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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?

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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 $dx^{\rho}dx^{\sigma}$

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  • $\begingroup$ Thank for you response. What is the physical sense of $\rho$, $\sigma$, $\nu$? $\endgroup$ – Aleksandrochka Apr 15 at 19:12
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    $\begingroup$ those are indexes that indicate which components of vectors and matrices you are considering. $\endgroup$ – AoZora Apr 15 at 19:20
  • $\begingroup$ What is $\eta_{\rho\sigma}$? Could you clear the question. How you can get the right side of the last formula? $\endgroup$ – Aleksandrochka Apr 15 at 19:27
  • $\begingroup$ $\eta_{\mu\nu}$ is the Minkowski metric: is the matrix you use to take the (pseudo)scalar product between vectors in special relativity; it is a diagonal matrix (4x4) with elements -1,1,1,1. Using this matrix the norm of vectors and all the scalars are Lorentz invariant. What I shown in my answer is that the two equations you presented are equivalent. These formulas themselves indicate that the norm of the vector $dx$ is invariant under lorentz transformations. If you want to review special relativity: damtp.cam.ac.uk/research/gr/members/gibbons/… $\endgroup$ – AoZora Apr 15 at 19:47
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I hope this is useful for you. This has to do with how to measure distances in Minkowski space. This is done through the metric, an amount that in principle depends on the point in space but in this case, special relativity, is constant. These equations that you have written tell us precisely about that. In the first, we see that the way to measure distances when we move from one frame of reference to another is "similar". The second is a bit more explicit, we see the way we transform the metric when we move from one frame of reference to another, this is done through the Lorentz transformations. The Lorentz transformations ($\Lambda_{\mu}^{\nu}=\frac{dx'^{\mu}}{dx^{\nu}}$) are those used in the Minkowsi space to change coordinates. Thus, $\eta_{\mu\nu}$ appears on both sides of the equation because it is invariant under the Lorentz transformations.

$\rho,\sigma, \nu$, they serve to label the coordinates

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