Talk:Ursell number

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A question about the limit for which linear wave theory applies: it's down as 3 / (32 π2). I went back to the references Dingemans (1997) and Stokes (1847) and it seems to come from ka << (kh) 3. Assuming H = 2a and k = 2 \pi / \lambda, this gives a linearity limit of 8 π2. Could you tell me if I have got this right or am I missing something? This is my first time inside Wikipedia so please excuse me if I'm not adhering to any of the rules!

--cheers! --Ally --13:52, 7 July 2009 (UTC)

Hi Alexandra. See Eq. (2.421a) on page 179 of Dingemans (1997). The ratio of the second-order amplitude A2 to the first-order amplitude a is:
\frac{A_2}{a}=ka \frac{3-\sigma^2}{4\sigma^3}   with   \sigma=\tanh\,(kh),
i.e. the factor in front of cos 2χ.
In shallow water, kh≪1, we have asymptotically σ ≈ kh≪1, and further that the wave height H=2a according to 2nd-order Stokes theory, so

  \frac{A_2}{a} \approx ka \frac{3}{4 (kh)^3}
                = \frac{3 a}{4 k^2\; h^3}
                = \frac{3 H}{8 \left( \frac{2\pi}{\lambda} \right)^2\; h^3}
                = \frac{3}{32 \pi^2} \frac{H \lambda^2}{h^3}
                = \frac{3}{32 \pi^2} U.
As a result, linear theory is applicable if A2a, which is the case if U≪32π2/3. Best regards, Crowsnest (talk) 16:43, 7 July 2009 (UTC)

hi Crowsnest Thanks for your prompt reply: I haven't have time to look at it yet, but will head to the library this week with the page printed off to get my head around it. -cheers -Ally-- 14:03, 20 July 2009 (UTC) —Preceding unsigned comment added by AlexandraPrice (talkcontribs)

hello again Crowsnest. Thanks for the pointers: I've been through the maths and I understand where I was getting stuck. From \frac{A_2}{a}=ka \frac{3-\sigma^2}{4\sigma^3} it follows that  \frac{3}{4}ka << (kh) ^3 , and from this it follows that linearity is defined by  U << \frac{32 \pi^2}{3} .

Stokes referred to ka << (kh) 3, presumably because he was talking the rough order of magnitude, and from this it follows that linearity is defined by  U << 8 \pi^2 . - cheers - Ally -- 09:13, 27 July 2009 (UTC)