Need mental picture

plxmny

Hello Gals,

I know what a scalar is.
I know what a vector is.
I know what a linear transformation is.

But what in the name of sweet aunt petunia is a rank 3 tensor?

Love,
Plx Mny

Demystifier

I believe that it is misleading to try to visualize higher-rank tensors. Think of vector as collections of 3 or 4 numbers, not as arrows. Then the algebraic generalization to matrices, rank-3 tensors, etc, is trivial.

pmb_phy

It is often recommended that you don't try to visualize a 3rd rank tensor, or any tensor for that matter, but to think of it merely as a multi-linear function which maps vectors and 1-forms to to the set of real numbers. The higher the rank the tougher the visualization becomes. In fact I never try to visualize a tensor myself.

Pete

plxmny

Hmm. I take it you've never read David Hume.

Anyways, I thought a little about this on my lunch break. I came up with "a
linear combination of linear combinations". Doesn't seem like a concept worth worrying about.

pmb_phy

There are always people who anyone never read. So what does David Hume have to do with this and why didn't you explain it when you posted his name? That's an irritating habit for you to fall into.

Pete

pervect

A rank 3 tensor inputs three generalized vectors (i.e. either a vector or their dual vector), and spits out a scalar.

One can also think of it as inputting 2 generalized vectors (or a rank 2 tensor), and outputting a vector, or inputting 1 generalized vector, and outputing 2 vectors (or a rank 2 tensor).

Xezlec

I was wondering this same thing. I can't even visualize linear transformations. I keep wondering: does not being able to visualize it make it impossible to do things like GR if you're a "geometric thinker" like me as opposed to a "formula thinker" like... pretty much everyone else?

I'm really bad. My eyes spin in circles whenever I see a sum in sigma notation, and I have to write it out with the ellipsis before I understand what it's saying!

plxmny

Well I am just learning this stuff and it's pretty clear that there is a lot of obfuscation going on. So I am taking very small bites.

I can definitely visualize a linear transformation. Maps a vector to another vector. Period.

pmb_phy

That holds true for any kind of transformation and not specifically to a linear one. A linear transformation is of the form

Y = aX + B

I.e. all linear transformations have this form.

Pete

Thrice

I know what a rank 10 tensor is
I know what a rank 11 tensor is...

Xezlec

As pmb_phy mentioned, that's just a "transformation". That I can visualize. It's the "linear" part that gets me.

Xezlec

That's not right. A linear transformation is not allowed to offset the vector (your B), and it can do a lot more than just scale the vector by a scalar "a". It can skew it, rotate it, and other things.

christianjb

Many tensors can be visualized as ellipsoids.

e.g. the inertia tensor, or the polarizability tensor.

pmb_phy

Thank you for that correction Xezlec. You're correct of course.

Pete

Xezlec

That's interesting. That got me thinking and exploring Wikipedia. I guess the best way to visualize a general linear transformation is to visualize the three basis vectors of a 3D coordinate system and think of just changing/moving any/all of them in any way, with their tails remaining stuck together at the origin. Then I can picture the effect on any object living in the space "attached" to those vectors.

Then, a rank-3 tensor is like picturing those three arbitrarily changed/moved vectors in a different coordinate system, and applying a different linear transformation to each one.

alan_stafford

m=(((1,2),(2,4)),((2,3),(5,6))) is a rank 3 tensor with dimension 2, a vector of matrices.
A tensor is a nested list.

An example Eigenmath http://eigenmath.net/ script with a rank 3 tensor is:

--Maxwell equations in tensor form.
--See the book Gravitation p. 81.
--
-- F + F + F = 0
-- ab,c bc,a ca,b
--
-- ab a
-- F = 4 pi J
-- ,b
--
--For this demo, use circular polarized light.
--
EX = sin(t+z)
EY = cos(t+z)
EZ = 0
BX = cos(t+z)
BY = -sin(t+z)
BZ = 0
FDD = (( 0, -EX, -EY, -EZ),
( EX, 0, BZ, -BY),
( EY, -BZ, 0, BX),
( EZ, BY, -BX, 0)) --See p. 74. Here, DD means "down down" indices.
X = (t,x,y,z) --Coordinate system
FDDD = d(FDD,X) --Gradient of F
T1 = transpose(transpose(FDDD,2,3),1,2) --Transpose bca to abc
T2 = transpose(transpose(FDDD,1,2),2,3) --Transpose cab to abc
check(FDDD + T1 + T2 = 0)
guu = ((-1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1))
FDDU = contract(outer(FDDD,guu),3,4) --Easier to make FDDU than FUUD.
check(contract(FDDU,2,3) = 0) --For light J is zero.
"OK"

The gradient of a rank 2 tensor (matrix) in a coordinate system (vector), is a rank 3 tensor. (GAMUDD) in example below.
Another example is the gradient of the metric in general relativity which is the connection.
An example of a 4th rank tensor is the Riemann curvature of spacetime. RUDDD in http://eigenmath.net/examples/bondi-metric.txt .

drsjs

Now some one try to visualize contravariant tensors! I tried to twenty years ago and then decided to do grad school in engineering. At least with fluids you don't get beyond three dimensions!