Understanding Pressure as a Scalar Quantity: Exploring Tensors and Rank

[ovais]

ok the normals stresses are neither scalars nor vectors but a second rank tensors!So what about pressure?Since it is obtain by just dividing the sum of three tensors(of rank 2)by -3, so does pressure is also a tensor of rank 2?And ofcourse i will go through these topic of tensors I Actually today buy a book of tensor analysis but it has only complex mathematic.My thanks are due to you all for presenting things to me so easy.

 

[Studiot]

Furthermore if they are not equal it implies motion.

 

[ovais]

so studiot do you mean the normal stress x,x y,y z,z will have to be equal for a static system?Also let me know once I found the pressure by dividing the sum of normal stresses by -3, then to calculate the force on the cube with what area i should multiply the pressure.do i take area of one face or of three faces?

 

[DrDu]

I hope not! E.g. I am sitting here on a chair and this implies normal stresses on its legs in vertical direction, but there are no equal stesses along the horizontal axes. The chair gets a little bit compressed when I sit down (normal strain) but reaches an equilibrium position.

 

[ovais]

please clarify me about pressure which is found by dividind tensors is a tensor or not?

 

[Studiot]

I didn't introduce continuum mechanics and its control volume to the thread. I was trying to avoid this as I don't consider such depth or tensors to be appropriate for 12 graders.

However, I apologise if I didn't make myself quite clear.
The stresses you guys are describing are the difference between the stresses acting upon opposing faces of the control volume. They are not absolute stresses. If these differences are not zero there is a net (resultant) force acting on the control volume, which implies motion of the control volume in the direction of that force.

What did you think of my comments about class discussions?

 

[ovais]

yes as far 12 graders are conserned they even haven't hear the word tensor.I am asking all this for my self i am studying in university.

 

[DrDu]

Pressure is not a tensor but a scalar. However, a symmetric second rank tensor like the stress tensor can be decomposed into a tensor whose trace is zero and a second rank unit tensor times a scalar.
In the case at hand, the scalar in question is minus the pressure.

 

[Studiot]

So do you understand the difference between pressure and stress, since both have the same units?

 

[Studiot]

You might like to look at post#5 of this thread. It describes what Dr Du is talking about.

https://www.physicsforums.com/showthread.php?t=371071&highlight=stress+tensor

 

[ovais]

So do you understand the difference between pressure and stress, since both have the same units?

what is understand is pressure only a component of stress which is a 2nd rank tensor and its isentropic trace is what we call pressure.M i right?

 

[Dale]

ok the normals stresses are neither scalars nor vectors but a second rank tensors!

No, they are COMPONENTS of a 2nd rank tensor. They are most definitely not tensors themselves.

 

[ovais]

No, they are COMPONENTS of a 2nd rank tensor. They are most definitely not tensors themselves.

if they are not tensors then what they are?scalara or vectors?do the normal stresses have direction?

 

[ovais]

And yes i understaand they themselves are not tensors but the components of tensors and i think drdu and studiot also mean the same but may be they write it tensors for my understanding.but now please clarify me what should we call these?scalars or vectors or anything new?

 

[Dale]

They are neither scalars nor vectors. They are components of a 2nd rank tensor. I don't know what more you want.

Do you understand the difference between a vector (a coordinate-independent geometric object) and its components (a set of numbers used to represent the vector in some specific coordinate system)? If so, then that is the same relationship between a tensor and its components.

 

[ovais]

ok now i understand what you mean a component of tensor.so trully they are not vectors or scalars.ok now on adding and dividing them by -3 we get pressure so dosent pressure also come in same category i mean we should call pressure to be neither scalar nor vector.

 

[Studiot]

I am not happy with the use of the term 'components' for Tensors. Vectors (in physics) have components, Tensors do not.

IMHO the term is best reserved for elements which can form a basis of a vector space.

I prefer the term 'elements'.

As to the nature of these elements. I offer a similar line of argument which showed pressures are not vectors.

Stresses are not vectors. Stress resultants are vectors. Vectors (in physics) have a definite line of action, stresses do not.

 

[ovais]

now i am confused in two things.1-components & elements.let me ask you by an example.if we say a car has a velocity(3i,4j)then 3i & 4j are its component.aren't they.and it feels to me that the components of velocity(a vector quantity) are vectors.is 3i not a vector?now where is the element here?this however is not a tensor.2 you call stess is not vector but resultant of stress is vector.i don't get this.please help.

 

[Studiot]

In a physical system at every point in space (x,y,z) we can observe physical quantities acting.

Some of these can be represented by simple numbers.
There are two types, we call these both scalars.

We can understand the two types best by considering a specific region of space.
Within the region
Extensive properties are additive. We add the numerical value at each point in the region to get one number that represents the property in that region.
For example the mass of our region is the sum of the mass values at every point.
Intensive properties
Are an average. We average the numerical value at every point to get a value to represent the property for the region.
For example temperature. We average, not add the temperatures at every point to get the region temperature.

We can write down the property as a function of position f(x,y,z)
This is known as a scalar field.

Some physical quantities are more complicated.

The simplest is where we assign or observe a vector at every point. Examples are velocity vectors in fluid flow, field vectors in electric and magnetic fields etc.

In our 3D coordinate system any vector can be written

$$v = \alpha x + \beta y + \gamma z$$

Where the greek letters are numbers and roman letters are vectors.

So at every point there is a vector v which can be written in the above way. Each of the
$$\alpha x$$; $$\beta y$$; $$\gamma z$$
are vectors in their own right and v is a linear combination of them. They are the basis vectors.

We call these components.

The important fact about components is that they are (linearly) independant. We can vary anyone without affecting one of the others.

We can write down a vector valued function v = f(x,y,z)
This is called a vector field.

But there are yet more complicated physical quantities, called Tensors, that we can assign or observe at any point in space.

The parts of tensors affect each other, unlike vector components, so we call these parts elements rahter than components.

Examples are the stress and strain tensors, the inertia tensor.

Again we can form a Tensor valued function whcih describes the distribution of the tensors in our 3D space.

T = f(x,y,z)

Since tensors are represented by 2 dimensional matrices the information of the interaction between the elements. However the elements may not form a basis - and don't in the case of the stress tensor.

You cannot in general change the stress on one axis, without affecting the stresses on the others.

this is a brief intorduction to physical quantities, without all the mathematical notation which tends to obscure the physical meaning.

Another way to look at it is to note that with vectors, there is no essential difference between its components. We could interchange x, y and z without altering things.
With tensors, on the other hand, we cannot do this. the elements are not all the same. We cannot interchange shear and normal stresses at will.

 

[Dale]

I am not happy with the use of the term 'components' for Tensors.
...
I prefer the term 'elements'.

I am fine with that, but "components" is in common usage even if it is sloppy so ovais should be aware. In any case, the point I was making is that elements/components of a tensor are not themselves either scalars or vectors.

 

[Dale]

now i am confused in two things.1-components & elements.let me ask you by an example.if we say a car has a velocity(3i,4j)then 3i & 4j are its component.aren't they.and it feels to me that the components of velocity(a vector quantity) are vectors.is 3i not a vector?now where is the element here?this however is not a tensor.2 you call stess is not vector but resultant of stress is vector.i don't get this.please help.

OK, your writing is a little sloppy here, you should not write (3i,4j). Assuming that i and j are vectors then the car would have a velocity of 3i + 4j which is also a vector. This follows from the normal rules of vector spaces. If i and j form a complete set of basis vectors then it is possible to use them to define a coordinate system where (A,B) would be the coordinates of any vector v = Ai + Bj. A and B are the elements (I will use Studiot's clean terminology) of v. A and B are not themselves either vectors or scalars.

Btw, technically (and for the purposes of this thread) vectors are rank 1 tensors and scalars are rank 0 tensors. Often the word "scalar" is used to refer to something that is just a real number rather than a tensor of rank 0. This is more common but sloppy terminology that you will see.

 

[DrDu]

Another way to look at it is to note that with vectors, there is no essential difference between its components. We could interchange x, y and z without altering things.
With tensors, on the other hand, we cannot do this. the elements are not all the same. We cannot interchange shear and normal stresses at will.

...which leads directly to the question about irreducible tensors and the difference between pressure and the traceless part of the stress tensor. Does the latter even have a special name?

 

[Dale]

Btw ovais, I think that this post is the key and is as far as you should go with 12th graders. The discussion about tensors is not going to be helpful.

(1) The vector normal to the surface, and (2) the force that creates the pressure. One way to think of it is that P = F/A implies F = PA. But we can write the second using both scalars and vectors so that it applies in both cases.

From the equation F=PA it is clear that P must be a scalar. I wouldn't go further because that would require IMO the introduction of tensors and continuum mechanics.

 

[dulrich]

I am fine with that, but "components" is in common usage even if it is sloppy so ovais should be aware. In any case, the point I was making is that elements/components of a tensor are not themselves either scalars or vectors.

I know that this is semantics and preferences (elements versus components), but its possible to talk about 5 and 3 as "components" of 8 because they are pieces that add back to the original. So I think it's okay to call these elements components.

Also, tensors do form a vector space (in the mathematical sense: they obey all the rules for vector addition, scalar multiplication, etc.) The basis for the tensors are called dyads:

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

So, even in the more restrictive sense of "vector components" I think it is appropriate to use the term components.

My two cents...

EDIT: Sorry for the delayed post -- i didn't notice the conversation had shifted back to the OP.

 

[ovais]

so whatever we call elements or components,they are somewhat different from the components of a vector.we can not chance the value of one component without affecting the other as is being said by studiot.Do these elements themselves poses a physical property?stress tensor is a tensor of rank 2.now if i chose anyone of the diagonal element or component out of the 3 diagonal element, will this component component poses some physical meaning?is this the normal force on one of the face of the cube?

 

[Studiot]

Does the latter even have a special name?

I do believe DrDu you are talking about the deviator and mean stress tensors, as I referenced in post#45.

ovais,

Did you look at the reference I gave in post#45? I don't really want to type all that out again, but it should answer your questions.

Incidentally some of the elements in the stress tensor are shear stresses and some are normal stresses.
Pressure does not appear directly anywhere, but would correspond to $$\sigma$$_m in the mean stress tensor. Note again that in this case all the entries in the leading diagonal are equal and the rest are zero. This is what I meant in post#37

A further note:
The extra quantities or information in a tensor over that in a vector is to do with rotation.

Have we given up on 12 grade now?

 

[Studiot]

Also, tensors do form a vector space (in the mathematical sense: they obey all the rules for vector addition, scalar multiplication, etc.) The basis for the tensors are called dyads:

I think you are confusing tensors themselves and the elements of the tensor matrix.

It is true that the sheaf of (geometric or physical) vectors at any point in space {x,y,z} forms a complete vector space. Much of differential geometry is based upon this fact.

It is also true that the set of 3x3 matrices form a complete vector space at any point.
The set of all second rank tensors is isomorphic to this space and is therefore a vector space.

However the set of all possible stress tensors (or their 3x3 matrices) is not necessarily isomorphic to these spaces so I would be interested in your proof that this set forms a subspace of the above.

 

[dulrich]

I think you are confusing tensors themselves and the elements of the tensor matrix.

It is true that the sheaf of (geometric or physical) vectors at any point in space {x,y,z} forms a complete vector space. Much of differential geometry is based upon this fact.

It is also true that the set of 3x3 matrices form a complete vector space at any point.
The set of all second rank tensors is isomorphic to this space and is therefore a vector space.

However the set of all possible stress tensors (or their 3x3 matrices) is not necessarily isomorphic to these spaces so I would be interested in your proof that this set forms a subspace of the above.

To be honest, I'm not sure what you are getting at. My background is with differential geometry, so maybe I'm missing something. I looked at the Wikipedia article suggested by DaleSpam in post #27. In particular:

http://en.wikipedia.org/wiki/Stress_(mechanics)#Transformation_rule_of_the_stress_tensor

Which states:

It can be shown that the stress tensor is a contravariant second order tensor...

It doesn't seem necessary to show that the set of all possible stress tensors form a subspace of the second rank tensors. Isn't it sufficient merely to show that the stress tensor is an element of the space?

 

[Studiot]

Isn't it sufficient merely to show that the stress tensor is an element of the space?

Sufficient for what?

All stress tensors are 3x3 matrices, but not all 3x3 matrices represent stress tensors.

Therefore the set of stress tensors is a subset of the set of all 3x3 matrices, but not necessarily a subspace in its own right.

I am trying to clarify things from a physics point of view, rather than obscure them from a mathematical one. I consider the different definitions of the word 'vector' adopted by different branches of science to be possibly the most unfortunate clash in science - it seems to lead to huge amounts of miscommunication.

 

[dulrich]

Sufficient for what?

I was trying to address this statement in post #62:

However the set of all possible stress tensors (or their 3x3 matrices) is not necessarily isomorphic to these spaces so I would be interested in your proof that this set forms a subspace of the above.

But as I said, I didn't really understand your point.

I am trying to clarify things from a physics point of view, rather than obscure them from a mathematical one.

I appreciate this, quite a bit actually.

I consider the different definitions of the word 'vector' adopted by different branches of science to be possibly the most unfortunate clash in science - it seems to lead to huge amounts of miscommunication.

I don't want to hijack the thread (although that's already been accomplished ), but can you expand on this statement or point me to somewhere I can learn more about it? Thanks.

 

[Studiot]

OK, last point first.

Vector is used in the sense of 'agent or 'carrier' in biological / medical / anthropological sciences.

Computer scientists use the word vector is some specialised database theory.

Physicists use those pointy arrow things with magnitude and direction we all know and love.

Mathematicians have extracted, extended and formalised the idea of linearity in the concept of a vector space (which I was trying to avoid introducing to this thread).

As far as I can tell, the first two have nothing to do with magnitude and direction or linearity.

There are many vector spaces which have nothing to do with magnitude and direction , for instance the vector space of integrals of continuous functions, the integrals being regarded as vectors in this space.

The only operations guaranteed by a vector space are addition of vectors and multiplication of vectors by a scalar. Vector multiplication may or may not be defined on a particular vector space.

Further the internal workings of the 'vectors' may or may not be linear. The internal workings of physics type pointy vectors are guaranteed to be linear. So although you can add the vectors to each other in a linear fashion the calculations needed to produce the individual vectors can be anthing but linear - as in the case of integrals.

The components of physics pointy vectors are linearly independant. The elements of the stress tensor matrix are not.

Hope this helps, I have no more time here tonight as it is now half past one in the morning.

Cheers

 

[Dale]

I consider the different definitions of the word 'vector' adopted by different branches of science to be possibly the most unfortunate clash in science - it seems to lead to huge amounts of miscommunication.

Yes, vectors are a well-known source of miscommunication.

Tower voice: Flight 2-0-9'er cleared for vector 324.
Roger Murdock: We have clearance, Clarence.
Captain Oveur: Roger, Roger. What's our vector, Victor?
Tower voice: Tower's radio clearance, over!
Captain Oveur: That's Clarence Oveur. Over.
Tower voice: Over.
Captain Oveur: Roger.
Roger Murdock: Huh?
Tower voice: Roger, over!
Roger Murdock: What?
Captain Oveur: Huh?
Victor Basta: Who?

-Airplane

 

[dulrich]

Physicists use those pointy arrow things with magnitude and direction we all know and love.

Mathematicians have extracted, extended and formalised the idea of linearity in the concept of a vector space (which I was trying to avoid introducing to this thread).

I see. Thanks for the clarification.

 

[Antiphon]

Back to the philosophy of pressure for a moment.

You may define a scalar pressure. The analysis begins with the stress tensor but if the medium doesn't support a shear then a scalar pressure is well defined.

The situation is the same as in electromagentics. The permeability and permittivity of free space are scalars but this is only because space is isotropic. In fact these quantities are tensors in the general case. In certain crystals there are different electric fields along different directions from say a point free charge in the crystal. We say D = epsilon* E but in this case the D and E vectors point in different directions.

 

[ovais]

guys i read all of your posts and it seems our discusion regarding pressure as a trace of stress tensor has extended to other things related to the base of tensors and vectors.This will lead to hide what i need to ask right now. See what I am now asking?Upto know I get understand that pressure is the diagonalize part of the stress tensor and that stree tensor is a rank 2 tensor represented by a 3x3 matrix.the 9 elements of the matrix are such that they are not indipendent of one another.