Pressure & Stress: Definition and Relationship

[Rasalhague]

If force on a surface in general can be defined as a vector field having, at each point on the surface, the value A\,S^{ij}\,n_j, where S^{ij} is the stress tensor, A the area of the surface, and n_j a unit vector normal to the surface (this being a generalisation of the equation for normal force, given at the PF info-page "pressure"), what restrictions are there on the kind of surface we can have? Must it be connected (in one piece)? I'm guessing maybe not. For a given scalar field of pressure, would the force on one side of the surface of a set of disconnected puddles be the same as the force on one side of one connected puddle with the same surface area? I suppose if we can apply the equation F = PA to a surface like the lower one in the diagram on the right here ( http://en.wikipedia.org/wiki/Barometer ) of a barometer, it must be permissible for there to be at least one hole in the surface.

This diagram ( http://en.wikipedia.org/wiki/File:Freebodydiagram3_pn.svg ) shows the forces acting on a block. Take the first picture, the one of the block resting on a horizontal surface. I think the tacit assumption is that the area over which these forces are defined is the area of the block resting on the surface. I keep thinking, well if we just take a much bigger area, can't we arbitrarily define any force as arbitrarily bigger, rendering the concept meaningless? But the elementary examples I'm finding tend to begin with a known force, derived by some means other than F=PA. So on the one hand, we have force defined in terms of pressure, as if a less fundamental property than pressure, since pressure is defined locally by the stress tensor, whereas force depends on an arbitrary choice of surface and therefore isn't entirely a local phenomenon (in this point of view), on the other hand, we have force defined as the time-derivative of momentum, without reference to pressure, and practical problems in elementary texts seem to treat force as more fundamental. (Not knowing anything practical about quantum mechanics, I'm not sure if the idea of "fundamental forces", EM, weak, strong, has any bearing on these definitions in elementary Newtonian mechanics...) Or maybe it's just arbitrary perspective which definition depends on the other, of they're all interrelated. Any comment on that welcome!

In the diagram cited above, if we take as our surface the area covered by the block, oriented upwards, a force of -mg + mg/2 is exerted by the matter on the negative side of the surface (since half of the normal force of electrostatic repulsion is due to the matter in the block, and half to the matter the block is resting on). So the mean pressure will be mg(1/2 - 1)/A = -mg/2A. Is that right? Since the block is in equilibrium, the pressure on the same surface with the opposite orientation should be mg/2A. And sure enough, by Newton 3, we have a gravitational force of mg, and subtract mg/2 by the same reasoning as above, and divide by area: mg/2A. Is that the idea?

 

[Matterwave]

I only read some of this thread...so if the points I'm going to bring up, has been brought up, then sorry.

First, \bold{F}=\frac{d\bold{p}}{dt}, Newton's second law, is confusing if you use it to define a force since you can no longer define forces for static situations. The F in that law is actually the NET force (summing up all the different forces on an object). If the NET force on an object were 0, it doesn't mean that there are no forces acting on the object.

Second, force fields (which sound pretty cool btw) are used, e.g. in gravity or E&M to denote the forces that would be on an ideal test particle of infinitesimal size, which is why you get one vector for each point in space. Alternatively, you can imagine that this is just the center of mass application of the force (for spherical test particles). If I push you, then the force of my hand is definitely NOT concentrated on one point, but spread out over my palm.

So, say I push you with my palm. My palm has an area of A and applies force F. The average pressure is then just F/A. If you look at only a small part of my palm, it is only providing a small fraction of the total force, so as A goes to 0, F goes to 0. This works well as long as you remain in the macroscopic range wherein my palm, for all practical purposes, is a continuum.

Microscopically, my palm is made up of atoms, and their electrons repel the other electrons via the Coulomb force. There's no reason to deal with the microscopic aspects, if you're only interested in macroscopic variables.

 

[Rasalhague]

force fields (which sound pretty cool btw) are used, e.g. in gravity or E&M to denote the forces that would be on an ideal test particle of infinitesimal size, which is why you get one vector for each point in space.

Are these fields only defined with respect to some specified surface? (Supposing this is "force" in the sense "traction vector (stress vector) times area".) Does the force on an ideal test particle of infinite size depend on an arbitrary choice of surface (area and orientation) with respect to which the force is defined? If it depends only on the orientation of the surface and not on area (as your mention of infinitesimal size suggests), this quantity seems more like S^{ij} \, n_j, a traction vector (stress vector). (Where S^{ij} is the stress tensor, and n_j a unit vector indicating an orientation.) But if these fields don't depend on a specified orientation either, how do they relate to force as defined in terms of stress and pressure?

 

[Matterwave]

Test particles aren't real...don't think too much into them.

If I have the force field:
\frac{\bold{F_g}}{m}=-\frac{GM}{r^2}\hat{r}

It just tells me the forces a small test particle of mass m would feel at those distances from the source.

I mean, there's no point to thinking about pressure or stress on that infinitesimal test particle.

Don't over think this...

 

[Rasalhague]

Oh, right ho. Come to think of it, the force given by the action of the stress tensor on a surface vector is only part of the force on a particle, since it only tells you the force due to entities (for want of a better word) on one side of the surface. So the equation in #25 can't be a general definition of force, so we don't necessarily need to specify a surface to define a force.

 

[Rasalhague]

If I have the force field:
\frac{\bold{F_g}}{m}=-\frac{GM}{r^2}\hat{r}

It just tells me the forces a small test particle of mass m would feel at those distances from the source.

Are the units of a gravitational force field the same as acceleration, then, while the units of other force fields are units of force?

 

[Matterwave]

Strictly speaking, what I have there isn't a "force" field, it's the gravitational field.

This is perfectly analogous to the Electric field E=F/q

It's easier to work with these since you don't have to know the mass (or charge) of the test particle...uhm...I suppose if you just want to move the m to the right side, that's fine also.

 

[Rasalhague]

First, \bold{F}=\frac{d\bold{p}}{dt}, Newton's second law, is confusing if you use it to define a force since you can no longer define forces for static situations. The F in that law is actually the NET force (summing up all the different forces on an object). If the NET force on an object were 0, it doesn't mean that there are no forces acting on the object.

The kind of "force" which the stress tensor gives, when supplied with a surface, seems to be the net force minus any component of it exerted any anything on the positive side of the surface, so with that understanding, one could still talk about force in terms of how the momentum of particles on the negative side would change if there was nothing exerting a counter force from the positive side. At least that's what I think the idea is; I could be mistaken.

 

[Rasalhague]

This seems crazy, or perhaps is a caution found in texts whose scope doesn't include tensors. Vector division happens all the time. We divide 4\bold{i} by the unit vector \bold{\hat{i}} to get its magnitude, 4. From the equation \bold{F}=\bold{\sigma}\bold{A}, where \bold{F} and \bold{A} are vectors and \bold{\sigma} a second-rank tensor, we have \bold{F}\bold{A}^{-1}=\bold{F}/\bold{A}=\bold{\sigma}\bold{A}\bold{A}^{-1}=\bold{\sigma}.

From diazona's definition in #13 here ( https://www.physicsforums.com/showthread.php?t=376048 ), it might be concluded that

\frac{\textbf{a}}{\textbf{a}}=\frac{A_1}{A_1}\textbf{i}+\frac{A_2}{A_2}\textbf{j}+\frac{A_3}{A_3}\textbf{k}=\textbf{i}+\textbf{j}+\textbf{k},

so that, taking 0/0 to be 0 in this context (if that's what's done),

\frac{4\textbf{i}}{\textbf{i}}=\frac{4}{1}\textbf{i}=4\textbf{i}.

Perhaps these are two completely different conventions for defining "division by a vector". Could you spell out the rules or point me in the direction of a source that does?

 

[Mapes]

From diazona's definition in #13 here ( https://www.physicsforums.com/showthread.php?t=376048 ), it might be concluded that

\frac{\textbf{a}}{\textbf{a}}=\frac{A_1}{A_1}\textbf{i}+\frac{A_2}{A_2}\textbf{j}+\frac{A_3}{A_3}\textbf{k}=\textbf{i}+\textbf{j}+\textbf{k},

I have no idea what you're doing here, or how this connects with diazona's post, where \Phi was a scalar.

 

[Rasalhague]

I have no idea what you're doing here, or how this connects with diazona's post, where \Phi was a scalar.

Good point, you were dividing a vector by a vector, whereas diazona had a scalar in the numerator and a vector only in the denominator. My question about general rules and formal definitions remains.

 

[Rasalhague]

For example,

\forall\textbf{a} \, \exists\textbf{a}^{-1}:\textbf{a}\textbf{a}^{-1}=1?

\forall\textbf{a} \, \exists\textbf{a}^{-1}:\textbf{a}^{-1}\textbf{a}=1?

\textbf{a}\textbf{b}^{-1}=\textbf{b}^{-1}\textbf{a}?

\textbf{a}\textbf{c}^{-1}=\textbf{b}\textbf{c}^{-1}\Rightarrow \; \textbf{a}=\textbf{b}?

My books, and that Wolfram Mathworld article, warned about the nonuniqueness of multiplicative inverses, so presumably the last equation doesn't hold, but without knowing the axioms, I can but guess. In #22, \textbf{i}\textbf{i}^{-1}=1[/tex], a scalar, but \textbf{F}\textbf{A}^{-1}=\sigma, a type-(2,0) tensor. What is the general rule? How would you define this operation formally?

 

[Mapes]

I'd test these by multiplying the vector/tensor on the same side:

\textbf{a}=\textbf{a}\textbf{a}^{-1}\textbf{a}=1\textbf{a}=\textbf{a}

\textbf{a}\textbf{b}^{-1}\textbf{b}=\textbf{a}\neq\textbf{b}^{-1}\textbf{a}\textbf{b}\mathrm{~(Not~necessarily,~that~is.)}

And so on. Try it with the last two equations to see why you end up with a scalar with one and a tensor with the other.

 

[Rasalhague]

I'd test these by multiplying the vector/tensor on the same side:

\textbf{a}=\textbf{a}\textbf{a}^{-1}\textbf{a}=1\textbf{a}=\textbf{a}

\textbf{a}\textbf{b}^{-1}\textbf{b}=\textbf{a}\neq\textbf{b}^{-1}\textbf{a}\textbf{b}\mathrm{~(Not~necessarily,~that~is.)}

And so on. Try it with the last two equations to see why you end up with a scalar with one and a tensor with the other.

The first case follows from the definition by substituting \mathbf{i}. The second is a 2nd order tensor because \mathbf{FA}^{-1} is a vector-valued function of vectors? I suppose this would also be a vector: \mathbf{FA}^{-1}\mathbf{B}, where \mathbf{B} is a vector other than \mathbf{A}.

 

[Rasalhague]

Aha, if \mathbf{A} \in V, where V is the underlying set of a vector space, then \mathbf{A}^{-1} \in V^*, its dual space, because \mathbf{A}^{-1} is a scalar-valued function of vectors... well, so long as it's linear.

 

[Rasalhague]

aa^{-1}(a+b)=a+aa^{-1}b=a(1+a^{-1}b)

so

a^{-1}a(1+a^{-1}b)=1+a^{-1}b.

And

aa^{-1}(a+b)=a^{-1}a(a+b)=a^{-1}(aa+ab)

so

a^{-1}(aa+ab)a^{-1}=(1+ba^{-1})

But what does it mean to add a scalar to a a type-(1,1) 2nd order tensor, if that is the nature of a^{-1}b and ba^{-1}? Another warning I've read is that the number of indices in a vector equation has to match on each term. Perhaps this rule is violated in a system that allows this operation of "vector division", given that my books don't recognise such a thing. But if the matching-indices rule is kept, perhaps this means that "vector division" doesn't distribute over vector addition.

Is see that HallsofIvy writes in this thread ( https://www.physicsforums.com/showthread.php?t=99883 ) that "division of vectors is not normally defined", which tallies with my experience, and what the internet at large seems to be saying. So if you do have a formal definition, could you please spell it out in full, with whatever axioms you use, and whatever modifications you make (if any) to the normal rules of vector and tensor algebra?

"1.5.3. 'Division' of vectors. The solution of equations usually leads to the operation of division, an operation which in the case of vectors is not unique. This difficulty appears even in the case of the scalar product [...] In fact, thinking of division as the inverse of multiplication, let \mathbf{a}\cdot \mathbf{x}=m \enspace (\mathbf{a}\neq \mathbf{0}), where x is an unknoqn vector. This operation has infinitely many solutions, since it merely determines the projection of x onto the direction of the given vector a. Hence the operation of division is best avoided altogether in vector algebra" (Boriskeno & Taparov: Vector and Tensor Analysis with Applications).

"Dividing by a vector is nonsense" (Griffel: Linear Algebra and its Applications, Vol 1, p.21).

 

[Mapes]

So if you do have a formal definition, could you please spell it out in full, with whatever axioms you use, and whatever modifications you make (if any) to the normal rules of vector and tensor algebra?

As an engineer, not a mathematician, I'll use any math that works. What is (perhaps cavalierly, to mathematicians) written \bold{F}/\bold{A}=\bold{F}\bold{A}^{-1}=\bold{\sigma} is simply meant to reflect the fact that \bold{F}=\bold{\sigma}\bold{A}. It is assumed from the context that \bold{F} and \bold{A} are vectors and \bold{\sigma} is a matrix.

Without these clues of the context of linear elasticity, we have to be cautious that the operations make sense. In your example of (1+ba^{-1}), I'd have to conclude that b and a are collinear vectors and that you're taking the dot product between them. Otherwise, I don't think the expression makes sense. In other words, we can't be overly general; we need to restrict the properties of our variables, which happens automatically in science and engineering.

So I agree with HallsofIvy in that divisions of vectors are not well defined in general. But in carefully defined applications, the process works fine. I think Griffel is being too cautious in describing it as "nonsense."

 

[Rasalhague]

Okay, thanks for your help.