Is Pressure a Scalar Quantity or a Vector?

[Ahsan Khan]

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.

 

[Ahsan Khan]

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.

 

[Ahsan Khan]

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.

 

[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?

 

[Ahsan Khan]

second thing i want to know after i answered for first is that, once i know the force in Newton in each side then how we calculate different stresses on one of.the surface.of definite direction of cube.thanks

 

[Studiot]

In order to pursue this form of analysis you need to realize that cubes have six faces not three.

You need to consider the forces on all six faces, plus anybody forces acting and any accelerations imposed on the source object of the cube.

This is the most general situation and leads to the equations of fluid and solid continuum mechanics.

Several simplifying assumptions can often be made, but we haven't defined whether our cube is part of the water in a stagnant pond or an element of the skin of a high velocity spacecraft .

 

[Ahsan Khan]

In order to pursue this form of analysis you need to realize that cubes have six faces not three.

You need to consider the forces on all six faces, plus anybody forces acting and any accelerations imposed on the source object of the cube.

I too already realized the six faces of the cube.Why i repeatedly talks three faces, since the opposite faces need not mentioning.whenever we apply any force on any plane surface we just see the force and and one of the area.

 

[Ahsan Khan]

further whatever be the sides there must be some way to find stresses on the surfaces of the cube so as to obtain stress tensor. i am looking for the meyhod to do so.and i am still waiting for my two queires of earlier posts.I like to mention to studiot that my system is in equilibrium under no acceleration.

 

[Studiot]

In order to answer either of your questions about stress states in this cube you need to consider a 'free body'

For the cube itself this means including the opposite faces.

For some selected point within the cube this means considering a plane cutting through the cube so that it includes the selected point. Three of the cube faces plus the cutting plane then form the free body.

Do you really need to do this in three dimensions to gain an understanding?

 

[Ahsan Khan]

Studiot in your last post i don't think i get something new,what i studied and understand about stresses is that they are different at different points in a system.to find a stress tensor at a point you need to imagine an small cube at the very point now to find tensor at the point of interest we need to find stresses(three on each surfaces) on the three surfaces.rest i mentioned in earlier posts.

 

[Ahsan Khan]

so i will like to say chose any point you like and with respect to the cube so formed answer my two questions.they will answer quench all my delima with stress tensor and thus to understand pressure.Regards

 

[mechprog]

If you don't mind intrusions at such a later stage...
Pressure has a great role to play not only in mechanics but also in thermodynamics (and probably greater) and thermodynamicists do not have much concern for vectors and tensors- you know. Take pressure as scalar or not mechanics has a better substitute- stress, but that will be inefficient in thermodynamics.
If I were to define pressure I would do so by
\bold{\sigma_n} = p \bold{\hat{n}}
just like we do so
\vec{v}=v\hat{\epsilon_s}
in case of speed and velocity.
And you know stresses are better dealt with tensors (usually cartesian), so if need insight into that there are many books that deal with them together (like Borg's Matrix tensor methods in continuum mechanics or freely available http://www.math.odu.edu/~jhh/counter2.html" )

 

[Studiot]

imagine an small cube at the very point now

How can you have a cube at a point?
A cube is a three dimensional object.

You can have your point of interest in the centre of the cube, at one of the vertices (this is conventional) or somewhere else.

I already asked you to consider this but you did not answer.

I really don't know where you are coming from on this, since your original question arose from a discussion with 12th graders, but you are studying (what?) tensors at university.

I know of four approaches to derive the formula you apparently seek.

The simplest is known as the Engineers' method and involves direct calculation with forces and moments and some geometry/trigonometry (direction cosines). It uses significant simplifications. The free body diagram is used (cube and cutting plane as previously described)

The next method is the simple continuum mechanics method. This involves simple manipulation of partial differentials and Cauchy's method, but does not need tensors. It can be extended to allow for body forces. Again this uses the cube.

The full continuum mechanics method does not work on the cube, but works directly with points and displacements. It allows body forces and accelerations to be included, the simpler formula appear as special cases by setting these equal to zero. Volume and Surface integrals are used, along with Greens theorem.

Finally there are energy methods involving Gauss' theorem and more intricate partial differential manipulation.

 

[Dale]

The full continuum mechanics method does not work on the cube, but works directly with points and displacements. It allows body forces and accelerations to be included, the simpler formula appear as special cases by setting these equal to zero. Volume and Surface integrals are used, along with Greens theorem.

I know the engineer's method, but would be very interested in this one.