Question

@ikram002p Ok how about a little tensor notation?

Answer 1

haha would be fabulous but consider im beginner so start easy :P

Answer 2

I dont understand your question @Kainui ? Can you be more clearer on it?

Answer 3

Ok so I'll start out with regular notation and then convert it over to tensors. =)

Let's say we have a path in space that is time dependent, so we have x(t) and y(t).

Now let's take their derivatives to find velocity:
\[\frac{d x}{dt}, \frac{d y}{dt}\]

So I'll just do this silly thing where I use the chain rule like this:
\[\frac{d x}{dt} = \frac{\partial x}{\partial x}\frac{dx}{dt}\]\[\frac{d x}{dt} = \frac{\partial x}{\partial y}\frac{dy}{dt}\]
\[\frac{d y}{dt} = \frac{\partial y}{\partial x}\frac{ d x}{dt}\]\[\frac{d y}{dt} = \frac{\partial y}{\partial y}\frac{dy}{dt}\]

So nothing really happened here since

\[\frac{\partial x}{\partial x}=1\] and \[\frac{\partial y}{\partial x}=0\] So those four things above simplify to this:

\[\frac{d x}{dt} = 1\frac{dx}{dt}\]\[\frac{d x}{dt} =0\frac{dy}{dt}\]
\[\frac{d y}{dt} = 0\frac{ d x}{dt}\]\[\frac{d y}{dt} =1\frac{dy}{dt}\]

Does this make sense haha, kinda looks long and confusing but I just copied the same thing basically.

Answer 4

yes , makes sense :)

Answer 5

So in matrices what I wrote was:
\[\Large\left[ \begin{array}c \frac{\partial x}{\partial x} & \frac{\partial x}{\partial y}\\\frac{\partial y}{\partial x} & \frac{\partial y}{\partial y}\\\end{array} \right] \left[ \begin{array}c \frac{dx}{dt}\\ \frac{dy}{dt}\\\end{array} \right] = \left[ \begin{array}c \frac{dx}{dt}\\ \frac{dy}{dt}\\\end{array} \right]\] which was really just an over complicated way of writing the identity matrix.

\[\Large\left[ \begin{array}c 1 & 0\\0& 1\\\end{array} \right] \left[ \begin{array}c \frac{dx}{dt}\\ \frac{dy}{dt}\\\end{array} \right] = \left[ \begin{array}c \frac{dx}{dt}\\ \frac{dy}{dt}\\\end{array} \right]\]

So really all I am saying here is that we can look at the chain rule in this sense as just multiplying by an identity matrix! That's it! We didn't even really bring in tensor notation at any point.

Answer 6

yeah was writing down first matrices then mentally thought of identity an hola it was in second line :P

Answer 7

So now for tensor notation, we can condense a lot of things down, so instead of writing x and y, we write using super scripts (not exponents!)

\(x=z^1\)
\(y=z^2\)

So now we can write these with an index so that our equations are much more general:

\(z^i\)

just plug in i=1 or i=2 to get that specific variable, but now of course we can plug in i=3, i=4, or whatever, so we can actually have as many variables as we want here.

Let's take the derivative like we did in the first post:

\[\frac{d z^i}{dt}\]

This represents 2 equations, not 1, so tensor notation will end up looking really small and representing quite a few things at once, so be careful, since it's about to get more:

Now let's look at the chain rule:

\[\frac{dz^i}{dt} = \frac{\partial z^i}{\partial z^1}\frac{dz^1}{dt}+\frac{\partial z^i}{\partial z^2}\frac{dz^2}{dt} \] So here if we plug in i=1 we recover:
\[\frac{dz^1}{dt} = \frac{\partial z^1}{\partial z^1}\frac{dz^1}{dt}+\frac{\partial z^1}{\partial z^2}\frac{dz^2}{dt} \] which is
\[\frac{dx}{dt} = \frac{\partial x}{\partial x}\frac{dx}{dt}+\frac{\partial x}{\partial y}\frac{dy}{dt} \]

So let's take this expression and compactify further with "Einstein summation notation" which just means whenever we see a repeated index, we sum over all possible values, so we can replace:
\[\frac{\partial z^i}{\partial z^j}\frac{dz^j}{dt}=\frac{\partial z^i}{\partial z^1}\frac{dz^1}{dt}+\frac{\partial z^i}{\partial z^2}\frac{dz^2}{dt} \]

So now we have the chain rule in tensor notation:
\[\frac{dz^i}{dt}=\frac{\partial z^i}{\partial z^j} \frac{dz^j}{dt}\]

Answer 8

Oh god @Kainui your figures aint tired?

Answer 9

fingers *

Answer 10

This isn't "THE chain rule" it just just one instance of it where the thing we get doesn't change what we have.

So we can see that this formula \[\frac{dz^i}{dt} = \frac{\partial z^i}{\partial z^j} \frac{dz^j}{dt}\] means that double indexed object must be the identity matrix, but in tensor notation we call this the Kronecker delta! \[\frac{\partial z^i}{\partial z^j} = \delta^i_j\] and it equals 1 when \(i=j\) and 0 when \(i \ne j\) which is just another perspective on the identity matrix. =P

\[\frac{dz^i}{dt} =\delta^i_j\frac{dz^j}{dt}\]

So in tensor notation the identity matrix just renames indices, so if you read this right to left, we have j being renamed into i, that's about it for this silly thing, but you can do a lot of cool stuff with matrices in tensor notation that's a pain in regular matrix notation.

Also yeah my fingers are kinda tired haha. @Here_to_Help15

Answer 11

I knew it lol see @ikram002p told you his fingers gonna fall ;)

Answer 12

there's definitely a better way to teach and introduce the subject, but I just wanted to give a fairly straight forward answer since ikram only has an hour before bed and wanted me to teach her something, so here it is hahaha.

Answer 13

hm? this was intended as an introduction to tensors?

Answer 14

lol this is a strong intro lol so dont judge :)

Answer 15

I think i am learning things my brain didnt even think of lol

Answer 16

Hahaha yeah, it's sorta just off the cuff, there's a lot being left out here since tensors have a lot of subtleties to them I haven't even mentioned. As far as it goes, it's more or less mechanically correct.

Answer 17

haha i can use to this notation from now on !

Answer 18

@Kainui you deserve 10 medals for this introduction( for each finger, unless you have more then just 10) :)

Answer 19

ok now do u have quick application ?

Answer 20

@Here_to_Help15 dont be sure he is using out of 10 fingers in writing =P

Answer 21

Taking derivatives with tensors in non Euclidean spaces ends up changing a lot of stuff, but as far as playing with matrices it can be real nice, for example in matrix notation if you multiply:

AB=C

in tensor notation this can be written as:

\(A^i_jB^j_k=C^i_k\)

However this is completely commutative since we are looking at the entries in the summation, the indices are what tells us what row and column:

\(A^i_jB^j_k=B^j_kA^i_j\)

Answer 22

I guarantee he is using all 10

Answer 23

Since you are likely not comfortable with Einstein summation notation yet, if we have a 2x2 matrix

\(A^i_jB^j_k = A^i_1B^1_k+A^i_2B^2_k=C^i_k\)

So if you want to look at a specific entry of the matrix C, you have two indices as you would expect, one for the row and one for the column. The proof of commutivity is more obvious maybe now:

\(A^i_jB^j_k = A^i_1B^1_k+A^i_2B^2_k = B^1_kA^i_1+B^2_kA^i_2 = B^j_kA^i_j\)

Answer 24

There's a lot of cool things in tensors if anyone's interested in learning more, I'm by no means an expert but there's a lot of interesting things I can teach. I haven't even said what a tensor is yet haha.

Answer 25

its kind of geometric relation of scalar and vectors right ?
what cool definition and simple examples u have ?

Answer 26

Well if we want to represent a vector, we could write it as:

\(\bar V = V^i \bar Z_i = V^1 \bar Z_1+V^2 \bar Z_2+V^3 \bar Z_3\)

You can see that \(V^i\) are the components of the vector with respect to the basis elements \(\bar Z_i\).

The thing that's important there is no matter what way you look at this vector, that vector is always the same, it's just a geometric object sitting in space, a directed line segment. So one of the key things that tensors must do is when you have an object with no indices, it must represent an invariant object, a geometric object that doesn't depend on your choice of coordinates.

The way you do this is by insuring that all of your objects obey a certain transformation rule by a matrix called a Jacobian. In tensor notation this matrix is:

\[J^{i'}_i=\frac{\partial Z^{i'}}{\partial Z^i}\] The primed coordinate system could represent \(r=Z^{1'}\) and \(\theta = Z^{2'}\) and \(x=Z^1\) and \(y=Z^2\)

So to make sure that the vector components are tensors, they must obey this rule:

\[V^iJ^{i'}_i= V^{i'}\]

And this would be the components in the new coordinate system! So while the components are different for a different coordinate system, if you contract these away you end up with the same object, that same vector!

I'm obviously leaving out a lot of details, but there's a lot of stuff to digest and get used to since it's weird working with tensors at first because of how condensed they are.

Answer 27

When you see a tensor it has a number of indices, could be 0, 1,2, 3, etc... and this is called the rank of the tensor. So a tensor of rank 0 is a scalar, rank 1 is a vector, rank 2 is a matrix, and rank 3 and higher don't really have names, they're just tensors.

Tensors can exist in any number of dimensions, so for instance a tensor of rank 1 in 3 dimensions is a vector with 3 components in it. A tensor of rank 2 in 3 dimensions is a 3x3 matrix. Similarly a tensor of rank 1 in 2 dimensions has 2 components and a tensor of rank 2 in 2 dimensions is a 2x2 matrix.

The number of entries in a tensor is then \(dim^{rank}\) however this isn't entirely true in general. You an have non square matrices right? Same goes for tensors, one example is the Shift Tensor which can be used to project vectors from 3D space onto a 2D surface.

Answer 28

Ok so maybe you remember from vector calculus these identities:

f is a scalar field, the curl of the gradient is 0:
\[ \bar \nabla \times (\bar \nabla f )= \bar 0 \]

F is a vector field, the divergence of the curl is 0:

\[\ \bar \nabla \cdot (\bar \nabla \times\bar F) = 0 \]

In tensor notation these equations are actually the same thing. So there's one fun example.

\[\epsilon^{ijk} \nabla_i \nabla_j f=0 \\ \epsilon^{ijk} \nabla_i \nabla_j f_k = 0\]

Answer 29

got it !

Answer 30

OMG @Kainui how are you keeping your fingers connected to your hands???

Answer 31

ok kai thanks for tonight i wrote them now :P
will see other stuff some other time gn :)

Answer 32

Good night!