Useful identities

Identities (1) and (3) are proved in general curvilinear coordinates—see Ref. [8]. For a less general (but more straightforward) treatment, these identities are also proved in index notation for orthonormal coordinates. Note that the outer product is defined as \((\vec{u} \otimes \vec{v}) \cdot \vec{w} = (\vec{v} \cdot \vec{w}) \vec{u}\).

1. \(\divergence(\vec{u}\otimes\vec{v})=(\gradient \vec{u})\vec{v}+ (\divergence \vec{v})\vec{u}\). This product rule is shown below in general curvilinear coordinates, where \(\vec{g}_k\) is a contravariant basis vector, and where \(_{,k}\) denotes differentiation with respect to the the \(k^\mathrm{th}\) coordinate in the basis, and where \(\vec{u} = u^k \vec{g}_k\) means \(\sum_{k=1}^3 u^k \vec{g}_k = u^1 \vec{g}_1 + u^2 \vec{g}_2 + u^3 \vec{g}_3\). Evaluating the left-hand side of the identity to be proved yields \begin{align*} \divergence(\vec{u}\otimes\vec{v})&= (\vec{u}\otimes\vec{v})_{,k}\cdot \vec{g}^k\tag{definition of divergence of tensor}\\ &= (\vec{u}_{,k}\otimes \vec{v} + \vec{u}\otimes \vec{v}_{,k}) \cdot \vec{g}^k \tag{product rule}\\ &= (\vec{u}_{,k}\otimes \vec{v}) \cdot \vec{g}^k + (\vec{u}\otimes \vec{v}_{,k}) \cdot \vec{g}^k \tag{distributive property}\\ &=(\vec{u}_{,k}\otimes\vec{g}^k) \cdot \vec{v} + (\vec{v}_{,k}\cdot \vec{g}^k) \vec{u} \tag{definition of outer product}\\ &=(\gradient\vec{u}) \cdot \vec{v} + (\divergence\vec{v})\vec{u}\,, \end{align*} where the last line holds by the definitions of the gradient and divergence of a vector. Note: This was a homework problem in Ref. [8].

   Below is a less general version of the proof, included for readers unfamiliar with Ricci calculus. In orthonormal index notation, the left-hand side of the identity to be proved in one line, by the product rule: \begin{align*} (u_iv_j)_{,j} &= u_{i,j} v_j + u_i v_{j,j}\,. \end{align*}

2. \(\langle \partial (fg)/\partial t\rangle = 0\) for time-harmonic \(f(t)\) and \(g(t)\). To show this, suppose \(f(t) = \cos \omega t\) and \(g(t) = \cos(\omega t + \phi)\). Then, since \([\cos(a+b) + \cos(a-b)]/2 = \cos a\cos b\), \[f(t)g(t) = \tfrac{1}{2}[\cos(2\omega t + \phi) + \cos \phi]\,.\] Taking the time derivative of the product yields \begin{align}\label{eq:id:fg} \frac{d}{dt}[f(t)g(t)] = -\omega \sin(2\omega t + \phi). \end{align} The time-average of \(h(t)\) having angular frequency \(\omega = 2\pi f = 2\pi/\tau\), where \(\tau\) is the period, is defined as \[\langle h(t)\rangle = \frac{1}{\tau} \int_{0}^{\tau} h(t) \,dt\,.\] Thus the time average of Eq. \eqref{eq:id:fg} is \begin{align*} \left\langle\frac{d}{dt}[f(t)g(t)]\right\rangle &= -\langle\omega \sin(2\omega t + \phi)\rangle\\ &=-\frac{\omega}{\tau} \int_0^\tau \sin (2\omega t +\phi) \,dt\\ &= \frac{\omega}{\tau} \frac{1}{2\omega}\cos (2\omega t + \phi)\bigg\rvert_{0}^\tau\\ &= \frac{1}{2\tau} [\cos (4\pi + \phi) - \cos\phi] = 0 \end{align*} since \(\cos(4\pi + \phi) = \cos \phi\). Thus, at least for time-harmonic functions, \(\langle{d}[f(t)g(t)]/{dt}\rangle = 0\). Since the theory on this website only involves time-harmonic functions, this property will be exploited when deriving the acoustic radiation stress tensor.

3. \(\divergence(\phi\dyad{T})= \phi\,\divergence\dyad{T}+ \dyad{T}\cdot (\gradient \phi)\), where rank-2 tensors on this website are denoted by being underlined. The left-hand side of the identity is manipulated: \begin{align*} \divergence(\phi\dyad{T}) &= (\phi\dyad{T})_{,k} \cdot \vec{g}^k\tag{definition of divergence of tensor}\\ &= \phi\dyad{T}_{,k} \cdot \vec{g}^k + \phi_{,k}\dyad{T}\cdot \vec{g}^k \tag{product rule}\\ &= \phi\dyad{T}_{,k} \cdot \vec{g}^k + \dyad{T} \cdot (\phi_{,k} \vec{g}^k) \tag{commutative property of multiplication}\\ &= \phi\,\divergence\dyad{T}+ \dyad{T} \cdot (\gradient \phi)\,.\tag{definition of div of tensor and vector} \end{align*}

   For a less general version of this proof, appeal to orthonormal index notation. As in Item (1), the statement is proved in one line by the product rule: \begin{align} (\phi T_{ij})_{,j} = \phi T_{ij,j} + T_{ij} \phi_{,j} \end{align}

4. For two time-harmonic functions \(f\) and \(g\) represented by the real parts of the complex-valued functions \(\tilde{f}(t) = f_0 e^{-i(\omega t + \phi_f)} = \tilde{f}_\omega e^{-i\omega t}\) and \(\tilde{g}(t) = g_0 e^{-i(\omega t + \phi_g)} = \tilde{g}_\omega e^{-i\omega t}\) (where \(\tilde{f}_\omega = f_0e^{-i\phi_F}\) and \(\tilde{g}_\omega = g_0 e^{-i\phi_G}\)), the time average of their product, \(\langle fg\rangle\), is given by \(\langle \Re (\tilde{f}) \,\,\Re (\tilde{g}) \rangle = \frac{1}{2} \Re (\tilde{f}_\omega \, \tilde{g}^*_\omega) = \frac{1}{2} \Re (\tilde{f}_\omega^*\, \tilde{g}_\omega)\), where "\(\Re\)" denotes "real part".

   To show this, note that \(\Re(\tilde{f}) = f_0 \cos(\omega t + \phi_f)\) and \(\Re(\tilde{g}) = g_0 \cos(\omega t + \phi_g)\). thus \begin{align} \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= \langle f_0 \cos(\omega t + \phi_f) \, g_0 \cos(\omega t + \phi_g)\rangle \notag\\ &= f_0 g_0 \langle\cos(\omega t + \phi_f) \, \cos(\omega t + \phi_g)\rangle\,. \label{eq:id:avg:simplify:1} \end{align} Since \(\cos A \cos B = \cos(A+B) + \sin A\sin B\), Eq. \eqref{eq:id:avg:simplify:1} becomes (by letting \(A = \omega t + \phi_f\) and \(B= \omega t + \phi_g\)) \begin{align} \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= f_0 g_0 \langle \cos(2\omega t + \phi_f + \phi_g) + \sin(\omega t + \phi_f)\sin(\omega t + \phi_g)\rangle\,.\label{eq:id:avg:simplify:2} \end{align} Noting that \(\sin A \sin B = \tfrac{1}{2} [\cos(A-B) - \cos (A+B)]\), Eq. \eqref{eq:id:avg:simplify:2} becomes \begin{align} \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= f_0 g_0 \langle \cos(2\omega t + \phi_f + \phi_g) - \tfrac{1}{2} \cos(2\omega t + \phi_f + \phi_g) + \tfrac{1}{2}\cos(\phi_f - \phi_g)\rangle \notag\\ &= f_0 g_0 \langle \tfrac{1}{2} \cos(2\omega t + \phi_f + \phi_g) + \tfrac{1}{2}\cos(\phi_f - \phi_g)\rangle\,.\label{eq:id:avg:simplify:3} \end{align} The time-averaging operation amounts to an integral, which is a linear operation. Thus Eq. \eqref{eq:id:avg:simplify:3} becomes \begin{align*} \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= \tfrac{1}{2} f_0 g_0 \langle \cos(2\omega t + \phi_f + \phi_g)\rangle + \tfrac{1}{2} f_0 g_0 \langle\cos(\phi_f - \phi_g)\rangle\,. \end{align*} The first term on the left-hand side is 0. Meanwhile, the second term does not depend on time, and therefore its time average is itself: \begin{align} \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= \tfrac{1}{2} f_0 g_0 \cos(\phi_f - \phi_g)\,.\label{eq:id:avg:simplify} \end{align} Noting that \(f_0 g_0 \cos(\phi_f - \phi_g) \) is \(\Re [f_0 g_0 e^{-i(\phi_f - \phi_g)}]\), which by the relations \(\tilde{f}_\omega = f_0e^{-i\phi_F}\) and \(\tilde{g}_\omega = g_0 e^{-i\phi_G}\) is \(\Re(\tilde{f}_\omega \, \tilde{g}_\omega^*)\), Eq. \eqref{eq:id:avg:simplify} becomes \begin{align} \langle fg \rangle = \langle \Re (\tilde{f}) \, \Re (\tilde{g}) \rangle &= \tfrac{1}{2} \Re(\tilde{f}_\omega\tilde{g}_\omega^*) = \tfrac{1}{2} \Re(\tilde{f}_\omega^*\tilde{g}_\omega) \,,\label{eq:id:avg} \end{align} where the final equality holds by noting that \(\cos (\phi_f - \phi_g) = \cos(\phi_g - \phi_f)\).

   A consequence of this relation is that the time-averaged intensity of a time-harmonic acoustic field is \[\langle \vec{I}\rangle = \langle p \vec{v} \rangle = \frac{1}{2} \Re(\tilde{p}_\omega \tilde{\vec{v}}_\omega^* ) = \frac{1}{2} \Re(\tilde{p}_\omega ^*\tilde{\vec{v}}_\omega )\,,\] as can be seen by letting \(\Re(\tilde{f}) = p\) and \(\Re(\tilde{\vec{g}})= \vec{v}\) in Eq. \eqref{eq:id:avg}

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