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## Relating the canonical energy momentum tensor to the Lagrangian gradient.

Posted by peeterjoot on September 12, 2009

[Click here for a PDF of this sequence of posts with nicer formatting]

In [4] many tensor quantities are not written in index form, but instead using a vector notation. In particular, the symmetric energy momentum tensor is expressed as

\begin{aligned}T(a) = -\frac{\epsilon_0}{2} F a F \end{aligned} \quad\quad\quad(25)

where the usual tensor form following by taking dot products with $\gamma^\mu$ and substituting $a = \gamma^\nu$. The conservation equation for the canonical energy momentum tensor of (23) can be put into a similar vector form

\begin{aligned}T(a) &= \gamma_\alpha \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} (a \cdot \nabla) A^\beta - a \mathcal{L} \\ 0 &= \nabla \cdot T(a) \end{aligned} \quad\quad\quad(26)

The adjoint $\bar{T}$ of the tensor can be calculated from the definition

\begin{aligned}\nabla \cdot T(a) = a \cdot \bar{T}(\nabla) \end{aligned} \quad\quad\quad(28)

Somewhat unintuitively, this is a function of the gradient. Playing around with factoring out the displacement vector $a$ from (26) that the energy momentum adjoint essentially provides an expansion of the gradient of the Lagrangian. To prepare, let’s introduce some helper notation

\begin{aligned}\Pi_\beta \equiv \gamma_\alpha \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \end{aligned} \quad\quad\quad(29)

With this our Noether current equation becomes

\begin{aligned}\nabla \cdot T(a) &= \left\langle{{ \nabla T(a) }}\right\rangle \\ &= \left\langle{{ \nabla (\Pi_\beta (a \cdot \nabla) A^\beta - a \nabla \mathcal{L} ) }}\right\rangle \\ &= \left\langle{{ \nabla \left(\frac{1}{{2}} \Pi_\beta (a (\nabla A^\beta) + (\nabla A^\beta) a) - a \mathcal{L} \right) }}\right\rangle \\ \end{aligned}

Cyclic permutation of the vector products $\left\langle{{a b c}}\right\rangle = \left\langle{{ c a b}}\right\rangle$ can be used in the scalar selection. This is a little more tractable with some helper notation for the $A^\beta$ gradients, say $v^\beta = \nabla A^\beta$. Because of the operator nature of the gradient once the vector order is permuted we have to allow for the gradient to act left or right or both, so arrows are used to disambiguate this where appropriate.

\begin{aligned}\nabla \cdot T(a) &= \left\langle{{ \nabla \left(\frac{1}{{2}} \Pi_\beta a v^\beta +\Pi_\beta v^\beta a \right) - \nabla \mathcal{L} a }}\right\rangle \\ &= \left\langle{{ \left( \frac{1}{{2}} v^\beta \stackrel{ \leftrightarrow }\nabla \Pi_\beta \frac{1}{{2}} \nabla (\Pi_\beta v^\beta)- \nabla \mathcal{L} \right) a }}\right\rangle \\ &=a \cdot \left( \frac{1}{{2}} {\left\langle{{ v^\beta \stackrel{ \leftrightarrow }\nabla \Pi_\beta + \nabla (\Pi_\beta v^\beta) }}\right\rangle}_{1} - \nabla \mathcal{L} \right) \end{aligned}

This dotted with quantity is the adjoint of the canonical energy momentum tensor

\begin{aligned}\bar{T}(\nabla) &=\frac{1}{{2}} {\left\langle{{ v^\beta \stackrel{ \leftrightarrow }\nabla \Pi_\beta + \nabla (\Pi_\beta v^\beta) }}\right\rangle}_{1} - \nabla \mathcal{L} \end{aligned} \quad\quad\quad(30)

This can however, be expanded further. First tackling the
bidirectional gradient vector term we can utilize the property that the reverse of a vector leaves the vector unchanged. This gives us

\begin{aligned}{\left\langle{{ v^\beta \stackrel{ \leftrightarrow }\nabla \Pi_\beta }}\right\rangle}_{1}&={\left\langle{{ v^\beta (\stackrel{ \rightarrow }\nabla \Pi_\beta) }}\right\rangle}_{1}+{\left\langle{{ (v^\beta \stackrel{ \leftarrow }\nabla) \Pi_\beta }}\right\rangle}_{1} \\ &={\left\langle{{ v^\beta (\stackrel{ \rightarrow }\nabla \Pi_\beta) }}\right\rangle}_{1}+{\left\langle{{ \Pi_\beta (\stackrel{ \rightarrow }\nabla v^\beta) }}\right\rangle}_{1} \\ \end{aligned}

In the remaining term, using the Hestenes overdot notation clarify the scope of the operator, we have

\begin{aligned}\bar{T}(\nabla) &=\frac{1}{{2}} \left({\left\langle{{ v^\beta (\nabla \Pi_\beta) }}\right\rangle}_{1}+{\left\langle{{ \Pi_\beta (\nabla v^\beta) }}\right\rangle}_{1} +{\left\langle{{ (\nabla \Pi_\beta) v^\beta }}\right\rangle}_{1} + {\left\langle{{ \nabla' \Pi_\beta {v^\beta}'}}\right\rangle}_{1} \right)- \nabla \mathcal{L} \\ \end{aligned}

The grouping of the first and third terms above simplifies nicely

\begin{aligned}\frac{1}{{2}}{\left\langle{{ v^\beta (\nabla \Pi_\beta) }}\right\rangle}_{1} +\frac{1}{{2}} {\left\langle{{ (\nabla \Pi_\beta) v^\beta }}\right\rangle}_{1} &=v^\beta (\nabla \cdot \Pi_\beta) +\frac{1}{{2}} {\left\langle{{ v^\beta (\nabla \wedge \Pi_\beta) }}\right\rangle}_{1} +{\left\langle{{ (\nabla \wedge \Pi_\beta) v^\beta }}\right\rangle}_{1} \\ \end{aligned}

Since $a (b \wedge c) + (b \wedge c) a = 2 a \wedge b \wedge c$, which is purely a trivector, the vector grade selection above is zero. This leaves the adjoint reduced to

\begin{aligned}\bar{T}(\nabla) &=v^\beta (\nabla \cdot \Pi_\beta) +\frac{1}{{2}} \left({\left\langle{{ \Pi_\beta (\nabla v^\beta) }}\right\rangle}_{1} + {\left\langle{{ \nabla' \Pi_\beta {v^\beta}'}}\right\rangle}_{1} \right)- \nabla \mathcal{L} \\ \end{aligned}

For the remainder vector grade selection operators we have something that is of the following form

\begin{aligned}\frac{1}{{2}} {\left\langle{{ a b c + b a c }}\right\rangle}_{1} = (a \cdot b ) c \end{aligned}

And we are finally able to put the adjoint into a form that has no remaining grade selection operators

\begin{aligned}\bar{T}(\nabla)&= (\nabla A^\beta) (\nabla \cdot \Pi_\beta) +(\Pi_\beta \cdot \nabla) (\nabla A^\beta) -\nabla \mathcal{L} \\ &= (\nabla A^\beta) (\stackrel{ \rightarrow }\nabla \cdot \Pi_\beta) +(\nabla A^\beta) (\stackrel{ \leftarrow }\nabla \cdot \Pi_\beta) -\nabla \mathcal{L} \\ &= (\nabla A^\beta) (\stackrel{ \leftrightarrow }\nabla \cdot \Pi_\beta) -\nabla \mathcal{L} \end{aligned}

Recapping, we have for the tensor and its adjoint

\begin{aligned}0 &= \nabla \cdot T(a) = a \cdot \bar{T}(\nabla) \\ \Pi_\beta &\equiv \gamma_\alpha \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \\ T(a) &= \Pi_\beta (a \cdot \nabla) A^\beta - a \nabla \mathcal{L} \\ \bar{T}(\nabla) &= (\nabla A^\beta) (\stackrel{ \leftrightarrow }\nabla \cdot \Pi_\beta) - \nabla \mathcal{L} \end{aligned} \quad\quad\quad(31)

For the adjoint, since $a \cdot \bar{T}(\nabla) = 0$ for all $a$, we must also have $\bar{T}(\nabla) = 0$, which means the adjoint of the canonical energy momentum tensor really provides not much more than a recipe for computing the Lagrangian gradient

\begin{aligned}\nabla \mathcal{L} &= (\nabla A^\beta) (\stackrel{ \leftrightarrow }\nabla \cdot \Pi_\beta) \end{aligned} \quad\quad\quad(35)

Having seen the adjoint notation, it was natural to see what this was for a multiple scalar field variable Lagrangian, even if it is not intrinsically useful. Observe that the identity (35), obtained so laboriously, is not more than syntactic sugar for the chain rule expansion of the Lagrangian partials (plus application of the Euler-Lagrange field equations). We could obtain this directly if desired much more easily than by factoring out $a$ from $\nabla \cdot T(a) = 0$.

\begin{aligned}\partial_\mu \mathcal{L}&=\frac{\partial {\mathcal{L}}}{\partial {A^\beta}} \partial_\mu A^\beta+\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\mu \partial_\alpha A^\beta \\ &=\left( \partial_\alpha \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) \partial_\mu A^\beta+\frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \partial_\alpha \partial_\mu A^\beta \\ &=\partial_\alpha\left(\left( \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) \partial_\mu A^\beta\right) \\ \end{aligned}

Summing over $\mu$ for the gradient, this reproduces (35), with much less work

\begin{aligned}\nabla \mathcal{L} &= \gamma^\mu \partial_\mu \mathcal{L} \\ &=\partial_\alpha\left(\left( \frac{\partial {\mathcal{L}}}{\partial {(\partial_\alpha A^\beta)}} \right) (\nabla A^\beta)\right) \\ &=(\Pi_\beta \cdot \stackrel{ \leftrightarrow }\nabla) (\nabla A^\beta) \end{aligned}

Observe that the Euler-Lagrange field equations are implied in this relationship, so perhaps it has some utility. Also note that while it is simpler to directly compute this, without having started with the canonical energy momentum tensor, we would not know how the two of these were related.

# References

[4] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.