A gauge-symmetrization method for energy-momentum tensors in high-order electromagnetic field theories

The Lagrangian density of a general electromagnetic system is written as

$$\mathcal{L}_{F}=\mathcal{L}_{F}\left(x^{\mu},DA_{\mu},\cdots,D^{\left(n+1\right)}A_{\mu}\right),$$

(3)

where $A=\left(\varphi,-\bm{A}\right)$ is the 4-potential and $D=\left(\left(1/c\right)\partial_{t},\bm{\nabla}\right)$ is the derivative operator over spacetime. The EL equation of the Lagrangian density is

$$E_{A}^{\mu}\left(\mathcal{L}_{F}\right)=0,$$

(4)

where the Euler operator $E_{A}^{\mu}$ of $A_{\mu}$ is defined by

$$E_{A}^{\mu}\equiv\sum_{i=1}^{n+1}\left(-1\right)^{i}D_{\mu_{1}}\cdots D_{\mu_{i}}\frac{\partial}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{i}}A_{\mu}\right)}.$$

(5)

Note that the Lagrangian density $\mathcal{L}_{F}$ depends on derivatives of $A$ with respect to the spacetime coordinates up to the $(n+1)$-th order. It includes the standard Maxwell system, i.e.,

$$\mathcal{L}_{F}=-\frac{1}{16\pi}F_{\mu\nu}F^{\mu\nu},$$

(6)

as a special case, where

$$F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}.$$

(7)

is the Faraday tensor. In Eq. (6), $\mathcal{L}_{F}$ depends only on first-order derivatives of $A$. High-order electromagnetic field theories appear in the study of gyrokinetic systems (Qin, 2005; Qin et al., 2007; Fan et al., 2020) for magnetized plasmas and radiation reaction for classical charged particles (Bopp, 1940; Podolsky, 1942). Physics requires that the EL equation (4) is gauge symmetric, i.e, invariant under the gauge transformation $A_{\mu}\mapsto A_{\mu}+\partial_{\mu}f$. In the present study, we assume that $\mathcal{L}_{F}$ is explicitly gauge symmetric in the form of

$$\mathcal{L}_{F}=\mathcal{L}_{F}\left(x^{\mu},F_{\mu\nu},DF_{\mu\nu},\cdots,D^{\left(n\right)}F_{\mu\nu}\right).$$

(8)

From the variational principle, $\delta\mathcal{A}=\delta\int\mathcal{L}_{F}d^{4}x=0$, we have

	$\displaystyle 0=\delta\int\mathcal{L}_{F}d^{4}x=\int\left\{E_{F}^{\mu\nu}\left(\mathcal{L}_{F}\right)\delta F_{\mu\nu}\right\}d^{4}x=\int\left\{E_{F}^{\mu\nu}\left(\mathcal{L}_{F}\right)\left(\partial_{\mu}\delta A_{\nu}-\partial_{\nu}\delta A_{\mu}\right)\right\}d^{4}x$
	$\displaystyle=-\int\partial_{\mu}\left\{2E_{F}^{\left[\mu\nu\right]}\left(\mathcal{L}_{F}\right)\right\}\delta A_{\nu}d^{4}x,$		(9)

where the boundary term has been dropped, and $E_{F}^{\mu\nu}$ denotes the Euler operator for the Faraday tensor $F_{\mu\nu}$ defined by

$$E_{F}^{\mu\nu}\left(\mathcal{L}_{F}\right)=\frac{\partial\mathcal{L}_{F}}{\partial F_{\mu\nu}}+\sum_{i=1}^{n}\left(-1\right)^{i}D_{\mu_{1}}\cdots D_{\mu_{i}}\frac{\partial\mathcal{L}_{F}}{\partial\partial_{\mu_{1}}\cdots\partial_{\mu_{i}}F_{\mu\nu}}\,.$$

(10)

In Eq. (9), superscript $\left[\mu\nu\right]$ represents anti-symmetrization with respect to $\mu$ and $\nu$, i.e.,

$$E_{F}^{\left[\mu\nu\right]}\left(\mathcal{L}_{F}\right)\equiv\frac{1}{2}\left[E_{F}^{\mu\nu}\left(\mathcal{L}_{F}\right)-E_{F}^{\nu\mu}\left(\mathcal{L}_{F}\right)\right].$$

(11)

Due to the arbitrariness of $\delta A_{\nu}$ in Eq. (9), the equation of motion for the system is

$$\partial_{\mu}\mathcal{D}^{\mu\nu}=0,$$

(12)

where

$$\mathcal{D}^{\mu\nu}\equiv-8\pi E_{F}^{\left[\mu\nu\right]}\left(\mathcal{L}_{F}\right)$$

(13)

is the electric displacement tensor.

In Sec. III, we will consider electromagnetic systems coupled with charged particles, and the Lagrangian density $\mathcal{L}$ will depend on 4-potential $A_{\mu}$ i.e.,

$$\mathcal{L}=\mathcal{L}\left(x^{\mu},A_{\mu},F_{\mu\nu},\cdots,D^{\left(n\right)}F_{\mu\nu}\right).$$

Equation (12) then becomes

$$\partial_{\mu}\mathcal{D}^{\mu\nu}=\frac{4\pi}{c}J_{f}^{\nu},$$

(14)

where

$$J_{f}^{\nu}\equiv-c\frac{\partial\mathcal{L}_{F}}{\partial A_{\nu}}$$

(15)

is the free 4-current.

For the standard Maxwell system (6) without free 4-current, the electric displacement tensor is the Faraday tensor, i.e., $\mathcal{D}^{\mu\nu}=F^{\mu\nu}$, and Eq. (12) reduces to Maxwell’s equation

$$\partial_{\mu}F^{\mu\nu}=0.$$

(16)

A continuous symmetry of the action $\mathcal{A}$ is a group of transformation

$$\left(x^{\mu},A^{\nu}\right)\mapsto\left(\tilde{x}^{\mu},\tilde{A}^{\nu}\right)=g_{\epsilon}\cdot\left(x^{\mu},A^{\nu}\right),$$

(17)

such that

$$\int\mathcal{L}_{F}\left(\tilde{x}^{\mu},\tilde{F}_{\mu\nu},\tilde{D}\tilde{F}_{\mu\nu},\cdots,\tilde{D}^{\left(n\right)}\tilde{F}_{\mu\nu}\right)d^{4}\tilde{x}=\int\mathcal{L}_{F}\left(x^{\mu},F_{\mu\nu},DF_{\mu\nu},\cdots,D^{\left(n\right)}F_{\mu\nu}\right)d^{4}x,$$

(18)

where $g_{\epsilon}$ constitutes a continuous group of the transformations parameterized by $\epsilon$ (Olver, 1993). The infinitesimal generator of the transformation group is

$$\bm{v}\coloneqq\frac{d}{d\epsilon}|_{0}g_{\epsilon}\cdot\left(x^{\mu},A^{\nu}\right)=\xi^{\mu}\frac{\partial}{\partial x^{\mu}}+\phi_{\mu}\frac{\partial}{\partial A_{\mu}}.$$

(19)

By rewriting the symmetry condition (18) as

$$\frac{d}{d\epsilon}|_{0}\int\mathcal{L}_{F}\left(\tilde{x}^{\mu},\tilde{F}_{\mu\nu},\tilde{D}\tilde{F}_{\mu\nu},\cdots,\tilde{D}^{\left(n\right)}\tilde{F}_{\mu\nu}\right)d^{4}\tilde{x}=0,$$

(20)

we can derive the following infinitesimal version of the symmetry condition,

$$\text{pr}^{\left(n+1\right)}\bm{v}\left(\mathcal{L}_{F}\right)+\mathcal{L}_{F}D_{\mu}\xi^{\mu}=0,$$

(21)

where $\text{pr}^{\left(n+1\right)}\bm{v}$ is the prolongation of $\bm{v}$. The standard prolongation formula for $\text{pr}^{\left(n+1\right)}\bm{v}$ can be found in Ref. (Olver, 1993). In the present study, we rewrite the prolongation formula with respect to $F_{\mu\nu}$, instead of $A_{\mu}$, as

	$\displaystyle\text{pr}^{\left(n+1\right)}\bm{v}=\bm{v}+\left[G_{\sigma\rho}+\xi^{\sigma}D_{\sigma}F_{\sigma\rho}\right]\frac{\partial\mathcal{L}_{F}}{\partial F_{\sigma\rho}}$
	$\displaystyle+\sum_{i=1}^{n}\left[D_{\mu_{1}}\cdots D_{\mu_{i}}G_{\sigma\rho}+\xi^{\sigma}D_{\sigma}D_{\mu_{1}}\cdots D_{\mu_{i}}F_{\sigma\rho}\right]\frac{\partial\mathcal{L}_{F}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{i}}F_{\sigma\rho}\right)},$		(22)

where

$$G_{\sigma\rho}=\partial_{\sigma}Q_{\rho}-\partial_{\rho}Q_{\sigma}\equiv 2\partial_{[\sigma}Q_{\rho]}$$

(23)

and

$$Q_{\nu}=\phi_{\nu}-\xi^{\sigma}D_{\sigma}A_{\nu},$$

(24)

is a characteristic of the Lie algebra.

Combining Eqs. (12) and (21) generates the conservation law corresponding to the symmetry,

$$D_{\mu}\left\{\mathcal{L}_{F}\xi^{\mu}-\frac{1}{4\pi}\mathcal{D}^{\mu\nu}\left(\mathcal{L}_{F}\right)Q_{\nu}+\mathbb{P}^{\mu}\right\}=0,$$

(25)

where

$$\mathbb{P}^{\mu}=\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}\left(D_{\mu_{j+1}}\cdots D_{\mu_{i}}G_{\sigma\rho}\right)\left[D_{\mu_{1}}\cdots D_{\mu_{j-1}}\frac{\partial\mathcal{L}_{F}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\sigma\rho}\right)}\right].$$

(26)

The conservation law given by Eq. (25) is not gauge-symmetric in general.

Now we assume the high-order electromagnetic field theory admits the spacetime translation symmetry, i.e.,

$$\frac{\partial\mathcal{L}_{F}}{\partial x^{\mu}}=0,$$

(27)

and derive the corresponding energy-momentum conservation law. Because of Eq. (27), the action is invariant under the spacetime translation

$$\left(x^{\mu},A_{\nu}\right)\mapsto\left(x^{\mu}+\epsilon X_{0}^{\mu},A_{\nu}\right),$$

(28)

where $X_{0}^{\mu}$ is 4D constant vector field. The infinitesimal generator $\bm{v}$, characteristic $Q^{\nu}$, and $G_{\sigma\rho}$ in Eq. (23) are

	$\displaystyle\bm{v}=X_{0}^{\mu}\frac{\partial}{\partial x^{\mu}},$		(29)
	$\displaystyle Q^{\nu}=-X_{0}^{\nu}\partial_{\nu}A_{\sigma},$		(30)
	$\displaystyle G_{\sigma\rho}=-X_{0}^{\nu}\partial_{\nu}F_{\sigma\rho}.$		(31)

The Lagrangian density satisfies the infinitesimal criterion because

$$X_{0}^{\mu}\frac{\partial\mathcal{L}_{F}}{\partial x^{\mu}}=0,$$

(32)

which implies a conservation law. Substituting Eqs. (29)-(31) into to Eq. (25), we obtain the canonical energy-momentum conservation law according the standard Noether procedure,

	$\displaystyle D_{\mu}T_{\text{N}}^{\mu\nu}=0,$		(33)
	$\displaystyle T_{\text{N}}^{\mu\nu}=\mathcal{L}_{F}\eta^{\mu\nu}+\frac{1}{4\pi}\mathcal{D}^{\mu\sigma}\partial^{\nu}A_{\sigma}-\Sigma^{\mu\nu},$		(34)
	$\displaystyle\Sigma^{\mu\nu}=\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}\left(D_{\mu_{j+1}}\cdots D_{\mu_{i}}\partial^{\nu}F_{\sigma\rho}\right)\left[D_{\mu_{1}}\cdots D_{\mu_{j-1}}\frac{\partial\mathcal{L}_{F}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\sigma\rho}\right)}\right].$		(35)

In Eq. (33), $T_{\text{N}}^{\mu\nu}$ is the canonical EMT derived from the standard Noether procedure.

Obviously, $T_{\text{N}}^{\mu\nu}$ depends on the gauge as expected. In the expression of $T_{\text{N}}^{\mu\nu}$ given by Eq. (34), the gauge dependence comes from the second term, and the first and third terms are gauge symmetric. Now we show how to gauge-symmetrize $T_{\text{N}}^{\mu\nu}$. Note that because electric displacement tensor $\mathcal{D}^{\sigma\mu}$ is anti-symmetric, the following equations hold,

	$\displaystyle D_{\mu}\left(D_{\sigma}\mathcal{F}^{\sigma\mu\nu}\right)=0,$		(36)
	$\displaystyle\mathcal{F}^{\sigma\mu\nu}\equiv\frac{1}{4\pi}\mathcal{D}^{\sigma\mu}A^{\nu}.$		(37)

Here, $\mathcal{F}^{\sigma\mu\nu}$ is a super-potential that is anti-symmetric with respect to the first two indices. For easy reference, we will call $\mathcal{F}^{\sigma\mu\nu}$ displacement-potential tensor. The divergence of $\mathcal{F}^{\sigma\mu\nu}$ defines a divergence-free tensor, i.e.,

$$T_{0}^{\mu\nu}\equiv D_{\sigma}\mathcal{F}^{\sigma\mu\nu}=-\frac{1}{4\pi}\mathcal{D}^{\mu\sigma}\partial_{\sigma}A^{\nu},$$

(38)

where used is made of Eq. (12). When $T_{0}^{\mu\nu}$ is added to $T_{\text{N}}^{\mu\nu},$ the gauge dependence is removed, i.e.,

	$\displaystyle D_{\mu}T_{\text{GS}}^{\mu\nu}=0,$		(39)
	$\displaystyle T_{\text{GS}}^{\mu\nu}\equiv T_{\text{N}}^{\mu\nu}+T_{0}^{\mu\nu}=\mathcal{L}_{F}\eta^{\mu\nu}+\frac{1}{4\pi}\mathcal{D}^{\mu\sigma}F_{\;\sigma}^{\nu}-\Sigma^{\mu\nu},$		(40)

where $T_{\text{GS}}^{\mu\nu}$ is the gauge-symmetric EMT.

It is worthwhile to mention that we derived the gauge-symmetrized EMT $T_{\text{GS}}^{\mu\nu}$ from the expression of $T_{\text{N}}^{\mu\nu}$ in Eq. (34), which is calculated from the prolongation with respect to $F_{\mu\nu}$. On the other hand, had we started from Eq. (3) and calculated the EMT from the prolongation with respect to $A_{\mu}$, we would have obtained a canonical EMT in the form of

$$T_{\text{N}}^{\mu\nu}=\mathcal{L}_{F}\eta^{\mu\nu}-\hat{\Sigma}^{\mu\nu},$$

(41)

where

$$\hat{\Sigma}^{\mu\nu}=\sum_{i=1}^{n+1}\sum_{j=1}^{i}\left(-1\right)^{j+1}D_{\mu_{j+1}}\cdots D_{\mu_{i}}\left(\partial_{\nu}A_{\sigma}\right)\left[D_{\mu_{1}}\cdots D_{\mu_{j-1}}\frac{\partial\mathcal{L}_{F}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}A_{\sigma}\right)}\right].$$

(42)

However, different from the situation in Eq. (34), every term in Eq. (42) is gauge dependent, making the gauge symmetrization difficult, if not impossible.

For the standard Maxwell electromagnetic system specified by Eq. (6), the electric displacement tensor reduces to the Faraday tensor $F^{\sigma\mu}$, and the displacement-potential tensor $\mathcal{F}^{\sigma\mu\nu}$ reduces to $F^{\sigma\mu}A^{\nu}/4\pi$, coinciding with the tensor used by Blaschke et al. for the U$\left(1\right)$ gauge theory (Blaschke et al., 2016).

As described above, the method proposed in the present study employs displacement-potential tensor $\mathcal{F}^{\sigma\mu\nu}$ to gauge-symmetrize the EMT, while the BR method use the super-potential $\mathcal{S}^{\sigma\mu\nu}$ to symmetrize the EMT. In this subsection, we discuss the difference between the displacement-potential tensor $\mathcal{F}^{\sigma\mu\nu}$ in Eq. (36) and the BR super-potential $\mathcal{S}^{\sigma\mu\nu}$. To calculate $\mathcal{S}^{\sigma\mu\nu}$, we need to first derive the 4D angular momentum conservation laws generated by the Lorentz symmetry. Assume that system is invariant under rotational transformation in 4D spacetime

$$\left(x^{\mu},A^{\nu}\right)\mapsto\left(\tilde{x}^{\mu},\tilde{A}^{\nu}\right)=\left(\Lambda_{\epsilon}^{\mu\sigma}x_{\sigma},\Lambda_{\epsilon}^{\nu\sigma}A_{\sigma}\right),$$

(43)

where $\left\{\Lambda_{\epsilon}^{\mu\sigma}\right\}$ is one-parameter subgroup of the Lorentz group. The infinitesimal generator $\bm{v}$, the characteristic $Q_{\rho}$, and the term $G_{s\rho}$ are calculated respectively by Eqs. (19), (24), and (23) as

	$\displaystyle\bm{v}=\frac{d}{d\epsilon}|_{0}\left(\Lambda_{\epsilon}^{\mu\sigma}x_{\sigma},\Lambda_{\epsilon}^{\nu\sigma}A_{\sigma}\right)=\left(\Omega^{\mu\sigma}x_{\sigma},\Omega^{\mu\sigma}A_{\sigma}\right),$		(44)
	$\displaystyle Q_{\rho}=\phi_{\rho}-\xi^{\alpha}D_{\alpha}A_{\rho}=\Omega_{\rho\alpha}A^{\alpha}-\Omega_{\alpha\beta}x^{\beta}D^{\alpha}A_{\rho},$		(45)
	$\displaystyle G_{s\rho}=\Omega_{\rho\alpha}F_{s}^{\;\alpha}-\Omega_{s\alpha}F_{\rho}^{\;\alpha}-\Omega_{\alpha\beta}x^{\beta}\partial^{\alpha}F_{s\rho},$		(46)

where the anti-symmetric tensor $\Omega^{\mu\sigma}=\left[d\Lambda_{\epsilon}^{\mu\sigma}/d\epsilon\right]_{0}$ is the Lie algebra element of the Lorentz group. Substituting Eqs. (44)-(46) into Eq. (25), we obtain the angular momentum conservation law in 4D spacetime,

$$\Omega_{\nu\sigma}D_{\mu}\left\{x^{\sigma}T_{\text{N}}^{\mu\nu}+2E_{F}^{\left[\mu\nu\right]}\left(\mathcal{L}_{F}\right)A^{\sigma}+L^{\mu\nu\sigma}\right\}=0,$$

(47)

which can be rewritten as

$$D_{\mu}\left\{\left[x^{\sigma}T_{\text{N}}^{\mu\nu}-x^{\nu}T_{\text{N}}^{\mu\sigma}\right]+S^{\mu\nu\sigma}\right\}=0.$$

(48)

In above equations,

	$\displaystyle L^{\mu\nu\sigma}=\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}D_{\mu_{j+1}}\cdots D_{\mu_{i}}\left(F_{\rho}^{\;\sigma}\right)D_{\mu_{1}}\cdots D_{\mu_{j-1}}\times$
	$\displaystyle\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\rho\nu}\right)}-\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\nu\rho}\right)}\right]$
	$\displaystyle-\left[\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}D_{\mu_{j+1}}\cdots D_{\mu_{i}}\left(x^{\sigma}\partial^{\nu}F_{s\rho}\right)D_{\mu_{1}}\cdots D_{\mu_{j-1}}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{s\rho}\right)}-x^{\sigma}\Sigma^{\mu\nu}\right]$		(49)
	$\displaystyle\frac{1}{2}S^{\mu\nu\sigma}=E_{F}^{\left[\mu\nu\right]}\left(\mathcal{L}_{F}\right)A^{\sigma}-E_{F}^{\left[\mu\sigma\right]}\left(\mathcal{L}_{F}\right)A^{\nu}+\Delta^{\mu\nu\sigma},$		(50)
	$\displaystyle\Delta^{\mu\nu\sigma}\equiv 2L^{\mu\left[\nu\sigma\right]}$
	$\displaystyle=\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}D_{\mu_{j+1}}\cdots D_{\mu_{i}}\left\{\left(F_{\rho}^{\;\sigma}\right)D_{\mu_{1}}\cdots D_{\mu_{j-1}}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\rho\nu}\right)}\right.\right.$
	$\displaystyle\left.\vphantom{\left(F_{\rho}^{\;\sigma}\right)}-\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\nu\rho}\right)}\right]-\left(F_{\rho}^{\;\nu}\right)D_{\mu_{1}}\cdots D_{\mu_{j-1}}$
	$\displaystyle\left.\vphantom{\left(F_{\rho}^{\;\sigma}\right)}\times\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\rho\sigma}\right)}-\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{\sigma\rho}\right)}\right]\right\}$
	$\displaystyle-\sum_{i=1}^{n}\sum_{j=1}^{i}\left(-1\right)^{j+1}D_{\mu_{j+1}}\cdots D_{\mu_{i}}\left(x^{\sigma}\partial^{\nu}F_{s\rho}-x^{\nu}\partial^{\sigma}F_{s\rho}\right)D_{\mu_{1}}\cdots D_{\mu_{j-1}}\times$
	$\displaystyle\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu_{1}}\cdots\partial_{\mu_{j-1}}\partial_{\mu}\partial_{\mu_{j+1}}\cdots\partial_{\mu_{i}}F_{s\rho}\right)}+x^{\sigma}\Sigma^{\mu\nu}-x^{\nu}\Sigma^{\mu\sigma},$		(51)

where the superscript $\left[\mu\nu\right]$ denotes anti-symmetrization with respect to $\mu$ and $\nu$.

The anti-symmetric BR super-potentail $\mathcal{S}^{\sigma\mu\nu}$ is defined from the tensor $S^{\sigma\mu\nu}$ in Eq. (48) as (Belinfante, 1939, 1940; Rosenfeld, 1940)

$\displaystyle\mathcal{S}^{\sigma\mu\nu}$

$\displaystyle\equiv\frac{1}{2}\left[S^{\sigma\nu\mu}-S^{\mu\nu\sigma}-S^{\nu\mu\sigma}\right].$

(52)

It is clear from Eqs. (49)-(51) that $\mathcal{S}^{\sigma\mu\nu}$ and $\mathcal{F}^{\sigma\mu\nu}$ are related as follows,

$$\mathcal{S}^{\sigma\mu\nu}=\mathcal{F}^{\sigma\mu\nu}+\frac{1}{2}\left[\Delta^{\sigma\nu\mu}-\Delta^{\mu\nu\sigma}-\Delta^{\nu\mu\sigma}\right].$$

(53)

Equation (52) shows that in general $\mathcal{F}^{\sigma\mu\nu}$ is different from $\mathcal{S}^{\sigma\mu\nu}$ when $\Delta^{\sigma\nu\mu}$ is non-vanishing. For a first-order field theory, such as the standard Maxwell system (6), $n=1$ and the last three terms vanish such that $\mathcal{S}^{\sigma\mu\nu}=\mathcal{F}^{\sigma\mu\nu}$. In this situation, adding $T_{0}^{\mu\nu}\equiv D_{\sigma}\mathcal{F}^{\sigma\mu\nu}$ to $T_{\text{N}}^{\mu\nu}$ will render it both symmetric and gauge-symmetric, and the method developed here can be used as a simpler procedure to calculate the BR super-potential $\mathcal{S}^{\sigma\mu\nu}$ without the necessity to calculate the angular momentum tensor in 4D spacetime.

As an example of high-order electromagnetic field theory, we consider the Podolsky system (Bopp, 1940; Podolsky, 1942), which was proposed to study the radiation reaction of classical charged particles. The Podolsky Lagrangian density is

$$\mathcal{L}_{\text{Po}}=\frac{1}{8\pi}\left\{\bm{E}^{2}-\bm{B}^{2}+a^{2}\left[\left(\bm{\nabla}\cdot\bm{E}\right)^{2}-\left(\bm{\nabla}\times\bm{B}-\frac{1}{c}\partial_{t}\bm{E}\right)^{2}\right]\right\}$$

(54)

or in a manifestly covariant form

$$\mathcal{L}_{\text{Po}}=-\frac{1}{16\pi}F_{\sigma\rho}F^{\sigma\rho}-\frac{a^{2}}{8\pi}\partial_{\sigma}F^{\sigma\lambda}\partial^{\rho}F_{\rho\lambda}.$$

(55)

We substitute the Lagrangian density (55) into Eq. (34) to obtain the canonical EMT

	$\displaystyle 4\pi T_{\text{N}}^{\mu\nu}=\left(-\frac{1}{4}F_{\sigma\rho}F^{\sigma\rho}-\frac{a^{2}}{2}\partial_{\sigma}F^{\sigma\lambda}\partial^{\rho}F_{\rho\lambda}\right)\eta^{\mu\nu}$
	$\displaystyle+\left[F^{\mu\sigma}-a^{2}\left(\partial^{\mu}\partial_{\lambda}F^{\lambda\sigma}-\partial^{\sigma}\partial_{\lambda}F^{\lambda\mu}\right)\right]\partial^{\nu}A_{\sigma}+a^{2}\left(\partial^{\nu}F_{\;\rho}^{\mu}\right)\left(\partial_{\sigma}F^{\sigma\rho}\right),$		(56)

where use is made of the following equations,

	$\displaystyle\frac{\partial}{\partial\left(\partial_{\sigma}F_{\mu\nu}\right)}\left[\partial_{\alpha}F^{\alpha\lambda}\partial^{\rho}F_{\rho\lambda}\right]=2\eta^{\sigma\mu}\partial_{\lambda}F^{\lambda\nu},$		(57)
	$\displaystyle D_{\sigma}\frac{\partial}{\partial\left(\partial_{\sigma}F_{\mu\nu}\right)}\left[\partial_{\alpha}F^{\alpha\lambda}\partial^{\rho}F_{\rho\lambda}\right]=2\partial^{\mu}\partial_{\sigma}F^{\sigma\nu},$		(58)
	$\displaystyle E_{F}^{\mu\sigma}=\frac{\partial\mathcal{L}_{\text{Po}}}{\partial F_{\mu\sigma}}-D_{\rho}\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\partial_{\rho}F_{\mu\sigma}}=-\frac{1}{8\pi}F^{\mu\sigma}+\frac{a^{2}}{4\pi}\partial^{\mu}\partial_{\lambda}F^{\lambda\sigma},$		(59)
	$\displaystyle 2E_{F}^{\left[\mu\sigma\right]}=-\frac{1}{4\pi}F^{\mu\sigma}+\frac{a^{2}}{4\pi}\left(\partial^{\mu}\partial_{\lambda}F^{\lambda\sigma}-\partial^{\sigma}\partial_{\lambda}F^{\lambda\mu}\right),$		(60)
	$\displaystyle\Sigma^{\mu\nu}=\left(\partial^{\nu}F_{\sigma\rho}\right)\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\left(\partial_{\mu}F_{\sigma\rho}\right)}=-\frac{a^{2}}{4\pi}\partial^{\nu}F_{\sigma\rho}\left[\eta^{\mu\sigma}\partial_{\lambda}F^{\lambda\rho}\right]=-\frac{a^{2}}{4\pi}\left(\partial^{\nu}F_{\;\rho}^{\mu}\right)\left(\partial_{\sigma}F^{\sigma\rho}\right).$		(61)

The displacement-potential tensor is

$$\mathcal{F}^{\mu\nu\sigma}\equiv\frac{1}{4\pi}\mathcal{D}^{\sigma\mu}A^{\nu}=\frac{1}{4\pi}\left[-F^{\mu\sigma}+a^{2}\left(\partial^{\mu}\partial_{\lambda}F^{\lambda\sigma}-\partial^{\sigma}\partial_{\lambda}F^{\lambda\mu}\right)\right]A^{\nu},$$

(62)

and

$$4\pi\partial_{\sigma}\mathcal{F}^{\mu\nu\sigma}=\left[-F^{\mu\sigma}+a^{2}\left(\partial^{\mu}\partial_{\lambda}F^{\lambda\sigma}-\partial^{\sigma}\partial_{\lambda}F^{\lambda\mu}\right)\right]\partial_{\sigma}A^{\nu}.$$

(63)

Adding Eq. (63) to Eq. (56), we obtain the gauge-symmetric EMT,

	$\displaystyle 4\pi T_{\text{GS}}^{\mu\nu}=\left[F^{\mu\sigma}F_{\nu\sigma}-\frac{1}{4}\left(F_{\sigma\rho}F^{\sigma\rho}\right)\eta^{\mu\nu}\right]-\frac{a^{2}}{2}\left(\partial_{\sigma}F^{\sigma\lambda}\partial^{\rho}F_{\rho\lambda}\right)\eta^{\mu\nu}$
	$\displaystyle-a^{2}F_{\;\sigma}^{\nu}\left(\partial^{\mu}\partial_{\rho}F^{\rho\sigma}\right)+a^{2}F_{\;\sigma}^{\nu}\left(\partial^{\sigma}\partial_{\rho}F^{\rho\mu}\right)+a^{2}\left(\partial^{\nu}F_{\;\rho}^{\mu}\right)\left(\partial_{\sigma}F^{\sigma\rho}\right).$		(64)

It is easy to see that $T_{\text{GS}}^{\mu\nu}$ for the Podolsky system is not symmetric, i.e., $T_{\text{GS}}^{\mu\nu}\neq T_{\text{GS}}^{\nu\mu}$.

To calculate the BR EMT $T_{\text{BR}}^{\mu\nu}$ for the Podolsky system, we evaluate the $\Delta^{\mu\nu\sigma}$ term in Eq. (52). Using Eq. (51), we have

	$\displaystyle\Delta^{\mu\nu\sigma}$	$\displaystyle=F_{\rho}^{\;\sigma}\left[\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\left(\partial_{\mu}F_{\rho\nu}\right)}-\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\left(\partial_{\mu}F_{\nu\rho}\right)}\right]-F_{\rho}^{\;\nu}\left[\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\left(\partial_{\mu}F_{\rho\sigma}\right)}-\frac{\partial\mathcal{L}_{\text{Po}}}{\partial\left(\partial_{\mu}F_{\sigma\rho}\right)}\right]$
		$\displaystyle=-\frac{a^{2}}{4\pi}\left[F^{\mu\sigma}\partial_{\rho}F^{\rho\nu}-\eta^{\mu\nu}\left(F_{\rho}^{\;\sigma}\partial_{\lambda}F^{\lambda\rho}\right)-F^{\mu\nu}\partial_{\rho}F^{\rho\sigma}+\eta^{\mu\sigma}F_{\rho}^{\;\nu}\left(\partial_{\lambda}F^{\lambda\rho}\right)\right].$		(65)

Substituting Eqs. (52) and (65) into Eq. (1), we obtain the BR EMT as

	$\displaystyle 4\pi T_{\text{BR}}^{\mu\nu}=4\pi T_{\text{GS}}^{\mu\nu}+2\pi\left[\Delta^{\sigma\nu\mu}-\Delta^{\mu\nu\sigma}-\Delta^{\nu\mu\sigma}\right]$
	$\displaystyle=\left[F^{\mu\sigma}F_{\nu\sigma}-\frac{1}{4}\left(F_{\sigma\rho}F^{\sigma\rho}\right)\eta^{\mu\nu}\right]+\frac{a^{2}}{2}\left[\left(\partial_{\sigma}F^{\sigma\rho}\right)\left(\partial^{\lambda}F_{\lambda\rho}\right)-2F_{\rho}^{\;\sigma}\left(\partial_{\sigma}\partial_{\lambda}F^{\lambda\rho}\right)\right]\eta^{\mu\nu}$
	$\displaystyle+a^{2}\left[F^{\nu\sigma}\left(\partial_{\sigma}\partial_{\rho}F^{\rho\mu}\right)+F^{\mu\sigma}\left(\partial_{\sigma}\partial_{\rho}F^{\rho\nu}\right)-\left(\partial_{\sigma}F^{\sigma\mu}\right)\left(\partial_{\rho}F^{\rho\nu}\right)\right.$
	$\displaystyle\left.\vphantom{\left(\partial_{\rho}F^{\rho\nu}\right)}-F_{\;\sigma}^{\nu}\left(\partial^{\mu}\partial_{\rho}F^{\rho\sigma}\right)-F_{\;\sigma}^{\mu}\left(\partial^{\nu}\partial_{\rho}F^{\rho\sigma}\right)\right].$		(66)

It is easy to verify that $T_{\text{BR}}^{\mu\nu}$ for the Podolsky system is both symmetric and gauge-symmetric.

The Lagrangian density of a generic classical electromagnetic field-charge particle system assumes the form of

	$\displaystyle\mathcal{L}$	$\displaystyle=\sum_{a}\mathcal{L}_{a}+\mathcal{L}_{F},$		(69)
	$\displaystyle\mathcal{L}_{a}$	$\displaystyle=L_{a}\delta_{a},\quad L_{a}=L_{a}\left(x^{\mu},\bm{X}_{a},\dot{\bm{X}}_{a};\varphi,\bm{A},\bm{E},\bm{B},D\bm{E},D\bm{B},\cdots,D^{n}\bm{E},D^{n}\bm{B}\right)$		(70)

where the subscript $a$ labels particles, $\bm{X}_{a}$ is its trajectory, $\mathcal{L}_{a}$ is its Lagrangian density, and $\delta_{a}\equiv\delta\left(\bm{x}-\bm{X}_{a}\right)$. Here, $\delta(x)$ is the Dirac $\delta$-function.

In the “3+1” form, the equations of motion for the electromagnetic field are

	$\displaystyle\nabla\cdot\bm{E}_{\bm{E}}\left(\mathcal{L}\right)=-\frac{\partial\mathcal{L}}{\partial\varphi},$		(71)
	$\displaystyle-\frac{1}{c}\frac{\partial}{\partial t}\left[\bm{E}_{\bm{E}}\left(\mathcal{L}\right)\right]-\bm{\nabla}\times\left[\bm{E}_{\bm{B}}\left(\mathcal{L}\right)\right]=\frac{\partial\mathcal{L}}{\partial\bm{A}}.$		(72)

In this study, it is assumed that the Lagrangian density $\mathcal{L}$ is linear in terms of $\varphi$ and $\bm{A}$, and Eqs. (71) and (72) are thus gauge symmetric. Specifically, we assume that $\mathcal{L}$ depends on $\varphi$ and $\bm{A}$ only through the term $-q_{a}\delta_{a}(\varphi+\bm{A}\cdot\dot{\bm{X}}_{a}/c)$, i.e., the Lagrangian density can be written as

$$\mathcal{L}=\sum_{a}\left[-\varphi+\frac{1}{c}\bm{A}\cdot\dot{\bm{X}}_{a}\right]q_{a}\delta_{a}+\text{GSP}\left(\mathcal{L}\right),$$

(73)

where “$\text{GSP}\left(\mathcal{L}\right)$” denotes the gauge-symmetric parts of the Lagrangian density $\mathcal{L}$. The right hand side of Eqs. (71) and (72) are the “3+1” form of Eq. (15), the free charge density $\rho_{f}$ and current density $\bm{j}_{f}$, respectively. Using Eq. (73), we have

$$\rho_{f}=-\frac{\partial\mathcal{L}}{\partial\varphi}=\sum_{a}q_{a}\delta_{a},\quad\bm{j}_{f}=c\frac{\partial\mathcal{L}}{\partial\bm{A}}=\sum_{a}q_{a}\dot{\bm{X}}_{a}\delta_{a}.$$

(74)