Is Pressure a Scalar Quantity or a Vector?

[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.

 

[Ahsan Khan]

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.

 

[Ahsan Khan]

means we can't change the value of one element without affecting the other elements.and as tensors are yet more complex they are neither scalar nor vector.but as their matrix is formed by two type of vectors area vector and force.I want to know if you have a cube with forces acting on it.how will you form each element of the tensor matrix?I am asking this to know what actually the elements of tensors made of?and what is the speciality diagonal elements have?

 

[Ahsan Khan]

knowing how each element is formed and understanding if the elements themselves have any physical meaning, we can move toward to know if the elements themselves are scalar, vector,tensor or none in case the elements themselves have no physical meaning.so i will more thankful if anybody teach me how to form these 9 elements of the stress matrix out of forces and the cube surfaces.regards

 

[Ahsan Khan]

just as in vetors if somewhere it is written that the velocity of car is(3i+4j-k) m/s then know then all its component have some meaning,3i shows car has a velocity of 3m/s in x direction,4j means its velocity in y direction is 4m/s.so if elements in the matrix have any physical meaning then we have look what they are.I understand that in tensor quantities the elements of matrix are not as simple as the components of a vector.

 

[Ahsan Khan]

but as these elemnts together when arranged in a matrix they represent stress tensor hence each element has something related to stress.I again will ask you people to makea tensor matrix for stress tensor.I want to see how each element of the matrix is formed and thus could know what each element intrept.I hope if some body do this then the discussion will end succesfully in 3 or 4 more posts.thanks for continue support I have taken a lot of you precious time.

 

[Pythagorean]

P=f/a is false in general. The actual relation involves calculus, which simplifies to p=f/a under appropriate assumptions.

 

[Ahsan Khan]

In facts studiot has already told me that some of the elements in the stress tensor matrix are shear stresses and some are normal stresses and that pressure is not directly represented in the matrix but is found by mean of diagonal ellements.What I want know is HOW we form these elements of stress tennsor?we initially have forces and surface areas of cube, now what to do with these to form these elements to form stress tensor matrix.Only then i can digest what element is what.regards

 

[Pythagorean]

In facts studiot has already told me that some of the elements in the stress tensor matrix are shear stresses and some are normal stresses and that pressure is not directly represented in the matrix but is found by mean of diagonal ellements.What I want know is HOW we form these elements of stress tennsor?we initially have forces and surface areas of cube, now what to do with these to form these elements to form stress tensor matrix.Only then i can digest what element is what.regards

The surfaces have a direction and the forces have a direction. Pressure exists between forces and surfaces that are aligned in space. Since the position of the elements in a tensor represents the combination of the spatial dimensions of those two quantities, the diagonals represent a match between dimensions (x-x, y-y, z-z)

the shear stresses are mismatchs (x-y, x-z, y-x, etc)

 

[DrDu]

Well, to actually find the stress tensor experimentally you have to consider the forces on 3 area elements or faces (in principle infinitely small ones if the forces are inhomogeneous as Pythagorean pointed out already), appropriately perpendicular to the x, y and z axes, respectively. The area elements are then specified by the vector normal on them with its length being the area. I will call these vectors \mathbf{n}_i. The i enumerates the three faces.

Then you find the forces acting on each of these faces (one vector for each face) \mathbf{F}_i
The stress tensor \epsilon (3x3 matrix) is then obtained by solving the equations
\epsilon \mathbf{n}_i=\mathbf{F}_i
where i ranges from 1 to 3.
When the 3 faces are perpendicular to the coordinate axes and have unit surface, the solution is especially easy as then e.g. \mathbf{n}_1=(1,0,0)^T so that the tree vectors n_i form a unit matrix.
Then \epsilon=(\mathbf{F}_1,\mathbf{F}_2,\mathbf{F}_3).

 

[Studiot]

Well, to actually find the stress tensor experimentally you have to consider the forces on 3 area elements or faces

You need to specify the point at which this stress system is supposed to act.

The centre of the cube?
A corner of the cube?
The centre of one face?

 

[DrDu]

Studiot, I got the impression from the discussion that you do know much more about stresses than I do. So maybe you can go on with the explanation.
I also wanted to ask you if you know an example where the deviator is used in praxis.

 

[DrDu]

You need to specify the point at which this stress system is supposed to act.

The centre of the cube?
A corner of the cube?
The centre of one face?

Now that I am thinking about it. Does this make any difference in the case of an infinitesimally small cube?

 

[Ahsan Khan]

You need to specify the point at which this stress system is supposed to act.

The centre of the cube?
A corner of the cube?
The centre of one face?

Studiot if say i need the stress tensor matrix for any point you like on the cube due to the force.and now its upto you, to chose any point I just want to see a fully solved procedure for making a tensor matrix, i want to see how do you use surfce and force to form the elements of matrix.

 

[Ahsan Khan]

i do net get how each element of the matrix are formed.Do we multiply the the x compnent of the force F1 acting on surface1 on the cube with the x component area to get and similarly multiply the y component of F1 with y component of area.is it like this?tell me how each component of the matrix is obtained.thanks a ton

 

[Ahsan Khan]

pythagorean has explained somewhat the way i want but i want to know it in more clear way.

 

[dulrich]

I think DrDu answered this in post #78 (for an infinitesimally small cube). The diagonal elements are the components of the force normal to the surface in question (i.e., F_x = \epsilon_{xx} n_x), while the off-diagonal elements are the components perpendicular to the surface (i.e., F_y = \epsilon_{xy} n_y), .

 

[Ahsan Khan]

now let be make things transparent.We know that the stress distribution is a system due to applied force is not uniform.Means when a force is applied then the stress at diffrent points of the system is not the same both in magnitude and direction.so we must talk of stress at a particul point on or within the system.now the chosen point is considered as an infinitesimal cube.Now what i know is that there is a force F Newton acting on the system (which definitely has some direction)

 

[Ahsan Khan]

and within the system i chose a point P where i have to find the stress tensor.I have made an infintsimal small cube having its sides in three diamentions.I thus have three surfces.and each of these surfaces have three directions one normal to it and two parllel to it.Now there will be some force on all the three faces.and each of these force further has three components along the three direction.

 

[Ahsan Khan]

Now one thing I want two know is that since we have imposed on the surface of the system only one force(in only one direction)F, now as we consider a single point P anywhere within the system and the point is considerd a small cube, so what about the forces on the three sides of the cube.Will the force(with some direction they make with the respective surface) have same magnitude on all the three sides?

 

[Ahsan Khan]

Whatever your answer may be for my previous post.Let me move further by just focusing for one of the face of the cube(though we need to work for all the three faces to obtain stress tensor matrix).Assume i first chose a surface along x axis.also suppose the force acting at side in three directions be fi, fj fk respectively in Newton .and I know the surface area vector of this small cube as ai+bj+ck.now how to find ther normal and shear stresses on this surface.

 

[Ahsan Khan]

obviosly ther must be three stresses one normal and two shear and thus on all three surfaces there should be a total of 9, now please clarify me two things.1.Will there be only one force to be considered on the whole cube and we have to resolve this force in three directions and thus specify one force on each cube normal to it or will there be three forces on each side of the cube and that in each direction the force it self has tree directions?