0
$\begingroup$

For $x^\mu$ with $\mu_0=t, \mu_i = x^i$ and $\eta_{\mu v}$ the metric tensor with diagonal $(-1,1,1,1)$ and zeros elsewhere, the condition for equivalence of inertial frames is stated as for some "coordinate transformation" $x \to x'$, (1) $\eta_{\mu v} dx'^\mu dx'^v = \eta_{\mu v}dx^\mu dx^v$ or equivalently (2) $\eta_{\mu v} \frac{\partial x'^\mu}{\partial x^\rho}\frac{\partial x'^v}{\partial x^\sigma} = \eta_{\rho\sigma}$.

I can understand (1) as saying that the Lorentzian metric/spacetime interval is preserved. I don't understand how (2) says the same thing or what's going on there mathematically. It can't be partial derivatives of a function taking $x$ to $x'$ since then the equality obviously wouldn't hold in general, but then I'm not sure what they are partial derivatives of, if anything. Any clarification is appreciated

$\endgroup$
2
  • 1
    $\begingroup$ Only Lorentz transformations preserve the Minkowski metric. So the mappings $x\to x’$ are very specific functions, not arbitrary functions. $\endgroup$
    – Ghoster
    Commented Jan 28 at 5:17
  • 1
    $\begingroup$ It’s not correct to write $\mu_0=t$. You can write $x^0=t$, or $x^\mu=(t,x,y,z)$. $\endgroup$
    – Ghoster
    Commented Jan 28 at 5:19

1 Answer 1

1
$\begingroup$

$x$ and $x'$ are coordinates on spacetime. Since each event in spacetime is labelled by a particular value in a given coordinate system, you can change between these two coordinate systems. Furthermore, you can write the coordinates $x'$ as functions of the coordinates $x$: $x' = x'(x)$.

This last point of view is what is in place when we take the partial derivatives you saw. You understand each $x'^{\mu}$ as a function of the different $x^\nu$ and take the partial derivative.

Now, condition (2) does not hold in general, as you pointed out. However, as pointed out in the comments, it doesn't need to hold in general: condition two is restricting the possible transformations from $x$ to $x'$ only to the so-called Poincaré transformations. The reason is that not all frames are inertial. For example, a general change of coordinates can take you from the usual coordinates for an inertial observer to the coordinates used by an accelerated observer, which is clearly non-inertial. Condition (2), and equivalently condition (1), forbid these general changes of reference frame and restrict you to inertial reference frames.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.