Vector & Tensor Fundamentals

Hi RMT!

I'm with you so far, that is very clear. I know how to multiply a vector by a scalar /ttiforum/images/graemlins/smile.gif I don't know the math involved to calculating a dot product or cross product. I knew them once and I still remember what they look geometrically - but whever I use them I just call a function, I don't have to multiply it out...

But I guess when you multiply a vector by a vector, the longhand version will help in understanding a 3x3x3 matrix by a 3-vector so I will read up on what is going on behind the scenes when I use cross product.
 
Scalar (DOT) Product & Vector (CROSS) Product

Hey bogz,
I don't know the math involved to calculating a dot product or cross product. I knew them once and I still remember what they look geometrically - but whever I use them I just call a function, I don't have to multiply it out...
Click to expand...
I understand. Many people learn them at one time, and often will quickly forget what these operations physically signify. But in understanding physics in general, and my equation that is an attempt at extending physics, I tend to think it is important to stay grounded in what specific mathematical operations (especially multiplication) means as we extend the original thoughts. And I thought this would be a good "time" in this thread to try to explain the SCALAR (dot) PRODUCT and the VECTOR (cross) PRODUCT and their significance to the physics of Massive SpaceTime engineering dynamics.


Dot Product of 2 Vectors Results in a SCALAR Quantity
We recall that any scalar quantity does not consider any specific direction in space as being preferred over any other. But what does this mean with the multiplication of two VECTOR quantities, which do consider direction in their measure? The physical significance of the dot product is that we are resolving the component of one of the two vectors (say the component of vector A) into the coordinate system of another vector (vector B), and then multiplying the component magnitude of A with the total magnitude of vector B. The dot product is commutative, so we could also describe this same operation as taking the component of vector B that acts in the direction of vector A and multiplying it by the total magnitude of vector A (just the opposite of what I just described). We state this mathematically as:

A . B = ABcos(theta) where "theta" is the angle measured between the vectors A & B.

So the numerical value of "dotting" these two vectors togther (recall each vector has an x, y, and z component) is nothing more than the total magnitude of A and B multiplied by the cosine of the angle between them. When we say "magnitude" of A and B we realize that we mean the following equation from Pythagoras:

Magnitude(A) = SQRT(Ax^2 + Ay^2 + Az^2) and similar equation for vector B's components.

The result of the DOT product is a number, and nothing more... no (x-y-z) direction is associated with this number. An example of this would be "Air Temperature". No matter which direction a thermometer faces at a single point, we measure the same temperature. Other good examples would be the scalar quantities we know as the simple measurements of Energy, and even Power. We only associate Energy with the number of Joules or Foot-Pounds associated with an energetic process. We do not claim that the energy has a specific direction associated with it.

However, we can consider the direction of how two vectors combine with each other, which means we can consider the directional tendencies of temperature, energy, and power, but we need a different mathematical operation to describe the resulting direction of how two vectors combine with each other. We call that mathematical operation...

Cross Product of 2 Vectors Results in a VECTOR (and Orthonormal) Metric
Note immediately that where the dot product results in a quantity , the vector product results in a metric. We say that a metric is a measurement that has both a magnitude and a direction. So we see that a metric shares the definition of a vector, but it is a special kind of vector because it is the result of the combination of two other (orthogonal) vectors. Therefore, the VECTOR METRIC is a fundamentally different beast than the SCALAR QUANTITY which results from the dot product.

The importance of the CROSS product as an operation over and above the DOT product is seen when we consider that the cross product EXTENDS DIMENSIONALITY FROM 2-D to 3-D. It does this because it projects the 2 dimensions defined by the 2 vectors into a third dimension that is mutually orthogonal to the the dimensions of the two original vectors. We use the "right hand rule" to visualize how this works:

1) Hold your right hand open flat, with fingers extended in front of you.
2) The direction of your fingers represents the direction of the first vector (A) you are going to cross with another vector (B).
3) As you begin to curl your fingers inward you are literally rotating the direction of vector (A) towards the direction of the vector (B).
4) This represents the crossing of A into B, and the projected dimension of this crossing is represented by the direction of your thumb. The result of AxB points in the direction of your thumb.
5) While the result of AxB points in the direction of your thumb, the resulting vector has a magnitude of

MAGNITUDE (AxB) = ABsin(theta) where theta is again the angle between vectors A and B

6) Note that if we perform (BxA) that your thumb will point in the opposite direction. So in terms of polarity (what might be called "spin" in QM) we would say:

direction of (AxB) = -direction of (BxA)

The best example to use for the physical significance of the CROSS PRODUCT is how it applies to rotation of a body as a result of a FORCE applied at some DISTANCE from a center of rotation. We call the result of this multiplication a TORQUE or a MOMENT. This is one of the most fundamental physical applications of the vector cross product, because when we cross a distance, or position vector, with a force that is applied at the end of this position vector, we will physically cause a body to rotate about the axis that is at right angles to both the position vector and the force vector.

The way I teach this to my aerospace students is by holding an airplane model and explaining how aerodynamic pitching moments created by the airplane cause the airplane to rotate about its pitch axis... in other words to pitch the airplane's nose up or down.

This mathematical operation that describes how two physical elements (a Force at a Displacement from a central point) combine with each other to create rotation about a mutually orthogonal axis is an extremely important mathematical model for a fundamental physical entity. And it is my assertation that the cross product is fundamental to understanding how the 3 elements of Mass, Space, and Time interact with one another.

Think of this: GRAVITY acts along a 1-dimensional axis within our 3-dimensional universe. This means that gravity is omnidirectional, and therefore must be subject to interactions with the two directions that are at right angles to it. Therefore, if we wish to figure out how to "nullify" gravity acting along its 1-dimensional axis, we must come to understand how Mass, Space, and Time interact with one another along the lines defined by gravity, and how they interact with each other in the two directions perpendicular to the axis of gravity.

The math is there. Yet our current treatments of Force, Energy, and Information are not yet dealing with them as full 3-D, integrated, dimensional quantities. Much of the "laws of physics" deal in scalars. Only when we generlize them all to vectors and tensors, using the full power of the cross product, will we ever be able to say we "fully understand" gravity and electromagnetism.

But that is just my opinion for now... I am working on being able to prove it mathematically (with tensors of course!)

RMT
 
Re: Scalar (DOT) Product & Vector (CROSS) Product

Hi Fletch,
Ray is there more
Click to expand...
There sure is. I sorta let this thread die because there were no questions or responses. If you have no questions so far, then perhaps we can move on to the concept of a rank 2 tensor, and use the example of mechanical stress. From there we can talk about covariance and contravariance.

I'll post more this weekend,
New/Improved RMT
 
Rank 2 Tensor (One Step Beyond a Vector)

OK, so it looks like Fletch (and possibly other lurkers) are happy with the basics of scalars and vectors, and the mathematical operations thereon, and are ready to expand into higher dimensional extensions of the vector...the generalized tensor. This can be a very involved topic, and one will no doubt see how involved it is once we get to our discussion of covariance and contravariance in how tensors operate. For now, we will introduce these multi-dimensional vectors with an example that should make it fairly easy to understand: The concept of mechanical stress within a solid body, which is a rank 2 tensor (i.e. it has two indices that denote its components)

Let me point out here at the beginning that I will ONLY be presenting the conventional interpretation and usage of tensors which are related to the directionality of 3-D Space. However, if you can comprehend this conventional usage where tensor components align with the 3 dimensions of Space, you should be able to extend it to handle my Massive SpaceTime theory which also treats Mass and Time as 3-D vector constructs.

First it is important to understand the units of stress: Force per unit Area (F/A). These are the same units used for fluid pressure. Examples of these units in both the Imperial and Metric unit systems would be Pounds per Square Inch and Newtons per Square Meter, respectively. In fact, in differential calculus treatments of fluid flowfields, one can also define a pressure tensor which is similar to the solid body stress tensor.

In examining the units, we can immediately identify that in "F/A" we know that the "F" is a Force, and we know that a Force is already a vector quantity (it has both magnitude and a directional orientation). Using an (x,y,z) coordinate system we can then decompose the resulting force into components acting in these 3 directions (Fx, Fy, Fz). To understand how we extend this Rank 1 tensor into the Rank 2 tensor called stress, we must now consider that each of these Force components (Fx, Fy, Fz) acts on some internal surface area within the solid object. Let us define this surface area as a plane oriented as either parallel to or coincident with the "y-z" axes of the (x,y,z) coordinate system. If you need diagrams to gain a visual understanding you can find some HERE (along with an explanation of stress that I am attempting to simplify here).

We are now ready to define THREE of the NINE components of the stress tensor, and we will do so by considering one component of the resultant Force vector (Fx) as it affects the surface area defined by the "y-z" plane. In mechanics we break down these 3 components of stress into two categories: Normal stress (acting normal to the defined plane, in this case "y-z") and tangential or Shear stress (acting tangentially to the defined plane).

Using the greek letter Sigma as our variable for stress, we can first define the Normal stress component on the "y-z" plane due to the force component (Fx) as:

Sigma(x-x) = (Fx)/Area(y-z)

Now this force component (Fx) induces stress in the tangential directions as well, so we can also define the other two components of stress caused by (Fx), and we call these the shear stress components:

Sigma(x-y) = (Fx)/Area(y-z)
and
Sigma(x-z) = (Fx)/Area(y-z)

The link I provided above will also help you understand what the two subscripts to sigma represent. For this case we can describe our 3 components of stress caused by (Fx) as follows:

Sigma(x-x) = Stress caused by force in the "x" direction acting normal to the (y-z) plane, which is otherwise known as the "x" direction.
Sigma(x-y) = Stress caused by force in the "x" direction acting tangentially to the "y-z" plane in the "y" direction.
Sigma(x-z) = Stress caused by force in the "x" direction acting tangentially to the "y-z" plane in the "z" direction.

So the above is just one "vector" out of three "vectors" which comprise the total stress tensor. We can derive similar sets of 3 stress components for the force components (Fy) and (Fz) We can now combine all NINE of these stress tensor components into a square matrix, as follows:

The Stress Tensor Matrix
{Sigma(x-x)___Sigma(x-y)___Sigma(x-z)}
{Sigma(y-x)___Sigma(y-y)___Sigma(y-z)}
{Sigma(z-x)___Sigma(z-y)___Sigma(z-z)}

So now we can see what we mean by a Rank 2 tensor having 2 indices. In the case of the stress tensor we see the first index which tells us which axial force component we are considering, and then the second index which tells us which direction of stress we are considering due to that axial force. Each index has a range of values of (x,y,z), and when you permutate each index over this range of three values, you end up defining the NINE elements of the stress tensor we see above.

Are we OK so far? /ttiforum/images/graemlins/smile.gif
New/Improved RMT
 
Re: Rank 2 Tensor (One Step Beyond a Vector)

Thanks for the info Ray. I understand tensors a bit more now. Enough to visualize a rank 3 tensor /ttiforum/images/graemlins/smile.gif

One thing I am confused about though more related to the topic of stress not tensors,

Sigma(x-x) = (Fx)/Area(y-z)

Sigma(x-y) = (Fx)/Area(y-z)
and
Sigma(x-z) = (Fx)/Area(y-z)
Click to expand...

The right side of the equals sign is the same for all 3. Do they all have the same value in this case?

And when will you explain what each component in your I=ms^3 equation means? =D
 
Re: Rank 2 Tensor (One Step Beyond a Vector)

I see... This is a nice jump on my physics studies for the next school year. I'll be in AP Physics, so this stuff is kind of important. (and interesting)
 
Re: Rank 2 Tensor (One Step Beyond a Vector)

Hello bogz,
One thing I am confused about though more related to the topic of stress not tensors,
In reply to:
--------------------------------------------------------------------------------
Sigma(x-x) = (Fx)/Area(y-z)

Sigma(x-y) = (Fx)/Area(y-z)
and
Sigma(x-z) = (Fx)/Area(y-z)
--------------------------------------------------------------------------------
The right side of the equals sign is the same for all 3. Do they all have the same value in this case?
Click to expand...

Yes, I can see your confusion. I did not mean to use these as actual equations. Replace the "=" with a "~" symbol. What these equations are trying to express can only be understood along with the words wrapped around them, or by the differential equation form for Traction that is given on the web page that I referred to above. All of the above 3 components of stress are related to how the internal force in the "x" direction of a solid body act over the same representative surface area defined by the "y-z" plane that the force (Fx) acts normal to. Again refer to how we describe the Sigma(x-x) as the NORMAL stress (which either acts as Tension or Compression depending on the direction of the Fx force vector). We call it NORMAL stress because it acts normal to the y-z plane. The other two values of stress (Sigma(x-y) and (Sigma(x-z)) represent the SHEARING stress acting tangentially to the y-z plane. Sigma(x-y) is the SHEARING stress in the "y" direction of the y-z plane, and Sigma(x-z) is the SHEARING stress in the "z" direction of the y-z plane. Get it? It does take a while to grasp the concept, and the cube visual on the other web page can help.

Another way to visualize these three components of stress is to imagine a square sheet of rubber (call this the y-z plane) with its four corners attached to a board and a string attached in the middle of the square. Now when you pull on the string you exert a force on the rubber (call this the "x" direction) the rubber stretches. The NORMAL stress is due to the tensile force acting to pull on the y-z plane. There are also the internal SHEARING tensions acting in both the y and z directions inside the sheet of rubber. So those represent the shearing stresses. Now to visualize the full stress tensor you only need to image two more squares of rubber to represent the (x-y) and (x-z) planes and strings exerting forces normal to those planes. This entire tensor models the internal distribution of forces through a solid body.
And when will you explain what each component in your I=ms^3 equation means?
Click to expand...
I likely will not explain that here on this forum. And I will not explain the details of them until I have finished all of my mathematical analysis and derivation of my tensor model such that it can be peer reviewed and published. /ttiforum/images/graemlins/smile.gif But I will reply to your post in the other thread and give you some hints...

RMT
 
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