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

## Conservation of the field momentum.

This is a follow up to the unanswered questions I had yesterday related to the apparent time dependent terms in the previous expansion of \( P^i \) for a scalar field.

It turns out that examining the reasons that we can say that the field momentum is conserved also sheds some light on the question. \( P^i \) is not an a-priori conserved quantity, but we may use the charge conservation argument to justify this despite it not having a four-vector nature (i.e. with zero four divergence.)

The momentum \( P^i \) that we have defined is related to the conserved quantity \( T^{0\mu} \), the energy-momentum tensor, which satisfies \( 0 = \partial_\mu T^{0\mu} \) by Noether’s theorem (this was the conserved quantity associated with a spacetime translation.)

That tensor was

\begin{equation}\label{eqn:momentum:120}

T^{\mu\nu} = \partial^\mu \phi \partial^\nu \phi – g^{\mu\nu} \LL,

\end{equation}

and can be used to define the momenta

\begin{equation}\label{eqn:momentum:140}

\begin{aligned}

\int d^3 x T^{0k}

&= \int d^3 x \partial^0 \phi \partial^k \phi \\

&= \int d^3 x \pi \partial^k \phi.

\end{aligned}

\end{equation}

Charge \( Q^i = \int d^3 x j^0 \) was conserved with respect to a limiting surface argument, and we can make a similar “beer can integral” argument for \( P^i \), integrating over a large time interval \( t \in [-T, T] \) as sketched in fig. 1. That is

\begin{equation}\label{eqn:momentum:160}

\begin{aligned}

0

&=

\partial_\mu \int d^4 x T^{0\mu} \\

&=

\partial_0 \int d^4 x T^{00}

+

\partial_k \int d^4 x T^{0k} \\

&=

\partial_0 \int_{-T}^T dt \int d^3 x T^{00}

+

\partial_k \int_{-T}^T dt \int d^3 x T^{0k} \\

&=

\partial_0 \int_{-T}^T dt \int d^3 x T^{00}

+

\partial_k \int_{-T}^T dt

\inv{2} \int \frac{d^3 p }{(2 \pi)^3} p^k

\lr{

a_\Bp^\dagger a_\Bp

+ a_\Bp a_\Bp^\dagger

– a_\Bp a_{-\Bp} e^{- 2 i \omega_\Bp t}

– a_\Bp^\dagger a_{-\Bp}^\dagger e^{2 i \omega_\Bp t}

} \\

&=

\int d^3 x \evalrange{T^{00}}{-T}{T}

+

T \partial_k

\int \frac{d^3 p }{(2 \pi)^3} p^k

\lr{

a_\Bp^\dagger a_\Bp

+ a_\Bp a_\Bp^\dagger

}

-\inv{2}

\partial_k \int_{-T}^T dt

\int \frac{d^3 p }{(2 \pi)^3} p^k

\lr{

a_\Bp a_{-\Bp} e^{- 2 i \omega_\Bp t}

+ a_\Bp^\dagger a_{-\Bp}^\dagger e^{2 i \omega_\Bp t}

}.

\end{aligned}

\end{equation}

The first integral can be said to vanish if the field energy goes to zero at the time boundaries, and the last integral reduces to

\begin{equation}\label{eqn:momentum:180}

\begin{aligned}

-\inv{2}

\partial_k \int_{-T}^T dt

\int \frac{d^3 p }{(2 \pi)^3} p^k

\lr{

a_\Bp a_{-\Bp} e^{- 2 i \omega_\Bp t}

+ a_\Bp^\dagger a_{-\Bp}^\dagger e^{2 i \omega_\Bp t}

}

&=

-\int \frac{d^3 p }{2 (2 \pi)^3} p^k

\lr{

a_\Bp a_{-\Bp} \frac{\sin( -2 \omega_\Bp T )}{-2 \omega_\Bp}

+ a_\Bp^\dagger a_{-\Bp}^\dagger \frac{\sin( 2 \omega_\Bp T )}{2 \omega_\Bp}

} \\

&=

-\int \frac{d^3 p }{2 (2 \pi)^3} p^k

\lr{

a_\Bp a_{-\Bp} + a_\Bp^\dagger a_{-\Bp}^\dagger

}

\frac{\sin( 2 \omega_\Bp T )}{2 \omega_\Bp}

.

\end{aligned}

\end{equation}

The \( \sin \) term can be interpretted as a sinc like function of \( \omega_\Bp \) which vanishes for large \( \Bp \). It’s not entirely sinc like for a massive field as \( \omega_\Bp = \sqrt{ \Bp^2 + m^2 } \), which never hits zero, as shown in fig. 2.

Vanishing for large \( \Bp \) doesn’t help the whole integral vanish, but we can resort to the Riemann-Lebesque lemma [1] instead and interpret this integral as one with a plain old high frequency oscillation that is presumed to vanish (i.e. the rest is well behaved enough that it can be labelled as \( L_1 \) integrable.)

We see that only the non-time dependent portion of \( \mathbf{P} \) matters from a conserved quantity point of view, and having killed off all the time dependent terms, we are left with a conservation relationship for the momenta \( \spacegrad \cdot \BP = 0 \), where \( \BP \) in normal order is just

\begin{equation}\label{eqn:momentum:200}

: \BP : = \int \frac{d^3 p}{(2 \pi)^3} \Bp a_\Bp^\dagger a_\Bp.

\end{equation}

# References

[1] Wikipedia contributors. Riemann-lebesgue lemma — Wikipedia, the free encyclopedia, 2018. URL https://en.wikipedia.org/w/index.php?title=Riemann%E2%80%93Lebesgue_lemma&oldid=856778941. [Online; accessed 29-October-2018].