Parallax method, 61-Cygni and the Hipporcas mission

November 14, 2018November 24, 2018 / 153armstrong / Leave a comment

It is trivial for most astronomy textbooks to illustrate the parallax method as follows:

This is absolutely fascinating, but it was really hard to find actual images of stars in books that illustrate this.

This is the proper motion of 61-Cygni, a binary star system over a span of couple of years.

61 Cygni showing proper motion at one year intervals

But Bessel discovered that in addition to this proper motion, 61-Cygni also wobbled a little bit from side to side because of the parallax during observation.

The following is a plot of the motion of 61 Cygni – A which beautifully elucidates the proper motion and the effect of parallax (i.e the wiggle of the blue line with respect to the mean free path)

In addition, if you would like to actually play around with data for yourself, the The Hipparcos Space Astrometry Mission might interest you a lot. The mission was Launched in August 1989 and successfully observed the celestial sphere for 3.5 years before operations ceased in March 1993 employ

The documentation and the catalogue are fairly clear , instructive and easy to use. Have fun!

Video – 61-Cygni and it’s importance
More proper motions of stars animations can be found here.
1838: Friedrich Bessel Measures Distance to a Star

Beautiful Proofs(#3): Area under a sine curve !

June 7, 2017June 9, 2017 / 153armstrong / Leave a comment

So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

Screenshot from 2017-06-07 00:19:11

$\sum sin(\theta)d\theta =$ Diameter of the circle/ The distance covered along the x axis starting from $0$ and ending up at $\pi$ .

And therefore by the same logic, it is extremely intuitive to see why:

$\int\limits_{0}^{2\pi} sin/cos(x) dx = 0$

Because if a dude starts at $0$ and ends at $0/ 2\pi/ 4\pi \hdots$ , the effective distance that he covers is 0.

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

Tricks that I wish I knew in High School : Trigonometry (#1)

June 4, 2017June 6, 2017 / 153armstrong / Leave a comment

I really wish that in High School the math curriculum would dig a little deeper into Complex Numbers because frankly Algebra in the Real Domain is not that elegant as it is in the Complex Domain.

To illustrate this let’s consider this dreaded formula that is often asked to be proved/ used in some other problems:

$cos(nx)cos(mx) =$ ?

Now in the complex domain:

$cos(x) = \frac{e^{ix} + e^{-ix}}{2}$

And therefore:

$cos(mx) = \frac{e^{imx} + e^{-imx}}{2}$

$cos(nx) = \frac{e^{inx} + e^{-inx}}{2}$

$cos(mx)cos(nx) = \left( \frac{e^{imx} + e^{-imx}}{2} \right) \left( \frac{e^{inx} + e^{-inx}}{2} \right)$

$cos(mx)cos(nx) = \frac{1}{4} \left( e^{i(m+n)x} + e^{-i(m+n)x} + e^{i(m-n)x} + e^{-i(m-n)x} \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( \left( \frac{e^{i(m+n)x} + e^{-i(m+n)x}}{2} \right) + \left( \frac{e^{i(m-n)x} + e^{-i(m-n)x}}{2} \right) \right)$

$cos(mx)cos(nx) = \frac{1}{2} \left( cos(m+n)x + cos(m-n)x \right)$
And similarly for its variants like $cos(mx)sin(nx)$ and $sin(mx)sin(nx)$ as well.

****

Now if you are in High School, that’s probably all that you will see. But if you have college friends and you took a peak what they rambled about in their notebooks, then you might this expression (for $m \neq n$ ):

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx) dx \\$

But you as a high schooler already know a formula for this expression:

$I = \int\limits_{-\pi}^{\pi} \left( cos(m+n)x + cos(m-n)x \right)dx \\$

$I = \int\limits_{-\pi}^{\pi} cos(\lambda_1 x) dx + \int\limits_{-\pi}^{\pi} cos(\lambda_2 x) dx \\$

where $\lambda_1$ , $\lambda_2$ are merely some numbers. Now you plot some of these values for lambda i.e ( $\lambda = 1,2, \hdots$ ) and notice that since integration is the area under the curve, the areas cancel out for any real number.

and so on….. Therefore:

$I = \int\limits_{-\pi}^{\pi} cos(mx)cos(nx)dx = 0$

This is an important result from the view point of Fourier Series!

Why is the area under one hump of a sine curve exactly 2?

May 7, 2017June 6, 2017 / 153armstrong / 1 Comment

Girls' Angle

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

If you’ve wondered about this too, perhaps you’ll find the following of interest.

View original post 162 more words

Solving the Laplacian in Spherical Coordinates (#1)

May 5, 2017June 6, 2017 / 153armstrong / Leave a comment

In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.

sph_coor

$x = rsin\theta cos\phi$
$y = rsin\theta cos\phi$
$z = rcos\theta$

where

$0 \leq r < \infty$
$0 \leq \theta \leq \pi$
$0 \leq \phi < 2\pi$

The Laplacian in Spherical coordinates in its ultimate glory is written as follows:

$\nabla ^{2}f ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial f}{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}f}{\partial \phi ^{2}}} = 0$

To solve it we use the method of separation of variables.

$f = R(r)\Theta(\theta)\Phi(\phi)$

Plugging in the value of $f$ into the Laplacian, we get that :

$\frac{\Theta \Phi}{r^2} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right) + \frac{R \Phi}{r^2 sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{\Theta R}{r^2 sin^2 \theta} \frac{d^2 \Phi}{d\phi^2} = 0$

Dividing throughout by $R\Theta\Phi$ and multiplying throughout by $r^2$ , further simplifies into:

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{h(r)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{g(\theta,\phi)} = 0$

It can be observed that the first expression in the differential equation is merely a function of $r$ and the remaining a function of $\theta$ and $\phi$ only. Therefore, we equate the first expression to be $\lambda = l(l+1)$ and the second to be $-\lambda = -l(l+1)$ . The reason for choosing the peculiar value of $l(l+1)$ is explained in another post.

$\underbrace{ \frac{1}{R} \frac{d}{dr} \left( r^2\frac{dR}{dr} \right)}_{l(l+1)} + \underbrace{\frac{1}{\Theta sin \theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{d\theta} \right) + \frac{1}{\Phi sin^2 \theta} \frac{d^2 \Phi}{d\phi^2}}_{-l(l+1)} = 0$ (1)

The first expression in (1) the Euler-Cauchy equation in $r$ .

$\frac{d}{dr} \left( r^2\frac{dR}{dr} \right) = l(l+1)R$

The general solution of this has been in discussed in a previous post and it can be written as:

$R(r) = C_1 r^l + \frac{C_2}{r^{l+1}}$

The second expression in (1) takes the form as follows:

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta + \frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2} = 0$

The following observation can be made similar to the previous analysis

$\underbrace{\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta }_{m^2} + \underbrace{\frac{1}{\Phi} \frac{d^2 \Phi}{d\phi^2}}_{-m^2} = 0$ (2)

The first expression in the above equation (2) is the Associated Legendre Differential equation.

$\frac{sin \theta}{\Theta} \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ l(l+1)sin^2 \theta = m^2$

$sin \theta \frac{d}{d \theta} \left( sin \theta \frac{d\Theta}{dr} \right)+ \Theta \left( l(l+1)sin^2 \theta - m^2 \right) = 0$

The general solution to this differential equation can be given as:
$\Theta(\theta) = C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta)$

The solution to the second term in the equation (2) is a trivial one:

$\frac{d^2 \Phi}{d\phi^2} = m^2 \Phi$
$\Phi(\phi) = C_5 e^{im\phi} + C_6 e^{-im\phi}$

Therefore the general solution to the Laplacian in Spherical coordinates is given by:

$R\Theta\Phi = \left(C_1 r^l + \frac{C_2}{r^{l+1}} \right) \left(C_3 P_l^m(cos\theta) + C_4 Q_l^m(cos\theta \right) \left(C_5 e^{im\phi} + C_6 e^{-im\phi}\right)$

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Trigonometry