Waves in Several Dimensions

Waves in multiple dimensions have a direction of motion that is perpendicular to the crests of the waves. 1 cycle of the wave is represented by the distance between 2 adjacent crests. We just showed = 2**c/ so * = 2**c. This relates the radian frequency and the wavelength with the wave's speed.

The wave height of as a function of space = Acos(ax+by+cz-dt) ax+by+cz-dt is the phase of the wave = an angle measured in radians. So as t increases the wave moves a distance dt in the positive coordinate direction.

Affine Functions in 2D + 3D

Ax + By + C = an affine function of 2 variables. Adding a 3rd variable it become a affine function of 3 variables, Ax + By + Cz + D.

Lets consider 2D functions. Let's set c = 0 for Ax + By + C So we get Ax + By. What's the cture of this function?

To do this we'll use level lines. Level lines are curves that have a single value for any given input value of the function. So Ax + By = 0 is a line that is parallel to the level lines of Ax + By = const. We will consider two cases: B = 0 and B not = 0.

If B = 0, then our function is just Ax = C, where C is a constant. So the level lines are horizontal lines.

If B not = 0, we can solve for y, obtaining y = (C-Ax)/B = C/B - (A/B)x. This is the slope equation, y = mx + b, m is the slope and b is the y-intercept. The level lines all have a slope of -(A/B) so they are parallel, with their spacing between each other depending on the constant C.

An easier way to describe the level lines of Ax + By = C, is to use the knowledge of dot products. Let's set C=0 so our function is Ax + By = 0. So the dot product [A,B]*[x,y] = 0. This means [A,B] is perpendicular to [x,y], by definition of dot product. So the level lines [x,y] are perpendicular to the vector[A,B]. The curve Ax + By = 0 is a line through the origin that is perpendicular to [A,B].

Given a point on the curve Ax + By = C and another point on the same curve Ax* + By* = C, the difference between them is A(x - x*) + B(y - y*) = 0. This tells us that any curve Ax + By = C is parallel to the curve Ax + By = 0, which goes through the origin.

The distance between Ax + By = C and Ax + By = 0 is the signed distance. Let's look at Ax + By = 1. The x-intercept when y=0 is Ax=1 or x=1/A. Similarly, the y-intercept occurs when x=0 so By = 1 or y = 1/B. The y and x axis' and the line Ax + By = 1 bound a right angle triangle. The signed distance line from the origin cuts the right triangle into two right triangle that are similar triangles to the original, which we can see by geometry. The hypotenuese and 1/A sides of the large right triangle are similar to the hypotenuese and signed distance side of the smaller triangle. The larger hypotenuese is sqrt( B^2 + A^2 ) by Pythagorus. So the signed distance is 1/ sqrt( B^2 + A^2 ). Since any other line Ax + By = C is parallel to it and have been a shift vertically proportional to the constant C, they are all equally spaced from each other, given C is always an integer. So Ax + By = C has a signed distance C times greater than Ax + By = 1 so Ax + By = C has signed distance of C / sqrt( B^2 + A^2 ).

The signed distance between two lines Ax + By = C1 and Ax + By = C2 is just the difference between the two: (C2 - C1) / sqrt( B^2 + A^2 ).

For a 2D wave with wave height = Acos( Ax + By ), where is the wavelength vector, is perpendicular to the level lines and is a multiple of [A,B], so = k[A,B]. The distance from level 0 and level 2 is 2/ sqrt( B^2 + A^2 ) which is the wavelength, since the phase difference is 2. so = 2/ sqrt( B^2 + A^2 ) = k[A,B], so k = 2 /( B^2 + A^2 ) and = 2 /( B^2 + A^2 )[A,B]. We also get [A,B] = /k, which at time t=0, the wave height is Acos( Ax + By), so it relates (a,b) to , the wavelength.

So how does the wave height change with time? wave height = Acos(ax+by+cz-t). The reason is clear if we look at t. =2**f from the definition of radian frequency, has units (2* * (cycles/unit time)) so t is 2**cycles giving us the angle change in radians for each unit of time t.