The math is sound. The method is not.
We consider an axisymmetric pressure distribution \(p(r, \theta)=\bar{p} H(r-l)\) (radial dependence only) and show that its angular Fourier series contains only the \(m=0\) mode. Consequently, the polar Fourier representation \(\tilde{p}(k, \beta)\) in wavenumber space is independent of the angular variable \(\beta\), and the Hankel transform reduces to the order-zero Bessel transform.
Statement and setup
We consider a radially symmetric (axisymmetric) pressure field of the form \[ p(r,\theta)=\bar p\,H(r-l), \] where \(H\) is the Heaviside function and \(\bar p\) is a constant amplitude. Axisymmetry means that \(p\) depends on \(r\) only (no \(\theta\)-dependence).
We use the standard angular Fourier series at each fixed radius \(r\): \[ p(r,\theta)=\sum_{m\in\mathbb Z} P_m(r)\,e^{im\theta}, \qquad P_m(r)=\frac{1}{2\pi}\int_0^{2\pi} p(r,\theta)\,e^{-im\theta}\,d\theta. \]
For each angular mode \(m\), the (order-\(m\)) Hankel transform is defined by \[ \hat p_m(k)=\int_0^\infty r\,J_m(kr)\,P_m(r)\,dr, \] where \(J_m\) is the Bessel function of the first kind.
A common polar Fourier representation in wavenumber space is then \[ \tilde p(k,\beta)=\sum_{m\in\mathbb Z}\hat p_m(k)\,e^{im\beta}, \] where \(\beta\) is the polar angle of the wavevector (Fourier-space angle).
Claim. If \(p\) is axisymmetric (independent of \(\theta\)), then \[ P_0(r)=\bar p\,H(r-l),\qquad P_m(r)=0\quad(m\neq 0), \] so only the \(m=0\) mode is nonzero. Equivalently, \(\tilde p(k,\beta)\) is independent of \(\beta\).
Proof (only the \(m=0\) mode survives)
Because \(p(r,\theta)=\bar p\,H(r-l)\) does not depend on \(\theta\), its angular Fourier coefficients are \[ \begin{aligned} P_m(r) &=\frac{1}{2\pi}\int_0^{2\pi} p(r,\theta)\,e^{-im\theta}\,d\theta \\ &=\frac{1}{2\pi}\int_0^{2\pi} \bar p\,H(r-l)\,e^{-im\theta}\,d\theta \\ &=\bar p\,H(r-l)\cdot \frac{1}{2\pi}\int_0^{2\pi} e^{-im\theta}\,d\theta. \end{aligned} \]
Using the orthogonality of complex exponentials, \[ \int_0^{2\pi} e^{-im\theta}\,d\theta= \begin{cases} 2\pi, & m=0,\\ 0, & m\neq 0, \end{cases} \] we obtain \[ P_m(r)= \begin{cases} \bar p\,H(r-l), & m=0,\\ 0, & m\neq 0. \end{cases} \]
Therefore, all angular modes vanish except \(m=0\). Consequently, the polar Fourier expansion in wavenumber space collapses to \[ \tilde p(k,\beta)=\hat p_0(k), \] which is independent of \(\beta\).
Remarks on terminology (\(\theta\) vs. \(\beta\))
- Axisymmetry is a property of the physical-space angle \(\theta\): \(p(r,\theta)\) is axisymmetric if it is independent of \(\theta\).
- The consequence is that the Fourier-space angle \(\beta\) drops out: \(\tilde p(k,\beta)\) becomes independent of \(\beta\) because only \(m=0\) contributes.