What is a Tensor? Definition & Applications

[abercrombiems02]

Can someone give me a dummies definition of what a tensor basically is and what its applications are? Thanks

 

[Janitor]

This very website had a nice little discussion of your topic not long ago.

https://www.physicsforums.com/showthread.php?p=262170#post262170

 

[Tom McCurdy]

http://en.wikipedia.org/wiki/Tensor

 

[abercrombiems02]

can someone give me an example of an applied problem where it would be necessary to use a tensor to find a solution? I still really don't see why tensors would be useful

 

[Feynman]

Look ,
http://scienceworld.wolfram.com/physics/Navier-StokesEquations.html

 

[mathwonk]

i am going to go out on a limb here and try a very elementary explanation. Fortunately there are plenty of knowledgeable people here to correct my mistakes.

A first order tensor is a vector, i.e. it is something linear, like a linear polynomial

say ax+by + cz.

A second order tensor is something that is bilinear, like a degree two polynomial

say ax^2 + by^2 + cZ^2 + dxy + exz + fyz.

And so on.


Given a function like f(x,yz), we can re - expand it as a Taylor series about each point, and pick off the linear term, thus assigning to each point a linear polynomial, i.e. a vector field.

On the other hand we could assign the second derivative, i.e. the second order term of the Taylor series at each point, thus obtaining a second degree tensor field.

and so on.


Hence any problem that requires second derivatives (or higher) for its solution is essentially one that uses tensors, i.e. multilinear as opposed to linear objects.

Curvature is such a concept, and curvature is an intrinsic part of the theory of relativity, since mass produces curvature in space time.

Ok, I drop out here and ask for help from the experts.

 

[pmb_phy]

No. None of that has anything to do with tensors. At least not according to any definition that I've seen. For a definition see

http://www.geocities.com/physics_world/ma/intro_tensor.htm

Pete

 

[mathwonk]

pmb phy, With all respect, I think you may be misunderstanding what you have seen.

For instance on the first page of the site you just referred us to, equation (1) displays the "metric tensor" in exactly the form I gave for a second order (or rank) tensor, namely it has an expression as a homogeneous polynomial of degree 2.

Later on, this same site describes a zero rank tensor as a scalar valued function that keeps the same value when coordinates are changed, and a first order tensor as one which transforms by the usual first order chain rule, when coordinates are changed, and a second order tensor as one which transforms by the chain rule for second derivatives when coordinates are changed, etc...

Perhaps the confusion is that I was referring to the appearance of a (symmetric) tensor in a given coordinate system, and your sources emphasize the way these representations change, under change of coordinates.

Unfortunately many sources emphasize the appearance or representation of tensors rather then their conceptual meaning. The essential content of a tensor (at a point) is its multilinearity.

Globalizing them, leads to a family or "field" of such objects at different points, and then to the necessity of changing coordinates, at which point the way in which they change appearance under coordinate change becomes of interest.

It seems odd to me at least to define them this way however. On the other hand, you are right that some features of tensors, or even vectors, are invisible except when one changes coordinates.

E.g. a vector and a covector at a point both look like an n tuple of numbers, but when you change coordinates one changes by the transpose of the matrix changing the other.

Of course conceptually they differ even at a point, as one is a tangent vector and one is a linear form acting on tangent vectors.

Do you buy any of this?

 

[pmb_phy]

For instance on the first page of the site you just referred us to, equation (1) displays the "metric tensor" in exactly the form I gave for a second order (or rank) tensor, namely it has an expression as a homogeneous polynomial of degree 2.

Equation 1 contains the components is the metric tensor. The components in that particular case were all zero except for g₁₁ and g₂ which equal 1.

Perhaps the confusion is that I was referring to the appearance of a (symmetric) tensor in a given coordinate system, and your sources emphasize the way these representations change, under change of coordinates.

I think you're confusing the tensor with the expression the components of the tensor appears in. A general tensor is a geometric object which is linear function of its variables which maps into scalars. For example: Let g be the metric tensor. Its a function of two vectors. The boldface notation represents the tensor itself and not components in a particular coordinate system. An example of this would be the magnitude of a vector.

$$A^2 = \boldf g \left \boldf A, \boldf B \right$$

When you represent the vector in a basis and using the linearlity of the tensor then you get the usual expression in terms of components.

Unfortunately many sources emphasize the appearance or representation of tensors rather then their conceptual meaning. The essential content of a tensor (at a point) is its multilinearity.

There are two ways of looking at tensors. I've been meaning to make a new web page to emphasize the geometric meaning but am unable to do so at this time. Plus I'm still thinking of the best way to do that.

E.g. a vector and a covector at a point both look like an n tuple of numbers, but when you change coordinates one changes by the transpose of the matrix changing the other.

The terms "covariant" and "contravariant" can have different meanings in the same context depending on their usage. For example: A little mentioned notion is that a single vector can have covariant and contravariant components. For details please see

http://www.geocities.com/physics_world/co_vs_contra.htm

Of course conceptually they differ even at a point, as one is a tangent vector and one is a linear form acting on tangent vectors.

Do you buy any of this?

Some of it.

Pete

 

[mathwonk]

Pete, let me try again. we may be getting closer together here.

An equation like summation gjk dxj dxk, as on the site you referenced, is a covariant tensor of second rank, because it is a second degree homogeneous polynomial in the expressions dxj, dxk, which are themselves covariant tensors of first rank. i.e. it is of rank 2, because there are two of them multiplied together. this results in a bilinear operator, which is linear in each variable separately.

now WHICH second rank tensor it is, is determined by what the coefficients are, or the "components" if you like, namely the gjk.

the usual metric tensor on the euclidean plane is dx1dx1 + dx2dx2, so the only non zero components are g11 = 1, g22 = 1.

But there are many other riemannian metrics given by other choices of the gjk.

I have tried to explain the conceptual idea of tensors at greater length in some other threads in this forum also with the word tensor in them. let me know what you think of them.

Brutally, if T is the tangent bundle and T^ the dual bundle of a manifold, then sections of T are contravariant tensors of rank one, and sections of T^ are covariant tensors of rank 1.

A section of (T^ tensor T^), where this is the bundle of tensor products of the dual spaces, is a covariant tensor of second rank, such as a metric.

if we consider only one tangent space isomorphic to R^n, its dual has basis dx1,...dxn, and the second tensor product of the duals has basis dxjdxk, for all j,k, (where the order matters).

Any second rank tensor can be expressed in terms of this basis and the coordinates or coefficients or componets are called gjk.

When we change coordinates, we get a new standard basis for the second tensor product and hence the coefficients of our basis expansion change. I/.e/; the gjk change into some other matrix valued function g'jk, in the way specified.

To define a tensor by saying how the components transform instead of what bundle it is a section of, is like defining a duck by the way it walks. Of course we all know the old saying: if it transforms like a tensor, then it is a tensor.

peace,

roy

 

[mathwonk]

Dear competing tensor advocates.

I just checked out the site recommended above by Tom McCurdy in post 3.

http://en.wikipedia.org/wiki/Tensor


It tries to discuss both viewpoints on tensors and even relate them. I recommend it also. It seems from that discussion that my viewpoint is the so called modern one.


"The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra."

I still like my duckwalk joke, though.

 

[pmb_phy]

An equation like summation gjk dxj dxk, as on the site you referenced, is a covariant tensor of second rank, because it is a second degree homogeneous polynomial in the expressions dxj, dxk, which are themselves covariant tensors of first rank.

Why do you call the summation a second rank tensor? It is not. g_jk is a tensor of rank two. The differentials dx_k are tensors of rank one. The summation is a contraction of a tensor of rank two with two tensors of rank one giving a tensor of rank zero.

i.e. it is of rank 2, because there are two of them multiplied together.

No. They are not multiplied together. They are summed. That is a huge difference.

if we consider only one tangent space isomorphic to R^n, its dual has basis dx1,...dxn, and the second tensor product of the duals has basis dxjdxk, for all j,k, (where the order matters).

I recommend learning how to use subscripts and superscripts on this forum. That way I can tell if you're using them or not. I don't see why you're referring to the differentials as a basis. A basis is a vector and not a component like dx^j.

Pete

 

[mathwonk]

Pete,

Let me try once again. I could be wrong of course, but I am not convinced by what you are saying. I am confident however that the miscommunication is due either to my ignorance, which hopefully is curable, or to a difference in language about tensors.


Now first of all you say that gjk is a tensor (subscript). To me this is like saying an n tuple of numbers is a vector. Just as an n tuple of numbers represents a vector, in terms of some basis, so also a matrix like (gjk) (subscripts) represents a tensor in terms of a basis.

Now if I denote by {...,ej,...} (sub) a basis of tangent vectors, then an n tuple of numbers like (,,,aj ...) represents

the vector: summation ajej.

Here it does not matter whether the indices are up or down, because we know what the objects are, namely, the ej are tangent vectors, so we know how they transform, whether I indicate it by sub or super scripts. Also the aj are numbers, and I have written a summation. Einstein's convention as I recall it, is that one can save on writing summation notation, if one uses oppositely placed scripts to signal summation automatically. I am not doing this.

Another convention is that subscripts are used like for tangent vectors ej, to denote that they are classical contravariant vectors (that is covariant in category theory). whereas superscripts are used like your elegantly written dx^j above to denote classically covariant vectors, i.e. covectors.

In modern terminology I believe this is achieved by saying that the ej are basic sections of the tangent bundle, whose variance is known to transform by the jacobian matrix, and saying the dx^j are (basic) sections of the cotangent bundle, or dual tangent bundle, whose sections are known to transform by the transpose of the jacobian matrix.

(These opposite transformation laws are used on the second site you referred me to, to define contravariant and covariant vectors.)

As to the word "basis", it is used by me in the sense of vector space theory. I.e. any space of objects closed under additon and scalar multiplication, is called a vector space. E.g. a tensor space is also an example of a vector space but I have refrained from using that term in that way on this post since it is not used that way by physicists it seems.

A basis for an abstract vector space is any collection of elements of the vector space, (e.g. if it is a basis of a tensor space, they will be tensors), such that every element of the space has a unique expression as a sum of scalar multiples of the given basis.


E.g. in R^n, if we denote the standard basis vectors (0,...0,1,0,...0) with a 1 in the jth place, by the symbol ej (sub), then the collection e1,...,en is a basis for the tangent space to R^n, simply because every tangent vector (a1,...,an) can be written uniquely as

summation: ajej.

(Here I have committed the apparent contradiction of referring to an n tuple of numbers (a1,...,an) as a tangent vector. But that is because R^n is the one vector space in the whole world, whose vectors really are n tuples of numbers. The concept of a basis is a way to represent elements of other vector spaces as elements of R^n, i.e. as n tuples of numbers.)

If I consider on the other hand the dual space (R^n)^ = linear maps from R^n to R, then this space is isomorphic to R^n, but the elements transform differently. One way to signal this would be to denote their coefficients by super scripts, but this is unnecessary, if we simply choose different symbols for them, such as dx^j.

These symbols are well chosen, because the basic elements of the dual space are the differentials of functions, and the simplest (coordinate) functions on R^n are the functions x^j.

(Thank you for your patience in bearing with what is no doubt extremely familiar to you. We may get somewhere yet however.)

Thus a covector like dx^j acts on a vector like ek by dot product in terms of their coordinate representations, or more intrinsically, by noting that dx^j(ek) = kronecker delta ?jk. Now I agree this pairing or contraction is signaled by the fact that the scripts of the dx's are up and those of the e's are down.

Now here is the source of the confusion for me in the equation (1) we were discussing.
But I will postpone it a little longer to clarify further my use of notation, and "basis".

In addition to the dual space T^ = (R^n)^, of R^n whose elements are apparently rank 1 covariant (old terminology) tensors, there is another space formed by taking the tensor product of this space with itself,

called T^ tensor T^,

whose elements are rank 2 covariant tensors. By definition, this space may be defined as the space of all bilinear maps from TxT to R. As such it is an abstract vector space, although it would be a sin to call its elements "vectors" in a physics forum, because to a physicist that word is reserved for rank 1 tensors.

Nonetheless T^ tensor T^ is a linear space, (that is a better word), and it has a basis, i.e. a set of elements such that all other elements can be written in terms of these.

E.g. such a basis is given by the tensor product dx^jdx^k of the two rank 1 tensors dx^j and dx^k. In gneral, if f,g are rank 1 covariant tensors, and if v,w is a pair of contravariant vectors, then the value of (f tensor g) on (v,w) is f(v)g(w).

Thus there is a tensor multiplication taking pairs of elements of T^ to one element of T^ tensor T^. Then it is a theorem, easily checked, that the set of products (dx^j)tensor(dx^k), for all pairs j,k, is a basis for the space of second rank covariant tensors.

In particular every such tensor can be written in terms of these. Thus a general second rank covariant tensor would be written as

summation gjk(sub if you like) dx^j dx^k, where I have omitted the tensor sign.

(Thus you are right there is a summation here of the tensors gjk dx^j dx^k,

but there is also a tensor product, of dx^j times dx^k.)

In particular, the standard scalar product on euclidean space would be written as

summation (kronecker delta) dx^j dx^k = dx^1dx^1 +...+ dx^ndx^n.


Such an object acts on pairs of tangent vectors and spits out a number. E.g. on the pair (a1,...,an),(b1,...,bn) = (summation ajej, summation bkek),

it spits out of course summation ajbj.

Thus I am interpreting the object in equation (1) at the referenced site as representing the second rank covariant tensor:

summation gjk dx^j dx^k

whose value on the pair of vectors (summation ajej, summation bkek),

is the number (matrix product):

(,,,aj,...) (gjk) (...,bk,...)T = summation (over j,k), gjk aj bk.

where here the T means transpose.


Now we can also consider the tensor product of the tangent space with itself and get the space (T tensor T) of second rank contravariant vectors, for which a basis is given by the products {ej tensor ek}.

Then we can consider that a covariant rank 2 vector like:

summ gjk dx^j dx^k

acts not on the pair (summation ajej, summation bkek),

but rather on the contravariant vector, their product:

summation ajbk (ej tensor ek).

Then the value in coordinates is given by:

summation(over j,k) ajbk gjk, a number.


Now I am trying to see how to possibly interpret equation (1) as you have done.

E.g. if I represent a contravariant rank 2 vector:

summation gjk ej tensor ek,

simply by the matrix gjk,

and represent the covariant tensor:

summation dx^j tensor dx^j,

by the same expression:

summation dx^j tensor dx^j,


then I suppose I could believe that the equation (1) represents the number

obtained by evaluating the covariant 2 tensor:

summation dx^j tensor dx^j,

on the contravariant 2 tensor:

summation gjk ej tensor ek.

This would violate two principles I hold sacred: first the symbols gjk are never used for contravariant tensors, but always for covariant tensors.

second and more important: the object being represented there is not a number as you say, but a tensor, the metric tensor. They say so right on the site. (let me check that and get back to you.)

Anyway I appreciate your sincere and patient attempt to communicate with me.

We are struggling since it seems you apparently speak primarily "indices" and i speak only "index free", so if someone with dual langauage capabilities would jump in, it might help, but maybe we are doing as well as can be expected.

OK I have been back your site, and to my mind it confirms what I have been saying.

E.g. it says there that the expression G (bold) = summation gjk dx^j dx^k is a tensor, whose components are the numbers gjk. (The matrix gjk is therefore not itself a tensor.)

From the basis point of view, this means that this tensor G is written as a linear combination of the basic tensors dx^j dx^k, using the components or coefficients gjk.

He does not say it there, but those basic tensors themselves are products of the rank 1 tensors dx^j and dx^k.

This is fun, and I hope we are helping each other. I know I appreciate your patience with my ignorance of common longstanding notation and practice in physics.

best regards,

roy

 

[pmb_phy]

I'm sorry, but due to major back problems it is impossibe for me to sit down and read such a long post. I'll have to spend some time reading and absorbing what you wrote bit by bit due to the short amounts of time I can sit in front of the computer.

In the meantime can you please find and post a reference to where you've seen/learned the definition(s) which you hold to be true.
Thanks

Pete

 

[mathwonk]

These are the definitions I believe to have been standard in mathematical treatments of differential geometry since the 1960's, for example in Michael Spivak's little book Calculus on Manifolds, or his large treatise Differential Geometry. I will look for an internet source, but i suspect the one on wikipedia would suffice.

http://en.wikipedia.org/wiki/Tensor

I will check it out more carefully.

I sympathize with the back problems as I also have them. Mine are helped by sitting only in an old captain's chair I inherited from my grandfather, but there must be others out there.

 

[mathwonk]

Pete,

Here is a quote from wikipedia:

"There are equivalent approaches to visualizing and working with tensors; that the content is actually the same may only become apparent with some familiarity with the material.

* The classical approach
The classical approach views tensors as multidimensional arrays that are n-dimensional generalizations of scalars, 1-dimensional vectors and 2-dimensional matrices. The "components" of the tensor are the indices of the array. ...


* The modern approach
The modern (component-free) approach views tensors initially as abstract objects, expressing some definite type of multi-linear concept. Their well-known properties can be derived from their definitions, as linear maps or more generally; and the rules for manipulations of tensors arise as an extension of linear algebra to multilinear algebra.

This treatment has largely replaced the component-based treatment for advanced study, in the way that the more modern component-free treatment of vectors replaces the traditional component-based treatment after the component-based treatment has been used to provide an elementary motivation for the concept of a vector. You could say that the slogan is 'tensors are elements of some tensor space'.

* The intermediate treatment of tensors article attempts to bridge the two extremes, and to show their relationships."


Here is the link for the intermediate article, but it is pretty sketchy.

http://en.wikipedia.org/wiki/Intermediate_treatment_of_tensors

One of the specific examples wikipedia cites of a tensor, is a homogeneous polynomial of degree two. There is also a short tutorial on this forum, in the thread "Math Newb wants to know what's a tensor", by chroot, where the scalar product is cited as an example of a rank 2 covariant tensor, not rank zero, (actually he calls it type (0,2), or rank 2 covariant and rank zero contravariant).

I also gave an explicit calculation in the thread "tensor product" that shows how to use the modern definition of a tensor to compute the tensor product as matrices.

If we get together on this, it will be a major success for both of us, but not worth a backache.

best regards,

roy

 

[pmb_phy]

Hi again

Yep. Back problems suck big time. Due to that sitting and typing too long I had to be rushed to the hospital in an ambulance. I've experienced levels of pain (from sitting here typing too long) that I never knew even existed, and I know pain since I've had 8 bone marrow biopsies. I have no intention of letting that happen again.

Meanwhile, if you happen to have A short course in General Relativity, J. Foster, J.D. Nightingale then see section 1.10. It will clarify the difference between contravariant vectors and 1-forms (aka covariant vectors). In Euclidean space you can get away with igoring the difference, but not in general. A 1-form maps vectors to real numbers. They are distinct objects from vectors (but they are related). That is why I have been emphasizing the placement of the indices. In some areas, such as mechanics, one uses only Cartesian tensors and this distinction never arises. A Cartesian tensor is an example of an affine tensor which is different than, but somewhat similar to, a tensor.

Note that it is common practice to use what Foster and Nightingale phrase as follows on page 45

From here onwards we shall adopt a much used convention which is to confuse a tensor with its components. This allows us to refer simply to the tensor T^ab, rather than the tensor with components T^ab.

g is literally the tensor while g_ab is literally the components. They are defined by g_ab = g(e_a,e_b).

Pete

 

[mathwonk]

Hi again Pete,

I admire your tenacity in this forum! It inspires me after my own last weeks "procedure" near the area I use for sitting.

I discovered that this whole hash about competing languages has already been discussed at unbelievable length in the thread "intro to differential forms, started by lethe, and subsequently largely deleted by him, over a flap about use of more informal language.

Check out posts #50,... in that thread to see some of the same discussions we have been having about up or down indices. Of course they were talking about the case of anticommutative covariant tensors, or differential p - forms, rather than general tensors.

It appears the original tutorial posted by lethe still exists at another site, namely

http://www.sciforums.com/showthread.php?t=20843&page=1&pp=20


He gives the full monty discussion there of the language I was advocating above. In particular, he does define general tensors on the way to defining skew commutative ones

And thanks for the reference to Foster and Nightingale.

Maybe I can still learn some relativity!

best wishes,

roy

 

[mathwonk]

PS

It seems one confusion is that although (the components of) a covariant tensor are subscripts, the basic co - tensors themselves are apparently written as superscripts.

i.e. gjk for the components as opposed to dx^j dx^k, for the basic guys.

E.g. the basic tensor dx^1dx^1 written in components,

would just be the matrix (gjk),

where all the gjk equal zero except g11 = 1.

Similarly, the coordinates (components) of a contravariant tensor are written as super, and the indices on the actual basic tensors are written as subs.

i.e. a^j as opposed to ej.

That way when we conjoin the components of a contra -tensor with the components of a co - tensor, it does mean to contract and get a number,

e.g. summ gjk h^(jk)



But when we conjoin the components gjk say of a co tensor, with the symbols for the basic co tensors, it only means to sum them up as numbers times co - tensors, so it is still a co tensor.

e.g. summ gjk dx^j dx^k, is a rank 2 cotensor.


So here are two versions of the same object

classical covariant 2 tensor: gjk

modern version of same covariant 2 tensor: summation gjk dx^j dx^k.


classical version of contravariant 2 tensor h^(jk)

modern version of same contravariant 2 tensor: summation h^(jk) ej ek.


Then (summation gjk dx^j dx^k) acts on: (summation h^(jk) ej ek),

by contracting their components: summation gjk h^(jk). this is a number.

How does this seem?

 

[mathwonk]

PPS:
In some sense the modern point of view has made the wonderful contribution of doubling the number of indices!

I.e. the modern way of writing that last contraction would be:

(summation gjk dx^j dx^k) (summation h^(rs) er es)

= big summation (gjk dx^j dx^k) (h^(rs) er es)

= big summation (gjk h^(rs)) (dx^j dx^k)(er es)

= big summation (gjk h^(rs)) (dx^j(er))(dx^k(es))

= big summation (gjk h^(rs)) kronecker^(jk),(rs)

[this last because dx^jdx^k (ej ek) = 1, and all other pairings

dx^jdx^k (er es) are zero]

= summation gjk h^(jk).

But no one would often do this I hope.



I admit that if one understands the indices on the components, there is never any need for the basic tensors, but it seems almost to throw out the baby and keep the bathwater, to an index - free guy like me.

I admit too the indices are too complicated for me. I even wrote an algebra book once, including treating tensors in a coordinate free way, and actually wrote out the tensor product of matrix as a consequence of these definitions, but it was a terrifying experience.

Just for laughs, I confess that to me the tensor product, Atensor(blank), is actually "the unique right exact functor on R - modules that commutes with direct sums and takes value A on the field R of real numbers", but I would not readily say that here, if we weren't good friends by now!

Perhaps that reveals why I am having such a hard time understanding classical tensors though.

Peace,

roy

 

[mathwonk]

By the way, in your nice post 17 above, you distinguish tensors from components by writing the tensors as bold, e.g. in the last line, you write "e suba (bold)" for a basic contravariant 1 tensor.

In that same spirit, on the site

http://www.geocities.com/physics_world/ma/intro_tensor.htm

the symbols dx^j in equation (1) should be bold, since they are the entirely analogous basic covariant 1-tensors.

The fact that they are not bold, leads to the confusion that this expression denotes a 0 - tensor instead of a 2 - tensor,

i.e. equation (1) is a sum of, scalar multiples of, pairwise products of, basic 1 tensors,

hence it is a homogeneous "polynomial" of degree 2 in the basic 1 tensors,

i.e. a 2-tensor.

Does that seem believable?

 

[pmb_phy]

In that same spirit, on the site

http://www.geocities.com/physics_world/ma/intro_tensor.htm

the symbols dx^j in equation (1) should be bold, since they are the entirely analogous basic covariant 1-tensors.

The components of tensors are in italics.

The fact that they are not bold, leads to the confusion that this expression denotes a 0 - tensor instead of a 2 - tensor,

That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero. Why do you keep calling the interval a dstensor?

i.e. equation (1) is a sum of, scalar multiples of, pairwise products of, basic 1 tensors,

hence it is a homogeneous "polynomial" of degree 2 in the basic 1 tensors,

i.e. a 2-tensor.

Does that seem believable?

Nope. Sorry.

Pete

 

[mathwonk]

Pete,

This is the basic point you are missing in my opinion. I cannot make it any clearer though than I have in post #19 above. If that doesn't do it, I have no more to add.

Of course I could always be wrong! (and I frequently am.)

have a good one, my friend.

roy

 

[mathwonk]

Pete,

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert. If it makes sense to you, you do not need to concern yourself with anything I have been saying. He also recommends the famous book Gravitation?, by Misner, Thorne, and Wheeler, which apparently also uses the modern approach.

These are notes of Sean Carroll at Institute for Theoretical Physics at Santa Barbara, UCAL.

http://xxx.lanl.gov/abs/gr-qc/9712019

best,

roy

PS: I was going to quit after 100 posts but this is kind of addictive! I must stop soon though, as I have to go to work.

 

[pmb_phy]

Pete,

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert.

Yep. I've seen them and have read part of them several years ago and during the last few years as reference. I've also read MTW as well as many other GR and tensor analysis texts.

And I understand what you were saying. What I was saying is that it is incorrect.

I've been trying to find the notes I used when I first studied GR but the site is not working right now. It will be under http://arcturus.mit.edu/~edbert when it does come up.

Pete

 

[mathwonk]

Hi Pete, another day another dollar.

Well, let's try again.

I believe that since x is a function, its differential dx is a covariant 1-tensor, i.e. a section of the cotangent bundle, because it acts on a tangent vector, via directional differentiation, and spits out a number.

Similarly, the tensor product of dx^j and dx^k is a covariant tensor of rank 2, because it acts on a pair of tangent vectors and spits out a number, namely the product of their jth and kth coordinates in the x - coordinate system. so dx^j dx^k is a covariant 2 tensor, i.e. a section of (T^) tensor(T^).


Now you are saying that gjk is also a 2 tensor. well, from what you have told me, it is called a 2 tensor as abuse of language. But what is the actual 2 tensor it is shorthand for? I understand it to be shorthand for the covariant 2 tensor:
summation gjk dx^j dx^k.

That would make it also a section of (T^)tensor(T^),

hence not a candidate for contraction against another such bird.

So here is the crucial point. You must contract tensors of opposite variance.

I,e, summation gjk dx^j dx^k is not actually a contraction.


The problem to me seems to be that if one thinks any expression with some indices up and others down is a contraction, one gets in trouble.

I.e. you are mixing two different languages here. the notation dx^j always stands for a section of the cotangent bundle, namely the differential of x^j. Hence dx^j really is a tensor, not just the components of one, i.e. it is not abuse of language to call dx^j a tensor.

if you want to contract two tensors, they have to be written in the same way, not one as an actual tensor and the other as components of a tensor.

If you want to contract gjk with some thing, it has to be with something like h^(jk), which would be the components of a contravariant 2 tensor like

summation h^(jk) ej ek.


The problem is that the components of a tensor transform opposite to the way the basis elements of the tensor transform.

Thus the basic covariant tensors are written with indices up, while their components are written with indices down.

Thus you do contract components having opposite indices, but a sum of components and basis elements with opposite placed indices is not a contraction.


Lets be more concrete.

What bundle do you think dx^j dx^k is a section of? and what bundle do you think gjk represents a section of?

You canno, contract them unless your answer is that they are sections of mutually dual bundles, but to me that would be hard to bring in line with any accepted notation or usage.

roy

 

[mathwonk]

Pete,

Let me give an example from your own post 17. I will write G for "bold g" and

ea for "e lower a".

Then you say that G is the tensor, and gab = G(ea,eb) are the components.

That is correct. But another way to write the tensor G is as:

G = summation gab dx^a dx^b.

In fact that is the meaning of the statement that "gab are the components of G".


Here is how it looks written out fully:

ea and eb are basic contravariant 1 tensors,

and dx^a and dx^b are the dual basic covariant 1 tensors.

I.e. dx^a(ea) = 1, and dx^a(eb) = 0 for any b different from a.


Any way of acting on the tensors ea,eb can be expressed in terms of these simplest possible actions. Thus any tensor G can be expressed as a linear combination of dx's.



I.e. there are some numbers gab such that G = summation gab dx^a dx^b.

Those numbers gab are computed by evaluating both sides

on the basic contravariant tensors (ea,eb).



One obtains:

G(ea,eb) = summation gjk dx^j dx^k (ea,ab) = gab,

because almost all terms dx^j dx^k (ea,ab) in the sum are zero

except dx^a dx^b (ea,ab) = 1.



I.e. because G and summation gab dx^a dx^b are both (the same) cotensor,

they can both be evaluated on the basic contra-tensors (ea,eb), and yield the same answer.


peace.

 

[mathwonk]

Dear Pete,

One last contribution. if e1,...,en, are basis vectors for R^n, i.e. e1 = (1,0,...,0),

e2 = (0,1,0,...,0), ...,en = (0,...,0,1), (where ej means e "subscript j")

Then any vector V can be written as a linear combination of these basis vectors with number coefficients.

In the notational convention taught in the books you have referred me to, for instance Sean Carroll, these number coefficients are written with superscripts,

thus to say the vector V has coefficients a^1,...,a^n, means we can write V as

V = summation a^j ej = (a1,...,an).

By your logic, because these have oppositely placed indices this would be a contraction, and hence a 0 - tensor, i.e. a number, whereas I hope we agree this is clearly a 1 - tensor, i.e. a plain old vector.

According to Carroll, the only significance iof these objects having oppositely placed indices is that the object obtained by summing them is an invariant object, independent of coordinates. In particular here it is a vector.

By duality with your example in post #17, where you find the coeeficients or components of G, by evaluating G on the basic vectors ej, here we could also find the coefficients, i.e. components of V, by evaluating V on the basic covectors dx^j, or if you like, by evaluating those covectors on V.

(Carroll uses a notation like theta^j for dx^j.)

I.e. V(dx^j) = dx^j(V) = a^j. THIS is a contraction, between a contratensor V and a cotensor dx^j, and yields a number, a^j.

Oh well, I'm sure we agree in some sense, since we both seem to be getting along successfully using the subject, but we just do not understand what each other are saying.


Hope to meet you some day and have a nice glass of wine. That always helps me clear my head and open my mind.


roy

 

[sal]

Pete asked me for another opinion on this one, so I'm going to stick my oar in...

First, I have to say that all you guys seem to be arguing about is notation, and IMHO the only "correct" notation is notation that's either completely universal and already known to the reader, or notation that's explained in the text. The only "incorrect" notation is notation that's so confusing, nonstandard, or self-contradictory that the reader can't figure it out.

Ain't none of that latter stuff here, that I can see.

Pete,
Let me give an example from your own post 17. I will write G for "bold g" and ea for

$$e_a$$.

Then you say that G is the tensor, and gab = G(ea,eb) are the components.

That is correct. But another way to write the tensor G is as:

G = summation gab dx^a dx^b.

Almost but not quite.

If you want to be nit-picky squeaky-clean correct, and you want to follow standard conventions here, you need to use a tensor product on the right, and you didn't do that. You should write

$$G = \sum g_{ab} dx^a \otimes dx^b$$

Otherwise you've got an ordinary product, which is a very different thing.

What you actually wrote was identical to what Pete had in equation (1) in http://www.geocities.com/physics_world/ma/intro_tensor.htm. That is not an equation between tensors. Rather, it is an equation in terms of infinitesimals. (See the end of this post for more on the line element.)

They're "physicist's infinitesimals", which have been in common use clear back to Einstein's papers. They're a shorthand for taking a limit. If you reduce the power to 1 and divide through by dt, poof, you have a derivative.

In fact that is the meaning of the statement that "gab are the components of G".

Yup, you can write it as a tensor equation in terms of the basis covectors, you can write it as a line element in terms of infinitesimals, you can write it as matrix (since it's rank 2), or you can just write it as a single bold-faced letter. They all mean the same thing, as long as your reader is on the same wavelength you're on!

Back on the subject of the basis covectors, you said someplace that they should be bold. Well, that's one convention. Not everybody does that. Furthermore, some people also use a tilde overtop to indicate that they're covectors, and not just infinitesimals, or even vector gradients.

And some people never use the word "covector" at all and would say that my post was riddled with nonsense as a result -- they'd call them "dual vectors" or even "basis 1-forms".

I'm going to snip the rest of the post I was replying to, and comment on a few other items from earlier in the thread.

I believe that since x is a function, its differential dx is a covariant 1-tensor, i.e. a section of the cotangent bundle, because it acts on a tangent vector, via directional differentiation, and spits out a number.

As I said, that's one way of using the symbol "dx", or more commonly, dx, or even
$$\tilde{d}x$$ .
It's not the only way to use dx; another very common use is as an infinitesimal, and in fact that's typically how the "line element" is written.

Going on,

Now you are saying that gjk is also a 2 tensor. well, from what you have told me, it is called a 2 tensor as abuse of language. But what is the actual 2 tensor it is shorthand for? I understand it to be shorthand for the covariant 2 tensor:

Calling it any kind of "2-tensor" is indeed an abuse of the language. A "2-tensor", most often, is a tensor on a 2-dimensional space; what you're talking about here is a rank 2 tensor.

Written as $$g_{ab}$$, it's actually a rank (0,2) tensor, but since it can be converted trivially into a rank (1,1) tensor or a rank (2,0) tensor, it's also perfectly reasonable to just refer to the whole "package" of 3 related tensors as a "rank 2 tensor".

I.e. you are mixing two different languages here. the notation dx^j always stands for a section of the cotangent bundle, namely the differential of x^j.

As I said, if that's a tensor you're talking about, then yeah, it's a section of the cotangent bundle. But just as often, it's an infinitesimal rather than a tensor.

In the land of physics, it's just not accurate to say it "always" stands for a 1-form.

Here is a good looking reference for notes on relativity that uses both indices and the conceptual approach, by a clear expert.

Check this reference again, and look at page 25, formula 1.95. It is exactly what Pete had. I quote:

"A more natural object is the line element, or infinitesimal interval:
$$ds^{2} = \eta _{\mu \nu}dx^{\mu} dx^{\nu}$$ "

So as I said to start with, Pete's talking about an infinitesimal line element, you're talking about covectors, and you're both using the same notation. It's incorrect to say the notation is only used for one or the other -- it's used for both, and if it's not clear from the context which is meant, you need to spell it out.

I hope this helps, at least a little, with the confusion...

 

[mathwonk]

Well it does help, because I thought "infinitesimals" went out with Newton. What do they mean to you? I also took Carroll's first chapter which you cite, as an imprecise conversational verson of the material before it gets precise.

At least I made it possible for you to understand what I meant by dx^j by defining it so you could tell I meant it is a differential.

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor, and also that gjk was a 2 tensor, and that therefore the combination summation gjk dx^j dx^k was a contraction to a 0 tensor.

Here is a quote from his post #22:

"That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero."

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors. do you agree with that? that is all I was puzzled by.

 

[sal]

Well it does help, because I thought "infinitesimals" went out with Newton.

No indeed! They're alive and well in the physics community.

Mathematicians sometimes refer to them as "Physicist's sloppy infinitesimals".

What do they mean to you?

In general they're shorthand for a limit process. Older (pre-1950) books on tensor calculus used them exclusively. It was only in the last half of the 20th century that it became really common to use the formal definition of a tangent vector as a partial (path) derivative and cotangent vector as the dual of that, rather than just talking about an "infinitesimal displacement".

For example, Einstein, "On the Electrodynamics of Moving Bodies" (1905), p44 in the Dover edition "The Principle of Relativity", in the course of deriving the Lorentz transform, says

"Hence, if x' be chosen infinitesimally small..."

and he goes on from there, using derivatives and "infinitesimals".

Synge and Schild, "Tensor Calculus" (c) 1949, p. 9 in the Dover edition, in section 1.3, discussing contravariant vectors and tensors, say

"...These two points define an infinitesimal displacement or _vector_ PQ..."

Synge and Schild is something of a classic though it's now considered rather out of date.

I don't know when the machinery for handling tangent vectors rigorously was invented, but most of the seminal physics in relativity was developed without it.

I also took Carroll's first chapter which you cite, as an imprecise conversational verson of the material before it gets precise.

That's the most common way of writing the line element that I've seen, and it's done in terms of infinitesimals.

With enough effort you can define rigorous "differential" functions in one dimension and work it out that way, but if you do, you're just wallpapering over the original meaning which was a relationship among infinitesimals.

At least I made it possible for you to understand what I meant by dx^j by defining it so you could tell I meant it is a differential.

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor, and also that gjk was a 2 tensor, and that therefore the combination summation gjk dx^j dx^k was a contraction to a 0 tensor.

Here is a quote from his post #22:

"That expression is a tensor of rank 0. If you notice, it is the contraction of a second rank covariant tensor with two rank 1 tensors. Such a contraction is always a tensor of rank zero."

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors. do you agree with that? that is all I was puzzled by.

I don't see how that can be correct.

The expression is a representation of a tensor. I would hesitate before saying $$g_{ab} dx^a dx^b$$ is actually a contraction, because that would make it a single number (just as Pete said) and I don't see how to apply that to anything.

Pete may have a different notion as to what the terms in the line element representation of the metric mean, but what I've described here is, I believe, fully consistent with what's on his website.

 

[mathwonk]

It has dawned on me that physicists may be willing to use something logically nonsensical just because Einstein did so, and achieved correct results.

I think if the modern version of differential geometry had been around in 1900 then Einstein would have used it instead.

I agree of course that physicists are really using their intelligence and intuition, rather than mathematiocal rigor, which is why they so seldom go astray.

I love the recent story of the puzzle as to how many rational cubic curves lie on a general quintic hypersurface in complex 4 space. The mathematicians, by brute force computation had one answer, and the physicists by relating the problem to one in quantum gravity or something, had a different prediction which popped out of a recursion formula and a differential eqaution they thought applicable.

Of course the physicists were actually right, and it led to a whole industry in enumerative algebraic geometry.

We mathematicians are merely trying to formulate precisely the intuitions physicicts seem blessed with because of their familiarity with nature. We are at a big disadvantage here.

But we do not seem to argue as hopelessly as some theoretical physicists do, because we do eventually make clear what we are saying.

My error in my previous long harangue, was not to ask precisely what Pete meant by his notation, and not to say precisely what I meant by it.

That was what I meant when I expressed confidence we would agree at some level, once we understood each other properly.

I have almost never heard a disagreement that was not found to be based on different interpretations of the same words being used.

 

[pmb_phy]

if you read some of Pete's posts however you will see that he himself said that in his cited equation (1) that dx^j dx^k was 2 tensor,..

Caution is required here. So long as you know that I said that dx^j are the components of a vector then we're all set. I don't mind the shorthand statement that dx^k is a vector though but I'm never sure what you mean by it.

So Pete never said his equation denoted infinitesimals, rather that it was a contraction of rank 2 tensors.

Ummm .. scuse me, but I did say that in
http://www.geocities.com/physics_world/ma/intro_tensor.htm
right above Eq. (1), i.e.

The arc length, dl, between two closely spaced points on a curve is given in Cartesian coordinates, by ...

unless you didn't know that the arc length was an infinitesimal? Usually one doesn't need to state that explicitly since the notation speaks for itself (hence the purpose of notation). dl is an infinitesimal and Eq. (1) gives the square of dl.

Use caution when using notation. dx^k is a differential but dx^k is not a differential. It is the gradient of a coordinate making it a basis 1-form.

Pete

 

[mathwonk]

Right you are Pete.

My apologies. I am not too up on infinitesimals and it must have gone right by me. In general I want to thank you (and the others here) for your extraordinary patience with me.

This is a great site for learning, and I will recommend it to my students too!

best regards,

roy

 

[sal]

It has dawned on me that physicists may be willing to use something logically nonsensical just because Einstein did so, and achieved correct results.

I think if the modern version of differential geometry had been around in 1900 then Einstein would have used it instead.

I'm sure you are correct about Einstein's willingness to use the modern forms. I wish I knew when the modern machinery was invented -- certainly, the notion that a sensible definition of a vector could be something like

$$\sum_a v^a \frac{\partial}{\partial x^a}$$

is not something that would ever have occurred to me, and I can't believe it's something that's just been floating around forever. It seems like it must have been an invention (or discovery) that happened at some particular moment, probably in the last 7 decades -- but when? I have seen no mention of its origin in any textbook.

Until that was invented, as far as I know there was no rigorous basis for tensor calculus -- the infinitesimals were all they had, and that's why they were used so heavily.