# What is a stress tensor

1. Jul 23, 2005

### chandran

what is a tensor. what is a stress tensor? what is a inertia tensor?

2. Jul 23, 2005

### marlon

3. Jul 23, 2005

### Antiphon

Marlon's writeup is good. Let me try to intuitize it a bit for your particular question.
(There is a special subset of tensors called dyads. I'm using those to get the
point across in this example.)

If you want to know how much of one vector points along the direction of another
vector you use a dot product.

But a tensor is something which can result in another vector when you take the dot
product of the tensor and the vector. The vector you get out can point in a different
direction too.

In the case of stress you want to know which way the forces push or pull on a small
portion of stuff inside a solid if you were to slice a tiny face of it. Since force is
a vector and direction is another vector the stress tensor says "give me a direction
you want to look in and I'll tell you which way the forces are pointing."

Tensors can do more than that mathematically but that should help you to
undertand why you need them. You need them whenever you have general
relationships between vectors which must transform those vectors into
different directions from the original vectors.

Another way to say it is that to change the amplitude of a vector you can
multiply it by a number. To change it's direction requires multiplying it with
a tensor. A vector cross-product is actually a tensor operation because it
operates on two vectors and results in a third which points in a new direction.

4. Jul 23, 2005

### Meir Achuz

In many materials the deformation vector (strain, S)) of the material is not in the direction of the applied stress vector (force per unit area, P). Then Hooke's law must be written as a matrix equation
$S_i=\Sigma\epsilon_{ij}P_j$. The $\epsilon_{ij}$
are the nine components of the "stress tensor". It is called a "tensor" because of the way it transforms under a rotation of coordinates.
The "tensor of inertia" is a tensor that relates angular momentum to angular velocity when they are not in the same direction. That equation is
$L_i=\Sigma I_{ij}\omega_j$.

5. Jul 23, 2005

### marlon

Well, ok, but do mention that the dot-product can be seen as a tensor itself; ie a (0,2)-tensor. You give in two vectors and this tensor gives back a scalar.
Then you also know that a vector product is a tensor...etc...

I am not sure this is quite accurate. i think you should replace the dot product by tensor product. I mean, the dot product is defined as a (0,2)-tensor so it must always yield a scalar. You are referring to a matrix product (which is also a tensor) or more generally a tensor-product.

What exactly are you talking about here ? A different direction to what ?

Well this is ofcourse true but this is certainly not a very general definition of a tensor. I mean, you can change the direction of a vector by simply multiplying it by a scalar : -1 for example. Indeed a scalar is also a tensor in itself, but the point is that i can apply your way of reasoning without using the word tensor ONCE. Hence this is not complete. Besides you are only giving 50% of the definition at best because you are totally omitting the required transformation properties. In your case Christoffel symbols would also be tensors but THEY ARE NOT. I explain this i my above journal entry

regards
marlon

6. Jul 25, 2005

### Antiphon

contained everything I restated. I merely tried to translate it down to the
level of the question being asked.

I'm an EE so I prefer to think of it as "impedance matching of ideas." Not
much will get through to the original poster unless he can relate to it in terms
he already understands. Later, he can come back and learn the full and correct
meanings. Of course I presume he needed than level of explaining, but what else can I assume without further explanation from him?

No vector operations can change the direction of a vector. For that and in
the general case (of arbitrary transformations) you need tensors.

Last edited: Jul 25, 2005