Now I'm trying to derive the hooke's law in general relativity,but I have many difficulties with it.

First,I try to search google/google scholar.But I found nothing about the hooke's law in general relativity.The only thing I got was this article:"Covariant formulation of Hooke's law".It gives the hooke's law in special relativity(flat spacetime): $$ F^a=kL^a $$ where $ F^a $ is the four-vector force,and $ L^a $ is the four-vector strain defined by $ L^\mu=(0,l_x,l_y,l_z) $ in the rest inertial coordinate of the spring(then we can transform it into any coordinate).

However,I don't know how to derive the hooke's law in curved spacetime.Should I just do the same work in the local rest inertial coordinate of the string?And I would like to make the definition of the four-vector strain not depend on any coordinate,such as the definition of the stress-energy tensor of the perfect fluid $ T^{ab} = (\rho+\frac{p}{c^2})U^aU^b+pg^{ab} $.

Thanks a lot for your help.


1 Answer 1


I have also thought about this question and have not seen an answer in the literature, but I will give you some thoughts. Unfortunately, the link you give is behind a pay wall.

First, strains do not transform as a 4-vector under rotations, boosts, or strains. Instead, they transform as part of a 2nd rank tensor $\Theta^{ab}$. The strains are group parameters like rotation angles and are measured in radians. We do a strain to an object. If we do a strain transformation to a vector (x,y,z), then the components get fractionally squashed/expanded or parallelopipeded as we have learned in a materials engineering class (where strains belonging to a group is never mentioned). In this answer to another question, I showed how rotations and strains are all the transformations of GL(3,R) and how to do them. Now, let's identify all the invertible 4x4 matrix transformations of a 4-vector (x,y,z,t), which is the group GL(4,R).

We can describe what these transformations do by just talking about the matrices $M$ that are very close to the identity matrix, where all elements in the matrix $\Theta$ are <<1. All these elements are in radians. $$ M=I+\Theta $$ $$ \Theta = \begin{bmatrix} 0 & \theta^{12} &-\theta^{13} & \theta^{14} \\ -\theta^{12} & 0 & \theta^{23} & \theta^{24} \\ \theta^{13} &-\theta^{23} & 0 & \theta^{34} \\ -\theta^{14} &-\theta^{24} &-\theta^{34} & 0 \\ \end{bmatrix}_{Antisymmetric} \ + \begin{bmatrix} \epsilon^{11} & \epsilon^{12} & \epsilon^{13} & \lambda^1 \\ \epsilon^{12} & \epsilon^{22} & \epsilon^{23} & \lambda^2 \\ \epsilon^{13} & \epsilon^{23} & \epsilon^{33} & \lambda^3 \\ \lambda^1 & \lambda^2 & \lambda^3 & \epsilon^{44} \\ \end{bmatrix}_{Symmetric} $$ By taking products of these matrices we build all the matrices $M$ in the group for any size elements in $\Theta$. $$ M=e^{\Theta}=I+\Theta+\dfrac{\Theta^2}{2!}+ \dfrac{\Theta^3}{3!}+ \dfrac{\Theta^4}{4!} +… $$

The $\theta$ are antisymmetric, make M orthogonal ($M^T =M^{-1}$), and leave the length ${x_1}^2+{x_2}^2+{x_3}^2+{x_4}^2$ invariant. Because lengths are invariant, the transformations are called rotations. The $\epsilon$ do not leave lengths invariant and are called strains. The $\lambda^1,\lambda^2,\lambda^3$ are the standard Lorentz Boost parameters which are space-time strains, though usually this is not pointed out.

Now the old (eg: for a,b=1,2,3 only) Hooke's law is that stress is proportional to strain for small strains $$ T_{Symmetric}^{ab}=C^{ab}_{cd}\Theta_{Symmetric}^{cd} $$ Presumably (guessing) this generalizes to the GL(4,R) covariant relation for a,b=1,2,3,4. $$ T^{ab}=C^{ab}_{cd}\Theta^{cd} $$ which may be the answer to you question. Notice that the stress energy tensor no longer has just symmetric components. This is because once we include strains, we can look at $T_{Symmetric}$ from a strained frame and we will find it has picked up antisymmetric components! This is because the commutator of two strains is a rotation (which is the reason a free falling cat can rotate itself by a sequence of strains to its body). $T_{Antisymmetric}$ can be identified as torque densities.

Masses in GR do strains. For example, the Schwarzschild metric is simultaneous stretch of t and squash of x (assuming x points toward the mass) by equal amounts ($\epsilon^{11}=-\epsilon^{44}=\frac{GM}{r}$). A gravity wave going in the z-direction does the squash/stretch strain ($\epsilon^{11}=-\epsilon^{22}=h_+$) or the parralelopiped strain which is called the other polarization ($\epsilon^{12}=\epsilon^{21}=h_X$).

These thoughts end here, uncompleted. Many things remain to be explained. For example, how do you do the space-time rotations ($\theta^{14},\theta^{24},\theta^{34}$)? Physically, what are Hooke's constants $C^{ab}_{cd}$ for mixed symmetries or when any of the indices are =4 ? Physically, what are the antisymmetric components of T ?

  • $\begingroup$ Thanks!You are right,I found this.By the way,do you know how to derive the spring's linear damping force in the curved spactime? It is really different from the hooke's law although they look similar to each other. $\endgroup$
    – lrh2000
    Commented Mar 5, 2016 at 10:33
  • $\begingroup$ @goodqt Thank you for the reference. The authors got the same 4-dim stress-strain relation as above (for symmetric $\Theta$ and $T$) and tried to identify what the constants C mean using the tools of differential geometry. I came at it from the idea of the linear group transformations and this caused the $\Theta$ and $T$ to pick up asym components. I will try and pursue what the C physically are in this point of view. Thank you again. Your paper reference will be useful. Also, sorry, I don't have any ideas about linear damping in curved space-time. $\endgroup$ Commented Mar 6, 2016 at 9:06

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.