Quantum Lorentz Transformations

Question

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?

Answer 1

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

Answer 2

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

Answer 3

You have to remember that the summation convention is being used, so that (for instance) $η_{μν} dx^μ dx^ν$ means $$\sum_{μ=0,1,2,3}{\sum_{ν=0,1,2,3}{η_{μν} dx^μ dx^ν}}.$$ If treat the line element as a proper distance $s$, then with $x^0 = t$ and $(x^1, x^2, x^3) = (x, y, z)$, the components are $$η_{00} = -c^2, \hspace 1em η_{11} = η_{22} = η_{33} = 1, \hspace 1em η_{μν} = 0 \hspace 1em (μ ≠ ν),$$ so that the respective line elements read: $$-c^2 dt'^2 + dx'^2 + dy'^2 + dz'^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2,$$ and that's what equation (2.3.1) is actually saying. There is no $μ$ or $ν$ anywhere in there, except as place-holders for the summation operator.

I won't write out (2.3.2) in full, but with the conventions spelled out explicitly like this, I think you can see more clearly that it's just an application of the chain rule, applied to the coordinate differentials $(dt', dx', dy', dz')$, starting from (2.3.1).

The Lorentz group is the symmetry group that preserves $x^2 + y^2 + z^2 - c^2 t^2$, with the coordinates in place of coordinate differentials. In contrast, it's the Poincaré group that preserves the line element with the coordinate differentials. In particular, the Poincaré group includes translations on the coordinates: $x^μ → x^μ + a^μ$ for constant $\left(a^0, a^1, a^2, a^3\right)$.

Weinberg, in fact, spoke wrong in another respect. There is no fundamental difference between the relativistic and non-relativistic cases - he only thought there was. This can be best seen by treating $s$ as proper time instead of as proper distance. Then, the corresponding line element becomes $$-c^2 ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2,$$ or just: $$c^2 (ds^2 - dt^2) + dx^2 + dy^2 + dz^2 = 0.$$

The Poincaré group can then be thought of as the group which preserves this line-element / constraint and which leaves $ds$ invariant. The only difference from non-relativistic theory is that $s$ and $t$ are distinct. In the limit, their difference $s - t$, the time-dilation, vanishes; thus leading the the (mistaken!) impression that there is nothing corresponding to time-dilation in non-relativistic theory.

Actually, there is. It just wasn't recognized until much later ... well into the 20th century. Though $s - t → 0$ in the non-relativistic limit, this is not the case for its scaled-up version $u = c^2 (s - t)$, which has a non-trivial and meaningful non-relativistic limit. In fact, if you substitute for $s$, then the line element becomes: $$dx^2 + dy^2 + dz^2 + 2 dt du + \frac{1}{c^2} du^2 = 0,$$ while the linear invariant (that is: $ds$) becomes: $$dt + \frac{1}{c^2} du.$$ The invariance group is also Poincaré ... trivially extended in that it also allows for translation on the $u$ coordinate.

This is one member of a family of invariants, parameterized by $α$: $$dx^2 + dy^2 + dz^2 + 2 dt du + α du^2 = 0, \hspace 1em dt + α du.$$ When $α > 0$, this yields the (trivially extended) Poincaré group, with light speed set to $c = \sqrt{1/α}$; while when $α = 0$, with the corresponding items being: $$dx^2 + dy^2 + dz^2 + 2 dt du = 0, \hspace 1em dt,$$ the symmetry group is the central extension of the Galilei group: also known as the Bargmann group. The $u$-translations corresponds to mass, in the very same way that the spatial-translations correspond to momentum, and the time-translations to (the negative of) energy.

For $α < 0$ the corresponding geometry provides a container for 4-dimensional Euclidean geometry, where $t$ is now a spatial dimension. In all cases, for all $α$, the ambient 5D geometry is a (4+1)-dimensional Minkowski space. They differ from one another only by which direction the invariant coordinate $s$ is oriented in.

So, there is a metric there - provided that (1) you add in the extra coordinate and dimension, (2) constrain to a 4D sub-space of the underlying ambient 5D-geometry and (3) impose the requirement that the coordinate differential $ds$ - for proper time - be invariant. In the non-relativistic case, where $α = 0$, the geometry is called the Bargmann geometry. Its relativistic version, where $α > 0$ has no name that I am aware of. But, as you can see, the Minkowski geometry is sitting there, right inside of it, as a 4D subspace. There's a (4+1)-Minkowski metric, for the line element, and a linear "metric" for the proper time, $ds = n_μ dx^μ$ where $n_1 = n_2 = n_3 = 0$, $n_4 = 1$ and $n_5 = α$, this time with $$\left(x^1, x^2, x^3, x^4, x^5\right) = (x, y, z, t, u).$$

In retrospect, you can see why Weinberg totally missed this: he kept using the ill-advised convention $c = 1$, which Sapir-Whorf's away the very ability to even express what's just been stated in the preceding paragraphs! That's an example (one of many) of the problem, which runs amok in the Physics literature.

And by the way, on that issue:

Sapir-Whorf Hypothesis:
https://en.wikipedia.org/wiki/Linguistic_relativity