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I am taking a fluid mechanics class and don't know very much physics. I was confused in class when the prof kept calling this derivative matrix of a fluid flow (A function from $\mathbb{R}^n\to\mathbb{R}^n$ representing fluid flow velocity) a tensor.

I confess I didn't get much out of reading the highly abstract math definition of tensor (multilinear, possibly antisymmetric etc.) and was wondering if I could get some insight on how physicists use and think of them? Is it just something multi-indexed?

Any suggestions of materials to read would also be appreciated!

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/32011/2451 , physics.stackexchange.com/q/20437/2451 , and links therein. $\endgroup$ – Qmechanic Apr 2 '17 at 22:52
  • $\begingroup$ @Qmechanic Thank you for linking, these didn't come up when I typed up my question. My quibble with the first is that I don't know what a matrix of second rank is. My quibble with the second is that I don't have a very good grasp of physics :). If it is okay, I will keep this post up for awhile and see if I get a good answer geared more towards my needs. $\endgroup$ – qbert Apr 2 '17 at 23:00
  • $\begingroup$ Check out Transport Phenomena, Bird, Stewart, and Lightfoot, in the appendices. $\endgroup$ – Chet Miller Apr 3 '17 at 20:04
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In physics we use the following (relevant to the tensor concept in three dimensional space, to make it simple):

scalars: these have a single value from the field of real numbers at each point is space . Example: temperature,

vectors: these have three values from the field of real numbers at each point in space and they obey vector algebra. They are useful to describe directional observations. example: force, velocity. (v_x,v_y,v_z) will describe the velocity and (v.v), the dot product of the vector is a single real number that is the speed. I suppose you are familiar with this.

In studying and measuring natural phenomena it was found that these were not enough in describing physical systems. Tensors were introduced, where for each point in space nine numbers are needed to describe the data. It is not enough to know the x component of the value under study, since it changes value ( at a fixed x) for different values the y axis and z axis.

All three obey particular coordinate transformation equations. Scalars do not change value, vectors and tensors follow the rules of the transformations.

A matrix representation for a tensor makes it simpler:

symmetric tensor

This example is a symmetric tensor , but good for an example. Looking at the columns it is like a vector, (x,y,z) components, looking at the row the same. It is used for physical quantities which differ in this "peculiar " manner and need all nine components to make sense. For example electric susceptibility in some crystal lattices.

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  • $\begingroup$ It is worth mentioning that what makes a scalar such is the fact of being independent of the reference frame chosen, not the fact of just being "real numbers". For example the coordinates of a vector, taken singularly, are also real numbers but they are by no means scalars (as they change with a change of reference). $\endgroup$ – gented Apr 3 '17 at 13:44
  • $\begingroup$ I don't know much about tensors, but doesn't the number of components depend on more than just the dimension of the vector space? I'd expect a type(1,1) tensor in 3-space to have nine components, but wouldn't a higher-order tensor have more than nine? $\endgroup$ – Solomon Slow Apr 4 '17 at 4:40
  • $\begingroup$ @jameslarge a higher order tensor in 3 dimensional space would have the clarification "higher order" in physics, rank 3 and larger in mathematics, though I do not know of a use in a physics model. Do you have an example in three dimensional space? The question is about physics, and the example is three dimensional space. $\endgroup$ – anna v Apr 4 '17 at 5:19
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You can find the rigorous definition anywhere, so here let me give a simple perspective: Tensors are building blocks of equations which are coordinate system independent.

For example, the width of a car when you see it from behind will be different than the "width" of the car when you rotate it 90 degrees. This is because this scalar "width" is not a product of a tensor. On the other hand the length of a rod will always be same regardless of the coordinate system in which you examine it, because length of an object is $l=\sqrt{\vec{l}\cdot\vec{l}}$, and the vector $\vec{l}$ is a tensor.

Like vectors, other tensors such as inertia tensor $I_{ab}$ are used in equations to make them coordinate system independent. We call them tensorial equations. For example, Einstein's GR equation is roughly $G_{\mu\nu}\sim T_{\mu\nu}$ which relates two second rank tensors, hence is valid at all coordinate systems.

Today, we know that tensorial equations are not enough to describe physics of QM, hence in addition we use spinor equations. They are generalized version of tensors.

The key point for both tensors and spinors is that they are all roughly the proper operators to work with if we want to respect some group transformation. And most of the time, this group is exactly the one which governs the coordinate transformations we want the equation to be independent of. For example, in Special Relativity, we want the equations to be independent of coordinate transformations, which are $SO(3,1)$ transformations, so we build our equations using $SO(3,1)$ tensors.

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In cartesian coordinates, a vector can represented as a summation of components times unit vectors:$$\vec{V}=V_x\vec{i}_x+V_y\vec{i}_y+V_z\vec{i}_z$$Similarly, a 2nd order tensor can be represented as a summation of components times pairs of unit vectors: $$\vec{T}=T_{xx}\vec{i}_x\vec{i}_x+T_{xy}\vec{i}_x\vec{i}_y+T_{xz}\vec{i}_x\vec{i}_z+T_{yx}\vec{i}_y\vec{i}_x+T_{yy}\vec{i}_y\vec{i}_y+T_{yz}\vec{i}_y\vec{i}_z+T_{zx}\vec{i}_z\vec{i}_x+T_{zy}\vec{i}_z\vec{i}_y+T_{zz}\vec{i}_z\vec{i}_z\tag{1}$$

Two vectors placed in juxtaposition like this, with no mathematical operation implied between them, is called a dyad. An example of a dyad is $\vec{a}\vec{b}$, where $\vec{a}$ and $\vec{b}$ are vectors. A dyad does not fulfill its mission in life until it is dotted with another vector. Then it maps the other vector into a new vector. Here are the two simple rules for dotting a dyad with a vector: $$\vec{c}\centerdot \vec{a}\vec{b}=(\vec{c}\centerdot \vec{a})\vec{b}=\vec{d}\tag{2}$$ $$\vec{a}\vec{b}\centerdot \vec{c}=\vec{a}(\vec{b}\centerdot \vec{c})=\vec{e}\tag{3}$$ In Eqn. 2, the dyad $\vec{a}\vec{b}$ maps the vector $\vec{c}$ into a new vector $\vec{d}$, and the vector $\vec{d}$ has the same direction as the vector $\vec{b}$. In Eqn. 3, the dyad $\vec{a}\vec{b}$ maps the vector $\vec{c}$ into a new vector $\vec{e}$, and the vector $\vec{e}$ has the same direction as the vector $\vec{a}$. A sum of components times dyads like Eqn. 1 is called a dyadic. Let's see what happens if we dot the dyadic sum in Eqn. 1 on its right side by a unit vector in an arbitrary direction $\vec{n}=n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z$:$$\vec{T}\centerdot \vec{n}=(T_{xx}\vec{i}_x\vec{i}_x+T_{xy}\vec{i}_x\vec{i}_y+T_{xz}\vec{i}_x\vec{i}_z+T_{yx}\vec{i}_y\vec{i}_x+T_{yy}\vec{i}_y\vec{i}_y+T_{yz}\vec{i}_y\vec{i}_z+T_{zx}\vec{i}_z\vec{i}_x+T_{zy}\vec{i}_z\vec{i}_y+T_{zz}\vec{i}_z\vec{i}_z)\centerdot (n_x\vec{i}_x+n_y\vec{i}_y+n_z\vec{i}_z)$$ $$\vec{T}\centerdot \vec{n}=(T_{xx}n_x+T_{xy}n_y+T_{xz}n_z)\vec{i_x}+(T_{yx}n_x+T_{yy}n_y+T_{yz}n_z)\vec{i_y}+(T_{zx}n_x+T_{zy}n_y+T_{zz}n_z)\vec{i_z}\tag{4}$$ You've been learning about the velocity gradient tensor. In terms of dyadic tensor notation, the velocity gradient tensor is written as $$\vec{\nabla} \vec{V}=\frac{\partial V_x}{\partial x}\vec{i}_x\vec{i}_x+\frac{\partial V_y}{\partial x}\vec{i}_x\vec{i}_y+\frac{\partial V_z}{\partial x}\vec{i}_x\vec{i}_z+\frac{\partial V_x}{\partial y}\vec{i}_y\vec{i}_x+\frac{\partial V_y}{\partial y}\vec{i}_y\vec{i}_y+\frac{\partial V_z}{\partial y}\vec{i}_y\vec{i}_z+\frac{\partial V_x}{\partial z}\vec{i}_z\vec{i}_x+\frac{\partial V_y}{\partial z}\vec{i}_z\vec{i}_y+\frac{\partial V_z}{\partial z}\vec{i}_z\vec{i}_z\tag{5}$$

Suppose you want to determine the difference in fluid velocity between two arbitrary neighboring points in the flow, at (x,y,z) and at (x+dx, y+dy, and z+dz). The differential position vector between these two points is $\vec{ds}=(dx\vec{i}_x+dy\vec{i}_y+dz\vec{i}_z)$. Here's a homework problem for you: Show that, if you pre-dot $\vec{\nabla} \vec{V}$ with $\vec{ds}$ to obtain $\vec{ds}\centerdot \vec{\nabla} \vec{V}$, the result is the velocity difference between the two points $d\vec{V}=(dV_x\vec{i}_x+dV_y\vec{i}_y+dV_z\vec{i}_z)$.

Hope this helps.

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    $\begingroup$ This seems pretty much everything tensors are NOT. The answer dangerously mixes the tensors and their representation in a chosen coordinate system. $\endgroup$ – gented Apr 3 '17 at 13:42
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    $\begingroup$ @Gennaro Tedesco Sorry you feel that way and that it doesn't coincide with your experience base. This is what I've used throughout my career as a fluid mechanics and rheology practitioner, and, with great success, I might add. The representation of this using other coordinate systems is easily handled. The development appears in the classic fluid mechanics, heat transfer, and mass transfer treatise Transport Phenomena by Bird, Stewart, and Lightfoot. This book has stood the test of time in engineering and material science for over 50 years. $\endgroup$ – Chet Miller Apr 3 '17 at 15:19
  • $\begingroup$ Another classic reference is Vector Analysis with an Introduction to Tensor Analysis by A. P. Wills. $\endgroup$ – Chet Miller Apr 3 '17 at 15:25
  • $\begingroup$ I'm not question your succesfull career, I'm just pointing out that a tensor is not its representation in this or that other coordinate system: the two are separate things. That the book has stood the test of time in <include any engineering topic> (where the space-time is always Euclidean) doesn't make it any more valid (or invalid). $\endgroup$ – gented Apr 3 '17 at 16:21
  • $\begingroup$ @GennaroTedesco The reality of the situation is that, if you are going to solve a real world problem involving fluid mechanics, solid mechanics, or material science, you are going to have to work in terms of the components of the tensors, and not just their symbolic conceptualization. This is what I believe the OP was asking about. $\endgroup$ – Chet Miller Apr 3 '17 at 16:30

protected by Qmechanic Apr 3 '17 at 6:22

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