What if the Lagrangian density $\mathscr{L}$, instead of a Lorentz scalar, were a Lorentz vector?

Question

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As an answer to this post, I made an impression that if $\mathscr{L}$ were not a Lorentz scalar in Eq.$(1)$ (see below), then Eq.$(1)$ would not be covariant. But now I think that is wrong! I state the reason below why I think so. If confirmed, I will delete the answer there.

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EDITED ON $14.04.23$

The Euler-Lagrange equation for a field $\phi_a$ ($a=1,2,3,...N$ counts that number of components that mix under Lorentz transformation) given by $$\frac{\partial \mathscr{L}}{\partial\phi_a}=\partial_\mu\left(\frac{\partial \mathscr{L}}{\partial(\partial_\mu\phi_a)}\right)\tag{1}$$ becomes $$\frac{\partial \mathscr{L}}{\partial\phi^\prime_a}=\partial_\mu^\prime\left(\frac{\partial \mathscr{L}}{\partial(\partial^\prime_\mu\phi^\prime_a)}\right)\tag{2}$$ where $\phi^\prime_a=S_{ab}\phi_b$ denotes the transformation of the field $\phi_a$ under Lorentz transformation. As far as the $\mu$ index is concerned, those are contracted and should not change. Also, since $\mathscr{L}$ is a Lorentz scalar, that too doesn't change. Therefore, the covariance of $(1)$ is manifest! So it satisfies the postulate of special relativity.

Now consider a different equation that has nothing to do with anything familiar! Write an equation by replacing $\mathscr{L}$ in $(2)$ with a quantity that has one uncontracted Lorentz index (say, $\mathscr{L}_\alpha$) : $$\frac{\partial \mathscr{L}_\alpha}{\partial\phi_a}=\partial_\mu\left(\frac{\partial \mathscr{L}_\alpha}{\partial(\partial_\mu\phi_a)}\right)\tag{3}$$

Question As far as covariance is concerned, wouldn't equations of the form $(3)$ also be covariant?

If yes, it would be incorrect to say: "if $\mathscr{L}$ were not a Lorentz scalar but a Lorentz vector $\mathscr{L}_\alpha$, equation of motions would not be covariant." In other words, the covariance of the equation of motion does not require $\mathscr{L}$ to be a Lorentz scalar. Is this correct?

Answer 1

I believe at least in flat spacetime, such things make sense. At the very least, Doran, Lasenby and Gull have studied even further generalized Lagrangians, namely multi-vector Lagrangians in the setting of geometric algebra. As for the specific case of vector (or tensor) valued Lagrangians in relativistic field theory, they do indeed give sensible covariant equations of motion, with tensor rank equal to or higher than the Lagrangian itself. As a simple example, lets consider a vector valued Lagrangian coupling a scalar field $\phi$ to a vector field $\psi^a$,

$$ \mathcal{L}^a = m \phi \psi^a + \partial^a \phi \partial_b \psi^b $$ This Lagrangian density is clearly a Lorentz vector. Now let us take the variation first with respect to $\phi$:

$$ \delta \mathcal{L}^a = m \delta \phi \psi^a + \partial^a \delta\phi \partial_b \psi^b =0 $$ Doing the usual partial integration of the second term and dropping surface terms gives $$ m \delta \phi \psi^a - \delta\phi\partial^a \partial_b \psi^b =0, $$ so our Euler Lagrange equation becomes $$ m \psi^a - \partial^a\partial_b \psi^b=0, $$ which is a perfectly sensible covariant equation of motion.

Similarly, let us take the variation with respect to $\psi^a$, $$ \delta \mathcal{L}^a = m \phi \delta\psi^a + \partial^a \phi \partial_b \delta\psi^b =0 $$ Doing the usual partial integration gives $$ m \phi \delta\psi^a - \delta\psi^b\partial_b \partial^a \phi =0 $$ We can write this as $$ (m \phi \delta^a_b - \partial_b \partial^a \phi )\delta\psi^b =0 $$ So our Euler-Lagrange equation is $$ m \phi \delta^a_b - \partial_b \partial^a \phi =0 $$ This is once again a covariant equation of motion.

As for the curved spacetime generalization, that I'm not sure if its still sensible, but perhaps a generalization in terms of tetrads or such could be done. Someone who's more familiar with Lagrangian field theories on curved spacetimes could chip in here.

Answer 2

The point of using a scalar Lagrangian is not necessarily obvious from classical physics in flat space-time. The point is that

• The Hamilton's principle of least action has proven to be an extremely useful tool to analyse the symmetries of a physical theory and thus also its conservation laws through Noether's theorem. For example, writing down a scalar Lagrangian with coordinate invariance immediately provides us with tools to analyze the conservation of momentum and energy (and what momentum and energy even are in the theory). A multi-component object makes this much less clear.
• Hamilton's action $S = \int_{\Omega} \mathcal{L} d^4 V$ is typically just a single number associated with the entire space-time domain $\Omega$. If you integrate a multi-component object over the space-time domain, you get a dangling index on $S$, $S^a = \int \mathcal{L}^a d^4 V$. This index must correspond to a "global" coordinate system or gauge group basis over the entire domain $\Omega$ somehow. This is in violation with local gauge invariance on which the entirety of the Standard model of particle physics and local diffeomorphism invariance on which General Relativity is built. (Local diffeomorphism covariance and/or invariance is the statement that your action principle and physical laws need have the same form irrespective of the coordinate system used, including curvi-linear coordinates, not just global Cartesian-Minkowski coordinates.) This is the point most relevant to the main body of your question.
• The vanishing variation $\delta S^a/\delta \phi = 0$ gives you more than one equation $\phi$ has to fulfill. Unless you make the functional $S^a$ (or the function $\mathcal{L}^a$) somehow degenerate, you will often get overconstrained systems. If you make $S^a$ degenerate, you will often find the corresponding Lagrangian can be "contracted'' or traced out into a scalar.
• Perhaps not as importantly, the path-integral would have great difficulties with a non-scalar action in expressions such as the transition amplitdue $\propto e^{i S/\hbar}$. How to do this with a multi-component object? And if you find a way, why don't you call the resulting scalar $\mathcal{S}(S^a)$ showing up in the transition amplitude $\propto e^{i\mathcal{S}(S^a)/\hbar}$ the action instead of the set $S^a$? This is just one other illustration how deeply the scalar Lagrangian and action are embedded into fundamental theories of physics.

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This is not to say that it is somehow forbidden to try and develop multi-component Lagrangian-like structures. It is interesting to imagine what could be the contexts where it is useful, such as when a medium defines some sort of global coordinates providing a basis for the dangling index $a$ etc. But in physics rather than mathematics you must justify the use of a generalized structure in a real physical context, not just postulate it arbitrarily. And unless the structure is justified by its usefulness, it is just met with a shoulder shrug.

This is because once you ask about multi-component Lagrangians, you do not need to stop there. You can start building theories that are not generated by any Lagrangian-like structure whatsoever. The field equations can just have objects in them that have the correct transformation properties. Why not, right? Or they can have objects in them that are a function of more than one space-time points. Or they can have objects in them that have a different meaning depending on the time of the day. Why not? The important question is why yes.

TL;DR: Modern fundamental physics such as general relativity, quantum mechanics, or quantum field theory are built around scalar actions and Lagrangians, and vector Lagrangians pose complications with no obvious advantage. Until there is no advantage in using them, you are not forbidden to try to develop use cases, but nobody will necessarily care.

Answer 3

You are correct that the covariance of the equations of motion does not require the Lagrangian density to be a Lorentz scalar, as shown in the example you provided. However, this is a rather unusual case and does not necessarily apply to physical theories.

In general, physical theories are formulated using Lagrangians that are Lorentz scalars. This is because Lorentz invariance is a fundamental symmetry of spacetime in special relativity, and physical observables should not depend on the choice of reference frame. By constructing a Lagrangian density that is a Lorentz scalar, we ensure that the action is invariant under Lorentz transformations, and the resulting equations of motion are covariant. This guarantees that the physical predictions of the theory are frame-independent.

In the example you provided with a Lagrangian density having one uncontracted Lorentz index, the equations of motion appear to be covariant. However, this does not necessarily mean that such a Lagrangian density would lead to a physically meaningful theory. It is possible that the resulting theory may not be consistent with the principles of special relativity or have other undesirable properties.

It is true that the covariance of the equations of motion does not require the Lagrangian density to be a Lorentz scalar, using a Lorentz scalar Lagrangian density is the standard approach in formulating physical theories consistent with special relativity. The example you provided demonstrates an unusual case and may not necessarily lead to a meaningful physical theory.

To answer your question, yes, your understanding is correct. Equations of the form (3) can also be covariant, even if the Lagrangian density is not a Lorentz scalar but a Lorentz vector. So, it would be incorrect to say that if the Lagrangian density were not a Lorentz scalar but a Lorentz vector, the equations of motion would not be covariant.

Answer 4

Lagrangian density is scalar because of the equation

$$S[\phi_i]=\int \mathcal{L}(\phi_i,\partial _\mu\phi_i, \mathbf{x}) d^4\mathbf{x}$$

Since the action is a scalar quantity you can't get a scalar by adding (integration) a bunch of vector,

Under Lorentz transformation or any other type of linear transformation of coordinate system, quantities such as scalars, vectors or higher order tensors remains invariant. (Things doesn't change just because one is trying to look at it from a different frame of reference), The components will differ in accordance to your basis, ( the length number may change if you started measuring in feet instead of meter, but length of object will never change, basis or coordinate system is like choosing meter or feet, while components are value associated with it ) but all such things remains invariant under any linear transformation, with coordinates transformation the new components comes out to be in accordance to your transformation rule, which is known as covariance.

So if there were some 'vector action' type thing with an associated vector lagrangian, sure we could have a 'vector lagrangian',

Edit:

Action is basically a functional that takes path(functions) $q_i(t)$ and gives a scalar quantity with it according to equation above,

The action principle says 'extremizing this action will give you a optimum path that a system will follow between two points in a configuration space,

However if one were to use a vector action, then it would be tedious to find extremums, as vectors are not like scalar that can be compared directly (loosely speaking, as scalars of different spaces can't be compared without any pull back or push forward map).(this part is to be taken loosely)

Answer 5

I'm quite stumped that this is not yet solved satisfactorily.

Yes, it is true that Euler-Lagrange equations allow us to consider Lagrangian densities that are Lorentz covariant, not needing them to be exactly a scalar.

However, a bit of thought shows that the Lagrangian densities can only be scalar, pseudo-scalar, or even ranked tensors of special type. Maybe a mix of the former two might be a way to generate matter-anti-matter assymmetry, but otherwise it can only be one or the other.

This is because Lagrangian densities are integrated over all of spacetime and exponentiated. If the Lagrangian densities has any unpaired Lorentz covariant index, then the exponentiation causes ever increasing complications. If the Lorentz covariant index is contracted away, then either it is contracted with a privileged tensor, or it is contracted with itself. The former case is rejected by the lack of privileged directions, and the latter case reduces back to scalar or pseudo-scalars. This is also related to how the action S = integrated Lagrangian density must not be dimension-full (after dividing by hbar).

Notice, however, it can be a square matrix, or a Lie algebra element. i.e. things that can be applied to themselves infinitely many times and still make sense. If it is to be a tensor, it must have equal numbers of contravariant indices and covariant indices with a natural way to pair them. It is however, not very clear what the physical interpretation of these things could be. Maybe someone in the future could make a sensible theory out of such exotic stuff. (Connecting with AwkwardWhale's answer, the even subalgebra of a geometric algebra is possible.)

Answer 6

An equation is said to be Lorentz covariant if it can be written in terms of Lorentz covariant quantities. A physical quantity is said to be Lorentz covariant if it transforms under a given representation of the Lorentz group. A Lorentz covariant scalar (e.g., the space-time interval) remains the same under Lorentz transformations and is said to be a Lorentz invariant.

Therefore, the covariance of the equation of motion does not require L to be a Lorentz scalar, but it does require L to be a Lorentz covariant quantity. For example, the wave equation ∂^2 ϕ ∂ x^2 = c^2 ∂^2 ϕ ∂ t^2 is Lorentz covariant, but ϕ is not a Lorentz scalar, it is a scalar field that transforms as a scalar under Lorentz transformations