View Full Version : tensor rank of pressure
Hello everybody, yesterday I stand to teach vectors and scalars to 12th standard students in a coaching.While giving examples of scalars I named mass , work , pressure etc.Then a student argued me that pressure should be a vector quantity since when you apply a push on wall that is force then the pressure would also acts in the direction where you are applying the force.So according to him pressure is a vector quantity.While in books I always read pressure a scalar.Actually I my self used to wonder how is pressure a scalar quantity while it seems associated with direction just as a force do, there I answer that since we do talk of pressure(while on solid surfaces) as pressure on a surface, which always acts normal to the surface no matter be the surface plane or curved thus it is immaterial to say the direction of the pressure since it is always calculated on the surface that is perpendicular to it or in opposite direction of area vector( just as we take are in guss law, which however has nothing to do with pressure but is helpful here to explain about area vector).Though I explain this, neither me nor he was fully satisfied, since still it has a direction though fixed.I search the net and get many useful points but still it is not getting clear as to how pressure a scalar quantity.Then I come across with tensors, I want to know-1: what is the tensor rank of pressure? 2:If its rank is zero (that is if is scalar) then please explain what physical quantities can be taken with non-zero rank. Thanks in advance
I am also not a specialist on this, but, at least in a solid, pressure is the trace of the stress tensor (or minus one third of it, to be precise) or, the isotropic part of it. Hence it transforms as a scalar. Liquids are special in so far as all other components of strain are zero.
If I may restate the question
Force is a vector. Force over an area is a pressure. How can pressure be a scalar?
I might add that area is properly expressed as a 2-form rather than a tensor of two upper indices such that the area is oriented; there is a unique choice for which direction the surface normal points.
Well, so to speak, pressure is the scalar product of normal surface vector (of unit area) times force averaged over all orientations of the surface.
thanks for your replies DrDu, I will say I am not clear when you say pressure is the trace of the stress tensor or the isotropic part of it. And how a part of something(stress tensor) non -scalar could become scalar. Thankx again
Well, the trace of a tensor T_ij is the sum over all T_ii, which is clearly a scalar.
E.g., I can define the tensor of the tensor product of the vectors r with itself T_{ij}=r_i r_j . It is important e.g. in the definition of the moment of inertia or of the electric multipole moments. Now the trace of this tensor tr(T)=\sum_i r_i r_i is simply the squared length of the vector r.
What does that mean in the case of pressure:
Consider a cube made out of an elastic material in a bench vice. The force on one of the two forces in the vice is F and the area of the surface is A. The force acting on the other surfaces is 0. Then the pressure is 1/3 (F/A+0+0).
ok the trace of tensor T_ij is the sum of all t_ii, can you please state the physical relationship between T_ij and T_ii using words other than trace?
T_ii are the diagonal elements of the tensor T_ij, i.e., the elements T_ij for which i=j.
Hence, if you think of the tensor as a 3x3 matrix, the trace (or spur) is the sum over all diagonal elements of the matrix.
ah i got it u explained very well.Now I will like to know what is that tensor whose trace you defined as pressure, i know it is stress tensor as you said i want to know how i visulaize the stress tensor?where can i feel its effect?I am familiar with pressure but stress tensor is new to me?
Q_Goest
Jul13-10, 09:42 AM
When pressure is referred to as a scalar quantity, what is meant is that pressure at any point within a fluid has no intrinsic direction to it. Consider the pressure of a fluid NOT acting against a surface. For example, consider what pressure water has at a depth of 10 feet in the middle of the ocean. If the water isn't acting against anything, there is no directional force, so we can't call it a vector. But there is certainly pressure at that depth in the ocean, just as there is pressure in a fluid at all locations within the fluid. So the pressure in a fluid is a scaler quantity. We call pressure a scalar quantity because without considering what surface the pressure is acting against, there is no force and no vector. (Note that "fluid" here means any liquid or gas.)
In contrast, pressure acting on a surface becomes a vector quantity because the interaction of the fluid against the solid surface creates a normal force. That normal force is just the pressure times the area (integrated of course) so that FORCE is a vector quantity.
did you mean the concept of pressure applies only to fluid?and the thing which acts on solid surfaces is pressure force(a vector) not pressure.Is that you mean?
Q_Goest
Jul13-10, 11:18 AM
When refering to pressure as a scalar quantity, the underlying assumption is that we're talking about a fluid. For a fluid acting on a solid, there is a force produced which is a vector quantity, but the concept of pressure in a fluid shouldn't be confused with the concept of that pressure acting at a surface.
so pressure in fluid is scalar and pressure acting on solid surface has direction normal to it?and in the bulk fluids have pressure in scalar form?
dulrich
Jul13-10, 11:48 AM
In contrast, pressure acting on a surface becomes a vector quantity because the interaction of the fluid against the solid surface creates a normal force. That normal force is just the pressure times the area (integrated of course) so that FORCE is a vector quantity.
Here is a quote from the wikipedia (http://en.wikipedia.org/wiki/Pressure#Definition) article: Pressure ... relates the vector surface element (a vector normal to the surface) with the normal force acting on it. The pressure is the scalar proportionality constant that relates the two normal vectors
This way of saying it works for both fluids and solids. The difference is that in a fluid the pressure is the same in all directions.
the statement that pressure is the scalar proportionality constant that relates two normal vectors.please tell me what are these two vectors and how is pressure(as a constant) relates these two vectors. thanks a ton
Studiot
Jul13-10, 12:14 PM
Interesting discussion, but won't all of this pass over the head of your 12th graders?
If you want a description to convince them, forget tensors and try this.
How do you combine (add) vectors?
How do you combine (add) scalars?
How do you combine (add) pressures?
Think of a sealed vessel half full of water. At any point in the water there is a pressure.
Now pump up the air pressure in the other half.
How much does the pressure increase at any point in the water?
This is scalar addition.
Another thing to bear in mind.
Only scalars can affect every point in this way. Vectors can only affect things in their line of action.
dulrich
Jul13-10, 12:19 PM
the statement that pressure is the scalar proportionality constant that relates two normal vectors.please tell me what are these two vectors and how is pressure(as a constant) relates these two vectors. thanks a ton
(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.
Studiot
Jul13-10, 12:21 PM
I have to observe that pressure can never be a vector, simply because it does not combine according to the laws of vector addition.
dear studiot your explanation works well when dealing with fluid in containers but what should I call the quantity F/A.Suppose i apply a force of 10 N normal to a plate .1 m2 in area.then we say we are applying a pressure of 100 N/m2 on the wall.Now if we say our pressure has nothing to do with direction, dosn't it seem wrong?
(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.
The vector normal to the surface-it that you write A and the force vector F?If this the case then P=F/A should be meaningless, since vectors do not follow division.we can give a meaning to it by your second relation F=PA.
Studiot
Jul13-10, 12:43 PM
No, the pressure does not have a direction.
The wall has a direction.
Take the wall away, what is then the direction of the pressure? Does it remain the same?
so if i imagine that just mass relates velocity and momentum as mv=p.where mass is thought as a scalar constant relating two vector quantities-velocity and momentum,scalar pressure relates area and force as two vectors as PA=F.Is this how one think?
Studiot
Jul13-10, 12:52 PM
Being in the UK I am not sure what standard 12 grade is but I guess it is some sort of middle school? So I don't know if your students have done any calculus.
You need to be very careful with pressure = force/area.
When you shrink your control area to zero and take the limit, what do you mean by area?
You need also to be careful with force = pressure x area
What force? Individual forces posess a line of action, they are not distributed over an area or in space.
oh 12th stanard here is equavalent to UK's High school and yes they have studied calculas but not advanced advanced calculus.and how can anyone talk of pressure without considering surface so my friend if i say my students to think of pressure(on a wall)without considering the direction of wall they will suspect me.and i think i am getting understand what pressure actually mean.i believe it must be scalar as you said.do u agree with my above posts?regards
dulrich
Jul13-10, 01:12 PM
so if i imagine that just mass relates velocity and momentum as mv=p.where mass is thought as a scalar constant relating two vector quantities-velocity and momentum,scalar pressure relates area and force as two vectors as PA=F.Is this how one think?This is what I was getting at. Using some better symbols: \vec{F} = P \vec{A}
Studiot
Jul13-10, 01:43 PM
Introducing momentum is not a good idea.
You need to be careful not to say or introduce anything which would actually be incorrect at higher level.
Unfortunately the curriculum in the UK no longer deals with basic concepts such as centre of pressure, and also what we can and cannot apply pressure to etc.
Remember that when you discuss 'force' in you mean the normal force. Pressure is not the tangential force/area for instance.
You also need to consider how the pressure is to be applied. In my experiment what would be the difference if the bottom half of the vessel was filled with ice not water?
DaleSpam
Jul13-10, 02:47 PM
Now I will like to know what is that tensor whose trace you defined as pressure, i know it is stress tensor as you said i want to know how i visulaize the stress tensor?where can i feel its effect?I am familiar with pressure but stress tensor is new to me?The Wikipedia page seems reasonably comprehensive, but I did not go into detail:
http://en.wikipedia.org/wiki/Stress_%28mechanics%29
If you have a material in static equilibrium then you can imagine cutting a small cube out of it. We label the faces of the cube x,y,z where the x face is normal to the x axis and so forth. On each face there is some force which we can further break up into x,y,z components. So the x,x component of the stress tensor would be the component of the force on the x face of the cube in the x direction, and the x,y component of the stress tensor would be the component of the force on the x face of the cube in the y direction. So the diagonal elements (x,x y,y z,z) are pressure and the off-diagonal elements (x,y x,z y,z) are shear stresses. Now, a fluid cannot support shear stresses without deforming, so in a static fluid all of the off-diagonal terms are 0 and all of the diagonal terms are equal. So in a static fluid the stress tensor can be reduced to a single number. That number is the pressure.
Nearly right, only that the diagonal elements are not called "pressures" but Nnormal stresses".
Only their mean is called pressure.
Studiot
Jul13-10, 03:51 PM
component of the force on the x face of the cube in the x direction
component of force
or
component of stress ?
Ovais,
I think you have the basis of a good classroom discussion or three
What is the difference between Force and Pressure?
What is the difference between Pressure and Stress?
Can you always add two vectors?
Can you always add two scalars?
What Force is represented by the expression Force = Pressure x Area?
DaleSpam
Jul13-10, 04:10 PM
Sure, it is normal stress and it is not the force on a given face in a given direction but the differential force divided by the differential area of the face. I was trying to use terminology that ovasis would recognize. For an accurate treatment he should get a good book on statics.
Nearly right, only that the diagonal elements are not called "pressures" but Nnormal stresses".
Only their mean is called pressure.
Dale spam has given a very good explanation.And he talks of(xx yy zz)as pressure.But as you extend that this diagonal elements are not pressures but infact normal stresses.now please let me know how we find the pressure using these normal stresses.also tell if the pressure found will be uniform on every side of the cube?
ok x,x (the normal stress) is the component of the stresses on face xx, and similarly y,y &z,z are component of stresses on yy and zz faces of the cube,.now how to find pressure on the cube?
I already explained this in post #6 and gave an example.
As x,x y,y z,z are normal stresses along the three faces. So you mean we have sum up them?I will be thankfull to know if these three NORMAL STRESSES scalar or vector?Sorry if I seem too argumentive.Thanks!
You have to sum them and divide by -3. So pressure is minus the arithmetical average of the three normal stresses.
The normal stresses are neither a vector nor a scalar. They are simply three components of a second rank tensor.
Btw., it might be interesting for you to look up the meanings of "contraction" of a tensor and "irreducible tensors".
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
Jul14-10, 05:59 AM
Furthermore if they are not equal it implies motion.
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?
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.
please clarify me about pressure which is found by dividind tensors is a tensor or not?
Studiot
Jul14-10, 06:26 AM
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?
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.
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
Jul14-10, 06:46 AM
So do you understand the difference between pressure and stress, since both have the same units?
Studiot
Jul14-10, 07:02 AM
You might like to look at post#5 of this thread. It describes what Dr Du is talking about.
http://www.physicsforums.com/showthread.php?t=371071&highlight=stress+tensor
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?
DaleSpam
Jul14-10, 07:18 AM
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.
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?
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?
DaleSpam
Jul14-10, 07:42 AM
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.
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 catagory i mean we should call pressure to be neither scalar nor vector.
Studiot
Jul14-10, 08:10 AM
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.
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 dont get this.please help.
Studiot
Jul14-10, 09:27 AM
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 any one 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.
DaleSpam
Jul14-10, 09:44 AM
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.
DaleSpam
Jul14-10, 09:58 AM
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 dont 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.
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?
DaleSpam
Jul14-10, 10:08 AM
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
Jul14-10, 10:21 AM
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... :smile:
EDIT: Sorry for the delayed post -- i didn't notice the conversation had shifted back to the OP.
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 any one 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
Jul14-10, 12:15 PM
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 \sigmam 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
Jul14-10, 02:22 PM
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
Jul14-10, 04:28 PM
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. :confused: 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_%28mechanics%29#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
Jul14-10, 06:38 PM
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
Jul14-10, 06:54 PM
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 :smile:), but can you expand on this statement or point me to somewhere I can learn more about it? Thanks.
Studiot
Jul14-10, 07:24 PM
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
DaleSpam
Jul14-10, 08:03 PM
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
Jul14-10, 11:06 PM
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
Jul14-10, 11:21 PM
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.
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.
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?
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
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.
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
Jul15-10, 01:10 AM
P=f/a is false in general. The actual relation involves calculus, which simplifies to p=f/a under appropriate assumptions.
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
Jul15-10, 01:53 AM
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)
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
Jul15-10, 03:07 AM
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?
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.
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?
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.
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
pythagorean has explained somewhat the way i want but i want to know it in more clear way.
dulrich
Jul15-10, 10:15 AM
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), .
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 definately has some direction)
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.
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?
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.
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?
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
Jul15-10, 12:02 PM
In order to pursue this form of analysis you need to realise that cubes have six faces not three.
You need to consider the forces on all six faces, plus any body 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.
[QUOTE=Studiot;2801366]In order to pursue this form of analysis you need to realise that cubes have six faces not three.
You need to consider the forces on all six faces, plus any body forces acting and any accelerations imposed on the source object of the cube.
I too already realised 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.
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
Jul15-10, 12:34 PM
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?
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 intrest we need to find stresses(three on each surfaces) on the three surfaces.rest i mentioned in earlier posts.
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
Jul15-10, 01:09 PM
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 Heinbockel's (http://www.math.odu.edu/~jhh/counter2.html))
Studiot
Jul15-10, 03:23 PM
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 apparantly 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.
DaleSpam
Jul15-10, 05:34 PM
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.
Ok now i consider the point at the centre of the cube.And I will prefer engineer's method.And I hope thise methods are somewhat lengthy but will give me what i need and I think the tradational engineer's method can be found somewhere in the net.Thank you all guys.I hope i can understand how to make stress tensor matrix after studying engineers method and also hope you people will continue your support in case i ask for further help!
Studiot
Jul16-10, 02:21 PM
It will take quite a bit of writing out. I will try to post something over the weekend.
DaleSpam
Jul16-10, 02:29 PM
No worries if you cannot.
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