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Shallow water wave approximation

  1. May 10, 2013 #1
    I'm working through a solved problem in a fluid mechanics textbook. In it, the group velocity of a dispersive wave is calculated as:
    $$c_g = \frac{1}{2}c\left (1 + 2kh\ \text{cosech} (2kh) \right)$$

    Where k is the angular wavenumber, and h is the depth of the water, which is fine. Now for shallow waves, we assume that $$kh \ll 1$$, which we can use to simplify the last part of the expression above:

    $$\text{cosech} (2kh) = \frac{1}{\sinh(2kh)}=\frac{2}{e^{2kh}-e^{-2kh}} \approx \frac{2}{1+2kh-(1-2kh)}$$

    resulting in $$c_g = c$$.

    I don't really understand how the $$1/\sinh(2kh)$$ in exponential form is approximated in the manner written above...how does:
    $$e^{2kh}=1+2kh$$
    Is this a mathematical approximation or is there some physical reasoning behind it?
    Thanks in advance
     
    Last edited: May 10, 2013
  2. jcsd
  3. May 10, 2013 #2

    Nugatory

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    It's an approximation. If kh<<1 then we can discard all of the terms of the series expansion except the first two.
     
  4. May 10, 2013 #3
    Thanks!
    Is it a Taylor expansion? I've managed to get this far in my self-study of physics without being able to expand a constant function (although I know how to expand a function of a variable around a point)...
     
  5. May 10, 2013 #4
    Thats right it is a Taylor expansion about 0 . What is a 'constant function'?
    I was hoping this discussion would be about the physics of shallow water waves! :cool:
     
    Last edited: May 10, 2013
  6. May 10, 2013 #5

    AlephZero

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    Yes it is a Taylor expansion.

    I'm not sure why you are bothered about something being a constant, but you can think of this as being the Taylor expansion of ##e^x## where ##x## iis a variable, and then substitude in the particular value of ##x = 2kh##.
     
  7. May 10, 2013 #6
    Thanks

    Ok, thanks for all the replies. On some of the points:

    A constant function: well I'm studying maths and physics without the benefit of a lecturer or tutor, so I sometimes get the terminology wrong. If you have a function f(x) = constant...can this not be described as a function despite being constant?

    Why I can do the expansion with a variable, but not a constant: I've learnt the Taylor series expansion for $$e^x$$ around the point zero, and I can see how to do this (and that in many cases the higher orders can be neglected (not sure under exactly which conditions). However, in this expression there is no variable so I'm not sure how the expansion goes...if I can just substitute $$x=2kh$$ then that's great.

    Supposed to be about shallow waves: well I wasn't sure whether this query was based on some element of fluid mechanics that I was unaware of, or whether it was a standard mathematical approximation. It turned out to be the latter, but I have many more questions about dispersive waves which I will no doubt come back with!
    Thanks again
     
  8. May 10, 2013 #7

    AlephZero

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    You can do both. The Taylor series expansion ##e^x = 1 + x + x^2/2! + \cdots## is valid for all values of ##x##, so in that sense you can think of ##x## as a variable.

    But if it is true for all values of ##x##, it is also true for any particular value that you want to use. If ##x = 0.1## for example you can say ##e^{0.1} = 1 + 0.1 + (0.1)^2/2! + \cdots## and you would probably call ##0.1## a "constant".

    In your original question, remember the formulas are true for any values of ##h## and '##k##, so long as ##hk \ll 1##, so in that sense ##h## amd ##k## are "variables" not "constants". But if you are thinking about one particular wave, i.e. one particular value of ##h## and ##k## and so you think ##hk## is a 'constant', that doesn't make any difference.

    The idea of a "constant function" ##f(x) = c## is a bit different. The value of ##f(x)## is constant, but ##x## is still a variable. Of course the Taylor series of ##f(x) = c## is just ##c##, and it doesn't involve ##x## at all (and it's not particularly useful either!)
     
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