Astronomy: Distance and Midpoint Between Two Celestial Objects

Stars and celestial objects have the equatorial coordinates ( α, δ, R ) where:

α = right ascension in hours, minutes, and seconds

δ = declination in degrees, minutes, and seconds

R = distance to the celestial about (in light years, astronomical units, etc)

Convert α and δ to decimal degrees:

α = (hours + minutes/60 + seconds/3600) * 15

δ = sign(δ) * ( |degrees| + |minutes/60| + |seconds/3600| )

However, calculating distance between two points in space requires that we have cartesian coordinates (x, y, z). The conversion formulas to go from equatorial to cartesian coordinates are:

x = R * cos δ * sin α

y = R * cos δ * cos α

z = R * sin δ

For completeness, here are the conversion formulas to from cartesian to equatorial coordinates:

R = √(x^2 + y^2 + z^2)

δ = asin(z/R)

α = atan(y/x) (pay attention to the quadrant!). Alternatively: α = arg(x + i*y)

The distance formula between two points in space is:

D = √( (x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2 )

Source: http://en.m.wikipedia.org/wiki/Equatorial_coordinate_system

Example

Find the distance between Sagittarius A* (center of the Milky Way Galaxy) and the Andromeda Galaxy (M31).

Equatorial Coordinates:

From Wolfram Alpha: ( http://m.wolframalpha.com )

Sagittarius A*:

R = 24,824 light years

α = 17h45m40s = 266.42°

δ = -29°0'28" = -29.008°

Andromeda Galaxy:

R = 2,571,000 light years

α = 42m40s = 10.68°

δ = 41°16'8" = 41.269°

Converting to the cartesian system:

Sagittarius A*:

x = -21,667.514038

y = -1,355.611275

z = -12,037.945401

Andromeda Galaxy:

x = 358,122.65578

y = 1,898,943.67736

z = 1,695,818.99789

Calculation:

D = √( (-21,667.514038 - 358,122.65578)^2 + (-1,355.611275 - 1,898,943.67736)^2

+ (-12,037.945401 - 1,695,818.99789)^2 ) ≈ 2,583,051.16059

The distance between Sagittarius A* and the Andromeda Galaxy is about 2.583 million light years.

Happy Thanksgiving, or Day of Thanks, wherever you are. Please be safe and use good judgement. We need all the positive we can get on Earth, and in the cosmos. Cheers!

Eddie

This blog is property of Edward Shore. 2014

A blog is that is all about mathematics and calculators, two of my passions in life.

## Wednesday, November 26, 2014

### Astronomy: Distance and Midpoint Between Two Celestial Objects

## Tuesday, November 25, 2014

### Complex Analysis: Line Integral

Contour Integrals

Let the complex function f(z) be defined as f(z) = u(x,y) + i*v(x,y)

Using the definition of the integral:

∫ f(z) dz = lim n → ∞ [ Σ f(z_k) * Δz_k from k = 1 to n ]

= lim n → ∞ [ Σ ( u(x_k, y_k) + i*v(x_k, y_k) ) * ( Δx_k + i*Δy_k ) from k = 1 to n ]

= lim n → ∞ Σ [ (u(x_k, y_k) * Δx_k - v(x_k, y_k) * Δy_k) + i*(v(x_k, y_k) * Δx_k

+ u(x_k, y_k) * Δy_k) ]

= ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )

In summary:

**∫ f(z) dz = ( ∫ u(x,y) dx - ∫ v(x,y) dy ) + i*( ∫ u(x,y) dy + ∫ v(x,y) dx )**

What is needed:

1. A contour curve f = y(x). This is where you get your end points.

2. We need f(z) to integrate. Separate f(z) into its parametric parts.

Let's look at a couple examples.

Examples

1. Contour: y = x + 1 from (3,4) to (4,5). Integrate ∫ z^2 dz.

Since y = x + 1, x = y - 1

z^2 = (x + i*y)^2 = (x^2 - y^2) + i*(2*x*y)

u = x^2 - y^2

v = 2*x*y

u = x^2 - y^2

u = x^2 - (x + 1)^2

u = -2*x - 1

∫ u dx from 3 to 4 = -8

v = 2 * x * y

v = 2 * (y - 1) * y

∫ v dy from 4 to 5 = 95/3

u = (y - 1)^2 - y^2

u = -2 * y + 1

∫ u dy from 4 to 5 = -8

v = 2 * x * y

v = 2 * x * (x + 1)

∫ v dx from 3 to 4 = 95/3

∫ z^2 dz from (3 + 4*i) to (4 + 5*i) = (-8 - 95/3) + i*(-8 + 95/3) ≈ -39.666667 + 23.666667*i

2. Contour: y = x^2 - 1 from (0, -1) to (2, 1). Integrate ∫ z^2 + 1 dz.

y = x^2 - 1, x = √(y + 1)

z^2 + 1 = (x + i*y)^2 + 1 = x^2 + 2*i*x*y - y^2 + 1

u = x^2 - y^2 + 1

v = 2 * x * y

u = x^2 - y^2 + 1

u = x^2 - (x^2 - 1)^2 + 1

∫ u dx from 0 to 2 = 8/5

v = 2 * x * y

v = 2 * √(y + 1) * y

∫ v dy from -1 to 1 = 8/15 * √2

u = x^2 - y^2 + 1

u = y + 1 - y^2 + 1

u = -y^2 + y + 2

∫ u dy from -1 to 1 = 10/3

v = 2 * x * y

v = 2 * x * (x^2 - 1)

∫ v dx from 0 to 2 = 4

∫ z^2 + 1 dz from -i to 2+i = (8/5 - 8/15 * √2) + i*(4 + 10/3) ≈ 0.845753 + 7.333333*i

This is the basics of line integrals.

Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. Addison-Wesley Publishing Company. 1994.

Take care, Eddie

This blog is property of Edward Shore. 2014

## Saturday, November 22, 2014

### Complex Analysis: Cauchy-Reimann Equations

Continuing my blogging adventures at the Last Drop Cafe in Claremont, CA:

Cauchy-Reimann Equations

Let the complex-valued function f(z), where z = x+i*y, be defined in the parametric form:

f(z) = u(x,y) + i*v(x,y) where u(x,y) = Re(f(z)) and v(x,y) = Im(f(z)).

Then by the definition of the derivative:

f(z) = lim x0 → x [(f(z) + x0) - f(z))/(x - x0)]

= lim x0 → x [(u(x +x0, y) - u(x, y) + i*v(x + x0, y) - i*v(x, y))/(x - x0)]

= du/dx + i*dv/dx (I)

Also:

f(z) = lim y0 → y [(f(z + y0) - f(z))/(i*y - i*y0)]

= lim y0 → y [(u(x, y + y0) - u(x, y) + i*v(x, y + y0) - i*v(x, y))/(i*(y - y0))]

= -i*(lim y0 → y [(u(x, y + y0) - u(x, y) + i*v(x, y + y0) - i*v(x, y))/(y - y0)])

= -i*du/dy + dv/dy (II)

Taking the real and imaginary parts of (I) and (II):

du/dx = dv/dy and -du/dy = dv/dx

The Cauchy-Riemann equations can be used to determine whether f(z) is differentiable and if so, where.

Two Quick Examples

1. z^2 = (x^2 - y^2) + 2*i*x*y

u=x^2 - y^2

v=2*i*x*y

du/dx = 2*x, dv/dy = 2*x

-du/dy = 2*y, dv/dx = 2*y (note the negative sign in front of du/dy!)

Since the Cauchy-Riemann equations hold, and without restriction, then z^2 is differentiable for all z. And:

df/dz = 2*x + 2*i*y = 2*(x+i*y) = 2*z

2. e^z = e^(x+i*y) = e^x*cos y + i*e^x*sin y

u=e^x*cos y

v=e^x*sin y

du/dx = e^x*cos y, dv/dy = e^x*cos y

-du/dy = e^x*sin y, dv/dx = e^x*sin y

Since the Cauchy-Riemann equations hold, and without restriction, then e^z is differentiable for all z. And:

df/dx = e^x*cos y + i*e^x*sin y = e^z

I used (I), but using (II) will garner the same result.

That is the Cauchy-Riemann equations in a nutshell!

Source: Wunsch, A. David. Complex Variables with Applications. 2nd Edition. Addison-Wesley Publishing Company. 1994.

As always, have a great day! Take care. Eddie

This blog is property of Edward Shore. 2014

### Complex Analysis: The Conjugate, the Modulus, and its Properties

Blogging today from Last Drop Cafè in Claremont, CA. I think I found a new favorite drink: mint mocha made with soy milk.

Here are some basics of the conjugate and modulus of complex numbers.

Let z = x+i*y, with i=√-1

Conjugate

Usually labeled "z-bar" (z with a line over it), the conjugate is also labeled conj(z) and z*.

conj(z) = x - i*y

Modulus or Absolute Value

Not surprisingly, the modulus, also called the absolute value of the complex number z is defined as:

|z| = √(x^2 + y^2)

Properties

Let's explore some properties of the conjugate and modulus.

z + conj(z) = (x + i*y) + (x - i*y) = 2*x

z - conj(z) = (x + i*y) - (x - i*y) = 2*i*y

(conj(z))^2 = (x - i*y)^2 = x^2 - 2*i*x*y + (i*y)^2 = x^2 - y^2 - 2*i*x*y

conj(z)^2 + z^2 = (x - i*y)^2 + (x + i*y)^2 = x^2 - 2*i*x*y - y^2 + x^2 + 2*i*x*y - y^2

= 2*(x^2 - y^2)

conj(z) * z = (x - i*y) * (x + i*y) = x^2 + i*x*y - i*x*y - i^2*y^2 = x^2 + y^2 = |z|^2

which easily leads to: |z| = √(z * conj(z)) and

|z1 * z2| = √(z1 * conj(z1) * z2 * conj(z2)) = √(z1 * conj(z1)) * √(z2 * conj(z2)) = |z1| * |z2|

and:

|z1/z2| = √((z1 * conj(z1))/(z2 * conj(z2))) = √((z1 * conj(z1))/√((z2 * conj(z2)) = |z1|/|z2|

Source: Wuncsh, David A. Complex Numbers with Applications. 2nd Edition. Addison-Wesley Publishing Company. Reading, MA. 1994

## Wednesday, November 12, 2014

### Casio fx-9750gII, fx-9860gII, Prizm: Prime and Complex Number Prime (Gaussian Prime)

**Real Number Primes**

“N”?->N

Int N->N

For 2->K To S

If Frac (N÷K)=0

Then

“NOT PRIME” /rtri

Stop

IfEnd

Next

“PRIME”

36 is NOT PRIME

**Complex Number Primes**

**(Gaussian Primes)**

a+bi

“Z”?->Z

(Abs Z)^2->T

If ReP Z=0 or ImP Z=0

Then

Goto 1

IfEnd

For 2->K To Abs Z

If Frac(T÷K)=0

Then Goto 2

IfEnd

Next

Goto 3

Lbl 1

If MOD(Abs Z, 4)≠3

Then Goto 2

IfEnd

Next

Goto 3

Lbl 2

“NOT PRIME” /rtri

Stop

Lbl 3

“PRIME”

7 is (complex) prime

-5+4i is (complex) prime

## Tuesday, November 11, 2014

### Darts: The Probability of Getting the Best Score

**Darts: The Probability of Getting the Best Score**

**Area of a Circular Ring**

**Simplified Darts Board**

**Standard Darts Board**

= (π * (R3^2 – R2^2 + R5^2 – R6^2)/(π * R^2)

= (R3^2 – R2^2 + R5^2 – R6^2)/R^2

= (R6^2 – R5^2)/R^2

= (R4^2 – R3^2)/R^2

= (π * R2^2)/(π * R^2)

= R2^2/R^2

**Scoring 60**

## Sunday, November 9, 2014

### Two New Videos: Derivatives on the HP Prime, Derivative and Integral of abs(x)

**Derivatives on the HP Prime Calculator**

Video: http://youtu.be/LXwLiPlKO2M

Notes:

Derivative Template: Template Key, template is on the top row, 4th column

Numeric Format: (d* function)/(d varaible = point), press Enter

In RPN Mode: You also need to press Shift+Comma (Eval)

Symbolic Format (CAS Mode): (d function)/(d variable)

*d represents the derivative symbol

diff command:

diff(function, variable)

**Derivative and Integral of abs(x)**

Video: http://youtu.be/8PikrlA5UlQ

A short derivation of the derivative and integral of absolute value function (abs(x)), and how the signum (sign(x)) function plays a part its calculation.

Graphs made with the online Desmos calculator. Please visit and explore at www.desmos.com

(This is NOT a paid endorsement)

Eddie

Blog Entry #403

This blog is property of Edward Shore. 2014

## Saturday, November 8, 2014

### My Morning at Honnold/Mudd Library: 2014-11-08 (Game Shows, Bravias Lattice, Prime Numbers)

**My Morning at Honnold/Mudd Library**

Honnold/Mudd Library - Claremont, CA |

**Game Shows and Winning**

^{nd}Edition Oxford Press: 2003

**Ideal Crystals and Bravais Lattices**

**Prime Number Goodies**

## Monday, November 3, 2014

### Derivative of abs(x) and Integrals of abs(x), abs(e^x), abs(e^(a*x) + e^(a*y))

**Derivative of abs(x) and Integrals of abs(x), abs(e^x), abs(e^(a*x) + e^(a*y)) **

For this blog entry, assume that are functions are in terms of x and a, x, y represent real numbers.

Definition of abs(x) (also symbolized as |x|)

Piecewise definition:

abs(x) = x if x > 0

abs(x) = 0 if x = 0

abs(x) = -x if x < 0

Also:

abs(x) = x * sign(x)

Where the sign(x), sometimes labeled sgn(x), is the sign or signum function. It is defined as:

sign(x) = 1 for x > 0

sign(x) = 0 for x = 0

sign(x) = -1 for x < 0

d/dx sign(x)

We can clearly demonstrate that d/dx sign(x) = 0 since:

d/dx sign(x) = 0 for x > 0

d/dx sign(x) = 0 for x = 0

d/dx sign(x) = 0 for x < 0

d/dx abs(x)

Using the chain rule:

d/dx abs(x)

= d/dx (x * sign(x))

= d/dx (x) * sign(x) + x * d/dx (sign(x))

= sign(x)

This can also be done with the piecewise representation:

d/dx abs(x) = 1 if x > 0

d/dx abs(x) = 0 if x = 0

d/dx abs(x) = -1 if x < 0

∫ abs(x) dx

∫ abs(x) dx

= ∫ x * sign(x) dx

Using integration by parts:

where u = sign(x), dv = x dx

Then: du = 0 dx, v = x^2/2

∫ abs(x) dx

= ∫ x * sign(x) dx

= x^2/2 * sign(x) - ∫ 0 dx

= x^2/2 * sign(x) + C

C is the arbitrary integration constant

With the piecewise representation:

∫ abs(x) dx = x^2/2 if x > 0

∫ abs(x) dx = 0 if x = 0

∫ abs(x) dx = -x^2/2 if x <0

Note that this is x^2/2 * sign(x).

∫ abs(e^(a*x)) dx

∫ abs(e^(a*x)) dx

= ∫ e^(a*x) * sign(e^(a*x)) dx

= 1/a * e^(a*x) * sign(e^(a*x)) - ∫ 0 dx

= 1/a * e^(a*x) * sign(e^(a*x)) + C

∫ abs(e^(a*x) + e^(a*y)) dx

∫ abs(e^(a*x) + e^(a*y)) dx

= ∫ (e^(a*x) + e^(a*y)) * sign(e^(a*x) + e^(a*y)) dx

= (e^(a*x)/a + x*e^(a*y)) * sign(e^(a*x) + e^(a*y)) - ∫ 0 dx

= (e^(a*x) + a*x*e^(a*y))/a * sign(e^(a*x) + e^(a*y)) + C

Eddie

This blog is property of Edward Shore. 2014

(Blog Entry # 401)

### Swiss Micros DM41X and Casio fx-CG 50: Minor Head Loss

Swiss Micros DM41X and Casio fx-CG 50: Minor Head Loss Introduction When a fluid, such as water, is flowing in a pipe system, ene...