Consider a thin ideal vortex segment of uniform strength G that lies along the z-axis between z1 and z2, and has a sense of rotation that points along the z-axis. Use the BioteSavart law to show that the induced velocity u at the location (R, φ, z) will be u(χ, t) = (Γ=4πR)(cosθ1 - cosθ2)eφ where

Question

Consider a thin ideal vortex segment of uniform strength Γ that lies along the z-axis between [latex]z_1[/latex] and [latex]z_2[/latex], and has a sense of rotation that points along the z-axis. Use the BioteSavart law to show that the induced velocity u at the location (R, φ, z) will be [latex]\mathbf{u}(\pmb{\chi},t)=(\Gamma / 4 \pi R)\left(\cos heta_1-\cos heta_2 ight) \mathbf{e} _{\varphi}[/latex] where the polar angles [latex] heta_1[/latex] and [latex] heta_2[/latex] are as shown in Figure 5.10.

Accepted Answer

Apply (5.17) to the geometry shown with x′ = (0, 0, z′):

[latex]d\mathbf{u}(\pmb{\chi},t)\cong \frac{1}{4 \pi}\underset{\Delta A^{\prime}}{\int}\left| \omega \left(\pmb{\chi}^{\prime}, t ight) ight|\mathbf{e}_\omega d^2 x^{\prime} imes \frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3} d l=\frac{\Gamma d l}{4 \pi}\mathbf{e}_\omega imes \frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3}, ext { or }[/latex]
[latex]\mathbf{u}(\pmb{\chi},t)=\underset{ ext {vortex }}{\int}d\mathbf{u}\cong \frac{\Gamma}{4 \pi}\underset{ ext {vortex }}{\int}\mathbf{e}_\omega imes \frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3} d l,[/latex] (5.17)

[latex]\frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3}=\frac{\left(x, y, z-z^{\prime} ight)}{\left[x^2+y^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}} ext { and }\mathbf{e}_z imes \frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3}=\frac{(-y, x, 0)}{\left[x^2+y^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}} .[/latex]

Switch to cylindrical coordinates where R² = x² + y², and [latex]\mathbf{e}_{\varphi}=-\mathbf{e}_x \sin \varphi+\mathbf{e}_y \cos \varphi[/latex]. This allows the geometrical results to be mildly simplified:

[latex]\mathbf{e}_z imes \frac{\left(\pmb{\chi}-\pmb{\chi}^{\prime} ight)}{\left|\pmb{\chi}-\pmb{\chi}^{\prime} ight|^3}=\frac{R\mathbf{e}_{\varphi}}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}}[/latex].

With dl = dz, the total induced velocity from the vortex segment is given by (5.17) integrated between [latex]z_1[/latex] and [latex]z_2[/latex]:

[latex]\mathbf{u}(\pmb{\chi},t)=\frac{\Gamma\mathbf{e}_{\varphi}}{4 \pi}\overset{z_2}{\underset{z_1}{\int}}\frac{R d z}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}}[/latex].

The integral can be evaluated by switching to a polar-angle integration variable:

[latex]\cos heta=\frac{z-z^{\prime}}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{1 / 2}}[/latex],

having a convenient differential:

[latex]d(\cos heta)=\left\{\frac{-1}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{1 / 2}}-\frac{1}{2} \frac{\left(z-z^{\prime} ight)}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}} \cdot 2\left(z-z^{\prime} ight)(-1) ight\} d z^{\prime}[/latex]
[latex]=-\left\{\frac{R^2}{\left[R^2+\left(z-z^{\prime} ight)^2 ight]^{3 / 2}} ight\} d z^{\prime}[/latex].

After making this variable change, the remaining integral is elementary:

[latex]\mathbf{u}(\pmb{\chi},t)=-\frac{\Gamma\mathbf{e}_{\varphi}}{4 \pi R}\overset{ heta_2}{\underset{ heta_1}{\int}}d(\cos heta)=\frac{\Gamma\mathbf{e}_{\varphi}}{4 \pi R}\left(\cos heta_1-\cos heta_2 ight)[/latex], (5.18)

where [latex] heta_1[/latex] and [latex] heta_2[/latex] are the polar angles from the vortex segment’s lower and upper ends, respectively:

[latex]\cos heta_1=\frac{z-z_1}{\left[R^2+\left(z-z_1 ight)^2 ight]^{1 / 2}} ext { and } \cos heta_2=\frac{z-z_2}{\left[R^2+\left(z-z_2 ight)^2 ight]^{1 / 2}}[/latex].

For an infinite line vortex [latex]\left( heta_1=0, ext { and } heta_2=\pi ight)[/latex], (5.18) produces [latex]\mathbf{u}(\pmb{\chi},t)=\Gamma\mathbf{e}_{\varphi} / 2 \pi R[/latex].

Question 5.4: Consider a thin ideal vortex segment of uniform strength G t...

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