Question about taking the difference between two square root expression

[zaybu]

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?

Thanks

 

[Simon Bridge]

Lets see - the equation would have been:

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

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

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

$$f(x) \approx f(0) + x{f^\prime}(0)$$

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.

 

[zaybu]

Lets see - the equation would have been:

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

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

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

$$f(x) \approx f(0) + x{f^\prime}(0)$$

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.

Thanks, I appreciate.

 

[Simon Bridge]

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

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

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

Otherwise you end up with:

$$f(v\delta t) \approx <br /> <br /> \frac{1}{c}\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}} -<br /> \frac{1}{c}\sqrt{H^2 + \frac{D^2}{4}} -<br /> <br /> \frac{1}{8}\frac{v\delta t}{c}\frac{D}{\sqrt{H^2 + \frac{1}{4}\frac{D^2}{4}}}<br />$$

I have not found any reason for the division by 4 in the equation for PD_sat-B(B:2) since it should just be the distance from B to the satellite (in position 2) by Pythagoras.

 

[zaybu]

Indeed, the 1/4 in the PD_sat-b(B:2) term is a typo error.

Thanks again