1-dimensions, because (going a little too fast here, and it's surely not the only route) the Clifford algebras over ℝ using either a (2,3) or a (4,1) metric are both isomorphic to a matrix algebra M₄(ℂ). [I'm slightly correcting vanhees71 mentioning the Poincaré group instead of the Lorentz group, which I think he intended here; time now for someone to slightly correct me, I expect.]

That's an interesting different point of view, but I really mean the Poincare group, because it's not the Lorentz group alone that's important to construct the unitary representations of the Standard Model but indeed the Poincare group. Of course, all arguments about the Lorentz group also apply, because it's a subgroup of the Poincare group.

Also it is important to stress that in Nature the symmetry group is not the full Poincare group but only the proper orthchronous subgroup, because today we know by independent (!) experiments that the weak interaction does not obey all the various discrete symmetries, P, C, T, and CP (while CPT seems to be intact as predicted by local relativistic QFT).

 

[vanhees71]

No it isn't. In his terminology $A^\mu(x)$ is a coordinate vector. It is $^*\!A^\alpha(x)$ that is a coordinate scalar in his terminology.

I'm not familar with this terminology. In the way I know it, $A^{\mu}(x)$ are components of a four-vector field and transform as such:
$$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} A^{\nu}(\hat{\Lambda}^{-1} x'),$$
where $x$ and $x'=\Lambda x$ are the column vectors built by the vector components of the space-time vector, $x^{\mu}$ and $x^{\prime \mu}$ respectively; $\hat{\Lambda}=({\Lambda^{\mu}}_{\nu})$ is an $\mathbb{R}^{4 \times 4}$-Lorentz-transformation matrix.

Vectors and vector fields themselves are, of course, invariant under any kind of basis transformation by definition!

 

[rubi]

And I fully agree with you that it doesn't work in general coordinates. That's why I treat the hydrogen atom in spherical coordinates differently, by using the "dirty smart trick" I described before. And in genuinely curved spacetime, even that kind of trick will not work. I agree with all that. In general, I cannot avoid the explicit use of tetrads, which makes the computations rather complicated. But in some cases I can make things simpler, so this is what I do in my paper - I make things simpler in cases in which it is possible.

It's nice that we agree. Using the chain rule instead of transformation laws should still work in curved spacetime, although I don't know what the generalization of your equation to curved spacetime would look like. My issue was just that your equation spoils the covariance (under general transformations) of the Dirac equation, but of course you can still perform calculations with it if you keep track of all the transformations that are involved. The nice thing about covariant equations is that you can just take the equation in cartesian coordinates and perform the replacement $\partial_\mu\rightarrow\nabla_\mu$ to get the general version. When physicists spot an equation that involves indices, they often take it for granted that it's covariant, so I struggle a bit to introduce non-covariant equations. I don't think this is beneficial for students, because they don't know what a non-coordinate spin frame is and that's quite advanced stuff that can't be explained easily, but it would be really important to teach them, so they understand why certain things work differently with your equation. However, it's of course nice if there are situations, where your equation makes things simpler.

 

[vanhees71]

One must of course also be careful with the replacement rule mentioned. Sometimes you cannot simply make $\partial_{\mu} \rightarrow \nabla_{\mu}$, but you have to carefully use the proper means to determine the correct differential operator for what you want to model. A pretty nice example is the quantization of the spinning top in Kleinert's path-integral book. There the correct way is of course to use the Hamiltonian, properly formulated with angular-momentum operators, while a naive "canonical quantization prescription" leads to wrong equations.

As far as I can see in Bjorken Drell vol. 1 they simply got the correct Dirac equation for the hydrogen atom in terms of spherical coordinates, because they used the proper definitions of the operators involved (energy, total angular momentum, orbital angular momentum, and spin).

 

[rubi]

One must of course also be careful with the replacement rule mentioned. Sometimes you cannot simply make $\partial_{\mu} \rightarrow \nabla_{\mu}$, but you have to carefully use the proper means to determine the correct differential operator for what you want to model. A pretty nice example is the quantization of the spinning top in Kleinert's path-integral book. There the correct way is of course to use the Hamiltonian, properly formulated with angular-momentum operators, while a naive "canonical quantization prescription" leads to wrong equations.

Well, okay, it's still possible that $\partial_\mu$ is supposed to be a Lie derivative or something like that instead of a covariant derivative, which you can't see easily from the cartesian coordinates formulation. My point is probably that covariant equations have a fully coordinate-independent formulation, i.e. you can write them in a way that doesn't involve a choice of coordinates in the first place and thus emphasizes the physical content of the equation rather than effects that really may be artifacts of the coordinate choice. Quantization is yet a different problem though. Even for a coordinate-free equation, there may be many inequivalent ways to quantize it.

As far as I can see in Bjorken Drell vol. 1 they simply got the correct Dirac equation for the hydrogen atom in terms of spherical coordinates, because they used the proper definitions of the operators involved (energy, total angular momentum, orbital angular momentum, and spin).

Well, the calculation in Bjorken Drell is very brief anyway. They essentially skipped the angular part and just wrote down the result, which is quite sad, because most of the connection coefficients are in the angular part, so there was no real part in the book where I could point to in order to illustrate the presence of the coefficients.

 

[vanhees71]

It's not only in quantizing classical models. Of course, I agree with you, that one should formulate everything in a covariant way in terms of coordinate-independent equations.

I don't remember where I've found the example (maybe in Misner Thorne, Wheeler, Gravitation?) that it's not clear how in a naive way you make the Maxwell Equations generally covariant by naively using the equations within SR in Cartesian coordinates and substituting the partial derivatives by covariant derivatives. It's of course unique when using the action principle, which has the great advantage that everything is written down in a manifestly covariant way as a scalar Lagrangian.

 

[rubi]

I don't remember where I've found the example (maybe in Misner Thorne, Wheeler, Gravitation?) that it's not clear how in a naive way you make the Maxwell Equations generally covariant by naively using the equations within SR in Cartesian coordinates and substituting the partial derivatives by covariant derivatives. It's of course unique when using the action principle, which has the great advantage that everything is written down in a manifestly covariant way as a scalar Lagrangian.

What's true is that there are in general many ways to generalize Maxwell's equation to curved spacetime, because you could for instance add a coupling to the Ricci scalar or other curvature invariants (any many more things) that vanish in the flat limit. But if you stay in flat spacetime, an equation will always have a unique coordinate-free formulation (of course up to equivalence, e.g. you can multiply both sides by 5). But if the equation is not covariant, the coordinate-free formulation will look require the introduction of unnatural background structure.

 

[Peter Morgan]

That's an interesting different point of view, but I really mean the Poincare group, because it's not the Lorentz group alone that's important to construct the unitary representations of the Standard Model but indeed the Poincare group.

Right. I wondered whether that was what you really intended. I was specifically channeling the Wightman axioms, per Haag, Local Quantum Physics, p. 57:

.
So the components of the field in that context are subject to transformations taken from a representation of the Lorentz group. The standard model of particle physics is the same, I think. I'd be interested in your choice of an example of a theory in which components of the field transform according to a representation of the Poincaré group?

 

[George Jones]

Or you can call it an element of a left module of the Dirac algebra, the latter being a Clifford algebra over the complex field using a (1,3) metric, which is isomorphic to a matrix algebra M₄(ℂ).

Representations of the Poincare group often are constructed by using infinite-dimensional function spaces that map into the above space, i.e., infinite-dimensional representations of the Lorentz group are used to construct infinite-dimensional representations of the Poincare group, where the action of the Poincare group is on the function space, and the action of the Lorentz group is on the codomain of the functions in the function space. Wave equations (e.g., the Dirac equations) are used to project onto subspaces of the function spaces that are relevant to a particular representation of the Poincare group.

 

[PeterDonis]

I don't think there are any relativists who use a different definition.

Just to note, this thread is in the quantum forum, not the relativity forum, so the question is really what definition quantum field theorists use.