Understanding Pressure as a Scalar Quantity: Exploring Tensors and Rank

In summary, pressure is a scalar quantity in a fluid, meaning that it has no intrinsic direction at any point within the fluid. However, when it acts on a solid surface, it becomes a vector quantity due to the normal force it creates. The stress tensor, which is related to pressure, is a tensor of rank 2 and its trace represents pressure. This can be seen in the example of a cube in a bench vice, where the pressure is 1/3 times the sum of the diagonal elements of the stress tensor.
  • #71
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?
 
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  • #72
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
 
  • #73
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.
 
  • #74
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.
 
  • #75
P=f/a is false in general. The actual relation involves calculus, which simplifies to p=f/a under appropriate assumptions.
 
  • #76
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
 
  • #77
ovais said:
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)
 
  • #78
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 [tex]\mathbf{n}_i [/tex]. The i enumerates the three faces.

Then you find the forces acting on each of these faces (one vector for each face) [tex]\mathbf{F}_i[/tex]
The stress tensor [tex]\epsilon [/tex] (3x3 matrix) is then obtained by solving the equations
[tex] \epsilon \mathbf{n}_i=\mathbf{F}_i [/tex]
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. [tex] \mathbf{n}_1=(1,0,0)^T[/tex] so that the tree vectors n_i form a unit matrix.
Then [tex] \epsilon=(\mathbf{F}_1,\mathbf{F}_2,\mathbf{F}_3) [/tex].
 
  • #79
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?
 
  • #80
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.
 
  • #81
Studiot said:
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?
 
  • #82
Studiot said:
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.
 
  • #83
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
 
  • #84
pythagorean has explained somewhat the way i want but i want to know it in more clear way.
 
  • #85
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., [itex]F_x = \epsilon_{xx} n_x[/itex]), while the off-diagonal elements are the components perpendicular to the surface (i.e., [itex]F_y = \epsilon_{xy} n_y[/itex]), .
 
  • #86
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)
 
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  • #87
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.
 
  • #88
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?
 
  • #89
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.
 
  • #90
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?
 
  • #91
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
 
  • #92
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 .
 
  • #93
Studiot said:
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.
 
  • #94
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.
 
  • #95
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?
 
  • #96
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.
 
  • #97
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
 
  • #98
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
[tex]\bold{\sigma_n} = p \bold{\hat{n}}[/tex]
just like we do so
[tex]\vec{v}=v\hat{\epsilon_s}[/tex]
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" [Broken])
 
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  • #99
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.
 
  • #100
Studiot said:
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.
 
  • #101
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!
 
  • #102
It will take quite a bit of writing out. I will try to post something over the weekend.
 
  • #103
No worries if you cannot.
 
<h2>1. What is pressure as a scalar quantity?</h2><p>Pressure is a measure of the force exerted on a surface per unit area. It is a scalar quantity because it only has magnitude and no direction.</p><h2>2. How is pressure related to tensors?</h2><p>Pressure can be represented as a tensor, which is a mathematical object that describes the relationship between physical quantities. In the case of pressure, it is a rank 0 tensor, meaning it only has one component and no direction.</p><h2>3. What is the difference between a scalar and a tensor?</h2><p>A scalar is a physical quantity that only has magnitude, while a tensor is a mathematical object that describes the relationship between multiple physical quantities. Scalars have a rank of 0, while tensors have a rank of 1 or higher.</p><h2>4. How can understanding tensors help us understand pressure?</h2><p>Understanding tensors can help us understand the underlying mathematical relationships between physical quantities, including pressure. By representing pressure as a tensor, we can better understand how it is affected by different factors and how it relates to other physical quantities.</p><h2>5. What are some real-world applications of understanding pressure as a scalar quantity?</h2><p>Understanding pressure as a scalar quantity is crucial in many fields, such as engineering, physics, and meteorology. It is used to design structures that can withstand different levels of pressure, to calculate fluid flow in pipes and channels, and to predict weather patterns and atmospheric pressure changes.</p>

1. What is pressure as a scalar quantity?

Pressure is a measure of the force exerted on a surface per unit area. It is a scalar quantity because it only has magnitude and no direction.

2. How is pressure related to tensors?

Pressure can be represented as a tensor, which is a mathematical object that describes the relationship between physical quantities. In the case of pressure, it is a rank 0 tensor, meaning it only has one component and no direction.

3. What is the difference between a scalar and a tensor?

A scalar is a physical quantity that only has magnitude, while a tensor is a mathematical object that describes the relationship between multiple physical quantities. Scalars have a rank of 0, while tensors have a rank of 1 or higher.

4. How can understanding tensors help us understand pressure?

Understanding tensors can help us understand the underlying mathematical relationships between physical quantities, including pressure. By representing pressure as a tensor, we can better understand how it is affected by different factors and how it relates to other physical quantities.

5. What are some real-world applications of understanding pressure as a scalar quantity?

Understanding pressure as a scalar quantity is crucial in many fields, such as engineering, physics, and meteorology. It is used to design structures that can withstand different levels of pressure, to calculate fluid flow in pipes and channels, and to predict weather patterns and atmospheric pressure changes.

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