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I am trying to derive Lorentz contraction purely through tensor calculations, without explicitly writing out the Lorentz transformation matrix. Here is what I have so far:

Let $i, j, k$ range over $1, 2, 3$, representing spatial indices, and $\lambda, \mu, \nu, \rho$ range over $0, 1, 2, 3$, representing spacetime indices. The Lorentz transformation $\Lambda$ satisfies the condition $$ \eta_{\mu \nu} = \Lambda^\lambda_{\ \mu} \Lambda^\rho_{\ \nu} \eta_{\lambda \rho}, $$ where $\eta = \text{diag}(1, -1, -1, -1)$ is the Minkowski metric.

Consider a small spatial interval in the $S$ frame: $$ dr^2 = \delta_{ij} \, dx^i dx^j. $$ After applying a Lorentz transformation to the coordinates, this becomes: $$ dr^2 = \delta_{ij} \Lambda^i_{\ \mu} \Lambda^j_{\ \nu} dx'^\mu dx'^\nu. $$ Similarly, in the $S'$ frame: $$ dr'^2 = \delta_{ij} \, dx'^i dx'^j. $$ From here, I expect to be able to show that $dr^2 < dr'^2$, which would imply Lorentz contraction. However, I am struggling to complete the derivation.

I would like to proceed by keeping the calculations purely in tensor form and without explicitly writing out the Lorentz boost matrix along the boost axis. How can I continue from here to demonstrate Lorentz contraction?

Wt.N
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1 Answers1

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In the unprimed frame we have, the squared differential spacetime interval between two events $\mathrm ds^2$ given as:

\begin{align*} \mathrm ds^2 &= -\mathrm dx^0\mathrm dx^0+\mathrm dr^2 \\&= \eta_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu \\&= \eta_{ij}\mathrm dx^i \mathrm dx^j + \eta_{00}\mathrm dx^0 \mathrm dx^0 & (i,j=1,2,3) \tag{1}\end{align*}

Now I think the key observation is that when we make a measurement of a length in that frame, $x^0$ (the $t$ coordinate) is constant, as we make a simultaneous measurement of both ends of an object, leading the $\mathrm dx^0$ term to vanish. Then from the above, we find that:

$$ \mathrm ds^2 = \mathrm dr^2\tag{2} $$

On the other hand, due to the invariance of the interval, we also have the relation to the "primed", relatively moving frame:

$$ \mathrm ds^2 = \mathrm ds'^2 = -\mathrm dx'^0\mathrm dx'^0 + \mathrm dr'^2\tag{3} $$

Putting $(2)$ and $(3)$ together we get:

$$ \boxed{\mathrm dr^2 = \mathrm dr'^2 - \mathrm dx'^0\mathrm dx'^0} $$

which is the result we expect. The frame taking a simultaneous measurement of the moving object, measures a smaller differential distance - smaller by an additive term $-\mathrm dt'^2$ where $t'$ is the time coordinate of the inertial frame where the object is in relative motion.

I don't know if that's what you were looking for exactly. On the one hand, I definitely haven't explicitly referred to the components of $\Lambda$ (I haven't even had to mention it because I just assumed the relevant transformation leaves the interval invariant). On the other hand, I don't know if this counts as a "purely tensorial" derivation because I do make use of specific components at several points. But I don't know if one has an alternative, because in deriving the Lorentz contraction from what I know, one always assumes we take two events that are simultaneous in one frame, and that somehow needs to be expressed in the equations.

Amit
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