Pressure & Stress: Definition and Relationship

[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, [itex]\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.