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- Thread starter nolanp2
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Here's my understanding. Tensors are essentially matrices, except they can exist in more than two dimensions. For example, a tensor with one index is a vector. If you were to write it out as a matrix, it would exist in one dimension. Tensors with two dimensions are just typical square matrices. A tensor of three indices can be thought of as a three-dimensional box of numbers. And it just goes from there.

Of course, I've never quite understood the difference between covarient and contravarient tensors. Let me know if you ever find out!

- #3

cristo

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Of course, I've never quite understood the difference between covarient and contravarient tensors. Let me know if you ever find out!

Do you know the difference between a covariant and a contravariant vector?

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https://www.physicsforums.com/showpost.php?p=1175763&postcount=7

https://www.physicsforums.com/showpost.php?p=1173978&postcount=12

Here is a short article to get you started:

http://arxiv.org/abs/gr-qc/9807044

(I am actually working on a VPython program to visualize tensors.)

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Do you know the difference between a covariant and a contravariant vector?

Well, I know that one has raised indices, and the other has lowered indices. And I know that raising or lowering both indices simultaneously means that that tensor becomes its inverse. So I can brainlessly do the operation. But I've never understood

- #6

quasar987

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I feel I cannot say more without rewriting the whole theory from scratch so I refer you to say Sharipov's ebook: http://uk.arxiv.org/abs/math/0403252/

- #7

mjsd

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how do you actually want to visualise it? As apples and oranges ? or as a collection of numbers/symbols/sea shells group together in some meaningful way? and that they follow particular rules when you try to juggle the content....?

I guess you may have a chance to visualise tensors if you limit yourself to tensors with very small ranks and very low dimensionality. Otherwise, it would be as difficult as to visualising a 10-dimensional hypercube for instance.

remember tensors are defined by how they transform, and certainly not all tensors naturally lead to some "everyday" visualizations.

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A tensor is a collection of values that alters a field.

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an mXm matrix will take a vector and transform it to produce another vector in the same "space" (ex. R^m)

now what if you had an array of several matrices? if you were to multiply this by a vector such that each component of the vector gave a certain "weight" to each of the component matrices in the tensor, thus producing a new matrix that could then be used to tranform a vector.

I believe that general relativity works in this way.

but most simply put a tensor is an array of matrices, and ghigher dimensional tensors are then arrays of lower dimensional tensors. similar in effect to a hypercube.

NOTE: I have no formal traiing or even informal self-study of tensors, I just gathered this from a couple of brief chapters on general relativity a while back, so what I said could easily be very wrong.

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I suppose that the name covariant cames from the fact that when you change one axis only the components related to this axis change while when working with contravariant components all the components change.

I hope I have made my self clear.

- #11

Hurkyl

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IMHO, one thing that would help is to try to understand what a tensor is **before** you try and understand what a tensor field is.

(much like it's easier to learn about vectors before you try to learn about a vector field)

If you've chosen a basis, the coordinates of a vector form an*nx1* array of numbers. The coordinates of a covector form a *1xn* array of numbers.

In coordinates, a tensor product is a generalization of a scalar product. For example, in three dimensions, if I tensor the covector [a b c] with something (could be anything: a scalar, a vector, a covector, a matrix, or some other kind of tensor)

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \mathbf{v}

= \left[\begin{array}{ccc}a\mathbf{v} & b\mathbf{v} & c\mathbf{v}\end{array}\right]

[/tex]

If**v** was, say, the vector with coordinates *x*, *y*, and *z*, this tensor product would be the matrix:

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes

\left[\begin{array}{c}x \\ y \\ z\end{array} \right]

=

\left[\begin{array}{ccc}

a \left[\begin{array}{c}x \\ y \\ z\end{array} \right]

& b \left[\begin{array}{c}x \\ y \\ z\end{array} \right]

& c \left[\begin{array}{c}x \\ y \\ z\end{array} \right]\end{array} \right]

=

\left[\begin{array}{c|c|c}ax & bx & cx \\

ay & by & cy \\

az & bz & cz

\end{array} \right][/tex]

Actually, we would drop the partitions in this case: they're more useful for other cases, such as if**v** happened to be the covector [p q r]:

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes

\left[\begin{array}{ccc}p & q & r\end{array} \right]

=

\left[\begin{array}{ccc|ccc|ccc}

ap & aq & ar & bp & bq & br & cp & cq & cr \end{array}\right]

[/tex]

Mind you, this is all in a coordinate representation: just like with vectors, we can do algebra with tensors without resorting to coordinates.

(much like it's easier to learn about vectors before you try to learn about a vector field)

If you've chosen a basis, the coordinates of a vector form an

In coordinates, a tensor product is a generalization of a scalar product. For example, in three dimensions, if I tensor the covector [a b c] with something (could be anything: a scalar, a vector, a covector, a matrix, or some other kind of tensor)

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes \mathbf{v}

= \left[\begin{array}{ccc}a\mathbf{v} & b\mathbf{v} & c\mathbf{v}\end{array}\right]

[/tex]

If

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes

\left[\begin{array}{c}x \\ y \\ z\end{array} \right]

=

\left[\begin{array}{ccc}

a \left[\begin{array}{c}x \\ y \\ z\end{array} \right]

& b \left[\begin{array}{c}x \\ y \\ z\end{array} \right]

& c \left[\begin{array}{c}x \\ y \\ z\end{array} \right]\end{array} \right]

=

\left[\begin{array}{c|c|c}ax & bx & cx \\

ay & by & cy \\

az & bz & cz

\end{array} \right][/tex]

Actually, we would drop the partitions in this case: they're more useful for other cases, such as if

[tex]

\left[\begin{array}{ccc}a & b & c\end{array} \right] \otimes

\left[\begin{array}{ccc}p & q & r\end{array} \right]

=

\left[\begin{array}{ccc|ccc|ccc}

ap & aq & ar & bp & bq & br & cp & cq & cr \end{array}\right]

[/tex]

Mind you, this is all in a coordinate representation: just like with vectors, we can do algebra with tensors without resorting to coordinates.

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This intended to help the fellows above who want to know about covariant and contravariant components. As magister above points out, if the axes of our co-ordinate system are not orthogonal, or are curved, there are two ways of relating a point to an axis. One way is to drop a perpendicular, the other way is to go parallel to another axis. So we have 2 numbers where we had one. These are the contravariant and covariant components. We used to describe a point in 2D with 2 co-ords, x and y, now we have a choice of x (contra), x (covar), y(contra), y (covar).

Now switching notation so super- and subscripts are the dimension ( eg 0=time, 1=x, 2=y etc), if we have 2 points in 2D (X,Y) we have 8 numbers ( n=0,1)

[tex]x^n, x_n , y^n, y_n [/tex]

Luckily the only quantities we can use in physics are products of contravariant and covariant components, so only two sensible choices exist to work with these numbers,

[tex] x_m y^m = x_0 y^0 + x_1 y^1 [/tex]

and

[tex] x^n y_n [/tex]

which are the same thing. All the intricacies of tensor analysis follow from the fact that physical quantities are scalars.

I haven't mentioned the metric tensor, to keep it simpler.

Now switching notation so super- and subscripts are the dimension ( eg 0=time, 1=x, 2=y etc), if we have 2 points in 2D (X,Y) we have 8 numbers ( n=0,1)

[tex]x^n, x_n , y^n, y_n [/tex]

Luckily the only quantities we can use in physics are products of contravariant and covariant components, so only two sensible choices exist to work with these numbers,

[tex] x_m y^m = x_0 y^0 + x_1 y^1 [/tex]

and

[tex] x^n y_n [/tex]

which are the same thing. All the intricacies of tensor analysis follow from the fact that physical quantities are scalars.

I haven't mentioned the metric tensor, to keep it simpler.

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I haven't mentioned the metric tensor, to keep it simpler.

hmm I'm intrigued

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Hi CPL - what I've posted is the first step in a journey of a thousand leagues.

B Reimann in the 1840's(roughly) investigated systems where the axes were not straight or orthogonal, realising that concepts like position coords are no longer useful, but that distances could be kept invariant.

I can recommend Stephani's book "General Relativity" as an introduction.

M.

B Reimann in the 1840's(roughly) investigated systems where the axes were not straight or orthogonal, realising that concepts like position coords are no longer useful, but that distances could be kept invariant.

I can recommend Stephani's book "General Relativity" as an introduction.

M.

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a geometric visualization (akin to visualizing a vector as an arrow and vector addition via the parallelogram rule)

or

an organizational typeset visualization (akin to visualizing a vector as a column matrix and vector addition via component-wise addition).

If you are looking for a geometric visualization, did you consult the arxiv.org paper?

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Hi Nolanp2. I don't think visualising tensors helps to understand them.

Thinking about geometry might help. You mention force fields, so I'll give you an example of how and why tensors are used. Consider the equation of motion of a charge in an electric field,

m.a = e(E + v x B ), i.e. mass times acceleration is electric charge times E + v x B, where v is the velocity vector and E and B are electric and magnetic field vectors. x is the vector cross-product.Note that this equation is really 3 equations, one each for x, y and z.

This formula is fine but it is not relativistic. It won't work if you put it in a moving frame of reference.

To make the equation obey special relativity, we go to four dimensions,

t,x,y,z, with stipulation that the time dimension has a negative sign, and

that we multiply all times by c=velocity of light to make them distances.

We can now rewrite the equation in tensor notation,

[tex] ma^{\mu} = e F^{\mu\nu}v_{\nu} [/tex]

which is four equations since [tex]\mu[/tex] can be t,x,y,or z. F is the EM field tensor ( see Wiki for instance).

But the point is that this equation is relativistically covariant, i.e it transforms properly between inertial frames. It's also very elegant.

To do a calculation from this formula, say for x ( mu=1), we expand the tensor to give,

[tex] ma^{1} = e F^{10}v_{0} + e F^{11}v_{1} + e F^{12}v_{2} + e F^{13}v_{3} [/tex]

Note that all the indexes are numbers, and we can plug in the values of the components to get an algebraic differential equation. The tensors are gone.

Thinking about geometry might help. You mention force fields, so I'll give you an example of how and why tensors are used. Consider the equation of motion of a charge in an electric field,

m.a = e(E + v x B ), i.e. mass times acceleration is electric charge times E + v x B, where v is the velocity vector and E and B are electric and magnetic field vectors. x is the vector cross-product.Note that this equation is really 3 equations, one each for x, y and z.

This formula is fine but it is not relativistic. It won't work if you put it in a moving frame of reference.

To make the equation obey special relativity, we go to four dimensions,

t,x,y,z, with stipulation that the time dimension has a negative sign, and

that we multiply all times by c=velocity of light to make them distances.

We can now rewrite the equation in tensor notation,

[tex] ma^{\mu} = e F^{\mu\nu}v_{\nu} [/tex]

which is four equations since [tex]\mu[/tex] can be t,x,y,or z. F is the EM field tensor ( see Wiki for instance).

But the point is that this equation is relativistically covariant, i.e it transforms properly between inertial frames. It's also very elegant.

To do a calculation from this formula, say for x ( mu=1), we expand the tensor to give,

[tex] ma^{1} = e F^{10}v_{0} + e F^{11}v_{1} + e F^{12}v_{2} + e F^{13}v_{3} [/tex]

Note that all the indexes are numbers, and we can plug in the values of the components to get an algebraic differential equation. The tensors are gone.

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I think this is just using the tensor summation convention. Expand it over all

values of i and k thus,

U = K(0,0)*x(0)*x(0) + K(0,1)*x(0)*x(1) + K(0,2)*x(0)*x(2) + ....

In fact this just matrix multiplication.

All you need now are the values ( components) of K and some x's.

M

values of i and k thus,

U = K(0,0)*x(0)*x(0) + K(0,1)*x(0)*x(1) + K(0,2)*x(0)*x(2) + ....

In fact this just matrix multiplication.

All you need now are the values ( components) of K and some x's.

M

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All you need now are the values ( components) of K and some x's.

M

so do the different k values represent different springs (for example) being stretched when one particle connected to them is moved or does it mean a spring is acting different when pulled in different directions? or does a second degree of freedom mean there's a second partcle and the two are connected via springs? in particular what would k_11 or k_22 represent?

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a geometric visualization (akin to visualizing a vector as an arrow and vector addition via the parallelogram rule)

or

an organizational typeset visualization (akin to visualizing a vector as a column matrix and vector addition via component-wise addition).

If you are looking for a geometric visualization, did you consult the arxiv.org paper?

it's a geometric interpretation of tensors i'm looking for, i looked at this article but didn't find much useful information on it, it seemed more related to arunma's comment

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It looks as if U is a potential energy of a some system. The different K values are spring constants and the x(i) i=1..s are displacements but I can't

work out the configuration right now. I'll have to think about it.

I've thought about and I think the x's represent displacements from some origin in

the different dimensions. K is the spring tensor, so Kii is the resistance in the

purely dimension i direction, and Kij is the resistance for a combined i, j dimension

direction. If you see what I mean.

The spring tensor is similar to a space metric in geometry, except the x's are infinitessimals.

work out the configuration right now. I'll have to think about it.

I've thought about and I think the x's represent displacements from some origin in

the different dimensions. K is the spring tensor, so Kii is the resistance in the

purely dimension i direction, and Kij is the resistance for a combined i, j dimension

direction. If you see what I mean.

The spring tensor is similar to a space metric in geometry, except the x's are infinitessimals.

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it's a geometric interpretation of tensors i'm looking for, i looked at this article but didn't find much useful information on it, it seemed more related to arunma's comment

Well... let me take a stab at what you may be looking for.

(And while writing this, I see that Mentz114 has the same interpretation.)

Since you provided the example of small oscillations in s-dimensions,

your object K is generally a symmetric tensor: [tex]K_{ab}=K_{(ab)}[/tex] (where [tex]K_{(ab)} \equiv \frac{1}{2!} (K_{ab}+K_{ba})[/tex]), which linearly maps a pair of vectors to the real-numbers. Thus, U is a scalar.

In the special case, [tex] K_{ab}=k g_{ab} [/tex], where k is a scalar constant and [tex]g_{ab}[/tex] is the metric tensor. In this case, it is like having a single spring. In the more general case, it is like having a system of coupled oscillators.

Geometrically, the symmetric tensor can be pictured as an ellipsoid [in s-dimensions]. (The moment of inertia tensor has a similar interpretation.)

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ok that makes sense, or at least when i think in two dimensions i can make sense of it! thanks to both of you for the help. actually robphy could you give a brief explanation of the last comment about the ellipsoid please? prefferably with s<4!

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Side detail: [tex] K_{ab}x^a x^b=x^a K_{ab} x^b =x_a K^a{}_{b}x^b = \vec x^\top K \vec x[/tex]

Consider the [level] surface:

[tex] 1=\vec x^\top K \vec x =

\left(

\begin{array}{ccc}

x & y & z

\end{array}

\right)

\left(

\begin{array}{ccc}

A & D & E\\

D & B & F \\

E & F & C

\end{array}

\right)

\left(

\begin{array}{c}

x \\

y \\

z

\end{array}

\right)

[/tex]

Upon multiplying this out, this is an ellipsoid with axes rotated away from the standard axes. (You can locate these axes by finding the eigenvectors of this matrix [which is associated with "diagonalizing this matrix"].) If the off-diagonal elements are zero, then the ellipsoid is lined up with the standard axes. Further, if these diagonal elements are equal (with the off-diagonal elements zero), then we have a sphere.

Consider the [level] surface:

[tex] 1=\vec x^\top K \vec x =

\left(

\begin{array}{ccc}

x & y & z

\end{array}

\right)

\left(

\begin{array}{ccc}

A & D & E\\

D & B & F \\

E & F & C

\end{array}

\right)

\left(

\begin{array}{c}

x \\

y \\

z

\end{array}

\right)

[/tex]

Upon multiplying this out, this is an ellipsoid with axes rotated away from the standard axes. (You can locate these axes by finding the eigenvectors of this matrix [which is associated with "diagonalizing this matrix"].) If the off-diagonal elements are zero, then the ellipsoid is lined up with the standard axes. Further, if these diagonal elements are equal (with the off-diagonal elements zero), then we have a sphere.

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