Q:

Quantum Lorentz Transformations

Now I am reading a Weinberg's book "Quantum theory of field". Vol.1 page: 55
Сould you explain me the following things?

Einstein's principle of relativity states the equivalence of certain 'inertial' frames of reference. It is distinguished from the Galilean principle of relativity, obeyed by Newtonian mechanics, by the transformation connecting
  coordinate systems in different inertial frames. If $x^\mu $ are the coordinates in one inertial frame (with $x^1$,$x^2$,$x^3$ Cartesian space coordinates, and $x ^ 0 = t$ a time coordinate, the speed of light being set equal to unity) then in any other inertial frame, the coordinates $x^\mu$ must satisfy:
  $$\eta_{\mu\nu}dx^{'\mu}dx^{'\nu} = \eta_{\mu\nu}dx^{\mu}dx^{\nu} \tag{2.3.1}$$
  or equivalently:
  $$\eta_{\mu\nu}\frac{dx^{'\mu}}{dx^{\rho}}\frac{dx^{'\nu}}{dx^{\sigma}} = \eta_{\rho\sigma}. \tag{2.3.2}$$

How can it be equivalent? What are $\rho$ , $\sigma$, $\nu$? Why do we have $\eta_{\mu\nu}$ at both sides in the first equation? 

A:

In the first equation you have $\eta_{\mu\nu}$ at both sides because you define the Lorentz transformations as those leaving the metric $\eta_{\mu\nu}$ invariant. So $\eta'_{\mu\nu}=\eta_{\mu\nu}$.
You can obtain the first equation from the second by multiplying times $dx^{\rho}dx^{\sigma}$:$$\eta_{\mu\nu}\frac{dx^{'\mu}}{dx^{\rho}}dx^{\rho}\frac{dx^{'\nu}}{dx^{\sigma}}dx^{\sigma} \equiv \eta_{\mu\nu}dx'^{\mu}dx'^{\nu}=\eta_{\rho\sigma}dx^{\rho}dx^{\sigma}$$
where the last step is just multiplying the right hand side of the start point equation times $dx^{\rho}dx^{\sigma}$

