Let’s say that! I wanted to specify the temperature in a city like New York. For instance. How would I do that well? I could simply use a number with a corresponding unit like 273 Kelvin, which is equivalent to zero degree Celsius and 32 degrees Fahrenheit. In other words, I could use a quantity which just has a magnitude. I could use a scalar to specify the temperature so to specify the temperature in New York. I just need the single number. This single component to specify the temperature, of course, there are no basis vectors involved in specifying this component because it’s just the magnitude temperature doesn’t really have a direction. As a result, One could say that there are zero basis vectors for this one component or zero basis vectors per component now staying with New York. Suppose that I wanted to know the displacement, not the distance the displacement from the JFK Airport, which I’ll label with point a to the top of the Empire State Building, which I’ll label with Point B. How would I specify this displacement? Well, in this case, I would need to give you. The magnitude of the displacement, which is about twenty two point five kilometers, as well as the direction of the displacement, which is, of course in the direction specified by the vector A B. Therefore, I would need a vector, a quantity that has both a magnitude and a direction to specify the displacement between the airport and the Empire State Building. One thing about this vector is that I can actually break it up into three components because I’m in three dimensions. So if I had a coordinate system where the X axis was going from west to east, the Y axis was going from south to north and the Z Axis was going from up to down. Then I could specify the displacement using three components, for instance. I could say that in order to go from A to B, You should go 19 kilometers west 12 kilometers north and about 0.45 kilometers of as a result one way. I could write the displacement. D is by writing it as negative 19 I plus 12 Jade plus 0.45 five K. If I use the IJK unit vectors in the X Y&Z directions, respectively, each of these three vectors that are added to give you. The total displacement are called the component vectors of this displacement. This means that the vector, which describes my displacement from. A to B has three components and one basis vector for each component. I’ve got the basis vector I. For the 19 kilometers west component, the basis vector J for the 12 kilometer north component and the basis vector K for the 0.45 kilometer up component. I’ve got one basis Vector corresponding to each component staying with the New York theme that we’ve got going suppose. I’m looking at a rectangular steel beam on Brooklyn Bridge. Let’s also suppose that there is a point O inside that steel beam. If I wanted to specify all the stresses that are acting on point. O, how would I go about doing that? Well, what I can do is that I can start by making three cross sections of the beam through this point. O, I can make a cross section that cuts the beam in half like this. I can make another cross section like this and a third cross section, which looks like this. I’ll also let the orientation of my XY and Z axes look like this. In this first cross section, the cross-sectional area is perpendicular to the X direction. In other words, the X axis makes a right angle with this shaded face of the beam and because the beam is under a whole bunch of stresses, we can say that the point. O, which is on the back. Half of the beam generally experiences a force per unit area from the front half of the beam, which can be broken up into three components. P xx P XY and p XZ. The first subscript in this P denotes the direction the area is perpendicular to. So in this case, the first subscript X means that the cross sectional area were looking at is perpendicular to the X direction. The second subscript denotes the direction the force is acting in. The idea is similar for the second cross section. The second cross section is perpendicular to the y-direction. Once again, we can say that the point. O, which is on the left half of the beam now generally experiences a force per unit area from the right half of the beam, which can be broken up into the components, Py X py y and Py Z And the third cross section, which is perpendicular to the Z direction, is similar point O experiences a force per unit area that can be broken down into the components. P ZX P Zy and PZ Z. We can actually combine all these force per unit area components into a three by three matrix as follows. Now you might be wondering why can’t. I just add the forces in the same direction and just end up with a force for you to area vector instead of a force per unit area matrix. For instance, why can’t I just straight up Add P xx and Py X. After all, they’re acting in the same x-direction. Well, it’s because even though the forces are acting in the same direction on the same point, the nature of those forces is different px. X acts to pull this point O forward and py X acts to shear or drag point O forward shearing and pulling actually caused the steel beam to be deformed in different ways. So that’s why. In addition to specifying the force, it’s necessary to specify the surface that the force acts on when looking at the stresses in the beam, so the way to specify the stresses on point O is to use this matrix P, where each component of P corresponds to a force per unit area on a particular surface that the point O inhabits and has a particular direction along which it acts. We can see that P has nine components and each component is specified by a magnitude and two basis vectors one vector for the cross-sectional area that’s being acted on and another vector for the force acting on that cross-sectional area. This is because as I just mentioned, we need to know both the direction of the surface area and the direction of the force. If we want to specify the nature of a stress component, it is not enough to use the force direction alone. So for instance, the component PXX corresponds to two basis vectors in the positive x-direction or two. I vectors while P. XY corresponds to one basis vector in the X direction for the area and a basis vector in the Y direction or J vector for the direction of the force. We can continue this logic for all the other components of P as well, For instance, P X Z would be specified by a basis vector in the X direction for the area and a basis vector in the Z direction for the force. Now all three of these mathematical objects that we talked about have something in common. They’re all something we call tensor’s. If we’re in an M dimensional space, a tensor of rank N is a mathematical object that has N indices M to the power n components, and it obeys certain transformation rules in the examples we talked about. We were dealing with a three-dimensional space and in most cases, that’s what we’ll be dealing with a three-dimensional space. One exception is in general relativity, where we’ll be dealing with a four dimensional space with time being an additional dimension. What about the rank of the tensor? Well, you can think of the rank of a tensor as the number of basis vectors you need in order to fully specify a component of the tensor, for instance. If we go back up to our scalar, we can see that we needed zero basis vectors to specify our scalar component. Therefore, we can say that a scalar is a tensor of Rank zero. What about a vector well for each component of the vector? I showed you that each component is specified by one basis Vector either. I J or K. Therefore, we can say that a vector is a tensor of Rank 1 and finally, what about our stresses on the beam well? I told you that! In order to specify each of the nine stress components, you needed two basis vectors per component, one basis vector for the area, the other basis vector for the direction of the force, and this makes P a tensor of Rank 2 also known as a stress tensor. Now, the definition of tensors that I wrote down here says that the number of components in my tensor equals M to the power N. Let’s verify this from the three tensors. We discussed earlier. My scalar has a rank of zero, so it has three to the power zero or one component, which is obviously true. My vector has a rank of one. So it has three to the power one or three components while. P, my stress tensor has a rank of two, so it has three to the power, two or nine components. You might have noticed that. I’ve hesitated quite a bit to call Kia Matrix. And that’s because there’s a fairly common misconception that rank two tensors and matrices are the same thing that’s not quite correct. A matrix is just an array of numbers, whereas a tensor has special transformation properties. It obeys certain transformation rules and it’s got of physical significance. Of course, we can use a matrix to represent a tensor, but a tensor actually has a deeper physical significance to it, so it would be rather inaccurate to say that a tensor and matrix are the same thing. Now, just for fun. Let’s also briefly mention tensors of rank three. You can represent tensors of rank three as three dimensional arrays, 3d arrays are not the same thing as tensors, just like matrices are not the same thing as tensors, but we can use 3d arrays to represent rank. Three tensor’s 3d arrays by the way are basically just matrices that are stacked on top of each other because this is a tensor of Rank 3 Each component of the tensor is specified by three separate basis vectors. In addition, a tensor of Rank 3 has 3 to the power 3 or 27 components, which again follows the end to the power and component rule in the tensor definition. I wrote above now. My goal with this video was to first get you warmed up to the idea of tensors and to give you a bit of intuition about what Tensors are. My belief Is that with this intuition, you’ll be better able to understand the more rigorous definitions of tensors because in the next video, that’s what. I’ll be going over specifically in the next video. I’m going to talk about these transformation rules that define a tensor that I didn’t really go over in this video anyway. That should do it for this lecture. I’d like to thank the following patrons for supporting me at the five-dollar level or higher. 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