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I am trying to work out the symmetric stress-energy tensor for a free massive vector field and show that it is conserved. The Lagrangian density and resulting EoM are:

$$L=-\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}+\frac{1}{2}m^2 A^{\beta}A_{\beta}$$ $$\partial_\mu F^{\mu\nu}+m^2 A^\nu=0$$

Where I am using the metric $\eta_{\mu\nu}=diag(+1,-1,-1,-1)$. Evaluating the usual expression for the canonical stress energy tensor:

$$T_C^{\mu \nu}=\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}-\eta^{\mu \nu}L$$ $$\frac{\partial L}{\partial \left(\partial_{\mu} A_{\lambda}\right)}=F^{\lambda\mu}$$ $$T_C^{\mu \nu}=F^{\lambda\mu}\eta^{\nu\beta}\partial_{\beta} A_{\lambda}+\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$

According to what I can find on this site and elsewhere, the symmetric stress energy should be (adding the mass term to the form found here):

$$T^{\mu\nu}=\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta}\eta^{\mu \nu}+F^{\mu\lambda}\eta_{\lambda\kappa}F^{\kappa\nu}-\frac{1}{2}m^2 A^{\beta}A_{\beta}\eta^{\mu \nu}$$

When I calculate the difference between these I get:

$$T^{\mu\nu}-T_C^{\mu \nu}=S^{\mu\nu}=-F^{\beta\nu}\partial_\beta A^\mu$$

It should be that $\partial_\mu S^{\mu\nu}=0$ so that $T^{\mu\nu}$ is conserved just as $T_C^{\mu\nu}$ is. But I calculate:

$$\partial_\mu S^{\mu\nu}=-\left( \partial_\mu F^{\beta\nu}\right) \partial_\beta A^\mu-F^{\beta\nu}\partial_\beta \left( \partial_\mu A^\mu\right)$$

The second term is zero since $\partial_\mu A^\mu=0$ by the EoM. But I can't figure out how to show that the first term is zero, nor can I see that $\partial_\mu T^{\mu\nu}=0$ when I try to calculate it directly. Where am I going wrong?

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    $\begingroup$ Minimally couple the Lagrangian and consider taking the derivative with respect to the metric, and then finally substituting the Minkowski metric. $\endgroup$
    – JamalS
    Mar 4, 2020 at 17:41
  • $\begingroup$ @JamalS Could you provide an answer showing how to do that? I am not familiar with how to take the derivative with respect to the metric. $\endgroup$ Mar 4, 2020 at 17:57
  • $\begingroup$ The conservation law is $\partial_\nu T^{\mu\nu} = 0$, not $\partial_\mu T^{\mu\nu} = 0$. $\endgroup$
    – Prahar
    Mar 5, 2020 at 22:22

2 Answers 2

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  1. The Lagrangian density for massive E&M is $${\cal L}~=~\sqrt{|g|}L, \qquad L~=~-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\mp \frac{1}{2}m^2 A^2, \tag{1}$$ for Minkowski sign convention $(\mp,\pm,\pm,\pm)$. The Euler-Lagrange (EL) equations read $$ d_{\mu}F^{\mu\nu}~\stackrel{(1)}{\approx}~\pm m^2A^{\nu} .\tag{2}$$

  2. The mass term breaks gauge-invariance but not translational invariance, so that the canonical stress-energy-momentum (SEM) tensor $$\mp T_C^{\mu}{}_{\nu}~:=~\frac{\partial L}{\partial(\partial_{\mu}A_{\lambda})}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L ~\stackrel{(1)}{=}~F^{\lambda\mu}\partial_{\nu}A_{\lambda}-\delta^{\mu}_{\nu}L\tag{3}$$ is conserved cf. Noether's first theorem

  3. The Hilbert/metric SEM tensor is $$ \mp T_{\mu\nu}~:=~\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}} ~\stackrel{(1)}{=}~- F_{\mu}{}^{\lambda}F_{\nu\lambda}\mp m^2A_{\mu}A_{\nu} - g_{\mu\nu}L.\tag{4}$$ In eq. (4) it is important that we think of $A$ as a co-vector/one-form rather than a vector.

  4. The difference becomes $$S^{\mu}{}_{\nu}~:=~T^{\mu}{}_{\nu}-T_C^{\mu}{}_{\nu} ~\stackrel{(3(+(4)}{=}~\mp F^{\mu\lambda}d_{\lambda}A_{\nu}+ m^2A^{\mu}A_{\nu} ~\stackrel{(2)}{\approx}~\mp d_{\lambda}(F^{\mu\lambda}A_{\nu}),\tag{5}$$ so that $$d_{\mu}S^{\mu}{}_{\nu}~\stackrel{(5)}{\approx}~\mp d_{\mu}d_{\lambda}(F^{\mu\lambda}A_{\nu})~=~0,\tag{6}$$ by (anti)symmetry of indices $\mu\leftrightarrow\lambda$. Hence the conservation law for the two SEM tensors (3) & (4) are equivalent, cf. OP's question.

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  • $\begingroup$ Thanks! To take the derivative with respect to the metric in finding the Hilbert SEM tensor, do you just have to lower all the indicies on the A field so that there is an instance of the metric for each contraction in the Lagrangian? $\endgroup$ Mar 5, 2020 at 18:21
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    $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Mar 5, 2020 at 18:51
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The Lagrangian density, based on your Lagrangian, $L$ is $$𝔏 = \sqrt{|g|} \left(-\frac 1 4 F^{μν} F_{μν} + \frac 1 2 m^2 A^μ A_μ\right).$$ Denoting the derivatives, respectively by the following densities: $$𝔊^{μν} = -\frac{∂𝔏}{∂F_{μν}}, \hspace 1em 𝔍^μ = \frac{∂𝔏}{∂A_μ},$$ which are, respectively, the tensor densities for the response field and current, we have, as the Euler-Lagrange equations: $$∂_ν 𝔊^{μν} = 𝔍^μ.$$ From your Lagrangian, we find: $$𝔊^{μν} = \sqrt{|g|} F^{μν}, \hspace 1em 𝔍^μ = \sqrt{|g|} m^2 A^μ.$$ The extra factor of 2 for $𝔊^{μν}$ arises from the double-counting that comes from $F_{μν} = -F_{νμ}$.

I'll denote the canonical stress tensor density by $𝔓^ρ_ν$. In terms of the Lagrangian density, it is defined by: $$𝔓^ρ_ν = \frac{∂𝔏}{∂\left(∂_ρ A_μ\right)} ∂_ν A_μ - δ^ρ_ν 𝔏.$$ The derivative with respect to the gradient $∂_ρ A_μ$ matches the response field: $$\frac{∂𝔏}{∂\left(∂_ρ A_μ\right)} = -𝔊^{ρμ}.$$ Therefore, $$𝔓^ρ_ν = -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏.$$

The stress tensor density, denoted here by $𝔗^ρ_ν$, is given in terms of the canonical stress tensor density by a Belinfante correction: $$𝔗^ρ_ν = 𝔓^ρ_ν + ∂_μ 𝔭^{ρμ}_ν.$$ The tensor density involved in this addition is anti-symmetric $𝔭^{ρμ}_ν = -𝔭^{μρ}_ν$, which ensures that both stress tensor densities have the same divergence. It is determined, ultimately, by the behavior of $A_μ$ under Lorentz transforms and works out to the following expression: $𝔭^{ρμ}_ν = 𝔊^{ρμ} A_ν$. Taking the divergence with respect to $μ$ and ... applying the Euler-Lagrange equation for the response field and source ... we get: $$\begin{align} ∂_μ 𝔭^{ρμ}_ν &= ∂_μ 𝔊^{ρμ} A_ν + 𝔊^{ρμ} ∂_μ A_ν \\ &= 𝔍^ρ A_ν + 𝔊^{ρμ} ∂_μ A_ν. \end{align}$$ Finally, upon substitution, we get: $$\begin{align} 𝔗^ρ_ν &= -𝔊^{ρμ} ∂_ν A_μ - δ^ρ_ν 𝔏 + 𝔍^ρ A_ν + 𝔊^{ρμ} ∂_μ A_ν \\ &= 𝔊^{ρμ} F_{μν} + 𝔍^ρ A_ν - δ^ρ_ν 𝔏, \end{align}$$ where the equation $F_{μν} = ∂_μ A_ν - ∂_ν A_μ$ is used.

Upon substitution of the tensor densities for the response and source fields and Lagrangian, with a little index-switching, we get: $$𝔗^ρ_ν = \sqrt{|g|} \left(F^{ρμ} F_{μν} + m^2 A^ρ A_ν - δ^ρ_ν \left(-\frac 1 4 F^{αβ} F_{αβ} + \frac 1 2 m^2 A^α A_α\right)\right).$$ When de-densitized and index-lowered with the metric $g_{μρ}$ to the symmetric stress tensor $T_{μν} = g_{μρ} 𝔗^ρ_ν/\sqrt{|g|}$, the result (with a little more index-switching) is: $$T_{μν} = F_{μα} g^{αβ} F_{βν} + m^2 A_μ A_ν - g_{μν} \left(-\frac 1 4 F^{αβ} F_{αβ} + \frac 1 2 m^2 A^α A_α\right).$$

Most of the answer present is independent of the details of your Lagrangian, and has been addressed generically. The continuity equation for the stress tensor density can be addressed generically by using the Euler-Lagrange field equation, along with the continuity equation for the source density (derived from it): $∂_μ 𝔍^μ = ∂_μ ∂_ν 𝔊^{μν} = 0$, that follows due to the anti-symmetry of $𝔊$: $$\begin{align} ∂_ρ 𝔗^ρ_ν &= ∂_ρ 𝔊^{ρμ} F_{μν} + 𝔊^{ρμ} ∂_ρ F_{μν} + ∂_ρ 𝔍^ρ A_ν + 𝔍^ρ ∂_ρ A_ν - ∂_ν 𝔏 \\ &= -𝔍^μ F_{μν} + 𝔊^{ρμ} ∂_ρ F_{μν} + 𝔍^ρ ∂_ρ A_ν - ∂_ν 𝔏 \\ &= -𝔍^μ \left(∂_μ A_ν - ∂_ν A_μ\right) + \frac 1 2 𝔊^{ρμ} \left(∂_ρ F_{μν} - ∂_μ F_{ρν}\right) + 𝔍^μ ∂_μ A_ν - ∂_ν 𝔏 \end{align}$$

Noting that $$∂_ρ F_{μν} - ∂_μ F_{ρν} = -∂_ν F_{ρμ},$$ which follows from $F_{μν} = ∂_μ A_ν - ∂_ν A_μ$, we get: $$\begin{align} ∂_ρ 𝔗^ρ_ν &= 𝔍^μ ∂_ν A_μ - \frac 1 2 𝔊^{ρμ} ∂_ν F_{ρμ} - ∂_ν 𝔏 \\ &= \frac{∂𝔏}{∂A_μ} ∂_ν A_μ + \frac 1 2 \frac{∂𝔏}{∂F_{ρμ}} ∂_ν F_{ρμ} - ∂_ν 𝔏. \end{align}$$ Again, noting the double-counting for the derivatives with respect to $F_{ρμ}$, the first two terms add up to the result of applying the chain rule to $∂_ν 𝔏$ - provided that $𝔏$ has no explicit coordinate dependence (such as: from external fields) - so that the divergence cancels.

You're using a constant metric, so there is no explicit coordinate-dependence in your Lagrangian. Otherwise, the dependence on the metric would have to be accounted for and fixed and - as usual - this means you have the Einstein-Hilbert stress tensor density $𝔍_{μν} = -2 ∂𝔏/∂g^{μν}$, this will match the tensor density we just derived, with $𝔍_{μν} = g_{μρ} 𝔍^ρ_ν$ and you'll need to compensate by adding an extra term for $g$: $𝔏 = (⋯) + k \sqrt{|g|} R$, where $R$ is the curvature scalar, for some constant $k$, take its derivative, as well, to get $k \sqrt{|g|} G_{μν}$, where $G_{μν}$ is the Einstein tensor, and you get the Einstein field equation for this stress tensor density.

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