Math Moments – Distance is Relative in Space-Time

This diagram is here as a reminder to read and understand the previous post.

It is obvious that because the speed of light is constant in every possible reference frame, then it requires distances and times to be different between events relative to the reference frame in which we measure them. So, if distances change from one reference frame to another, then how exactly do we make sense of the concept of distance in Space-Time?

We have seen in 3-D space, when a point (x, y, z) is not moving in time, we simply change our reference frame by “moving our head”. We can set our Nose direction to point directly toward the point (x, y, z), in the direction (Dx, Dy, Dz). Recall that the point:

     (x, y, z) = (Dx, Dy, Dz) * S

where S is the distance from our nose (0, 0, 0) to the point (x, y, z). We assume that rotating our head does not change the distance to the point.  This is what is called a “spherical metric”.

It is easier to picture this in 2-dimensions (when z = 0), so we will use only the ‘x’ and ‘y’ dimensions for our illustration. Recall:

    (x, y) = (Dx, Dy) * S

      meaning

      x = Dx * S
      y = Dy * S

Where the (Dx, Dy) direction is shown in the following diagram:

Recall that if we “dot” (x, y) with the direction (Dx, Dy):

   (Dx, Dy).(x, y) = (Dx, Dy).(Dx, Dy) * S
                             = (Dx*Dx + Dy*Dy) * S
                             = (Dx2 + Dy2) * S

If the metric (Dx2 + Dy2) = 1, then the Distance (S) is just this dot product.  We have looked at this in detail before.  This is called a “spherical” metric, because distances are the same in all directions, like the radius of a sphere.  With this metric, the distance is the same regardless of the orientation (angle) of the reference frame. Nothing new here.

Yet if the point (x, y, z, C*t) in Space-Time is moving, then the situation is different.  We cannot set a fixed “Nose direction” to look at a moving point.  We first must start moving with the same speed in the same direction as the point.  Now in this moving reference frame, the point is stationary, not moving, and then the “Nose Direction” can be set.

In space-time, the speed at which a point is moving is determined by the angle (At) it is off the C*t-axis. Going into another frame of reference means finding the “angle” of a line that will match the point’s speed. The angle At is often called the “boost” parameter and is directly related to the speed of the point traveling along this line. We will look at how to change the speeds (At) of reference frames relative to each other in space-time.

We can use the same concepts to find the meaning of Distance in Space-Time.  Consider the example of a moving flash bulb (bulb frame) between two stationary mirrors (mirror frame) that we examined in the last post.  For simplicity and clarity, we are only looking at the ‘x’ and ‘C*t’ dimensions and have put in grid lines parallel to the “x-axis” and “C*t-axis”.  Here the x-axis points in the same direction that the bulb is moving.

This does not restrict us because we always can rotate our frame of reference so that the ‘x’ direction is pointing in the same direction that the point is moving, we are ignoring the y = 0 and z = 0 coordinates.  Now as before, let’s look at the same events in the perspective of the “bulb frame”, with the “xb-axis” and “C*tb-axis” defining this frame.  In the following diagram we mark in the same “grid lines” parallel to the x-axis and C*t-axis, but now in the viewpoint of the “bulb” frame.

As we mentioned before, the grid lines in the mirror frame are not mutually perpendicular when seen in the bulb frame, they are stretched out “diamonds”.  Note however that the “light lines” (yellow) in both frames of reference fall along the same points (they are 45 degree lines in both frames of reference).  Another thing to note is that the angle (At) of the bulb from the C*t-axis to the C*tb-axis is the same as the angle (At) between the negative-x-axis and the negative-xb-axis, but unlike the spherical rotations, they are now rotating in opposite directions and squeezing closer together.

As the angle (At) is increased, the two “mirror” axes, as seen from the “bulb” frame of reference, would come closer together, squeezing the “mirror” reference grid together like one of those fold-out books, until they get to 45 degrees (the speed of light), where the mirror frame axii are completely squeezed together and the grid lines disappear completely. Stare at the diagram above and you might be able to imagine it. This might seem really weird at first, but it is a consequence of the speed of light being the same (at 45 degrees) in both frames.

What do you think the “bulb” frame of reference would look like in the “mirror” frame of reference?  See the diagram below:

Let’s now see if the math works out.  Let us look at a specific point (x, C*t) in the mirror frame and ask how it would look in the bulb frame. First let’s write the equation for the “C*tb” axis as seen in the mirror frame.  Let’s say it is in the direction (Dx, Dt). Recall that this is the same direction as the “bulb” world line is in the above world diagram. Then any point (xtb, C*ttb) at a distance St along the C*tb-axis can be written:

   (xtb, C*ttb) = (Dx, Dt)*St

Where (Dx, Dt) marks the direction of the bulb world line in the mirror frame, and St is a distance marked along the C*tb-axis, and is yet to be determined.  We can also find the equation of a point along the xb-axis in the mirror frame.

    (xxb, C*txb) = (Dt, Dx)*Sx

Note that (Dt, Dx) is the direction of the xb-axis which is at the same angle, At, above the x-axis (as required by the space-time geometry) and Sx is the (yet undetermined) distance marked along the xb-axis.

From our orienteering days, we have seen that we can arrive at the point (x, C*t) by “moving” along the C*tb-axis a distance St, and then turning right to “move” in the direction parallel to the xb-axis a distance Sx (as seen by the green lines in the diagram).  Note that the Sx and St are the coordinates in the “bulb frame” of reference. And so:

    (x, C*t) = (Dx, Dt)*St + (Dt, Dx)*Sx

We can now “dot” both sides with a direction that is perpendicular to the C*tb-axis (Dt, -Dx) to eliminate the St term and thus find Sx:

 (Dt, -Dx).(x, C*t) = (Dt, -Dx).(Dx, Dt)*St + (Dt, -Dx).(Dt, Dx)*Sx
                                =(Dt*Dx + –Dx, Dt)*St  + (Dt*Dt – Dx*Dx)*Sx
                                = 0*St + (Dt2 – Dx2)*Sx
                                = (Dt2 – Dx2)*Sx

and we can find the St coordinate by a similar process:

 (-Dx, Dt).(x, C*t) = (-Dx, Dt).(Dx, Dt)*St + (-Dx, Dt).(Dt, Dx)*Sx
                                =(-Dx*Dx + Dt, Dt)*St  + (-Dx*Dt + Dt*Dx)*Sx
                                = (Dt2 – Dx2)*St + 0*Sx
                                = (Dt2 – Dx2)*St

If we define the metric such that

        Dt2 – Dx2 = 1,

(called a hyperbolic metric) then the “magic” matrix that transforms the event (x, C*t) into the (Sx, St) event coordinates in the “bulb frame” is found by combining the three equations above in a matrix equation, as usual:

     [(Dt, -Dx), (-Dx, Dt)].(x, C*t) = (Sx, St)

And there we have it, the way to convert between the two frames of reference, this is also referred to as the “Lawrentz Transform”.

We should say a few words about the meaning of the hyperbolic metric Dt2 – Dx2 = 1.  If we draw a plot of this metric, we can see that Dx, and Dt are defined by coordinates of points on the unit hyperbola as shown:

There are special functions, the hyperbolic sine (Sinh) and cosine (Cosh), that are defined to find (Dx, Dt) from the angle At:

   (Dx, Dt) = (Sinh(At), Cosh(At))

We see that this is a new way of defining distances in Space-Time.  If the point (x, C*t), is in the direction (Dsx, Dst):

           (x, C*t) = (Dsx, Dst)*S   (where S is the spacetime distance)
then   1 = Dst2 – Dsx2
           S2 = S2*(Dst2 – Dsx2)
           S2 = (S*Dst)2 – (S*Dsx)2
           S2 = (C*t)2 – (x)2

And we have found a way to define distance in space-time.

Returning to our moving bulb example, we finish off by finding the direction (Dx, Dt) of the bulb, moving with a speed “v” in the“positive-x-direction” in space-time (which is also the direction of the “C*tb” axis in the mirror frame).  After a time ‘tb’, the bulb has traveled a distance of xb = v*tb thus:

   (v*tb, C*tb) = (Dx, Dt) * St

Where St is the space-time distance traveled by the bulb, which we now know to be:

    St2 = (C*tb)2 – (v*tb)2
          = (C2 – v2)*tb2

And so

St = Sqrt(C2 – v2)*tb

   And thus we have found:

Dx = v/Sqrt(C2 – v2)
Dt = C/Sqrt(C2 – v2)

Finally we have come full circle and filled in the unknowns.  Don’t worry if you are having trouble understanding it all. It still takes me a bit of time to completely wrap my mind around these concepts.  Just for reference, in science literature, Dt is often called the Greek letter “gamma” and the ratio ‘v/c’ is often called “beta”, thus Dx = gamma*beta and Dt = gamma.

I believe that that is sufficient for now. You might have many questions, but if you read and study this post a few times, you will be able to get the gist of how Special Relativity works to transform between Space-Time reference frames.

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Math Moments – Special Relativity

So far, we have been thinking of the time (C*t) dimension to be just another dimension in space, completely symmetric with the other three directions (x, y, z), yet we have already seen a problem with this concept.  The reasonable concept of time requires that a point cannot travel everywhere with infinite speed and be everywhere at the same time. Einstein postulated that the speed of light is the speed limit that prevents everything from happening all at once.  This concept creates limitations on the “time direction” (Dt) and even affects the “symmetric” geometry of space-time.

Let us look at a simple example that illustrates this concept.  In space-time, every point (x, y, z, C*t) is called an “event” with a position in space and in time.  Let us consider a flash bulb traveling in the positive ‘x’ direction with speed “v”.  When the bulb is exactly between two mirrors, it flashes.  We can call this flash our reference event, (0, 0, 0, 0).  Let us suppose that the two stationary mirrors are on both sides of the flash bulb, so that when the bulb flashes, the space-time coordinates of the two mirrors are (-xm, 0, 0, 0) and (xm, 0, 0, 0) respectively, where xm is the distance to each of the mirrors.  See the diagram below.  It is assumed that the light from the flash will propagate out in all directions (in a perfect sphere) at the speed of light.

It is simple to predict what will happen as time passes, the light from the flash will travel at the speed of light toward both mirrors. Since we assume by symmetry that the speed of light traveling in the positive ‘x’ direction is the same speed that it travels in the negative ‘x’ we imagine that it will arrive at both mirrors at the same time, tm, and then reflect back toward the center.  The diagram below is called the “world diagram” in the reference frame of the mirrors, it maps out what will happen at any time “t” in the future.  In this diagram we only show the ‘x’ axis, for simplicity, describing only what happens in the ‘x-direction’, which is plotted against the ‘C*t’ axis that shows what will happen in time.

In this diagram, the flash of light follows the yellow “light line” (where x = C*t in the ‘positive-x-direction’ and x = -C*t in the ‘negative-x-direction’).  In this diagram, then, Light lines always travel at a 45-degree angle off the ‘C*t’ axis.  Anything that is traveling at a speed slower than light (everything else) travels at an angle less than 45 degrees off the ‘C*t’ axis.  For example, the flash bulb that is traveling with a speed ‘v’ in the positive x-direction on this diagram travels along its green world line (x = v*t) and, of course, ‘v’ is less than ‘C’. In this diagram, the lines tell you where everything will be at any time ‘t’ in the future.

The world line for the mirrors, for example, go ‘straight up’, the mirrors do not move in space, they start at -xm and xm and just travel through time along the ‘C*t’ axis.  From the above diagram, can you see everything that will happen?  The light will hit both mirrors at the same time ‘tm’ and bounce back. The flash bulb will just keep moving off in the x-direction, and the mirrors will just stay where they are.  Staring at the diagram you can see it all.  You might even see that the flash bulb will flash again, when the light bouncing back off the mirror hits it.

The events are labeled on the diagram, the flash is at the origin, light bounces off the mirror in the positive direction at ‘O+’ and simultaneously bounces of the mirror in the negative direction at ‘O–‘,  ‘O’ is the midpoint between O+ and O–, and ‘P’ is when the light comes back to center.

Things start getting a bit weird, however, if you are looking at the world in the reference frame of the flash bulb.  In this frame of reference (you are traveling along with the bulb), the flash bulb is not moving, yet the two mirrors are moving at a relative speed ‘v’ in the negative-x-direction.  As before, the flash is triggered when the bulb is exactly in the mid-point between the mirrors.

In this reference frame, the moving mirrors are certainly not going to change the velocity of light, you will still see the light propagating outward in a perfect sphere at a 45 degree angle in your world diagram.  There is an obvious difference, however. You observe the light bouncing off the mirror that is traveling toward you (O+) before it bounces off the mirror that is traveling away from you (O–), the distances are different.  Please refer to the diagram above. In the frame of the flash bulb, the events O+ and O– do not happen at the same time.

How can this be? In the first case, the reference frame of the mirrors, the light bounces off both mirrors at exactly the same time, but in the frame of reference of the flash bulb, the light bounces off the approaching mirror first before it bounces off the receding one.  Which observation is right? Looking at the two world diagrams of each case, it is obvious that one diagram is not a simple rotation of the other.  Distances and directions have changed when the relative velocities change.  Lines between events that are perpendicular in one frame of reference are not perpendicular in the other.

So, we see that if there does exist a uniform speed limit ‘C’ in all frames of reference, that is the same regardless of direction or speed of the observer, then events that are simultaneous in one frame, must happen at different times in another, and the angles and distances observed between events in the space-time diagram must change from one reference frame to another.  This concept is what was going through the mind of Albert Einstein when he formulated the theory of Special Relativity.

In order for the math to work, to predict what will be observed in one reference frame from the perspective of the other, we find that the symmetry or “geometry” of space-time cannot be spherical and distances are not preserved between different “rotations” or frames of reference.  For spherical symmetry and geometry, Dx2 + Dy2 + Dz2 + Dt2 = 1 is true for all directions and times.

We now are starting to see that this doesn’t work in real space-time, where the speed of light is constant in all frames of reference and rotations of world lines cannot exceed 45 degrees off the ‘C*t’ axis.  There is a geometry that does allow this all to happen, it is called “hyperbolic” geometry where distances are measured as Dx2 + Dy2 + Dz2 – Dt2 and vary by the velocity of one frame of reference relative to another.  Mathematically, this is the true working version of space-time and is called Minkowski Space.  The math has been meticulously worked out. We will cover some of the details later. The interesting thing is that through more than a century of observations, the Theory of Special Relativity has been proven time and time again.  Scientists have still not been able to find an exception.  It is used in everything from GPS, to positioning of satellites, to radioactive decay of cosmic particles.  It works! But it takes a bit of getting used to.  Space and time are inextricably connected.

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Math Moments – Finding Directions in 4-D

As we enter dimensions beyond our physical experience, it might help our understanding to look back and see patterns in smaller dimensions that still apply in larger dimensions.  Is there such a pattern when finding directions (Dx, Dy, Dz, Dt) in 4 dimensions or larger.  One obvious pattern, as dimensions increase, is the way we measure distance:

   1-D:         Dx2 = 1
   2-D:         Dx2 + Dy2 = 1
   3-D:          Dx2 + Dy2 + Dz2 = 1
   4-D:          Dx2 + Dy2 + Dz2 + Dt2 = 1

The pattern for higher dimensions is obvious. We expect this pattern to continue to any number of dimensions; we just keep adding on perpendicular triangles.  See the diagram above.

There is one problem with setting directions in higher dimensions that is not so obvious. For example, in 1-D, Dx2 = 1 means that Dx = 1 or Dx = -1, this defines the directions to be along the positive or negative x-axis.  So far so good. In 2-D, however, when Dy2 = 1, it forces Dx2 = 0, which does not indicate either a positive or negative direction along the x-axis.  This ambiguity is due to the rotational symmetry of perpendicular lines, as we have mentioned before, you can “flip” negative and positive directions around a perpendicular direction, and nothing changes.

This ambiguity of direction exists whenever we add a new dimension.  In 3-D, for example, when Dz2 = 1, it forces Dx2 + Dy2 = 0 and there are ambiguities as to where to place the direction of both the x-axis and y-axis.  To a man standing exactly on the north pole, for example, every direction is south, there is no east and west.  I have also heard it said that no matter how you comb the fuzz flat on a tennis ball, there will always be a part (singularity).

To deal with these ambiguities when any of our direction parameters is zero, it is helpful to fix the angles of the x-axis, y-axis, z-axis, etc. so that directions will not be ambiguous when you pass through the “singularities” where one or more of the direction parameters are forced to be zero.  So a man standing on the north pole, where every direction is south, still knows which angle to face if he wants to point to Paris, France.  This makes defining directions consistent.

Let us look at a way to define directions as we are increasing dimensions. We defined the angle Axy to be the angle from the x-axis in the x-y plane.  Then we defined the angle Az to be the angle off of the x-y plane in the z direction.  Listing the direction parameters for the dimensions we know:

   2D: D2 = (D2x, D2y) = (Cos(Axy), Sin(Axy))
   3D: D3 = (D3x, D3y, D3z) = (D2x*Cos(Az), D2y*Cos(Az), Sin(Az))
                 = (D2x, D2y, 0)*Cos(Az) + (0, 0, 1)*Sin(Az)

We added the “2” and “3” subscripts to indicate the dimension.

It might take you a few minutes to see the pattern.  Expanding into 4-D, we define the angle At to be the angle off the x-y-z space toward the t-axis, then continuing the pattern, the direction in 4-D is:

    D4 = (D4x, D4y, D4z, D4t) = (D3x, D3y, D3z, 0)*Cos(At) + (0, 0, 0, 1)*Sin(At)

    So putting them together and stacking them we get:

         D4x = Cos(Axy)*Cos(Az)*Cos(At)
         D4y = Sin(Axy)*Cos(Az)*Cos(At)
         D4z =                   Sin(Az)*Cos(At)
         D4t =                                   Sin(At)

And we have found the Direction parameters in 4-D, and hopefully we can see the pattern that will allow us to use the same scheme to find directions in any dimension. An interesting side note: The angle in the t-direction, At, determines how close the 4-D direction is to the t-axis. When it gets close to 90 degrees, the direction is close to the t-axis and Cos(At) is close to zero, and D4x, D4y and D4z are close to zero, this restricts motion in the x-y-z space. The faster time is moving, the more restricted the motion is in 3-D space. How strange. We will see just how strange in the following post.

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Math Moments – Does Space-Time Exist?

We are now ready to expand our 3-D space into 4 Dimensions.  My ability to draw diagrams in 4-D is very limited, so we will not have any this time.

We need to pick a point in 4 dimensions that does not exist anywhere in our 3-D world.  At the start, we might be tempted to think that such a world is simply unphysical and impossible until we consider that the 4th dimension might be related to time.  Consider a point (x, y, z, t), where ‘t’ represents the time (watch) that is associated with the point (x, y, z).  Does this even make sense, points running around with timer watches?

     It might be easier to believe if we consider a single point that is able to move around in 3-D space.  When we track the same point (x, y, z) over time, we can imagine that it doesn’t have to be in the same position all the time relative to where it is right now.  If we want to track the movement of a point through time, then its associated timer watch (t) comes in handy.  We can think of the path of the point walking through time as walking through a path in 4-D “space-time”, where the point (x, y, z, t) describes the position of the point (x, y, z) at the time ‘t’.

Suppose that we choose the origin (0, 0, 0, 0) to be the position of the point when we set the timer to zero.  Then a “new point” in 4-D space-time could be (0, 0, 0, 1) which describes the same point in the same place 1 second later. We are getting closer to constructing our new 4-D space.  Now we need to draw line segments from every point in 3-D space to this new point in 4-D space-time and then need to locate the shortest of these line segments (like we did when expanding 2 dimensions to 3 dimensions).

A problem arises, however, because the time ‘t’ is not a distance, so it is confusing to find a measure for distance in this 4-D space.  What is x2 + y2 + z2 + t2?  One way to solve this problem is to somehow convert the time ‘t’ into a measure of distance:

    (Distance/Time)*Time = Distance

By choosing a constant speed ‘C’  (with units of Distance/Time) that is the same for every point in this 4-D space, we have a way to convert any time into a distance:

      C*t = the distance traveled in time ‘t’

So, if C is truly a constant speed for all points in space-time, then ‘C*t’ is the distance that any point with this constant speed ‘C’ would travel in a time ‘t’.  So let us redefine a point in 4-D space time to be described by:

   (x, y, z, C*t)

And we have a 4-D space of distances.  Then distance ‘S’ between the origin (0, 0, 0, 0) and any point (x, y, z, C*t) is found to be:

   S2 = x2 + y2 + z2 + (C*t)2

And, as usual, the direction ‘D’ of a line from the origin to any point can be expressed as the vector:

    D = (Dx, Dy, Dz, Dt)

      where

    D.D = Dx2 + Dy2 + Dz2 + Dt2 = 1

Here Dt expresses how fast time is changing as you travel a 4-D distance ‘S’ along the line.  As usual, lines from the origin in 4-D spacetime can be expressed as:

     (x, y, z, C*t) = (Dx, Dy, Dz, Dt)*S

                        or

     (x, y, z, C*t) = (Dx*S, Dy*S, Dz*S, Dt*S)

‘Dx*S’ expresses how far the point has traveled in the x-direction, as usual, and the other 3-D coordinates have similar meanings.  The new one is ‘Dt*S’ that describes how much time has elapsed after we have traveled a distance ‘S’ along this 4-D line.  Dt determines the speed of the timer.  Notice that the 4-D distance ‘S’ is not the same as the 3-D distance we will now call ‘S3’:

     S32 = (Dx*S)2 + (Dy*S)2 + (Dz*S)2
         = (Dx2 + Dy2 + Dz2)*S2

Since 1 – Dt2 = Dx2 + Dy2 + Dz2, we see a more interesting expression for S3:

   S32 = (1 – Dt2)*S2

This is interesting, and maybe a bit surprising, because it says that the distance that the point (x, y, z, C*t) travels in 3-D space is completely determined by the Dt direction parameter of the 4-D line.  If you look at it a bit closer, however, it starts to make sense.  Since 1 – Dt2 = Dx2 + Dy2 + Dz2 then when Dt = 1, the other direction parameters, Dx2, Dy2, and Dz2 must add to zero.  Thus (Dx, Dy, Dz, Dt) must be (0, 0, 0, 1), and the point is only moving in the ‘time’ direction, but not moving in any of the 3-D directions.  This 4-D direction describes a point that is not moving in 3-D space, S32 = 0.

If Dt is less than 1 then the point has room to move in 3-D space.  A more perplexing situation happens when Dt = 0.  Then in this direction, time stands still, no time is elapsing, but the point can go any distance it wants in the 3-D space.  Physically this makes very little sense.  The point could be everywhere all at once. Let’s look at what this means in terms of the speed ‘V’ of the point.  Speed is the 3-D distance it travels per the time elapsed ‘t’ (recall C*t = Dt*S):

    V = S3/t
    V = C*S3/(Dt*S)

    V2 = C2*S32/(Dt*S)2
         = C2*(1 – Dt2)*S2/(Dt2*S2)
         = C2*(1 – Dt2)/Dt2

And we see that the speed of the point is constant everywhere on the 4-D line and it is completely determined by how fast the point is traveling in the ‘time’ direction, Dt.  Things also start to be more understandable.  However, if Dt goes to zero, then the speed is infinitely fast (you can’t divide by zero).  Having things that can go at infinite speed is very problematic.  The point could literally be everywhere at once.  One fun definition of time is: “Time is something that keeps everything from happening all at once”.  It makes sense to have some sort of a speed limit for our points.

Einstein, a brilliant physicist, made the argument that the speed ‘C’ (a constant speed for all points in space-time) should be that speed limit.  If we assume this (C2 >= V2), then we can find restrictions on the direction parameter Dt:

     C2 >= C2*(1 – Dt2)/Dt2
      1 >= (1 – Dt2)/Dt2
      Dt2 >= ½

This puts some serious restrictions on the directions of lines that are possible to traverse in space-time:

    ½ <= Dt2 <= 1

Dt is between the square root of ½ and 1.  Another restriction that we see is that we cannot go backwards in time, Dt cannot cross over to become negative unless the point violates the speed limit and can go infinitely fast.

So, a 4-D space-time can exist mathematically, but for it to make sense physically, certain restrictions must be placed on the directions of lines we can draw in this space-time. Next time, we will look at how to find (Dx, Dy, Dz, Dt) in space-time and their associated angles.  We will introduce a new angle, At, that describes the angle of the “time” direction.  We will find that Dt = Sin(At).  At the speed limit ‘C’, Dt is the square root of ½.  From trig this means that At = 45 degrees.  The restrictions require that At is between 45 degrees and 90 degrees.  This “cone” of 4-D space-time is the only place where anything can exist when it comes from the origin.  It is commonly called the “light cone”. And the speed of light is the speed limit ‘C’.

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Math Moments – What do Dot Products Mean?

We have seen the power of the dot products between two direction vectors.  We have shown that the dot product between two perpendicular directions is zero.  We will review this in two dimensions, because it is easier to see. Using our head reference the nose direction in 2-D is N:(Dx, Dy) and the Left ear direction is L: (-Dy, Dx):

   N.N = (Dx, Dy).(Dx, Dy) = Dx2 + Dy2 = 1
   N.L = (Dx, Dy).(-Dy, Dx) = -Dx*Dy + Dy*Dx = 0

We have also seen that we can travel to any point in 2-D (x, y) by traveling in the N direction a distance SN and then turning left and traveling in the perpendicular L direction another distance SL:

    (x, y) = N*SN + L*SL

And then we used the power of the direction dot products to see what happens when we dot these direction vectors with any point (x, y) in the 2-D space:

   N.(x, y) = N.(N*SN + L*SL) = N.N*SN + N.L*SL = 1*SN + 0*SL
                  = SN

    L.(x, y) = L.(N*SN + L*SL) = L.N*SN + L.L*SL = 0*SN + 1*SL
                 = SL

Written in concise dot product notation:

      [N, L].(x, y) = (SN, SL)

This means that dotting these “head direction vectors” (N, L) with any point gives you back the coordinates of the point in the “head frame of reference” (SN, SL), the distance traveled in the Nose direction and the distance traveled in the Left direction to arrive at the point.  From the last post, in 3-D, we saw that SN is the distance from the Nose to the perpendicular View Plane where the point lives.

We can conclude that in any dimension, if we dot the Nose direction (N) with any point we get back the “nose coordinate” in the “nose frame of reference” (SN):

     N.(any point) = SN

It might take a few moments of thought for this to sink in, but when it does, we start to realize that we can convert any point into the “nose frame of reference” no matter where our nose is pointing.  It is like rotating the reference frame into any direction we want and converting the coordinates into this “nose frame of reference”.

We might ask what happens when we dot two direction vectors together that are not perpendicular.  Suppose we have two Nose directions (N1 and N2) in 2-D:

   N1 = (Dx1, Dy1)
   N2 = (Dx2, Dy2)

Since any point (x, y) can be expressed in the N1 frame of reference, (Dx2, Dy2) can also be written in the N1 frame:

  (x, y) = N1*SN + L1*SL
  N2 = (Dx2, Dy2) = N1*SN + L1*SL

Refer to the diagram above for clarification.  Now we can easily see what happens when we dot N1.N2:

    N1.N2 = N1.(N1*SN + L1*SL) = N1.N1*SN + N1.L1*SL
               = SN

Now from the diagram we also can see that SN is just the Cos(A) because the length of N2 is 1, where A is the angle between the two directions:

   N1.N2 = Cos(A)

So, when we dot two direction vectors together, we get the Cosine of the angle between them.  This is consistent with what we have seen, because when the two directions are the same, A = 0 degrees, and when they are perpendicular, A = 90 degrees. Cos(o) = 1 and Cos(90) = 0 this is consistent with what we have shown when dotting direction vectors together so far.

   I have earned some good money from knowing this.  On one occasion, a ship captain wanted me to find the distance between two points in the ocean given the longitude and latitude of each point.  Recall that the longitude and latitude is the same as the “head direction” as seen from the center of the earth. (Axy1, Az1 and Axy2, Az2).

I realized that if I could find the angle between these two “head” directions (A), I could find the distance between the two points along the surface of the earth.  Since I know that the circumference of the earth (CE), the distance between the points is CE*(A/360).  I found N1 (the direction vector to the first point) and N2 the direction to the second point, dotted them together to find the angle: N1.N2 = Cos(A), and took home a large check (he still uses it to navigate).  Could you fill in the details?

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Math Moments – How to Project 3-D on a 2-D Flatscreen

No matter where we turn our head in our 3-D world, we see the world projected on the 2-D “screen” in the back of our eyes.  These days, flat screens and cell phones are everywhere, they can display 2-D images on a flat surface that our brain interprets as being projections into a 3-D world.  From the last post, we are now equipped with math language to describe how this projection works.

We can imagine that everything we see in front of our face is displayed on a big flat screen that is always a certain fixed distance in front of our nose and moves around with our head (it could be transparent).  Let’s call this screen the “View Plane” and say that it is always placed a fixed distance (SV) perpendicular to our Nose direction (N).  This is not hard to imagine, I see lots of people carrying their cell phones around like this in front of their face all the time.

Your personal reference point (0, 0, 0), the origin, is always between your ears, the Nose direction (N), always points straight out in front of your face, the Left Ear direction (L) always points in the direction of your left ear, and the Top of head direction (T) always points through the top of your head.  This personal reference frame moves as you change the orientation of your head.

Since the View Plane in front of you is 2-D, every point on this screen can be located on a 2-D “xv-yv plane”.  Let’s draw such Cartesian coordinates on our View Plane. We choose the origin of the View Plane (0, 0) to be where our Nose direction (N) passes through the plane. We can choose the xv-axis on the plane to point to the right (opposite the L direction), and then the yv-axis would point in the Top of Head direction (T).

A single point (pixel) on this 2-D view screen (xv, yv), with a color (cv), can also be located in the 3-D world.  Start at the origin (0, 0, 0), the middle of your head, then travel a distance Sv in the N direction to arrive at the origin of the View Plane (0, 0), then from there travel along the xv-axis (in the negative L direction) a distance xv, and then along the yv-axis (in the T direction) a distance yv.  From our orienteering days, we already know how find this 3-D point in math language:

    (x, y, z) = N*Sv + (-L)*xv + T*yv
                   = [N, L, T].(Sv,-xv,yv)

And so, we can find the 3-D coordinates for any point on the 2-D View screen (xv, yv) when we know the distance to the view screen (Sv) and the orientation of the head (N).  In review, the middle of the head is always at the origin (0, 0, 0) and you are looking in the Nose direction N:(Dx, Dy, Dz) in the 3-D world, the direction vectors of the head reference frame are:

         N = (Dx, Dy, Dz)
         L = (-Dy/Dr, Dx/Dr, 0)
         T = (-Dx*Dz/Dr, -Dy*Dz/Dr, Dr)

         Dr2 = Dx2 + Dy2

In the Cartesian frame of the 3-D world around us the orientation of our head can be determined by two angles (Axy and Az).  As you remember, Axy is the angle we have rotated our head left or right in the x-axis in the x-y plane, and Az is the angle that we have tilted our head up or down off the x-y plane.  In review, the direction vector of the head is found from the following:

       Dx = Cos(Axy)*Cos(Az)
       Dy = Sin(Axy)*Cos(Az)
       Dz = Sin(Az)

The dot products of the [N, L, T] head reference vectors are:

   N.N = L.L = T.T = 1
   N.L = N.T = L.T = 0

since they are all perpendicular to each other.
We also find that:

     N.(x, y, z) = N.( N*Sv + L*(-xv) + T*yv )
                        = N.N*Sv + N.L*(-xv) + N.T*yv
                        = 1*Sv + 0*(-xv) + 0*yv
                        = Sv

     L.(x, y, z) = L.( N*Sv + L*(-xv) + T*yv )
                        = L.N*Sv + L.L*(-xv) + L.T*yv
                        = 0*Sv + 1*(-xv) + 0*yv
                        = -xv

     T.(x, y, z) = T.( N*Sv + L*(-xv) + T*yv )
                        = T.N*Sv + T.L*(-xv) + T.T*yv
                        = 0*Sv + 0*(-xv) + 1*yv
                        = yv

Another way of writing these stacked equations with dot products is simply:

      [N, L, T].(x, y, z) = (Sv, -xv, yv)

So, we can see that with this miracle matrix [N, L, T] can map a point in 3-D space onto a point (xv, yv) on a View Plane that is a distance Sv away.

Let’s finish this idea off.  Suppose we have any point in 3-D space (x, y, z), then

[N, L, T].(x, y, z) = (SN, -xN, yN)

maps (x, y, z) onto a point (xN, yN) on 2-D view plane that is a distance SN away from your nose.  Using simple geometry, this point can be mapped onto a point (xv, yv) in any parallel View Plane that is a distance Sv away by using:

    (xv, yv) = (xN, yN)*Sv/SN

And so, we can map a 3-D world onto a 2-D view plane.  Our brain perceives these 2-D images from colored points (pixels) from our flat screens as 3-D images.

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Math Moments – Navigating in 3-D Space

We can navigate in 3-D to any point (xR, yR, zR), in Cartesian Coordinates, by starting at the origin (0, 0, 0) and in the 2-D, x-y plane, walking in the x-direction (1, 0, 0) a distance xR, then turning left 90 degrees and walking a distance yR in the y-direction (0, 1, 0), still on the x-y plane.  This is the same as we did in 2-D except we put an extra ‘0’ in the third ‘z’ direction.  At this point we are still on the x-y plane at the point (xR, yR, 0).  The z-direction is straight up, so we would now need a ladder to climb up or a shovel to dig down, it is helpful to pretend we can fly (or dig).  From this point (xR, yR, 0) on the x-y “ground” plane, we fly straight up in the z-direction (0, 0, 1) a distance zR, and we successfully arrive at the point (xR, yR, zR).

We can express this journey with vectors, as we did in 2-D, by connecting (adding) lines together to form the path:

   (xR, yR, zR) = (1, 0, 0)*xR + (0, 1, 0)*yR + (0, 0, 1)*zR
                        = (xR, 0, 0) + (0, yR, 0) + (0, 0, zR)
                        = (xR, yR, zR)

        And using dot product notation, this can be written:

   (xR, yR, zR) = (1, 0, 0)*xR + (0, 1, 0)*yR + (0, 0, 1)*zR
                        = [(1, 0, 0), (0, 1, 0), (0, 0, 1)].(xR, yR, zR)

The [(1, 0, 0), (0, 1, 0), (0, 0, 1)] matrix is called the identity matrix.

We could also arrive faster by flying along a straight line connecting the two. Fly along a direction (Dx, Dy, Dz), from the origin directly to this point, (xR, yR, zR) a distance “R” away:

   (xR, yR, zR) = (Dx, Dy, Dz)*R

        Where the distance ‘R’ is found by:

    R2 = xR2 + yR2 + zR2

From the equation above for this line, we can easily find the direction vector from the origin to any point (xR, yR, zR), similar to what we did in 2-D.

      (Dx, Dy, Dz) = (xR/R, yR/R, zR/R)

Recall that the direction vector always has a distance of 1 unit, as seen in the diagram above.  In the diagram, we have attempted to show a 3-D path on a 2-D piece of paper.  Note that in the diagram, Dr is the distance from the origin to the point (Dx, Dy, 0) in the x-y plane. From the Pythagorean theorem, we find that the length:

      Dr2 = Dx2 + Dy2    and
      Dr2 + Dz2 = 1        thus combining these
      Dx2 + Dy2 + Dz2 = 1.

The total length of the distance vector is 1 unit, as required.  Everything is consistent.  Now we use the definition of the ‘Sin’ and ‘Cos’ and can see, from the diagram above, that:

     Cos(Axy) = Dx/Dr
     Sin(Axy) = Dy/Dr

      Cos(Az) = Dr/1
      Sin(Az) = Dz/1

    Thus

       Dx = Cos(Axy)*Dr = Cos(Axy)*Cos(Az)
       Dy = Sin(Axy)*Dr = Sin(Axy)*Cos(Az)
       Dz = Sin(Az)

And so, if we stand at the origin facing in the x-direction, and rotate our “head” an angle Axy to the left, and then tilt our head an angle Az off the x-y plane to look directly at the point (xR, yR, zR), we can find the direction vector pointing to this point using these angles:

   (Dx, Dy, Dz) = (Cos(Axy)*Cos(Az), Sin(Axy)*Cos(Az), Sin(Az))

       And then draw the line to this point:

    (xR, yR, zR) = (Dx, Dy, Dz)*R

Now that we are staring at a point in the sky, with our nose pointed directly at the point, we can ask about “reference frame” of our head.  What direction is our left ear pointing, for example.  Our left ear is pointing in the direction, still in the x-y plane, that is 90 to the left of the Axy angle we rotated our head (-Dy/Dr, Dx/Dr, 0). We divide by Dr because in 3-D we must make the length of the direction vector equal to 1.  If this seems vague, look back and review how to find 90 degree directions in the 2-D plane in the top diagram taken from previous posts.

When we tilt our head, we can ask what is the direction that the top of our head points, 90 degrees up from the angle Az. It perhaps takes a bit more imagination to see this, since we are facing in the x-y plane (Dx/Dr, Dy/Dr, 0) and have tilted our head up to look at Dz, we find that the “shadow” of our head is pointed behind us, in the x-y plane (-Dx/Dr, -Dy/Dr, 0), then looking at the elevation Dr (when rotated 90 degrees) we find that (-Dx*Dz/Dr, -Dy*Dz/Dr, Dr) is the direction of the top of our head.  So, in the reference frame of our head:

     Direction of our Nose:     (Dx, Dy, Dz)
     Direction of Left ear:         (-Dy/Dr, Dx/Dr, 0)
     Direction of Top of head:  (-Dx*Dz/Dr, -Dy*Dz/Dr, Dr)

Let us verify that these direction vectors all have a length of 1.  The Nose direction we have verified.  For the Left ear direction:

    L.L = (-Dy/Dr)2 + (Dx/Dr)2 + 02 = (Dy2 + Dx2)/Dr2 = 1

For the Top of head direction:

    T.T = (-Dx*Dz/Dr)2 + (-Dy*Dz/Dr)2 + Dr2 = (Dy2 + Dx2)*Dz2/Dr2 + Dr2
                                                                    = Dz2 + Dr2 = 1

Now let’s find what happens when we find the dot product of these direction vectors, we will call them the N, L, and T directions respectively.

N.N = (Dx, Dy, Dz).(Dx, Dy, Dz) = Dx2 + Dy2 + Dz2 = 1

N.L = (Dx, Dy, Dz).(-Dy/Dr, Dx/Dr, 0) = -Dx*Dy/Dr + Dy*Dx/Dr + 0
                                                                  = 0

N.T = (Dx, Dy, Dz).(-Dx*Dz/Dr, -Dy*Dz/Dr, Dr)
        = -Dx2*Dz/Dr + -Dy2*Dz/Dr + Dz*Dr
        = -(Dx2 + Dy2)*Dz/Dr + Dz*Dr
        = -Dr2*Dz/Dr + Dz*Dr
        = 0

L.T = (-Dy/Dr, Dx/Dr, 0).( -Dx*Dz/Dr, -Dy*Dz/Dr, Dr)
       = Dy*Dx*Dz/Dr2 + -Dx*Dy*Dz/Dr2 + 0
       = 0

We also verified that L.L = 1 and T.T = 1 above.  So, in general, if we dot any of these N, L, T direction vectors with themselves, we get ‘1’, but if we dot any of them with each other, we get ‘0’.  It turns out that this is a requirement for any three perpendicular direction vectors in a 3-D Cartesian reference frame.  I find the “head” reference frame to be more natural because we can relate to it better, so I use this reference frame for whatever direction I am looking.  It is also a great reference frame for video games.

We have already seen from perpendicular 2-D paths how we can convert between a reference frame to our “ground” x-y-z reference frame.  To understand this, let’s take a journey: Fly in the Nose direction a distance SN, then fly in the Left ear direction a distance SL, and then in the Top of head direction a distance ST.  Then we will arrive at the point:

    (x, y, z) = N*SN + L*SL + T*ST

       or in dot product notation:

     (x, y, z) = [N, L, T].(SN, SL, ST)

This amazing equation lets you convert any point in your “head frame” of reference (SN, SL, ST), when your Nose is pointing in a direction N:(Dx, Dy, Dz), to a point in the “ground” frame (x, y, z).  You might also want to do this the other way around.  Review how we did this in 2-D, by “dotting” each side by each of the direction vectors, we get:

    (SN, SL, ST) = [N, L, T].(x, y, z)

This is how video games convert a 3-D world, stored in “ground” (x, y, z) coordinates, into the perspective of the viewer’s head (SN, SL, ST) coordinates.  The details need to be worked out but using this equation we can map a 3-D world onto a 2-D screen in the viewer’s perspective, no matter the relative direction of the viewer’s head.  We will stop here, there is a lot to take in.

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Math Moments – Review and 3-Dimensional Expansion

All the concepts that we have discussed so far can be expanded into new dimensions.  We have only been looking at two dimensions so far, and yet we live in a world of 3 physical dimensions.  Although mathematically we don’t have to stop at 3 dimensions, we could expand the same concepts to work in as many dimensions as we want. First, let’s start by generating a 3-dimensional space from the 2-dimensional space we have been working in.

Do you remember how we generated new dimensions? Look back to the expanding dimensions post to refresh your memory. Take the 2-dimensional x-y plane where we have been working and find a point “P” that is NOT in it.  Now imagine line segments between every point in our 2-D space (x-y plane) and the new point P.  Find the point on the x-y plane (2-D space) that forms the line segment with the shortest distance to P and label it “O”.   Call this minimal distance “r”.

We can choose the point O to be the origin on the x-y plane and pick an x-direction for the x axis, which also determines the direction for the y-axis.  We can now use the same diagram above, the same procedure (and a bit of imagination) to generate a 3-D space from a 2-D space with one observation. In our original 1-D space there were only two directions from the point O, negative (O-) and positive (O+).  In the 2-D space, there are an infinite number of directions pointing away from the point O, and each direction is indistinguishable from any other direction.

We can use this rotational symmetry to generate our 3-D space.  Imagine that you printed the diagram above and on a sheet of paper and placed it so that the point O is on the origin of the x-y plane, the point P (the new point) is above the x-y plane, and make the x-axis go through the O+ point.  The sheet (plane) of our diagram would now be perpendicular (vertical) to the x-y plane.  The same argument holds that since the O-P segment is the shortest one that connects the x-y plane it is an axis of symmetry.

We call this new O-P direction the “z” axis.  Now imagine rotating this diagram around the z axis.  The O+ point would trace out a circle on the x-y plane, and touch every point on the plane that is a distance “r” from point O.  The distance from every one of these points on this circle would also be the same distance from P.  The segments P-O+ would sweep out the surface of a cone as we rotated the diagram around the z axis, and parallel lines would sweep out parallel planes along the z axis. We can rotate our sheet of paper around the z-axis to any angle and the cross section would still remain the same. Can you see it?

We can use the same trick as before, we can imagine that these parallel planes are mirrors facing each other that will reflect an infinite array of parallel planes to the x-y plane along the z axis, thanks to reflection symmetries.  We can also make an infinite set of parallel planes along the x axis, and along the y-axis by processes similar to those we did before in 2-D to make a cartesian “grid” of planes and fill in all the gaps.

Note that the line segment O-P does not move when we rotate the diagram around it.  It is stationary under rotation.  Now we have fully leveraged the rotational and reflectional symmetries of 3-D space.  We can pick any axis of rotation (a line), rotate the universe around it, and the universe will remain unchanged, only our vantage point changes.

As always, we can pick any point in 3-D space to be the reference point (origin). And any 2 perpendicular directions (an x and y axis) can be chosen to define a 2-D plane, and the third direction (z axis) perpendicular to this x-y plane is determined. These three directions define how we write points in this new 3-D space.

In cartesian coordinates,  the vector (x, y, z) can label every point and (Dx, Dy, Dz) can label every direction.

Lines in 3-D space can be defined by:

   (x, y, z) = (x0, y0, z0) + (Dx, Dy, Dz)*S

Where (x0, y0, z0) is the point of origin for the line, (Dx, Dy, Dz) is the direction of the line, and “S” is the distance along the line from the origin.  At this point, this notation should not be difficult or surprising.  The above is also an equation for a sphere if you hold S constant to be the radius and vary the direction.

There might be some questions on how you can determine directions in 3-D space. Similar to 2-D space, the direction vector always has a length of 1 unit:

    Dx2 + Dy2 + Dz2 = 1

A natural way to determine directions is by using your head.  Stand facing North (the x-axis), now rotate your head an angle “Axy” to the left (West towards the y-axis).  You can determine every direction on the x-y plane with this angle Axy (between -180 and 180 degrees).  Now tip your head up an angle “Az” off the x-y plane (between -90 and 90 degrees).  Varying these two angles, you can now look in any direction possible in 3-D space.

This is also how we locate positions on the spherical surface of the earth.  We place an imaginary observer in the middle of the earth and have it “face” 0 degrees longitude and 0 degrees latitude (the x-axis).  The Axy angles are degrees of longitude as the observer rotates its head around the north pole (z-axis), and the Az are degrees of latitude as the observer tips its head up and down.  Using this scheme the observer can look at any place on the surface of the earth given a longitude (Axy) and latitude (Az).

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Math Moments – The Rotation Matrix

On the Cartesian plane, if we walk along the x-direction (1, 0) for a distance of 4 units, then turn left 90 degrees (perpendicular) and walk a distance of 3 units, we arrive at the point (4, 3), which is a distance of 5 units from the origin (42 + 32 = 52), great.  Well, what happens if we do the same, but start off walking in any direction (Dx, Dy) 4 units and then turn left 90 degrees (-Dy, Dx) and walk 3 units; we are still 5 units from the origin.  We have just “rotated” the whole path to a new direction.  Where would we end up (x, y)?  We use our orienteering skills:

     (x, y) = (Dx, Dy)*4 + (-Dy, Dx)*3
     (x, y) = [(Dx, Dy), (-Dy, Dx)].(4, 3)  –  using the dot product

Now notice that the matrix [(Dx, Dy), (-Dy, Dx)] when dotted with the cartesian point (4, 3) rotates it to another point on the circle (radius 5).  See the diagram. This matrix is called the “rotation matrix”.

Now that we are getting a bit more comfortable with the dot product, let’s play around and see what happens when we dot direction vectors with each other?  We’ll start off by dotting a direction with itself:

     (Dx, Dy).(Dx, Dy) = Dx2 + Dy2 = 1

Any direction dotted with itself is just “1”.  What about dotting a direction with its perpendicular direction?

    (Dx, Dy).(-Dy, Dx) = Dx*(-Dy) + Dy*Dx = 0

It turns out that any direction dotted with its perpendicular direction is “0”.  Now how about dotting the position (x, y) above with the direction vector?

   (Dx, Dy).(x, y) = (Dx, Dy).((Dx, Dy)*4 + (-Dy, Dx)*3)
                             = (Dx, Dy).(Dx, Dy)*4 + (Dx, Dy).(-Dy, Dx)*3
                             = 1*4 + 0*3
                             = 4

And so, dotting a direction vector (Dx, Dy) with any position vector (x, y) gives you the distance (Sx) you would have to travel in that direction before making a 90 turn and traveling the distance (Sy) that would bring you to the position (x, y).  Now let’s dot the perpendicular direction with the position vector:

   (-Dy, Dx).(x, y) = (-Dy, Dx).((Dx, Dy)*4 + (-Dy, Dx)*3)
                             = (-Dy, Dx).(Dx, Dy)*4 + (-Dy, Dx).(-Dy, Dx)*3
                             = 0*4 + 1*3
                             = 3

You can see that dotting the perpendicular direction with a position also gives you the distance (Sy) you would have to travel in the perpendicular direction (-Dy, Dx) to get to the position (x, y).

And so can you see that in this case:

     [(Dx, Dy), (-Dy, Dx)].(x, y) = (4, 3)

So in general if

     (x, y) = [(Dx, Dy), (-Dy, Dx)].(Sx, Sy)

              then

     [(Dx, Dy), (-Dy, Dx)].(x, y) = (Sx, Sy)

What does this mean?  The rotation matrix […] when it “operates” on (is dotted with) any point (x, y), gives you the coordinates of the point relative to a new set of “x-y” axis that is pointing in a direction (Dx, Dy) relative to the old axis.  Or you can say that it rotates the point around the origin.  Either vantage point is valid, you can rotate the point around the origin or rotate the reference frame in the opposite way around the origin; either way, it is describing the same thing.

Remember the concept of “rotational symmetry”, that we can pick the direction of the x-y axis to be in any direction we want.  Now we have the tools to convert the cartesian coordinates of a set of points from any reference frame to a reference frame pointing in any direction we choose.

THIS IS A FUNDAMENTAL CONCEPT of the MATH Language: ROTATIONAL SYMMETRY means that the universe is the same no matter the direction of your frame of reference.  The Rotation matrix gives you a way of describing the same universe from any prospective of direction.  It is pure magic.

One more point, remember that

   (Dx, Dy) = (Cos(A), Sin(A))

Where “A” is the angle of the direction off x-axis.  Thus, the rotation matrix “R(A)” can be written as:

     R(A) = [(Cos(A), Sin(A)), (-Sin(A), Cos(A))]

And there you have it.

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Math Moments – Using Vectors in Orienteering

Soldiers and airplanes in unfamiliar places need to find their way around. They can do this with methods similar to what we have already discussed.  They receive instructions to go first in a certain direction (Dx1, Dy1) for a certain distance S1, and then to change course to another direction (Dx2, Dy2) and keep going for a distance S2.  This same pattern can be followed for many different directions and distances. This discipline is called orienteering.  Soldiers are trained to find directions with their compasses and measure distances with the number of steps they take.

With our new vector notation, we can say the same thing in math language:

     (x, y) = (Dx1, Dy1)*S1 + (Dx2, Dy2)*S2

Which says that the point of our destination (x, y) can be found by going in the direction (Dx1, Dy1) for a distance (S1) and then going a direction (Dx2, Dy2) for a distance (S2).  Notice that to find our destination (x, y), we just add these two vectors together.  It is just like piecing together lines to form a path.

Any destination (x, y) is determined by just adding together all the lines (vectors) that form the path in between.  In the example above, there were only 2 directions and distances (lines), but we could keep on going for as many vectors as we want, adding on lines, one line after another until we arrive at our destination.

It is helpful to define yet another notation that is even more compact.

     (x, y) = (Dx1, Dy1)*S1 + (Dx2, Dy2)*S2

   can be written

     (x, y) = [(Dx1, Dy1), (Dx2, Dy2)] . (S1, S2)

The dot “.” between the two vectors is called a “dot product”.  In the case above the dot product means you multiply the first direction vector in the square brackets [1,2] with the first distance in the round brackets (1,2) and do the same with the second direction vector and second distance and then add them both together.

This new notation might seem a bit mysterious at first, but it does make things clearer later on.  The dot product is one of the most fundamental operations of “linear algebra” which is the math that tells you where you end up when adding together a bunch of lines (vectors) to form a path.

We do something similar to write the (x, y) coordinates of a point in our cartesian plane.  Let’s look at the point (3, 4), we can get there by walking along the x-direction 3 units, then turning left 90 degrees and walking in the y-direction 4 units:

    x-direction: (Dx1, Dy1) = (1, 0)
    y-direction: (Dx2, Dy2) = (0, 1)

Then using the “dot product” notation we discussed above:

    (x, y) = [(Dx1, Dy1), (Dx2, Dy2)].(S1, S2)

    (x, y) = [(1, 0), (0, 1)].(3, 4)
               = (1, 0)*3 + (0, 1)*4
               = (3, 0) + (0, 4)
               = (3+0, 0+4)
               = (3, 4)

We find that we end up at the point (3, 4).  This is also one way in the cartesian coordinates that we can define the location of a point.  So, the [(1, 0), (0, 1)] vector when “dotted” with any two distances (Sx, Sy) brings you to the point (x, y) = (Sx, Sy).

     (x, y) = [(1, 0), (0, 1)].(Sx, Sy) = (Sx, Sy)

Do you get the point?  You are starting to see the “language” of the dot product.  In math language a vector of vectors is called a matrix and [(1, 0), (0, 1)] is called the “identity matrix”, because when you dot it with any vector you get back the same vector.

The dot product can be applied to any two vectors of the same size. This concept is used all over the place.  Here is an interesting example, we can use the dot product notation to write down the number 365 (in decimal) starting with the vector (3, 6, 5):

   (3, 6, 5).(100, 10, 1) = 3*100 + 6*10 + 5*1 = 365

This is how we write numbers (in base 10). It is read like this: 3 in the 100’s column, 6 in the 10’s column and 5 in the one’s column, added together, normally written “365”.

We could also use it to find the number of seconds from midnight to 04:35:20am from the vector (4, 35, 20):

   (4, 35, 20).(3600, 60, 1) = 4*3600 + 35*60 + 1*20
                                                  = 16520 sec

There are 3600 seconds in an hour, and 60 seconds in a minute.  I think you got the idea.

Standard