4
$\begingroup$

I often see people interpreting Hooke's Law $σ=Eε$ as,

"The deformation $ε$ that occurs when you subject a material to a stress $σ$."

This makes it sound like stress is an external stimulus that causes the material to deform. But from what I know, stress is an internal phenomenon, not an external one.

So technically, isn't it more correct to interpret Hooke's law as

"The stress σ that develops in a material given a deformation of $ε$"?

I would greatly appreciate it if someone could clear this up for me.

$\endgroup$

2 Answers 2

5
$\begingroup$

Unfortunately, both descriptions are misleading because both describe the relationship as a one-way cause-and-effect relationship. The correct way to describe it is

the stress is proportional to the strain

It doesn’t matter if the stress is given and the strain is obtained by Hooke’s law or if the strain is given and the stress is obtained by Hooke’s law. Either way they are proportional to each other.

Unfortunately, whether it is Newton’s laws, Ohm’s law, Hooke’s law, or Maxwell’s equations, the tendency to verbally express such equations in cause-and-effect language is fairly strong and common in many introductory physics courses. It is almost universally inappropriate. A real cause-and-effect relationship is given by an equation of the form $$f(t)=g(t-\Delta t)$$ In this equation $g$ is the cause and $f$ is the effect and $0<\Delta t$ so that the cause always precedes the effect. In an expression like Hooke’s law both stress and strain are happening at the same time and causes happen before effects. It is a simple proportionality, not a cause-and-effect relationship.

$\endgroup$
9
  • 1
    $\begingroup$ I don't think you need a time delay to express cause she effect. For example, with Newton's second law if a net force is applied to an object then it will cause an acceleration, but you don't have an acceleration causing a force. Similarly, just because one can be deduced from another does not mean cause and effect are not existent. $\endgroup$ Jan 3, 2021 at 2:56
  • 1
    $\begingroup$ @BioPhysicist that there is a time delay is usually part of the very definition of causality. Eg “In general, a process has many causes ... and all lie in its past” en.m.wikipedia.org/wiki/Causality $\endgroup$
    – Dale
    Jan 3, 2021 at 3:32
  • 1
    $\begingroup$ I agree... but at the same time I find it odd that for the scenario of me pushing a box across the floor that we can say I'm pushing on the box because it is accelerating. Yes, the application of the force and the acceleration occur simultaneously (for all intents and purposes here), but I don't think it is incorrect to say that the force causes the acceleration. Or perhaps we really do need to bring in the fact that no object is truly rigid? $\endgroup$ Jan 3, 2021 at 4:04
  • 1
    $\begingroup$ @BioPhysicist although it is commonly said and seems comfortable, strictly speaking it is wrong. If A and B always happen at the same time and place how can you decide in general which is the cause and which is the effect? $\endgroup$
    – Dale
    Jan 3, 2021 at 4:14
  • 2
    $\begingroup$ So instead of saying "gravity causes objects to fall" should we technically only be saying "objects that fall also experience gravity"? $\endgroup$ Jan 3, 2021 at 4:18
0
$\begingroup$

Hooke's Law—in the standard form as you've written it—says that the normal stress $\sigma$ and normal strain $\varepsilon$ are linearly coupled by a constant of proportionality $E$ (termed Young's modulus). This is a good approximation for a long rod of a stable, constant-temperature solid for small axial deformations over moderate time scales. (All these qualifiers are needed to eliminate the effects of stresses in other directions, temperature dependence, creep, etc.) Note that no specific cause or effect regarding the stress or strain is implied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.