- #1

chandran

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

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- Thread starter chandran
- Start date

- #1

chandran

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

- #2

marlon

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Scroll down to the end of this page and look at the 'what is a tensor' entry

https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=7 [Broken]

regards

marlon

https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=7 [Broken]

regards

marlon

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- #3

Antiphon

- 1,683

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(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

Meir Achuz

Science Advisor

Homework Helper

Gold Member

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[itex]S_i=\Sigma\epsilon_{ij}P_j[/itex]. The [itex]\epsilon_{ij}[/itex]

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

[itex]L_i=\Sigma I_{ij}\omega_j[/itex].

- #5

marlon

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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.Antiphon said:If you want to know how much of one vector points along the direction of another

vector you use a dot product.

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.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.

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 entryTo change it's direction requires multiplying it with

a tensor.

regards

marlon

- #6

Antiphon

- 1,683

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Of course all your comments are accurate Marlon, and your link already

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.

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?

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

No vector operations can change the direction of a vector. For that and in

the general case (of arbitrary transformations) you need tensors.

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