hobz Posted November 2, 2010 Posted November 2, 2010 What is a tensor and why is it useful? I have grabbed "Vectors, Tensors and the Basic Equations of Fluid Mechanics" by Rutherford Aris, but it is not a gentle introduction (some of the notation used is not explained at all!).
ydoaPs Posted November 2, 2010 Posted November 2, 2010 http://www.scienceforums.net/topic/49957-tensors-for-dummies/ <= theres a good pdf from NASA in that thread.
timo Posted November 2, 2010 Posted November 2, 2010 The easiest way to think of a tensor is probably as a linear function over a vector space (for those a bit more familiar with tensors note that I did not specify the range is or the number of arguments). I have no idea why functions are useful but I am quite sure they are.
ajb Posted November 2, 2010 Posted November 2, 2010 (edited) Fix a smooth manifold. A tensor is a special kind of geometric object. Definition: A geometric object (at a point) consists of with respect to any allowable coordinate system there is one and only one ordered system of functions called components with respect to the given coordinate system. a law which allows the representation of the components in an allowable coordinate system in terms of the components in any other allowable coordinate system, the corresponding coordinate transformations, the Jacobian matrix and their derivatives. Characteristics of a geometric object are the number of coordinates, the highest order of derivatives in the transformation law and the particular representation of the groupoid of coordinate transformations the object forms. Definition Tensors are first order geometric objects that form a linear representation. (For supermanifolds you will want to be a bit more relaxed on the representation, but this is another story.) So, the components of a tensor in a given coordinate system look like [math]T^{A_{1} A_{2} \cdots A_{p}}_{B_{1} B_{2} \cdots B_{q}}(x)[/math]. Then under a coordinate change like [math]x^{A} \rightarrow x^{A'}(x)[/math] the components transform as [math]T^{A_{1}^{\prime} A_{2}^{\prime} \cdots A_{p}^{\prime}}_{B_{1} ^{\prime}B_{2}^{\prime} \cdots B_{q}^{\prime}}= \left(\frac{\partial x^{B_{1}}}{\partial x^{B_{1}^{\prime}}} \right)\left(\frac{\partial x^{B_{2}}}{\partial x^{B_{2}^{\prime}}} \right)\cdots \left(\frac{\partial x^{B_{q}}}{\partial x^{B_{q}^{\prime}}} \right)T^{A_{1} A_{2} \cdots A_{p}}_{B_{1} B_{2} \cdots B_{q}} \left( \frac{\partial x^{A^{\prime}_{1} }}{\partial x^{A_{1}}} \right)\left( \frac{\partial x^{A^{\prime}_{2} }}{\partial x^{A_{2}}}\right) \cdots \left( \frac{\partial x^{A^{\prime}_{p} }}{\partial x^{A_{p}}} \right)[/math], where I have used the Einstein summation convention. (I have not worried about any ordering here as everything is commutative, again on supermanifolds you would need to be a little more careful.) Now, the important thing about these transformation laws is that they preserve the form of any tensor identity. Thus they are naturally suited to physics where nothing should really depend on the details of how you decide to present things. Other geometric objects, particularly densities, that is objects that pick up powers of the Jacobian are also very important in physics. To stress things, in this local description using coordinates the transformation law is the vital thing when describing tensors or more general geometric objects. You can do things globally without local coordinates using natural vector bundles. That is in terms of sections of a vector bundle built entirely from the data of a smooth manifold (or supermanifold). In more sophisticated language a natural vector bundle is a functor from the category of smooth manifolds to vector bundles such that local changes of coordinates become vector bundle automorphisms. The canonical example here is the tangent bundle whose sections are vector fields. Hope I have not confused you more than you already were. Edited November 2, 2010 by ajb
Bignose Posted November 3, 2010 Posted November 3, 2010 (edited) What is a tensor and why is it useful? I have grabbed "Vectors, Tensors and the Basic Equations of Fluid Mechanics" by Rutherford Aris, but it is not a gentle introduction (some of the notation used is not explained at all!). Wow. Despite the friendly-sounding title, this book is quite the mountain. If you finish and understand this book, you will have an excellent understanding of fluids, but the learning curve of this book is unbelievably steep. This is a book over the heads of most graduate students. Are you looking for an introduction to fluids or an introduction to tensors? I can recommend much more friendly books for either, that are still challenging without being overwhelming. To answer the question, vector (which is really a tensor of rank 1) and tensor quantities follow certain rules when coordinate changes happen. These rules are in place to ensure that the principle that nature doesn't have preferred coordinate system remains in effect. This is probably best explained by an example. Consider fluid flow in a (round) pipe. Because of the geometry, this is a problem that is very nicely described using the cylindrical coordinate system. However, nature doesn't know what a cylindrical coordinate system is. Nor what a Cartesian or spherical or bipolar or any of the other coordinate systems are. In the cylindrical coordinate system, we may describe the velocity at a certain point by [math] v_r, v_{\theta}, v_z[/math]. That velocity is the same as some other set of components in the Cartesian coordinate system [math] v_x, v_y, v_z[/math]. How you convert from one set to another is based on the rules of tensors. Shear stress in a fluid is a rank 2 tensor, and has components [math] \tau_{xx}, \tau_{xy}, \tau_{xz}, \tau_{yx}, \tau_{yy}, \tau_{yz}, \tau_{zx}, \tau_{zy}, \tau_{zz}[/math]. And, these components have some other values in another coordinate system. These components change from one to another coordinate system, following the rules of tensors. They change values so that no matter what coordinate system you choose to describe the problem with, in the end you describe the same thing. Edited November 3, 2010 by Bignose
hobz Posted November 8, 2010 Author Posted November 8, 2010 Wow. Despite the friendly-sounding title, this book is quite the mountain. If you finish and understand this book, you will have an excellent understanding of fluids, but the learning curve of this book is unbelievably steep. This is a book over the heads of most graduate students. Are you looking for an introduction to fluids or an introduction to tensors? I can recommend much more friendly books for either, that are still challenging without being overwhelming. To answer the question, vector (which is really a tensor of rank 1) and tensor quantities follow certain rules when coordinate changes happen. These rules are in place to ensure that the principle that nature doesn't have preferred coordinate system remains in effect. This is probably best explained by an example. Consider fluid flow in a (round) pipe. Because of the geometry, this is a problem that is very nicely described using the cylindrical coordinate system. However, nature doesn't know what a cylindrical coordinate system is. Nor what a Cartesian or spherical or bipolar or any of the other coordinate systems are. In the cylindrical coordinate system, we may describe the velocity at a certain point by [math] v_r, v_{\theta}, v_z[/math]. That velocity is the same as some other set of components in the Cartesian coordinate system [math] v_x, v_y, v_z[/math]. How you convert from one set to another is based on the rules of tensors. Shear stress in a fluid is a rank 2 tensor, and has components [math] \tau_{xx}, \tau_{xy}, \tau_{xz}, \tau_{yx}, \tau_{yy}, \tau_{yz}, \tau_{zx}, \tau_{zy}, \tau_{zz}[/math]. And, these components have some other values in another coordinate system. These components change from one to another coordinate system, following the rules of tensors. They change values so that no matter what coordinate system you choose to describe the problem with, in the end you describe the same thing. This is the most lucid explanation I have come across. I was really looking into tensors, but please do recommend books for fluid introduction as well. Thank you. Fix a smooth manifold. A tensor is a special kind of geometric object. Definition: A geometric object (at a point) consists of with respect to any allowable coordinate system there is one and only one ordered system of functions called components with respect to the given coordinate system. a law which allows the representation of the components in an allowable coordinate system in terms of the components in any other allowable coordinate system, the corresponding coordinate transformations, the Jacobian matrix and their derivatives. Characteristics of a geometric object are the number of coordinates, the highest order of derivatives in the transformation law and the particular representation of the groupoid of coordinate transformations the object forms. Definition Tensors are first order geometric objects that form a linear representation. (For supermanifolds you will want to be a bit more relaxed on the representation, but this is another story.) So, the components of a tensor in a given coordinate system look like [math]T^{A_{1} A_{2} \cdots A_{p}}_{B_{1} B_{2} \cdots B_{q}}(x)[/math]. Then under a coordinate change like [math]x^{A} \rightarrow x^{A'}(x)[/math] the components transform as [math]T^{A_{1}^{\prime} A_{2}^{\prime} \cdots A_{p}^{\prime}}_{B_{1} ^{\prime}B_{2}^{\prime} \cdots B_{q}^{\prime}}= \left(\frac{\partial x^{B_{1}}}{\partial x^{B_{1}^{\prime}}} \right)\left(\frac{\partial x^{B_{2}}}{\partial x^{B_{2}^{\prime}}} \right)\cdots \left(\frac{\partial x^{B_{q}}}{\partial x^{B_{q}^{\prime}}} \right)T^{A_{1} A_{2} \cdots A_{p}}_{B_{1} B_{2} \cdots B_{q}} \left( \frac{\partial x^{A^{\prime}_{1} }}{\partial x^{A_{1}}} \right)\left( \frac{\partial x^{A^{\prime}_{2} }}{\partial x^{A_{2}}}\right) \cdots \left( \frac{\partial x^{A^{\prime}_{p} }}{\partial x^{A_{p}}} \right)[/math], where I have used the Einstein summation convention. (I have not worried about any ordering here as everything is commutative, again on supermanifolds you would need to be a little more careful.) Now, the important thing about these transformation laws is that they preserve the form of any tensor identity. Thus they are naturally suited to physics where nothing should really depend on the details of how you decide to present things. Other geometric objects, particularly densities, that is objects that pick up powers of the Jacobian are also very important in physics. To stress things, in this local description using coordinates the transformation law is the vital thing when describing tensors or more general geometric objects. You can do things globally without local coordinates using natural vector bundles. That is in terms of sections of a vector bundle built entirely from the data of a smooth manifold (or supermanifold). In more sophisticated language a natural vector bundle is a functor from the category of smooth manifolds to vector bundles such that local changes of coordinates become vector bundle automorphisms. The canonical example here is the tangent bundle whose sections are vector fields. Hope I have not confused you more than you already were. Perhaps, but then again, I bet tensors (and their related background) are not taught as single-post forum replies I am having trouble with your notation (or the summation notation). What does it mean? Also," Tensors are first order geometric objects that form a linear representation", what does it mean to form a linear representation?
ajb Posted November 8, 2010 Posted November 8, 2010 I am having trouble with your notation (or the summation notation). What does it mean? If an index appears one up and once down you have a sum. For example [math]X^{A}\omega_{A}[/math] has a "secret" summation implied thus we really mean [math]X^{A}\omega_{A}= X^{1}\omega_{1} + X^{2}\omega_{2} + \cdots [/math] up to the dimension of the space the object are define on. Have a look at the Wikipedia entry. Also," Tensors are first order geometric objects that form a linear representation", what does it mean to form a linear representation? Tensors form a vector space is what we really mean here. You should think of this in terms of a representation of the groupoid of coordinate changes. Anyway, the point is as Bignose also states that tensors have a very nice transformation law when changing local coordinates. It is so nice that any equations between tensors (physical laws) remain of the same form.
hobz Posted November 8, 2010 Author Posted November 8, 2010 Have a look at the Wikipedia entry. According to this, "The upper indices are not exponents, but instead different axes. Thus, for example, x^2 should be read as "x-two", not "x squared", and corresponds to the traditional y-axis." What is meant by "traditional y-axis"? Anyway, the point is as Bignose also states that tensors have a very nice transformation law when changing local coordinates. It is so nice that any equations between tensors (physical laws) remain of the same form. Could you point out a simple example of this?
Bignose Posted November 8, 2010 Posted November 8, 2010 For tensors, I recommend Synge & Schild's Tensor Calculus. It is an older form of the subject, but I think it is significantly easier to grasp than some of the more modern treatments. S&S is still in print and a Dover book so it is a cheap own. I also like Sokolnikoff's Tensor Analysis, and it also has a chapter on fluid mechanics. I do think that this book is out of print as I bought it from a used bookseller. According to this, "The upper indices are not exponents, but instead different axes. Thus, for example, x^2 should be read as "x-two", not "x squared", and corresponds to the traditional y-axis."What is meant by "traditional y-axis"? traditional as in Cartesian x,y,z. A lot of the books will just use generalized coordinates to develop the subject. I.e. one coordinate system will just be labeled [math]x^i[/math], and a different will just be labeled [math]y^i[/math]. What x and y are, will be specific to a problem at hand. For example, if you are converting from Cartesian to cylindrical, and you let x be the Cartesian system, then [math]x^1 \rightarrow x, x^2 \rightarrow y, x^3 \rightarrow z[/math] where I am using the right arrow to mean "corresponds to". In this case [math]y^1 \rightarrow r, y^2 \rightarrow \theta, y^3 \rightarrow z[/math] And, in fact, there isn't any particular reason you have to "go in order"... [math]x^1 \rightarrow z, x^2 \rightarrow x, x^3 \rightarrow y[/math] is perfectly fine -- you just have to keep track of the indicies correctly. Could you point out a simple example of this? The Navier-Stokes equations of fluid mechanics: [math] \rho \frac{\partial\mathbf{v}}{\partial t} + \rho \mathbf{v}\cdot\nabla\mathbf{v} = -\nabla p + \mu\nabla^2\mathbf{v} + \rho \mathbf{b}[/math] This is the equation in vector form. It is valid for all Newtonian fluids. But, to apply it in any specific coordinate system, you will have to convert the components of the equation into the coordinate system you want to work in: http://en.wikipedia.org/wiki/Navier%E2%80%93Stokes_equations has them all typed out for Cartesian, cylindrical, and spherical equations (about half way down, and you'll see why I didn't re-type them here)
ajb Posted November 9, 2010 Posted November 9, 2010 According to this, "The upper indices are not exponents, but instead different axes. Thus, for example, x^2 should be read as "x-two", not "x squared", and corresponds to the traditional y-axis." What is meant by "traditional y-axis"? Bignose has covered this. I will stress that the indices do not necessarily refer to a Cartesian coordinate system. Could you point out a simple example of this? A nice example is that of vectors. As you know, the derivative has a nice transformation rule when you change coordinates. Namely [math]x^{A} \rightarrow \overline{x}^{A} = \overline{x}^{A}(x)[/math] [math]\frac{\partial }{\partial x^{A}} \rightarrow \frac{\partial }{\partial \overline{x}^{A}} = \frac{\partial x^{B}}{\partial \overline{x}^{A}} \frac{\partial}{\partial x^{B}} [/math]. (I have changes notation slightly, but I hope that won't add to your worries.) Now, we can define a vector field to be a first order differential operator acting on smooth functions over a manifold (sections of the structure sheaf). (There is a theorem that says "all first order differential operators acting on functions are derivatives".) In local coordinates we have [math]X = X^{A}(x)\frac{\partial}{\partial x^{A}}[/math]. This expression should be invariant under changes of coordinates. We know how the derivative changes, so we need something that will cancel that. Thus, [math]\overline{X}^{A} = X^{B}\frac{\partial \overline{x}^{A}}{\partial x^{B}}[/math]. This should be seen as the starting point for developing higher order tensors. In essence, they boil down to changing variables in the derivative.
hobz Posted November 9, 2010 Author Posted November 9, 2010 So is the tensor what takes a description in one coordinate system to the description in another? Or is it the description it self?
ajb Posted November 9, 2010 Posted November 9, 2010 So is the tensor what takes a description in one coordinate system to the description in another? Or is it the description it self? A bit of both, assuming I understand the question. You have a collection of functions specified in some coordinates system. This is the components of the tensor in a given coordinate system. By tensor, this maybe what is really meant. As it is, by definition (tautology) a tensor we know how it transforms when changing coordinates. So you may take that for granted. However, any geometric object requires you to state the transformation law.
pmb Posted April 30, 2012 Posted April 30, 2012 What is a tensor and why is it useful? I have grabbed "Vectors, Tensors and the Basic Equations of Fluid Mechanics" by Rutherford Aris, but it is not a gentle introduction (some of the notation used is not explained at all!). A while back when I decided to learn relativity it became clear to me that I needed to learn tensors. So I learned a good deal about them and decided to post the URL to the site to relativity sites and it seemed to help people who were attempting to learn relativity. The URL is located at http://home.comcast.net/~peter.m.brown/math_phy/math_phy.htm
imatfaal Posted April 30, 2012 Posted April 30, 2012 A while back when I decided to learn relativity it became clear to me that I needed to learn tensors. So I learned a good deal about them and decided to post the URL to the site to relativity sites and it seemed to help people who were attempting to learn relativity. The URL is located at http://home.comcast....hy/math_phy.htm Peter - none of the images do not display for me (firefox on linux)
studiot Posted April 30, 2012 Posted April 30, 2012 Flipping through this thread I couldn't make out your prime interest so here are some (probably irrelevent) comments. I found Aris a good book, but rather dated. Tensors are almost useless in real world fluid mechanics, but you can certainly spin a lot of theory with them. I would recommend the companion Dover book by Harley Flanders - Differential Forms with Applications to the Physical Sciences DFs are the modern 'alternative' to tensors in many applications and the gaining ground rapidly. HF contrasts both approaches. Another good book is Tensor Geometry by Dodson and Poston If you want to study mathematical fluids, but with a practical bent, you would have to go a long way to get a better book than Elementary Fluid Dynamics by Acheson. go well
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