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If a spring is at rest and lies along $X$ axis in a frame $O$ with a spring constant $k_{0}$ then its spring constant in a frame $O'$ which is moving with a speed $v$ at an angle $\theta$ with the $X$ axis (the motion is as observed from $O$) comes out to be $\displaystyle\frac{k_{0}\sqrt{1-v^2}}{1-v^2\cos^2\theta}$ . I have derived this formula considering some cases of a particle being held in equilibrium by the means of a spring against some particularly chosen electromagnetic forces acting on it.

Now if we consider the spring as a prismatic rod and define a quantity $N$ to be equal to $\displaystyle\frac{kL}{A}$, where $L$ is the unstressed length of the rod and $A$ is the cross-section area of the prismatic rod then it comes out that with the above-considered transformation for the spring constant and the usual transformation for area and length, the quantity $N$ remains frame invariant. But the considered ratio can be certainly interpreted as the Young's Modulus of the rod in a given frame. And as pointed out, it remains frame invariant. So Young's Modulus is a Lorentz Scalar.

I haven't previously encountered any such assertion so I am posting this question to check if I have made any mistake(s) in my procedure.

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    $\begingroup$ I would expect Young's modulus to be (one component of) a tensor, since a material with a nontrivial crystal structure may have strong planes and weak planes. Perhaps a materials person can comment. $\endgroup$
    – rob
    Commented Sep 27, 2016 at 14:51
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    $\begingroup$ @rob in continuum mechanics, for a general body, you definitely want to have a tensor description for the relation between stress and strain. $\endgroup$
    – Sanya
    Commented Oct 5, 2016 at 21:32
  • $\begingroup$ @Dvij I've found at least a few papers saying something about relativistic Hooke, see e.g. arxiv.org/pdf/1205.5299.pdf and links therein. However, I think to really talk of a Young Modulus as a Lorentz scalar, we might want to consider relativistic continuum mechanics which I've got no clue of. $\endgroup$
    – Sanya
    Commented Oct 5, 2016 at 21:34
  • $\begingroup$ @Sanya I'm not so fond of that paper on a superficial glance. They just seem to be stuffing a relativistic mass expression into nonrelativistic Hooke's law and don't really think about the Lorentz transformation of Hooke's law, a key thing to think about. $\endgroup$ Commented Jun 17, 2017 at 5:49
  • $\begingroup$ @WetSavannaAnimalakaRodVance I think a key thing to think about would actually be to consider the whole problem not in the framework of classical continuum mechanics - but no, I didn't want to advocate the paper, I just wanted to share something I had found because the question made me google a bit ... $\endgroup$
    – Sanya
    Commented Jun 17, 2017 at 10:38

2 Answers 2

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A mass $m$ on a spring of spring-constant $k$ will oscillate at (angular) frequency $\omega=\sqrt{k/m}$, so $k=m\omega^2=4\pi^2 m/\tau^2$, where $\tau$ is the period. So if you know $m$ and $\tau$ that gives you $k$.

An observer in an inertial frame moving past with speed $v$ will see $m$ increased by a factor $\gamma=1/\sqrt{1-v^2/c^2}$ but the time $\tau$ is also increased by $\gamma$, so $k'=k/\gamma$. That's not quite the same as the formula you give.

For a rod as you describe $N=kL/A$, (assuming there is a trivial crystal structure so stress and strain point in the same direction) the transformations of $L$ and $A$ depend on the direction the rod is pointing, as the Fitzgerald contraction acts on one or the other or both. If the rod is along the axis of motion then $L'=L/\gamma$ and $A'=A$, if it is transverse then $A'=A/\gamma$ and $L'=L$. So the value of $N'$ is in general different from $N$, and it is not a Lorentz invariant.

In fact the motion means that there's a difference between the transverse and longitudinal directions, so stress and strain will point in different directions, and $N$ is a tensor in its own 3D right, as Rob suggests (but not for the same reason).

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If stress-strain relationship is measured and plotted, then close to origin where the graph is linear, its slope gives Young's modulus, $E$. Now strain being ratio of two lengths in the same direction, strain measurement of all inertial observers will agree. However, stress is defined as ratio of force applied to cross-section area (normal to force), and the area is different for different inertial observers (in general). This means that inertial observers will disagree on what the applied stress is, and so disagree on what $E$ is.

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