# Pressure and stress

1. Feb 9, 2010

### Rasalhague

If pressure at some point on a surface is defined, as here ( http://en.wikipedia.org/wiki/Pressure ), as

$$\frac{\mathrm{d}||F_n||}{\mathrm{d}A}$$

(apparently meaning the rate of change wrt area of the magnitude of force normal to the specified surface), why is pressure at a point not, in general, 0? How does the force at a point in space depend on the size of an area, given that no area appears in the definition of force

$$\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t}.$$

Likewise, a stress vector (traction vector) for a given surface is defined here ( http://en.wikipedia.org/wiki/Stress_(physics) ) as

$$\frac{\mathrm{d} \textbf{F}}{\mathrm{d}A},$$

but, given the force acting at a point, how is the force at that point changed at all by changing the value of $A$ in the equation?

Alternatively, if the magnitude of the force perpendicular to the an area is denoted by $\Delta F$ (Fishbane et al.: Physics, 2nd ed., p. 436), and pressure p defined thus:

$$p = \lim_{\Delta A \rightarrow 0} \frac{\Delta F}{\Delta A} = \frac{\mathrm{d} F}{\mathrm{d} A},$$

why isn't pressure always infinite at every point, given that for any choice of point, as one choses a smaller area to divide it by, that choice should have no affect on the magnitude of a vector belonging to a force field, nor on its normal component, since what is changing is only the area we're choosing to divide the magnitude of the force by, and not the orientation of the surface.

2. Feb 9, 2010

### sophiecentaur

Where did "F = dp/dt" come from? The units seem all wrong! (Have you defined what the symbols stand for?)

The Units for stress are, as you say, the same as the units of pressure: force/area

As with all differential calculus, dy/dx is the limit of the ratio as dx goes to zero.
So the force per unit area is still finite because that is the limit when an ever decreasing force divided by an ever decreasing area. The force at a point is infinitessimal.

3. Feb 9, 2010

### Rasalhague

Sorry, I should have defined my symbols. By p, I meant momentum, which unfortunately is traditionally given the same letter as pressure. I'm using bold letters to represent vectors though.

$$\textbf{F} = \frac{\mathrm{d} \textbf{p}}{\mathrm{d} t} = \frac{\mathrm{d}}{\mathrm{d} t} \left ( m\textbf{v} \right ) = m \frac{\mathrm{d} \textbf{v}}{\mathrm{d} t} = m\textbf{a}$$

where m is mass, t is time, v velocity, and a acceleration (supposing that mass is constant with time). The units on either side are newtons = kg m s-2.

Last edited: Feb 9, 2010
4. Feb 9, 2010

### Rasalhague

Why does the force decrease as the area does?

In the definition of the derivative dy/dx, if y isn't actually a function of x, it doesn't change as x changes, so

$$\lim_{\Delta x \rightarrow 0} \frac{y \left ( x + \Delta x \right )-y\left ( x \right )}{\Delta x} = \lim_{\Delta x \rightarrow 0} \frac{0}{\Delta x} = 0.$$

Rightly or wrongly, I'm thinking of force as a vector field, with a vector defined at each point in space. If we take the magnitude of the component of each of the force vectors which is normal to a specified surface, we get a scalar field.

How then do we get the pressure at one point (if pressure is even defined at a single point)? If we take the value at some point of the scalar field of normal force, as described in the previous paragraph, then divide that by an area, then take the limit as we divide that same unchanging finite number by ever smaller areas, won't the limit be $+\infty$? Or if we use the definition

$$\frac{\mathrm{d}||F_n||}{\mathrm{d}A}$$

and the conventional definition of the derivative, it just seems that the only result can be 0.

5. Feb 9, 2010

### sophiecentaur

"Why does the force decrease as the area does?"
How can it not? If all the individual forces on each small area add up to the whole force.
Why should dF/dA go to zero? By that argument, any derivative would be zero and that is certainly not the case.

The slope of a uniform hill is the same at one point as it is overall - in just the same way.

"Pressure at a single point" will be the same as pressure over a small area around that point and can be the same as the pressure over the whole of the (level) bottom of a swimming pool.
I think that your F = dP/dt is not relevant to this argument even though the definition is correct. What does it do for the force and pressure question?

When you say that you look at Force as a vector field, I think that is where you are going astray. Pressure is the vector field - force is the integral over a surface of that field. That "zero" in your basic definition of the derivative is not correct so what follows isn't right either.

6. Feb 9, 2010

### Rasalhague

This seems like a different kind of quantity to what I'm used to seeing called force. I'm used to seeing force defined as a vector, namely the time derivative of a momentum vector. Adding forces (to give a net force or total force), in that sense, I think means adding vectors at one point, rather than adding the values of a scalar field at many points. The quantity you call whole force sounds more like a surface integral. Is this the same thing as momentum flux, and would it be written something like this?

$$\int_S \textbf{F} \cdot \hat{\textbf{n}} \; \mathrm{d}A.$$

In the standard limit definition,

$$\lim_{\Delta x \rightarrow 0} \frac{y \left ( x + \Delta x \right )-y\left ( x \right )}{\Delta x}$$

if y varies as a function of x, then in general $y \left ( x + \Delta x \right ) \neq y\left ( x \right )$, so I don't think what I wrote would imply that any derivative must be zero, only any derivative of a constant function. Apparently the force in the pressure equation isn't constant with respect to area. But I'm curious about how this kind of force is defined?

The derivative of that constant slope is zero. (The derivative of the hill's height as a function of position isn't zero, since the height isn't constant with respect to position.) Would you agree that $$\frac{\mathrm{d} F}{\mathrm{d} A} = 0$$ unless F varies as a function of A?

Well, that's what I'm wondering ;-)

Obvious I'm getting something wrong... Is this what you mean:

$$\int_S \textbf{(pressure)} \cdot \hat{\textbf{n}} \; \mathrm{d}A = (Force)$$

(where bold indicates a vector, italics a scalar).

7. Feb 9, 2010

### Rasalhague

I must say, all the sources I've looked at seem pretty unanimous that pressure is a scalar quantity...

8. Feb 9, 2010

### sophiecentaur

But I do get your drift. Perhaps I mean that the effect of the pressure is in the direction of the normal n. That must be where my field idea comes in. I don't see that Firce is a field, in any case. An appropriate field can produce a force on a charge / mass / current which is not the same as saying that force is a field. In each of theose cases, the force is proportional to another quantity.

"The derivative of that constant slope is zero. (The derivative of the hill's height as a function of position isn't zero, since the height isn't constant with respect to position.) Would you agree that LaTeX Code: \\frac{\\mathrm{d} F}{\\mathrm{d} A} = 0 unless F varies as a function of A?"

Of course I couldn't agree with that because the force increases as you increase the area. How else would hydraulic brakes work? In fact, isn't that a clincher argument?

Edit - your maths didn't copy properly but you see what I mean.

9. Feb 9, 2010

### sophiecentaur

Isn't this all to do with the difference between intensive and extensive quantities?

10. Feb 9, 2010

### sophiecentaur

Looking at all this again. Are we both heading in different directions?
Pressure under a solid increases as you decrease the area (force constant).
Hydrostatic force (fluids) decreases as you reduce the area ( pressure constant).
Could that be part of our problem?

11. Feb 9, 2010

### lightarrow

You hang an iron cube at the roof through a spring. When the system is stationary, the spring makes a force on the cube, exactly equal to its weight. What is, and which is the value of momentum? If you define force in that way, I see it difficult for a civil engineer to solve static problems to design a building

Hello sophiecentaur.

12. Feb 9, 2010

### sophiecentaur

Well hi there lightarrow.
Long time no argue!

I'm not sure I got the relevance of the momentum aspect either.

13. Feb 9, 2010

### Mapes

There are some terminology differences between fields. In solid mechanics, pressure is a scalar and is specifically defined as -1/3 multiplied by the trace of the stress tensor (i.e., $p=-\frac{1}{3}(\sigma_{11}+\sigma_{22}+\sigma_{33})$). What other people (including Wikipedia) call pressure (and declare to be a tensor), solid mechanics folks would describe as a surface stress: an infinitesimal traction vector acting on a infinitesimal surface area with its own unit normal vector.

14. Feb 9, 2010

### Rasalhague

In that case, what I need to work out is (1) how many concepts are involved (how many concepts go by the names of pressure, stress and force in physics); (2) what unique and distinct names shall I call them while learning the difference between them (just to keep track of them); (3) what (various) names are each of them called by in the contexts I'm likely to meet them; (4) what do each of the concepts mean.

In the Wikipedia article Stress, at least in this section ( http://en.wikipedia.org/wiki/Stress_(mechanics)#Stress_deviator_tensor ), the trace of the stress tensor divided by three is called "mean stress", and denoted $p$ and $\frac{1}{3}I_1$.

To begin with, I'm particularly interested to understanding the meaning of the kind(s) of "pressure" introduced in elementary physics texts and defined

$$\frac{\mathrm{d} F}{\mathrm{d} A} \text{ or } \frac{F}{A},$$

and the kind of pressure which the diagonal entries of the stress tensor are said to represent. I'd assumed that the latter kind of pressure was the same thing as the former kind(s), but now I've no idea!

Last edited: Feb 9, 2010
15. Feb 9, 2010

### Mapes

In elementary physics texts, it's likely that $\bold{F}$ and $\bold{A}$ are collinear, $\bold{F}$ is distributed, and $\bold{A}$ is flat, so the surface stress $\lim_{A\rightarrow 0} (\mathbf{F}/\mathbf{A})=F/A$ and is a scalar that is equivalent to pressure. More generally, $\bold{F}$ and $\bold{A}$ will point in different directions and the surface stress $\bold{\sigma}$ will be a second-rank tensor.

16. Feb 9, 2010

### Rasalhague

Your reason for disagreeing seems to be of the form:

(1) P (force is a function of area).
(2) Therefore not: "if not P then Q" (if force isn't a function of area, its derivative wrt area will be 0).

But there's no inconsistency between "if not P then Q" and "P", so I don't think (2) is a valid conclusion. (How's that for pernickity!)

Yeah, obviously I've misunderstood something fundamental. My money's on F, and maybe the nature of force generally.

To get the maths symbols to display, if they're on a seperate line, just type "tex" in square brackets [...], then after the code put "/tex" in another pair of square brackets. If the maths is in a paragraph, on the same line as normal text, it works better if you use "itex" and "/itex" instead. In the code itelf, just use a single \ before commands like "frac" (it looks like it inserted double slashes \\ when you tried to copy it).

$$\frac{\mathrm{d} F}{\mathrm{d} A}$$

17. Feb 9, 2010

### Rasalhague

And, if I've got this right, the diagonal elements of a matrix representing that tensor are the components of stress normal to each of the coordinate surfaces, i.e. the pressure (in the elementary physics textbook sense) on each coordinate surface.

I take it $\bold{F}$ is the component of force normal to the surface, and $\bold{A}$ is a vector normal to the surface and proportional to the area. What I don't understand is how, formally, a vector can be "distributed"; does this mean taking a surface integral of a vector field? But how do we do that if force isn't a vector field? Perhaps force isn't a vector in the same sense as acceleration or velocity or electric field strength, given that the velocity of each particle that makes up a car doesn't "distribute" over the car, giving a big car an inherently bigger velocity, whatever that would mean... I guess it's related to the fact that force depends on mass, but that suggests force would get smaller as volume got smaller, rather than area. Anyway, suffice to say, I'm a bit confused...

Last edited: Feb 9, 2010
18. Feb 9, 2010

### Rasalhague

Yes, I think it must be.

Vector fields have a value at each point in space, and since force is said to be a vector and is often derived from vector fields such as the gravitational field and the electric field, I assumed that force was meant to be a vector field too. What do you make of this?

http://en.wikipedia.org/wiki/Force_field_(physics)

The discussion page suggests that it might not be standard terminology.

19. Feb 9, 2010

### Rasalhague

That's the second time this week I've seen what looks like a vector in the denominator of a fraction ( https://www.physicsforums.com/showthread.php?t=376048 (#10, #17)). All my books on vectors warned me somewhere in the first few pages: "It is meaningless to divide by a vector!" Is the meaning of this notation analogous to the meaning of diazona's notation, explained in #13 of that thread (where its a scalar field that's being differentiated, rather than a vector)? Could you spell it out for me? E.g.

$$\frac{\mathrm{d} \textbf{F}}{\mathrm{d} \textbf{A}} \equiv \lim_{\left \| \textbf{A} \right \|\rightarrow 0} \frac{\textbf{F}}{\left \| \textbf{A} \right \|}\cdot \frac{\textbf{A}}{\left \| \textbf{A} \right \|}?$$

Last edited: Feb 9, 2010
20. Feb 10, 2010

### Rasalhague

Is this right?

Taking Blandford and Thorne's version of the stress tensor matrix (in Applications of Classical Physics) which is the transpose of the version in Wikipedia's Stress article, if we right-multiply it by a column vector representing a unit vector normal to one of the coordinate surfaces, we get the stress vector (traction vector) embodying the stress on that coordinate surface at that point. And in a Cartesian frame, a unit vector normal to a surface of constant $x^{(k)}$ will be the kth basis vector, so the components of this stress vector will then correspond to the jth column of the stress tensor matrix $S^{ij}$, for example,

$$\begin{bmatrix} S^{11} & S^{12} & S^{13} \\ S^{21} & S^{22} & S^{23} \\ S^{31} & S^{32} & S^{33} \end{bmatrix} \begin{bmatrix}1\\0\\0\end{bmatrix} = \begin{bmatrix}S^{11} \\ S^{21} \\ S^{23} \end{bmatrix}$$

or in slot-naming index notation (abstract index notation), and using the Einstein summation convention (summing over identical indices on different levels),

$$S^{ij} \; e_{(k)j} = \frac{\mathrm{d}F^i}{\mathrm{d}A},$$

where an $e_{(k)j}$ is the jth component of the kth basis vector. Or will the matrix calculation be the same in a general frame? I'll have to think about that... And if we take the dot product of this vector with a unit vector normal to the coordinate surface, we get

$$e_{(k)i} \; S^{ij} \; e_{(k)j} = e_{(k)i} \; \frac{\mathrm{d}F^i}{\mathrm{d}A} = S^{(ii)} = \frac{\mathrm{d}F}{\mathrm{d}A} = p,$$

the normal stress (that is, pressure) on the specified coordinate surface at that point. The bracketed indexes in $S^{(ii)}$ indicating one particular value, a single number. And for an arbitrary surface with unit normal vector $n^j$, the pressure (normal stress) is

$$n_i \; S^{ij} \; n_j = n_i \; \frac{\mathrm{d}F^i}{\mathrm{d}A} = \frac{\mathrm{d}F}{\mathrm{d}A}.$$

Last edited: Feb 10, 2010
21. Feb 10, 2010

### Rasalhague

PF's own page about pressure has this definition of force:

$$\bold{F}\,=\,\int_SP\,\hat{\bold{n}}\,dA\ \ \ \ \ \ (F = PA\ \ \text{for constant pressure on a flat surface}).$$

Which gels with what sophiecentaur's being saying (if I've understood) that force (unlike stress tensors, stress vectors and pressure) isn't a local property, since at a point, there is no area, and so no force.

But is this equation only valid where there's no shear stress? Or, to put it another way, I'm (tentatively) thinking this $\bold{F}$ must be a vector field $\bold{F}\left( P, \hat{\bold{n}},A, x,y,z \right)$ defined over the specified surface, everywhere normal to the surface, and each force vector having the identical magnitude

$$AP=\int_SP \, \mathrm{d}A.$$

Perhaps we could write:

$$\text{pressure} = \frac{\partial}{\partial A} \left ( \textbf{F}\left ( p, \hat{\textbf{n}},A,x,y,z \right ) \right ).$$

Last edited: Feb 10, 2010
22. Feb 10, 2010

### Mapes

All real forces are distributed; point forces are only an idealization. Does this clarify things?

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 $\bold{F}\bold{A}^{-1}=\bold{F}/\bold{A}=\bold{\sigma}\bold{A}\bold{A}^{-1}=\bold{\sigma}$.

23. Feb 10, 2010

### Rasalhague

Yeah, thanks. I hope so! Can we say that force is only defined with respect to a specified surface of some orientation and non-zero area, subject to a certain (vectorial) stress; and force is then defined at each point on the specified surface, having a constant magnitude, this being the area times the pressure? Or does this definition only work in the absence of sheer stress? Is that only a particular kind of force: normal force, wrt the surface?

The warning was given in the context of the dot product

$$\textbf{a}\cdot\textbf{b} = \left \| \textbf{a} \right \| \, \left \| \textbf{b} \right \| \, \cos{\theta_{\textbf{a},\textbf{b}} } = x$$

where for a given vector $\textbf{a}$, and a given x, there is not generally a unique vector $\textbf{b}$. Also:

http://mathworld.wolfram.com/VectorDivision.html

Last edited: Feb 10, 2010
24. Feb 10, 2010

### Mapes

Not quite; (1) stress is generally a tensor. (2) You can define a force at a point, but this is going to muddy the relationship with pressure, since all real forces are distributed. (3) There's no problem with incorporating shear stress, which exists if the force and the surface are not perpendicular.

25. Feb 10, 2010

### Rasalhague

Okay, so how's this? We let the stress tensor, $S^{ij}$, act on the unit vector normal to an arbitrary surface, $n_j$, resulting in a traction vector (also called a stress vector in the Wikipedia article Stress). Again using the Einstein summation convention:

$$S^{ij} \, n_j = \frac{\mathrm{d} F^i}{\mathrm{d} A}$$

This vector tells us all about the stress on a surface element at the point where it's defined, including pressure and sheer stress. Compared to the stress tensor, the traction vector has "forgotten" about the stress on all the other surface elements we might consider at that point (except that we can still deduce from it the equal and opposite stress on a surface element with the opposite orientation). The dot product of the traction vector with the unit normal vector gives us the pressure on this surface element:

$$n_i \, \frac{\mathrm{d} F^i}{\mathrm{d} A} = p\left ( \hat{\textbf{n}} \right ).$$

All of this can be done for each point where we define our surface.

The normal force is

$$\textbf{F}_\perp \left(p,\hat{\textbf{n}},A,x,y,z\right) = \int_S \; n_i \; \frac{\mathrm{d} F^i}{\mathrm{d} A} \; \hat{\textbf{n}} \; \mathrm{d}A = \int_S \; p \; \hat{\textbf{n}} \; \mathrm{d}A = A \; p\left(x,y,z\right) \; \hat{\textbf{n}}$$

$$\equiv \left(F_{\perp}\right)^i = \int_S \; n_j \; \frac{\mathrm{d} F^j}{\mathrm{d} A} \; n^i \; \mathrm{d}A = \int_S \; p \; n^i \; \mathrm{d}A = A \; p\left(x,y,z\right) \; n^i,$$

so I guess the whole force must be

$$\textbf{F} \left(p,\hat{\textbf{n}},A,x,y,z\right) = F^i = \int_S \; \frac{\mathrm{d} F^i}{\mathrm{d} A} \; \mathrm{d}A = \int_S \; S^{ij} \; n_j \; \mathrm{d}A = A \; S^{ij} \; n_j.$$

And both are vector fields defined on the specified arbitrary surface. Even though they have a value at each point on the surface, both are "distributed" entities in the sense that their magnitude at any one point depends partly on the area of the whole surface we've chosen, and the value of either vanishes if we reduce the surface area to nothing. At first, I thought the magnitude of the normal force would be constant over the surface, but I was forgetting that the pressure might vary over the surface as a function of position, so I suppose the magnitude of both vector fields (both normal and whole force) can vary over the surface, although they will everywhere by proportional to the unvarying area.

Last edited: Feb 10, 2010