Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question about taking the difference between two square root expression

  1. Nov 22, 2011 #1
    In Ramakrishna's paper, http://arxiv.org/ftp/arxiv/papers/1111/1111.1922.pdf , he derived equation (9), page 5.

    It is a difference of two square-roots using an approximation method. Can anyone help in how this is done?

  2. jcsd
  3. Nov 22, 2011 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Lets see - the equation would have been:

    H^2 + \frac{1}{4}\left (
    \frac{D}{2}-v\Delta t
    \right )^2
    } - \frac{1}{c}\sqrt{
    } \approx -\frac{1}{2}\frac{v \delta t}{c}\frac{D}{\sqrt{H^2 + \frac{D^2}{4}}}

    The text says this is "to first order in [itex]v\Delta t[/itex]" ... which means he took the Taylor Series to first order, putting [itex]x=v\Delta t[/itex] about [itex]x=0[/itex].

    Taking the LHS for f(x), the 1st order Taylor approx is:

    [tex]f(x) \approx f(0) + x{f^\prime}(0) [/tex]

    He just rewrote the delta-tee to emphasize it's small size.

    I havn't actually crunched the numbers but it looks about the right shape - the deriverative gets you the -1/2 and pulls the vδt outside the root - putting the root itself in the denominator.
  4. Nov 22, 2011 #3
    Thanks, I appreciate.
  5. Nov 22, 2011 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    I suspect that 1/4 in the first square root is not supposed to be there.
    If you leave it off, then the zeroth order term vanishes at x=0 and

    [tex]f^\prime (v\Delta t) = \frac{1}{c}\frac{1}{2}\left [ H^2 + \left ( \frac{D}{2} -v\Delta t \right )^2 \right ]^{-1/2} 2 \left ( \frac{D}{2} -v\Delta t \right ) (-1)[/tex]

    (Leaving it messy to show the working.) So the first order term, [itex](v\delta t) f^\prime(0)[/itex], comes out as shown.

    Otherwise you end up with:

    f(v\delta t) \approx

    \frac{1}{c}\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}} -
    \frac{1}{c}\sqrt{H^2 + \frac{D^2}{4}} -

    \frac{1}{8}\frac{v\delta t}{c}\frac{D}{\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}}}


    I have not found any reason for the division by 4 in the equation for PDsat-B(B:2) since it should just be the distance from B to the satellite (in position 2) by Pythagoras.
  6. Nov 22, 2011 #5
    Indeed, the 1/4 in the PDsat-b(B:2) term is a typo error.

    Thanks again
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook