1
$\begingroup$

This question rises from the comments on @G Smith's answer's to this question https://physics.stackexchange.com/a/603032/113699

Precisely I was trying to understand the Lorentz Transformations between Spherical Polar Coordinates. The point is that I have read that Lorentz transformations have to be Linear but if for spherical polar coordinates they will be Non-Linear.

These answers here say that Interval preserving transformations/ Lorentz Transformations need to be linear Interval preserving transformations are linear in special relativity and Kleppner derivation of Lorentz transformation ( the answer by Selena Routley) and Why is this non-linear transformation not a Lorentz transformations? It does preserve $x^2 + y^2 + z^2 - c^2t^2 = x'^2 + y'^2 + z'^2 - c^2t'^2 $ ( especially Knzhou's comment on A Hussain's answer where he also says we should do SR in cartesian coordinates and not polar coordinates).

But the answer by Void here Are Lorentz transformations linear transformations?

says that Lorentz Transformations should be Homogenous and not Linear.

So my question is-

Can be there Lorentz Transformations between spherical polar coordinates or cylindrical coordinates. If yes, then they will be Non-Linear. So can the Lorentz Transformations be Non-Linear too?

$\endgroup$
5
  • $\begingroup$ @CosmasZachos Sorry, I didn't understand the clue. How is rotation to going help. I want to write the Lorentz Transformations for boosts in spherical coordinates. In the 1st link G Smith says it's possible and perfectly okay to write Lorentz Transformations for boosts in spherical or cylindrical coordinates. But then my query is that I have read in Caroll, that Lorentz Transformations are linear. If I will write Lorentz Transformations for boosts in spherical coordinates, the transformations would be non linear. Can Lorentz Transformations be Non Linear. Could you please elaborate the hint $\endgroup$
    – Shashaank
    Dec 27, 2020 at 5:56
  • $\begingroup$ You were given the explicit answer in the answers you quote. $\endgroup$ Dec 27, 2020 at 16:14
  • $\begingroup$ @CosmasZachos yes. But those Lorentz transformations are not linear in the coordinates. Lorentz transformations need to be linear. The question I am asking is can Lorentz Transformations be NON LINEAR Are they allowed to be non linear... $\endgroup$
    – Shashaank
    Dec 27, 2020 at 16:22
  • $\begingroup$ ? A linear function of nonlinear functions can be nonlinear. $\endgroup$ Dec 27, 2020 at 16:39
  • $\begingroup$ @CosmasZachos so it doesn't mean that Lorentz Transformation need to be always linear. Is that correct, because the 1st three linked questions have answers where the answered insists that Lorentz Transformation need to be Linear ( Caroll also says that)... What's the final thing... Can Lorentz Transformations be NoN linear. They don't need to be necessarily linear $\endgroup$
    – Shashaank
    Dec 27, 2020 at 16:42

1 Answer 1

1
$\begingroup$

The Lorentz transformation is always linear in all co-ordinates, including polars.

I feel that the subtlety you may be forgetting here is that the Lorentz transformation always acts on the linear tangent spaces to a manifold at particular given points of discussion, not the manifold itself. Aside from in one special case ....

In special relativity, where we are always working in flat Minkowski spacetime, we can and almost always do use Cartesian co-ordinates to globally label the manifold in a one chart atlas. In this unusual situation, not only is the tangent space but also the manifold itself a linear (vector) space. So, in Cartesian co-ordinates, Lorentz transformations can, through this Cartesian co-incidence, be imparted to global co-ordinates to work out their images under Lorentz transformations. This is a special case and not a general property of Lorentz transformations or of manifolds.

If you choose polar (or other nonlinear images of Cartesians), even in Minkowski spacetime we don't have this luxury anymore and the Lorentz transformation $\Lambda:T_XM \to T_XM$ only has meaning as the linear transformation between the common tangent spaces $T_XM$ to the manifold $M$ constructed by two two relatively moving observers at the point $X$ where they are momentarily collocated. A tangent vector $A$ with contra components $A^j$ as seen by observer $\mathbf{A}$ naturally has components $B^k = \Lambda^k_{{}j} A^j$. And, as usual, if we want to construct the components of $\Lambda$ in polars $\bar{x}$ from the Cartesian components $x$, we invoke $\Lambda^k_{{}j}\mapsto \frac{\partial x^k}{\partial \bar{x}^\ell} \frac{\partial \bar{x}^m}{\partial x^j} \Lambda^\ell_{{}m}$, where the expressions of the form $\frac{\partial x^k}{\partial \bar{x}^\ell}$ are calculated from the derivatives of the polar in terms of Cartensian co-ordinate transformation in the wonted way.

I'm pretty sure i have made exactly this mistake too in the past, and it arises from the ubiquitous use of Cartesians in special relativity, which in turn makes us forget that Lorentz transformations do not usually apply to global co-ordinates. Of course, when you wish to do this latter, you concatenate $C\circ \Lambda \circ C^{-1}$ where $C$ is the nonlinear polar to Cartesian co-ordinate conversion. This is a nonlinear process and a simplification of the more general, nonlinear process of co-ordinate transformations between charts in a general manifold.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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