Question about the nonlinear Hooke's law

[patric44]

Homework Statement

why the non linear hoock's law is given by : F_spring = ku + u^3 in this question ? i guess a u^2 is missing ?!

Relevant Equations

F = -kx-eu^2+...

hi guys
i saw this problem online about using the MATLAB ode45 to solve the nonlinear Hoock's law and its specifically stated that
the nonlinear hoock's law is given by
$$F = k\;u + \epsilon\;u^{3}$$ , but when expanding the potential function in a Taylor series where you obtain the force from it
it seems that the force is given by :
$$F = -\frac{dV}{dx}_{x_0} *(x)-\frac{1}{2} \frac{d^{2}V}{dx^{2}}_{x_0}*(x)^2+...$$
$$F = -k\;x-\epsilon\;x^{2}$$
the problem :

my question is where is the x^2 term ?? why did he jump directly to the x^3 term ?

 

[etotheipi]

If the potential energy function is$$V(x) = V(0) + x \frac{dV}{dx} (0) + \frac{1}{2!} x^2 \frac{d^2 V}{dx^2}(0) + \frac{1}{3!} x^3 \frac{d^3 V}{dx^3}(0) + \frac{1}{4!} x^4 \frac{d^4 V}{dx^4}(0) + \dots $$the force is$$F_x = -\frac{dV(x)}{dx} = -\frac{dV}{dx}(0) - x\frac{d^2 V}{dx^2}(0) - \frac{1}{2} x^2 \frac{d^3 V}{dx^3}(0) - \frac{1}{6}x^3 \frac{d^4 V}{dx^4}(0) +\dots$$then why is it unreasonable that the coefficient of $x^2$ in $F_x$ is zero? All you need is that $\frac{d^3 V}{dx^3}(0)$ is zero. That doesn't mean that $\frac{d^2 V}{dx^2}(0)$ and $\frac{d^4 V}{dx^4}(0)$ can't be non-zero. And in fact, in the problem you attached, you're given that $\frac{d^2 V}{dx^2}(0) = k$ and $\frac{1}{6}\frac{d^4 V}{dx^4}(0) = \varepsilon$.

 

[patric44]

If the potential energy function is$$V(x) = V(0) + x \frac{dV}{dx} (0) + \frac{1}{2!} x^2 \frac{d^2 V}{dx^2}(0) + \frac{1}{3!} x^3 \frac{d^3 V}{dx^3}(0) + \dots $$the force is$$F_x = -\frac{dV(x)}{dx} = -\frac{dV}{dx}(0) - x\frac{d^2 V}{dx^2}(0) - \frac{1}{2} x^2 \frac{d^3 V}{dx^3}(0) + \dots$$then why is it unreasonable that the coefficient of $x^2$ in $F_x$ is zero? All you need is that $\frac{d^3 V}{dx^3}(0)$ is zero. That doesn't mean that $\frac{d^2 V}{dx^2}(0)$ and $\frac{d^4 V}{dx^4}(0)$ can't be non-zero.

but using the same argument i could say that hoock's law is given by :
$$F = -k\;x-\eta\;x^{7}$$
assuming that the other coefficient is zero ?
why did he assumed that the coefficent of x^2 equal zero , the problem specifically states that the hoock's law is generaly given by
$$F = ku+\epsilon\;u^3$$
they didnt include an x^2 term then sat its coefficent equal zero ?
is there some thing special about $\frac{d^{3}V}{dx^{3}}$ that make it always zero ?

 

[haruspex]

why did he assumed that the coefficent of x^2 equal zero

I suggest it is an observation about how materials actually behave, not an assumption.

 

[etotheipi]

Different springs might obey different potential functions. The more accurate the model, the more higher order terms will contribute to Hooke's law. For the purposes of this problem, the author probably thought that the behaviour of his spring was most simply modeled by a potential function for which the derived force has terms in $x$ and $x^3$ (the $x^3$ term being a correction term).

 

[patric44]

ok its clear for me now thanks a lot , but i have a another question , is there is a way to obtain the potential
curve of a specific spring ?
another question : does these higher order derivatives has some physical or even a mathematical meaning , i know that the first derivative specify if the point is maxima or minima and so on , does $\frac{d^{3}V}{dx^{3}}$ has any meaning what so ever related to the property of the spring ?

 

[etotheipi]

another question : does these higher order derivatives has some physical or even a mathematical meaning , i know that the first derivative specify if the point is maxima or minima and so on , does $\frac{d^{3}V}{dx^{3}}$ has any meaning what so ever related to the property of the spring ?

In all cases, it's a mathematical model, which hopefully fits the properties of the real spring close enough for purpose. To determine better and better models, you'll need to look at some research surrounding strain energy functions and I suspect you'll need to start looking quite closely at what happens on the molecular level. However, most probably, these models are experimentally determined.

I don't know enough about elasticity theory to know how the higher order terms vary between different materials and types of springs, but perhaps someone else will.

 

[patric44]

thank you so much for helping

 

[kuruman]

I think the motivation for omitting the $x^2$ term in the force is that it will mean a $x^3$ in the potential. It is reasonable to assume that an elastic potential is symmetric about $x=0$, i.e. $U(x)=U(-x)$. If one adds higher order terms to the potential to make it more realistic, these terms must be even powers of $x$. That's because the restoring elastic force, $F=-\frac{\partial U}{\partial x}$ will have only odd powers in $x$ and its magnitude will be symmetric about $x=0$. Of course if the magnitude of the force is not the same when the spring is compressed as opposed to stretched, then odd terms in the potential will be required. Normally, one would collect data and fit the force function to a polynomial.

My own view is that this problem is not about elasticity theory but about using ode45 and plot (whatever they are) to get some results that you can then interpret in terms of a concrete example.

 

[patric44]

In all cases, it's a mathematical model, which hopefully fits the properties of the real spring close enough for purpose. To determine better and better models, you'll need to look at some research surrounding strain energy functions and I suspect you'll need to start looking quite closely at what happens on the molecular level. However, most probably, these models are experimentally determined.

I don't know enough about elasticity theory to know how the higher order terms vary between different materials and types of springs, but perhaps someone else will.

I think the motivation for omitting the $x^2$ term in the force is that it will mean a $x^3$ in the potential. It is reasonable to assume that an elastic potential is symmetric about $x=0$, i.e. $U(x)=U(-x)$. If one adds higher order terms to the potential to make it more realistic, these terms must be even powers of $x$. That's because the restoring elastic force, $F=-\frac{\partial U}{\partial x}$ will have only odd powers in $x$ and its magnitude will be symmetric about $x=0$. Of course if the magnitude of the force is not the same when the spring is compressed as opposed to stretched, then odd terms in the potential will be required. Normally, one would collect data and fit the force function to a polynomial.

My own view is that this problem is not about elasticity theory but about using ode45 and plot (whatever they are) to get some results that you can then interpret in terms of a concrete example.

that's a really good explanation thanks a lot , i have a little bit of a stupid question when i plotted this using Matlab ode45 with an epsilon = 0.2 it seems like any ordinary oscillating system ! , shouldn't i see some non linear behavior or something ? i was expecting some alien looking graph or something due to that extra $x^{3}$ term .
my Matlab graph :

 

[kuruman]

The symmetric potential will display oscillatory motion regardless of whether it has terms higher than $x^2$ in it or not. There is nothing alien about it. The motion will not be simple harmonic, but how good is your eye distinguishing a pure sinusoid from what you plotted? That's perhaps part of the lesson to be learned here.

As another example of adding non-linear terms to linear forces consider the simple pendulum equation $$\ddot \theta+\omega^2~\sin\theta=0.$$ In the "small angle approximation" you get the pure sinusoidal solution, but what happens at angles larger than "small"? You can keep, say, the next two terms in the sine expansion and get $$\ddot \theta+\omega^2~\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}\right)=0.$$Note that here too you can use ode45 and plot to illustrate what this looks like. Also note that you should not expect it to look astonishingly different from a pure sinusoidal. Why? Because if you swing a pendulum at large amplitude, its motion looks to the eye pretty much like that of a pendulum swinging at a smaller amplitude.

 

[patric44]

The symmetric potential will display oscillatory motion regardless of whether it has terms higher than $x^2$ in it or not. There is nothing alien about it. The motion will not be simple harmonic, but how good is your eye distinguishing a pure sinusoid from what you plotted? That's perhaps part of the lesson to be learned here.

As another example of adding non-linear terms to linear forces consider the simple pendulum equation $$\ddot \theta+\omega^2~\sin\theta=0.$$ In the "small angle approximation" you get the pure sinusoidal solution, but what happens at angles larger than "small"? You can keep, say, the next two terms in the sine expansion and get $$\ddot \theta+\omega^2~\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}\right)=0.$$Note that here too you can use ode45 and plot to illustrate what this looks like. Also note that you should not expect it to look astonishingly different from a pure sinusoidal. Why? Because if you swing a pendulum at large amplitude, its motion looks to the eye pretty much like that of a pendulum swinging at a smaller amplitude.

your answers and comments are astonishingly clear and very informative thank you so much

 

[TeethWhitener]

Homework Statement:: why the non linear hoock's law is given by : F_spring = ku + u^3 in this question ? i guess a u^2 is missing ?!
Relevant Equations:: F = -kx-eu^2+...

my question is where is the x^2 term ?? why did he jump directly to the x^3 term ?

If the force is quadratic, then where is the minimum of the potential?

 

[haruspex]

If the force is quadratic, then where is the minimum of the potential?

It's not whether the force is quadratic but whether it has a quadratic term.
I do not see that it would create a problem for the minimum of the potential. If the force is $F=ax+bx^2$ then there's a local extremum of potential at x=0. dF/dx(0)=a, so if a>0 it is a minimum.
Maybe I misunderstand your point.

 

[etotheipi]

It's not whether the force is quadratic but whether it has a quadratic term.
I do not see that it would create a problem for the minimum of the potential. If the force is $F=ax+bx^2$ then there's a local extremum of potential at x=0. dF/dx(0)=a, so if a>0 it is a minimum.
Maybe I misunderstand your point.

I think you're right, but I interpreted it to mean that it seems very unlikely that such a force would represent a spring. If the force and potential energy is $$F_x = -ax - bx^2 \implies U(x) = \frac{a}{2}x^2 + \frac{b}{3}x^3$$then if the mass overcomes the potential barrier in the $-\hat{x}$ direction, it undergoes runaway contraction with a large force in the $-\hat{x}$ direction forever... which is quite unphysical!

 

[TeethWhitener]

It's not whether the force is quadratic but whether it has a quadratic term.
I do not see that it would create a problem for the minimum of the potential. If the force is $F=ax+bx^2$ then there's a local extremum of potential at x=0. dF/dx(0)=a, so if a>0 it is a minimum.
Maybe I misunderstand your point.

If $F(x)=ax+bx^2$, what is the minimum of $V(x)$? (Hint: it’s not $x=0$)

 

[haruspex]

I think you're right, but I interpreted it to mean that it seems very unlikely that such a force would represent a spring. If the force and potential energy is $$F_x = -ax - bx^2 \implies U(x) = \frac{a}{2}x^2 + \frac{b}{3}x^3$$then if the mass overcomes the potential barrier in the $-\hat{x}$ direction, it undergoes runaway contraction with a large force in the $-\hat{x}$ direction forever... which is quite unphysical!

But the presence of a quadratic term term does not rule out a cubic one. For what value of x<0 is $-x-x^2-x^3$ negative?

 

[TeethWhitener]

But the presence of a quadratic term term does not rule out a cubic one. For what value of x<0 is $-x-x^2-x^3$ negative?

@etotheipi got the gist of what I was saying. And the OP (whom I was addressing in my initial post) asked specifically about a force where the highest power was quadratic.

 

[haruspex]

If $F(x)=ax+bx^2$, what is the minimum of $V(x)$? (Hint: it’s not $x=0$)

I think you are saying it has a second minimum, but you are assuming there is no cubic term. See post #17.

 

[etotheipi]

But the presence of a quadratic term term does not rule out a cubic one. For what value of x<0 is $-x-x^2-x^3$ negative?

Yeah, I think as a rule of thumb potentials with an even highest order term (and consequently forces with an odd highest order term) will be candidates to model an elastic force. In the context of the OP, it would be unphysical to have a highest order term in the force of $x^2$, however there is nothing specifically wrong with having non-zero terms of both $x^2$ and $x^3$ in the force expression. So I assume the author just omitted an $x^2$ term for reasons of simplicity.

In short, which lower order terms you choose to keep in your model will come down to the specific physical properties of the spring under consideration, and how many terms you include in the model reflects how accurate you want the model to be.

 

[haruspex]

the OP (whom I was addressing in my initial post) asked specifically about a force where the highest power was quadratic.

Did he? Post #1:

my question is where is the x^2 term ?? why did he jump directly to the x^3 term ?

Similarly in other posts.

 

[haruspex]

Yeah, I think as a rule of thumb potentials with an even highest order term (and consequently forces with an odd highest order term) will be candidates to model an elastic force. In the context of the OP, it would be unphysical to have a highest order term in the force of $x^2$, however there is nothing specifically wrong with having non-zero terms of both $x^2$ and $x^3$ in the force expression. So I assume the author just omitted an $x^2$ term for reasons of simplicity.

In short, which lower order terms you choose to keep in your model will come down to the specific physical properties of the spring under consideration, and how many terms you include in the model reflects how accurate you want the model to be.

As long as there are no nontrivial solutions of F=0, and dF/dx has the right sign at x=0, there is no obvious reason the function cannot be valid.
But @kuruman's symmetry argument in post #9 is persuasive.

 

[TeethWhitener]

Did he?

Maybe I just misinterpreted.

I think you are saying it has a second minimum, but you are assuming there is no cubic term. See post #17.

If the highest order term in the force is quadratic, then the highest order term in the potential is cubic, and the potential therefore has no global minimum. It may have a local minimum which might be useful for small perturbations, but there will always be a range of initial values that sends the poor particle careening off into the wild blue yonder.

(It’s not much of a problem for classical mechanics, but in quantum mechanics, it turns into a very big problem.)

 

[haruspex]

If the highest order term in the force is quadratic, then the highest order term in the potential is cubic

Agreed, but nobody else is suggesting that the highest term in the force could be quadratic.

 

[patric44]

@TeethWhitener i guess you misunderstood my question
my question can be boiled down to this graph of the potential :

it was why not the left , and after all the arguments about the symmetry and some deeper physical reasons i am convinced that it should be the one on the right for this this specifically (symmetrically elastic spring) .

 

[etotheipi]

It’s not much of a problem for classical mechanics, but in quantum mechanics, it turns into a very big problem.

Out of interest, and if it's not too much of a deviation from the topic of the thread, I wondered if you could explain why potentials with no global minimum are more of a problem in QM than in CM? My very untrained intuition tells me that it can be problematic in both disciplines. Is it because in QM we're more often dealing with bound states?

 

[TeethWhitener]

Out of interest, and if it's not too much of a deviation from the topic of the thread, I wondered if you could explain why potentials with no global minimum are more of a problem in QM than in CM? My very untrained intuition tells me that it can be problematic in both disciplines. Is it because in QM we're more often dealing with bound states?

It’s probably off topic, but at a B-level, there’s no ground state. If the system does start off in a local potential well, it will have a nonzero probability of tunneling through the barrier, at which point its energy shoots off to infinity.

 

[etotheipi]

Ah, yes that makes sense, thanks! Also when you say its energy shoots off to infinity, are you talking about the kinetic energy (in that if $V \rightarrow -\infty$, then $T = E-V \rightarrow + \infty$, with $E = \text{constant}$)? In any case, the tunnelling seems like the key difference!

 

[TeethWhitener]

Ah, yes that makes sense, thanks! Also when you say its energy shoots off to infinity, are you talking about the kinetic energy (in that if $V \rightarrow -\infty$, then $T = E-V \rightarrow + \infty$, with $E = \text{constant}$)? In any case, the tunnelling seems like the key difference!

Yes sorry, I haven’t been putting enough care into my responses in this thread. The potential energy is unbounded from below and the (expectation value of the) kinetic energy is unbounded from above: you can show this using the virial theorem. But yes, as you pointed out: tunneling is what makes the quantum case different.