# Hyperelasticity and the relativistic Lagrangian of a 1-dimensional rod

1. Sep 7, 2007

### pervect

Staff Emeritus

I plan to write this in several sections, as time permits. The first first section will be a recap of what "hyperelasticity" is about, following sections are planned to develop this into a complete Lagrangian model of a one dimensional, hyperelastic, relativistic rod.

The hyperelastic model in relativity is discussed in some of the following links:

Greg Egan's webpage

Relatistic elastodynamics, Wernig Piilchler (a PHD thesis also linked on Greg Egan's webpage)

Nonrelativistic and relativistic continuum mechanics a paper by R Beig

The basic idea of hyperelasticity is this. Suppose we have an elastic rod, and we assume that the rod has the following characteristics:

1) It obeys Hooke's law, so that if the rod is stretched in length by some factor s (i.e. if the rod is stretched from one foot to two feet, the stretch factor is 2) that the force is always proportional to the stretch factor s, even for large values of stretch.

We will further assume that the rod does not change in width as it is stretched. This assumption is mainly for convenience, one could introduce Poisson's ratio if needed.

2) Energy is conserved, and evenly distributed throughout the rod. This assumption allows us to compute the energy density and stress-energy tensor of the stretched rod, quantities that will be needed later. In fact, as we will see, the key quantity needed is the proper energy density of the rod, i.e. the energy density in the rod as seen by an observer co-moving with some small section of the rod.

This calculation is carried out on Greg Egan's webpage, but a brief review might be helpful:

Consider a unit length of the rod, with the Hooke's law constant of the spring being k.

If it is stretched from unit length to some length s, the total energy in the rod goes from its initial rest energy $E_0$ to:

$$E = E_0 + \int_{S=1}^{S=s} -k \left( S - 1 \right) dS = E_0 + \frac{k}{2} \left( s - 1 \right)^2$$

because the total energy in the spring is equal to its initial energy plus the work done in stretching it.

We now wish to re-write this in terms of energy density, rather than energy. Since the original rod had a unit length, its energy density is equal to its total energy

$$\rho_0 = E_0$$

The energy density in the stretched rod is just:

$$\rho = \frac{\rho_0}{s} + \frac{k}{2} \left( s + \frac{1}{s} - 2 \right) \hspace{1.5 in} (1)$$

We divide the previous result by s, because that is the length of the stretched rod, so that the density in the stretched rod is the total energy / stretched length. We have also taken advantage of the fact that $E_0 = \rho_0$ in writing this expression.

These results follow from our assumption that the energy is evenly distributed throughout the spring.

The details of the dynamics of the actual energy transport are not modeled, it is simply assumed that the energy is always evenly distributed through the rod. It is important to restrict k to a small enough value so that the rod is not "too rigid". If the rod is too rigid, the speed of sound through the rod will become greater than 'c', i.e. there is no physical way our assumption of uniform energy density can actually be maintained. At some future point after the Lagrangian has been developed, this issue will be revisited. I'll mention the important result here, though - there is a maximum stretch factor s possible for any given spring constant k that limits the speed of energy transport to below that of light.

This concludes the introductory section, next up will be the topic of an overview of Lagrangians, Lagrangian densities, and the Lagrangian of a Newtonian elastic 1d rod.

Last edited: Sep 8, 2007
2. Sep 8, 2007

### pervect

Staff Emeritus
Part 2: The non-relativistic Lagrangian of a 1-d rod.

Before considering the relativistic case, let's consider the non-relativistic case using the Lagrangian formalism.

The Lagrangian of a system can in most cases (including the case of interest here) be expressed as L = T - V, where T is the kinetic energy and V is the potential energy.

Suppose we have a rod of uniform density $\rho_0$. We can describe the state of the rod by a function that maps the original position of a point on the rod at t=0 to its future position at some time t. That is, we want to find some function X(t,x0) that gives the position (x coordinate of the rod) as a function of two variables. The first variable, t is time, and the second variable, $x_0$ is the initial position of some point on the rod at t=0. Because this is a 1-d problem, we only need to specify one such function function.

x0 will be called a body coordinate, and the initial configuration of the rod at t=0 s called the body manifold (one also sees the term "material manifold" in the literature).

Given this function (or map) X(t,x0), we can compute the kinetic energy T as follows:

$$T = \frac{1}{2} \rho_0 \left( \frac{\partial X}{\partial t} \right)^2 d x_0$$

here $\rho_0$ is the initial density of the rod at t=0 (i.e. in the body manifold), assumed for simplicity to be a constant.

The variable of integration, x0, is a body coordinate. $\rho_0 dx_0$ is just the mass of a small section of a rod, and (1/2) v^2 dm is the kinetic energy of this small piece of the rod. Integration over x0 gives us the total energy of the rod at time t.

Similarly, we can write the total potential energy stored in the rod, using the results earlier derived in (1), as

$$V = \int \frac{k}{2} \left(s + \frac{1}{s} - 2 \right)\, s \, dx_0$$

Our initial volume element, dx0, is stretched to a volume s*dx0, so we multiply the energy density from (1) by the stretched volume element to compute the total stored energy.

We need, however, to compute the stretch factor, s, from X(t,x0). This is easily done - the stretch factor is just

$$s = \frac{\partial X}{\partial x_0}$$

i.e. if $X = 2 \, x_0$, s=2

Putting these results together, we can write:

$$L = \left( \frac{1}{2} \rho_0 \left( \frac{\partial X}{\partial t} \right)^2 - \frac{1}{2} k \left( \left( \frac{\partial X}{\partial x_0} \right)^2 - 2 \frac{\partial X}{\partial x_0} + 1 \right) \right) dx_0$$

We can simplify and cosmetically alter this result, however. The linear terms in $$\frac{\partial X}{\partial x_0}$$ will have no effect on the equations of motion, neither will constant or additive terms. They therefore can be eliminated.

We then wind up with

$$L = \left( \frac{1}{2} \rho_0 \left( \frac{\partial X}{\partial t} \right)^2 - \frac{1}{2} k \left( \frac{\partial X}{\partial x_0} \right)^2 \right) dx_0$$

It's further convenient to normalize this equation by defining $$v_c^2 = \frac{k}{\rho_0}$$, we can then write

$$L = \frac{\rho_0}{2} \left( \left( \frac{\partial X}{\partial t} \right)^2 - v_c^2 \left( \frac{\partial X}{\partial x_0} \right)^2 \right) dx_0$$

Now that we have the Lagrangian, how do we get the equations of motion? This is discussed in textbooks such as "Classical Mechanics" by Goldstein.

If for convenience we symbolically substitute
$$vt = \frac{\partial X}{\partial t} \hspace{.5 in} vx_0 = \frac{\partial X}{\partial x_0}$$

we can write:

$$\mathcal{L} = \frac{\rho_0}{2} \left( vt^2 - v_c^2 \: vx_0^2 \right)$$

where $\mathcal{L}$ is the Lagrangian density, i.e. we have dropped the integral sign and considered the Lagrangian of only a small piece of the rod.

Our equations of motion are then just (Goldstein, pg 549 - replace Goldstein's $\eta$ with X):

$$\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial vt} \right) + \frac{d}{d\,x_0} \left( \frac{\partial \mathcal{L}}{\partial vx_0} \right) = \frac{\partial \mathcal{L}}{\partial X}$$

where $\mathcal{L}$ is regarded as a function $\mathcal{L} (X,vt,vx_0,t,x_0)$. It is needed to know what variables $\mathcal{L}$ is a function of in order to perform the partial differentiation properly, and avoid the "second fundamental confusion" of calculus.

Carrying this step out, it is seen that we get the standard wave equation, the result we expect for a Newtonian rod:

$$\frac{\partial^2 X}{\partial t^2} - v_c^2 \: \frac{\partial^2 X}{\partial x_0^2} = 0$$

Reviewing our definitions, recall that $v_c$ was defined as $$\sqrt{k/\rho_0}$$. Further recall that k was the force of the spring per unit stretch factor: since the stretch factor is dimensionless, if we use non-geometric units, k has units of newtons. $\rho_0$ was the density of the spring per unit length, in non-geometric units it thus has units of kg/ meter. $v_c$ thus has units of velocity, and can be regarded as the longitudinal speed of sound in our bar.

Note that most of the derivation above was performed in geometric units, rather than standard units. However, the only affect on the end result of switching back to geometric units would be a trivial multiplication of the entire Lagrangian by c^2, which would make no difference to the equations of motion.

Basically the above result contains no surprises - the speed of longitudinal waves in a uniform bar with a linear force/displacement ratio is a constant, as it should be.

Last edited: Sep 9, 2007
3. Sep 8, 2007

### pervect

Staff Emeritus
Part 3: The relativistic Lagrangian

Now we get to the interesting part of the problem.

As discussed in

Nonrelativistic and relativistic continuum mechanics a paper by R Beig

the relativistic Lagrangian density is very simple - it is just $-\rho_0$, the energy density of the rod (rest energy + stored energy) in its rest frame.

Thus we replace the non-relativistic expression for the Lagrangian, (2) by the simpler

$$L = \int \rho(s) \: dV$$

where dV is the volume element in the lab frame, and where $\rho(s)$ is the total energy density (rest energy + stored energy) of the rod, which is a function of how much the rod is stretched.

$\rho(s)$ is given by equation (1) for the hyperelastic rod, and is a function of s, the relativistic "stretch factor".

The correct volume element can be derived as follows: the initial volume element in the body manifold is $dx_0$. This is then stretched by the relativistic stretch factor s to a proper volume of $s \, dx_0$. Finally, because the section of the rod is moving, it's volume in the lab frame is the proper volume multiplied by a factor of 1/gamma.

Substituting in for the volume element in the above integral, we get:

$$L = \int \rho(s) \, s \, \sqrt{1-\left( \frac{\partial X}{\partial t} \right)^2} dx_0$$

Substituting and simplifying for $\rho(s)$ using eq (1) we find:

$$L = \int \rho_0 \, \left( 1 + \frac{k}{2 \, \rho_0} \left(s^2 -2 s + 1 \right) \right) \sqrt{1-\left( \frac{\partial X}{\partial t} \right)^2} dx_0$$

which can be further re-written making use of the definition of $v_c$ as

$$L = \int \left( 1 + \frac{v_c^2}{2} \left(s^2 -2 s + 1 \right) \right) \sqrt{1-\left( \frac{\partial X}{\partial t} \right)^2} dx_0$$

Note that if we have an unstretched rigid rod, moving at some velocity $\frac{\partial X}{\partial t} = v$, the above Lagrangian density gives us the correct Lagrangian for a rigid body: $L = m \sqrt{1-v^2}$. Goldstein gives this formula for the relativistic Lagrangian of a rigid body on pg 321, for instance.

Finally, we have to discuss how we compute the relativistic stretch factor s. I may or may not expand on this point further in some future post, but I'll offer the following simple argument for the value of s. Consider, again, a relativistic rod, that initially starts out at length L, and accelerates so that it maintains a length of L as measured in the lab frame. If the rod did not stretch, its length would shrink to L / gamma, thus the rod in this example must be in tension - its stretch factor should in fact be equal to gamma. However, the non-relativistic stretch factor, ${\partial X}/{\partial x_0}$ for the rod is unity, so some correction for relativity in computing the stretch factor is necessary.

This example suggests the following for the relativistic stretch factor, which turns out to be correct:

$$s = \frac{\partial X}{\partial x_0} \frac{1}{\sqrt{1 - \left( \frac{\partial X}{\partial t} \right) ^2}}$$

Substituting and expressing the result as a density, we find that

$$\mathcal{L} = -\rho_0 \left( \left( 1 + \frac{v_c^2}{2} \right) \sqrt{1 - \left( \frac{\partial X}{\partial t} \right)^2} + v_c^2 \: \left( \frac{\partial X}{\partial x_0} \right) + \left( \frac{v_c^2}{2} \right) \left( \frac {\left( \frac{\partial X}{\partial x_0} \right)^2}{\sqrt{1 - \left( \frac{\partial X}{\partial t} \right)^2}}\right) \right)$$

We can again drop the middle term because it is linear in ${\partial X}/{\partial x_0}$. Furthermore, the Lagrangian is scale invariant - it is convenient to divide this expression by $1 + v_c^2/2$. This gives us the final result:

$$\mathcal{L} = -\rho_0 \left( \sqrt{1 - \left( \frac{\partial X}{\partial t} \right)^2} + \left( \frac{v_c^2}{2+v_c^2} \right) \left( \frac {\left( \frac{\partial X}{\partial x_0} \right)^2}{\sqrt{1 - \left( \frac{\partial X}{\partial t} \right)^2}}\right) \right)$$

Next: comparing the above with the Newtonian result, and doing useful things with it.

Last edited: Sep 9, 2007
4. Nov 1, 2011

### qnusmani

I am reading this thread with considerable interest. So far a relativistic langrangian density of a rod has been obtained. What about the equation of motion. This may find application in nuclear physics,

5. Nov 2, 2011

### pervect

Staff Emeritus
The equations of motion were presented earlier, for the Newtonian case. They're the standard Lagrange equations for a Lagrangian density.

Goldstein has a derivation, that shows how you can approximate a continuous system as a "lumped" system, then take the limit as the number of lumps approaches infinity. It turns out that you get partial differential equations in the limit. There's a reference to the appropriate section of Goldstein in post #2.

Look at part 2:

What you need to do for the relativistic case is replace the Newtonian Lagrangian density with the relativistic Lagrangian density. I was originally intending to do that, but never wound up fijishing it.

6. Nov 4, 2011

### qnusmani

I have worked out the relativistic equation of motion (wave equation). It is in the attached file. I found it difficult to type in latex - hence the doc file. Please let me know your opinion.
qnusmani

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• ###### Euler-Lagrange-Equation.doc
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7. Apr 30, 2012

### hoomanya

Hi,
I know this is an old thread but I am hoping that you will still read this.
I am write a code to model a 1D hyperelastic model. I need to come up with a test case and test it using a finite element code written in Matlab. I was wondering if you knew any test case results, and material properties. I would really appreciate it if you could let me know of any relevant papers or any ideas on what kind of a test case to use and how to check the results.
Regards,
H

8. May 1, 2012

### pervect

Staff Emeritus
9. May 16, 2012

### FieldTheorist

I have to admit, I'm a bit confused by this statement:

Why is this the Lagrangian? This should be the Hamiltonian of the system, no? The Lagrangian should be a Lorentz scalar; this quantity is by definition not a Lorentz scalar (energy changes under a Lorentz boost).

I would think that the correct Lagrangian density would be the Legendre transformation of this Hamiltonian density. Of course, I'm not sure there's a definitive guarantee that this quantity will be Lorentz invariant, either; but in either case, that is certainly not a Lagrangian that leads to Lorentz covariant equations of motion. (At least, it strongly appears not to be)