Definition of Second-Order Tensor by Jim Adrian

[jamesadrian]

TL;DR Summary

The search for a scientific (formal) definition of a second-order tensor.

A second-order tensor is comprised at least of a two-dimensional matrix, as an nth-order tensor is comprised at least of an n-dimensional matrix, but what else is in the formal definition.

A scientific definition needs to name the term being defined, and describe the meaning of that term only by reference to meanings previously established.

I am not looking for examples of applications. Every form of math, including the partial derivative which may be a part of the definition I seek, has somewhere a formal definition that does not simply help you get the idea through applications.

I have found it very difficult to obtain such a definition. I am asking about the second-order tensor rather than the word tensor because I think that would be too combersome, but perhaps it is not.

Thank you for your help.

Jim Adrian

 

[strangerep]

Perhaps you should summarize your understanding of the mathematical meaning of "tensor" and "tensor product", and then tell us which particular physics situation you've seen it used, and what specifically you don't understand about that usage.

I could, say, point you at Wikipedia's tensor product page explaining how to generalize 1D linear spaces to higher-dimensional linear spaces (which the Cartesian product does not).

Or I could explain the metric tensor $g_{\mu\nu}$ as used in GR to give a mapping from 2 vectors to a scalar representing distance.

But it's hard to know which explanation would be best, since your question is rather nonspecific.

 

[sysprog]

I think that "is comprised" is often better said as "is composed" or as "comprises" $-$ I like @strangerep 's moderately reticent answer regarding the question $-$ I'm not sure of exactly what you mean by "second-order tensor" $-$ isn't a tensor already second-order with respect to a vector? $\dots$ I hope that maybe one of the strong math guys here will weigh in $-$ @fresh_42 I think might shed some light . . .

 

[strangerep]

[...] I'm not sure of exactly what you mean by "second-order tensor"

Which "you" did you mean? The OP or myself?

isn't a tensor already second-order with respect to a vector? [...]

A tensor can be 0'th rank (scalar), 1st rank (vector), 2nd rank, 3rd rank, 4th rank (e.g., Riemann tensor), and so on. A vector is a tensor (i.e., a 1st-rank tensor).

(I'm reasonably sure I'm capable of explaining tensors, but my "moderately reticent" answer was more intended to tease out precisely where the OP was having difficulty.)

 

[pervect]

I believe you're looking for the definition of a rank 2 tensor - I'm not aware of anyone calling it a "second order" tensor, though it's possible that people have done so and I've missed it.

There are several possible definitions, one of the best definitions, applicable for tensors of any rank, is as a multi-linerar map.

To define a tensor as a multi-linear map, we need a vector space V, and the dual of that vector space, the vector space V*.

Then a general (m,n) tensor is a multi-linear map from m vectors belonging to V and n vectors belonging to V* to a scalar.

Multi-linear means that the map is linear for each input.

The rank of the tensor is the value of m+n, so a rank 2 tensor would have m=2, n=0, or m=1, n=1, or m=0,n=2.

So for instance, a multi-linear map from 2 vectors to a scalar would be a rank 2 tensor.

For more detail , try the relevant section of wiki. https://en.wikipedia.org/w/index.php?title=Tensor&oldid=976358362#As_multilinear_maps

You may be more fond of the clumsier definition in terms of multi-dimensional arrays, but the definition in terms of multi-linear maps is IMO much more concise and provide more insight. Wiki discusses the other defintion too - I don't care for it much, but you can read it if you like.

 

[sysprog]

Which "you" did you mean? The OP or myself?

I meant to address the OP.

A tensor can be 0'th rank (scalar), 1st rank (vector), 2nd rank, 3rd rank, 4th rank (e.g., Riemann tensor), and so on. A vector is a tensor (i.e., a 1st-rank tensor).

Yes. I tend to think of a tensor as a system of more than one vector; however, I recognize that what you said there is right.

(I'm reasonably sure I'm capable of explaining tensors, but my "moderately reticent" answer was more intended to tease out precisely where the OP was having difficulty.)

I'm not sure that I could well explain, but I think that regarding the elicitation, I had a similar intention.

 

[vanhees71]

It's most simple to first think about basis-independent (i.a., invariant) definitions. You start with some vector space, e.g., a finite-dimensional real vector space. Then a tensor of rank $n$ (I think "order" is inaccurate, though I've seen this expression in some textbooks) is a linear form mapping $n$ vectors to a real number, i.e., a function $T:V^n \rightarrow \mathbb{R}$ which is linear in each of its arguments, i.e.,
$$T(\lambda_1 \vec{v} + \lambda_2 \vec{w},\vec{v}_2,\ldots,\vec{v}_n)=\lambda_1 T(\vec{v},\vec{v}_2,\ldots,\vec{v}_n)+\lambda_2 T(\vec{w},\vec{v}_2,\ldots,\vec{v}_n)$$
for any vectors $\vec{v},\vec{w},\vec{v_2},\ldots,\vec{v}_n$ and real numbers $\lambda_1$, $\lambda_2$.

 

[PeterDonis]

A second-order tensor is comprised at least of a two-dimensional matrix

No. A matrix is one possible representation of a tensor, but it's not the same thing as a tensor.

 

[jamesadrian]

Here is Einstein's Equation:

8(π)G(Tμν)
__________
Gμν =
c4This equation is often referred to in the plural as Einstein's Equations. This is because the subscripts, μ (pronounced Myu) and ν (pronounced Noo) each may independently take on integer values from zero to 3. This means that Einstein's equation, as written above, is a compact representation of 16 equations, each of which quantifies some aspect of space time (three familiar spatial dimensions plus time as a dimension).

G, called the Einstein Tensor, describes the geometry of space and the way spacetime is curved. G is Newton's gravitational constant. T is also a tensor and measures matter content. It is the stress-energy-momentum tensor, which describes the total energy content of space, including the rest-mass energy, the kinetic energy of moving matter, the pressure in fluids, and the electromagnetic fields. In the denominator, c is the speed of light in a vacuum. The subscript numbers of T must be identical to those of G.

Apparently, there are several definitions of a tensor. This means that the definition most appropriate to helping the reader understand Einsteins's Equation is the one to select, I presume.Jim Adrian

 

[jamesadrian]

The rendering of the equation was messed up after I posted it. G sub mu v = 8(π)G(Tμν) divided by c raised to the 4th.

Jim Adrian

 

[PeterDonis]

Here is Einstein's Equation:

8(π)G(Tμν)
__________
Gμν =
c4This equation is often referred to in the plural as Einstein's Equations. This is because the subscripts, μ (pronounced Myu) and ν (pronounced Noo) each may independently take on integer values from zero to 3. This means that Einstein's equation, as written above, is a compact representation of 16 equations, each of which quantifies some aspect of space time (three familiar spatial dimensions plus time as a dimension).

G, called the Einstein Tensor, describes the geometry of space and the way spacetime is curved. G is Newton's gravitational constant. T is also a tensor and measures matter content. It is the stress-energy-momentum tensor, which describes the total energy content of space, including the rest-mass energy, the kinetic energy of moving matter, the pressure in fluids, and the electromagnetic fields. In the denominator, c is the speed of light in a vacuum. The subscript numbers of T must be identical to those of G.

We already know all this. There is no need to belabor it, especially in an "A" level thread.

Apparently, there are several definitions of a tensor.

There are several different ways of stating the definition of a tensor mathematically, but they are all equivalent.

I find it difficult to believe that you have looked very hard for definitions, since every GR textbook that I'm aware of defines what tensors are in a fairly early chapter.

 

[PeroK]

The rendering of the equation was messed up after I posted it. G sub mu v = 8(π)G(Tμν) divided by c raised to the 4th.

Jim Adrian

You could try:

https://arxiv.org/pdf/gr-qc/9712019.pdf

Or, this:

 

[jamesadrian]

We already know all this. There is no need to belabor it, especially in an "A" level thread.
There are several different ways of stating the definition of a tensor mathematically, but they are all equivalent.

I find it difficult to believe that you have looked very hard for definitions, since every GR textbook that I'm aware of defines what tensors are in a fairly early chapter.

You might be surprised by how many descriptions of the concept of a tensor are online that are not formal definitions but explanations based on examples. After more that thirty of of these, I posted my question.
Perhaps you could quote a good one.Jim Adrian

 

[jamesadrian]

No. A matrix is one possible representation of a tensor, but it's not the same thing as a tensor.

This is very interesting. I would appreciate any definition or a second-rank tensor that does not refer to a matrix.

Thank you for your help.Jim Adrian

 

[Nugatory]

Apparently, there are several definitions of a tensor.

Yes, and all are mathematically equivalent. However, the text you quoted is not any of them; it’s an explanation of the physical significance of two particular tensors that appear in the Einstein field equation (but omits several more - Ricci, Weyl, metric, Riemann - which are needed to understand how the Einstein tensor describes the curvature and geometry of spacetime).

Perhaps you could quote a good one.

@pervect gave one of the best in post #5 of this thread.

 

[PeterDonis]

You might be surprised by how many descriptions of the concept of a tensor are online

I didn't say "online". I said "textbooks". You should not be trying to learn science or math from random online sources, particularly not if you are interested in rigorous formal definitions.

I would appreciate any definition or a second-rank tensor that does not refer to a matrix.

Go read posts #5 and #7 of this thread.

 

[vanhees71]

This is very interesting. I would appreciate any definition or a second-rank tensor that does not refer to a matrix.

Thank you for your help.Jim Adrian

It's a bilinear form on a vector space. No matrix needed and an invariant definition!

 

[PeterDonis]

It's a bilinear form

I think "form" is too specific. Not all (0, 2) tensors are 2-forms.

 

[dextercioby]

I think "form" is too specific. Not all (0, 2) tensors are 2-forms.

Bilinear forms are a concept different from differential forms.

 

[PeterDonis]

Bilinear forms are a concept different from differential forms.

Ah, sorry, I had misread.

 

[pervect]

You might be surprised by how many descriptions of the concept of a tensor are online that are not formal definitions but explanations based on examples. After more that thirty of of these, I posted my question.
Perhaps you could quote a good one.Jim Adrian

I would second the recommendation that if you want a formal definition of a tensor, a textbook (and not an online presentation) would be the best source. Formal definitions are not necessarily "easy to understand", as frequently they require some other formal background knowledge.

I'm a bit unclear what you actually want. At one instant you're talking about wanting a formal definition, at the next moment you're looking for "easy to understand" and "online sources".

Also, you seem to be ignoring the formal definitions that were offered. But perhaps you didn't see them, it's easy to miss a response due to the way the forum software works.

The specifications of "easy to understand", "available online", and "rigorous and formal" are probably at least as incompatible as "good, cheap, and fast", and maybe even more incompatible.

 

[jamesadrian]

I would second the recommendation that if you want a formal definition of a tensor, a textbook (and not an online presentation) would be the best source. Formal definitions are not necessarily "easy to understand", as frequently they require some other formal background knowledge.

I'm a bit unclear what you actually want. At one instant you're talking about wanting a formal definition, at the next moment you're looking for "easy to understand" and "online sources".

Also, you seem to be ignoring the formal definitions that were offered. But perhaps you didn't see them, it's easy to miss a response due to the way the forum software works.

The specifications of "easy to understand", "available online", and "rigorous and formal" are probably at least as incompatible as "good, cheap, and fast", and maybe even more incompatible.

Perfect,

I don't recall asking for "easy to understand" and I don't insist on Internet sources, only inexpensive ones.

It would be nice if somebody said "A second-rank tensor is . . . . . . ." without including the term "second-rank tensor" in the . . . . . . . and placing in this description string, only terms that have scientific definitions that can be found, or are commonly know words like "the" and "set." I don't care how long it is.

Maybe the rules of formal definitions are not as well known as I assumed.

If somebody picks a definition, we can discuss whether it is a formal definition and other matters concerning it, but I did not expect my post to be at all controversial.Jim Adrian

 

[TeethWhitener]

It would be nice if somebody said "A second-rank tensor is . . . . . . ." without including the term "second-rank tensor" in the . . . . . . . and placing in this description string, only terms that have scientific definitions that can be found, or are commonly know words like "the" and "set." I don't care how long it is.

@pervect and @vanhees71 already did this in posts 5 and 7, respectively: A second rank tensor is a bilinear map from a vector space and/or its dual to the real numbers.

 

[PeterDonis]

It would be nice if somebody said "A second-rank tensor is . . . . . . ."

You mean you are unable to fill in the dots from what has already been said? Posts #5 and #7 in particular?

 

[pervect]

Perfect,

I don't recall asking for "easy to understand" and I don't insist on Internet sources, only inexpensive ones.

It would be nice if somebody said "A second-rank tensor is . . . . . . ." without including the term "second-rank tensor" in the . . . . . . . and placing in this description string, only terms that have scientific definitions that can be found, or are commonly know words like "the" and "set." I don't care how long it is.

Maybe the rules of formal definitions are not as well known as I assumed.

If somebody picks a definition, we can discuss whether it is a formal definition and other matters concerning it, but I did not expect my post to be at all controversial.Jim Adrian

Try Wald, "General Relativity", pg 20. Or read my post #5, and compare to Wald. You did read post #5, right?

 

[anuttarasammyak]

Summary:: The search for a scientific (formal) definition of a second-order tensor.
A second-order tensor is comprised at least of a two-dimensional matrix, as an nth-order tensor is comprised at least of an n-dimensional matrix, but what else is in the formal definition.

Say vectors are transformed from coordinate system S to S' as
$$a'^\mu=\Lambda^\mu_\xi a^\xi$$
$$b'^\nu=\Lambda^\nu_\gamma b^\gamma$$
The product is
$$a'^\mu b'^\nu= \Lambda^\mu_\xi \Lambda^\nu_\gamma a^\xi b^\gamma$$
If we write $T^{\mu\nu}=a^\mu b^\nu$
$$T'^{\mu\nu}= \Lambda^\mu_\xi \Lambda^\nu_\gamma T^{\xi \gamma}$$
Not only these products of vector components but anything that follows this transformation rule in coordinate system change, is tensor. For the third rank or order tensor we make triple product, for the fourth quatro and so on.

Example:　Angular momentum tensor for a particle
$$M^{\mu\nu}=\frac{1}{2}(x^\mu p^\nu - p^\mu x^\nu)$$
Though substract for antisymmetrization is applied, it clearly meets the above transformation rule. The three non zero distinguished numbers from components of $\mu,\nu$=1,2,3 give angular momentum.

Addendum attached thought on the difference of $\Lambda^\mu_\nu$ and $a'^\mu b'^\nu$.

 

[vanhees71]

I think "form" is too specific. Not all (0, 2) tensors are 2-forms.

A multilinear form is a mapping $V^n \rightarrow \mathbb{R}$ (assuming $\mathbb{R}$ being the field of scalars). I didn't mean to imply alternating forms here, of course.

 

[pervect]

I am not sure if the following will be helpful for a number of reasons, but I am wiling to devote the time to say the following and write a few words about how a matrix satisfies the definition I gave earlier of a rank 2 tensor.

In matrix notion, there are column vectors, and row vectors. Column vectors can be regarded as members of a vector space V, while row vectors can be regarded as a member of a dual space V*.

Let C be a column vector, R be a row vector, and A be a matrix.

Then in matrix notation, the expression $R \cdot A \cdot V$, A is a multi-linear map from a row vector R, and a column vector V, to a scalar, $R \cdot A \cdot V$

For instance, in 3 dimensions

$$ \begin{bmatrix} r_1 & r_2 & r_3 \end{bmatrix} \cdot \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \cdot \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

is a scalar.

As mentioned previously, we can consider the column vector V to be an member of a vector space V, and the row vector R to be a member of a vector space V* that is dual to V

Thus we see that the matrix A is a linear mapping from one element belonging to the vector space V* and another element belonging to the vector space V to a scalar. Thus the matrix A satisfies the definition given earlier of a tensor. In particular, it is a type (1,1) tensor, which makes it a rank 2 tensor.

Matrix notation is not really suitable (as far as I know, at least) to express type (2,0) or type (0,2) tensors, but matrices can be understood as an example of (1,1) tensors.

 

[PeroK]

If somebody picks a definition, we can discuss whether it is a formal definition and other matters concerning it, but I did not expect my post to be at all controversial.

YouTube decided to put this on my recommended list today.



It and/or the whole series of lectures might be worth a watch.

 

[jamesadrian]

I found a video that has some features of a definition in it, although is has a lot to say about examples. By the end you will see the features:



Transformation rules must be explicitly stated in a formal definition of a second rand tensor or any tensor.Jim Adrian

 

[PeterDonis]

Transformation rules must be explicitly stated in a formal definition of a second rand tensor or any tensor.

Who came up with this meaning of "formal definition"? Are you incapable of deducing the transformation rules from the rank of the tensor and which indexes are upper and which are lower?

 

[jamesadrian]

You mean you are unable to fill in the dots from what has already been said? Posts #5 and #7 in particular?

That is correct.

Jim Adrian

 

[PeterDonis]

That is correct.

Then this thread's level cannot be "A", since you do not have that level of background knowledge. Changing it to "I".

 

[PeterDonis]

That is correct.

Do you understand what vector spaces and their duals are? Do you understand how vectors and dual vectors (or covectors or 1-forms--nomenclature varies among different sources) transform under transformations of the coordinates?

 

[jamesadrian]

Then this thread's level cannot be "A", since you do not have that level of background knowledge. Changing it to "I".

You didn't ask Why.

Jim Adrian