Timeline for Physical reason for defining a Lorentz transformation as one that preserves the inner product of 4-vectors?
Current License: CC BY-SA 4.0
19 events
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| May 25, 2020 at 14:09 | comment | added | Vivek | @AndreasBlass True. Under differentiability assumptions you can show that you want Christoffels to be zero, which leads to linear transformations (cf. Rindler). Physically, this stems from homogeneity of space time in Cartesian coordinates. Or you could show it algebraically using a proof similar to Michael Artin's orientation preserving isometry of the plane proof (that these transformations have to be linear). | |
| May 25, 2020 at 0:28 | comment | added | Andreas Blass | You already have two answers showing that a linear transformation that preserves norms must also preserve inner products. I believe I've also seen a proof that a transformation preserving norms is necessarily linear under some mild hypothesis (like maybe continuity). Unfortunately, I don't recall where I saw that. | |
| May 24, 2020 at 20:57 | history | became hot network question | |||
| May 24, 2020 at 15:58 | comment | added | Shirish | @Vivek: Perfectly clear. Thanks a lot! | |
| May 24, 2020 at 15:56 | comment | added | Vivek | Indeed. You could do that. Then you'd be doing a general coordinate transformation. But that's not the definition of a Lorentz transformation (eg. cartesian to spherical polars is not a Lorentz transformation). In these general coordinates one sees weird behavior of free particles and one has to introduce pseudo forces to explain why they are accelerated without any physical forces etc. OTOH, Lorentz transformations preserve the flatness of the metric, so correspond to special choices of coordinate systems employed by inertial reference observers, so that free particles stay "free". | |
| May 24, 2020 at 15:39 | comment | added | Shirish | @Vivek : (cont'd)...So $v^T\eta w=v'^T\eta' w'=v^TT^T\eta'Tw=v^T(T^T\eta'T)w$, which means that $(T^{-1})^T\eta T^{-1}=\eta'$. Even if $\eta'\neq\eta$, I don't see why that would contradict $v^T\eta w=v'^T\eta' w'$ | |
| May 24, 2020 at 15:39 | comment | added | Shirish | @Vivek: So here's why I'm confused - if we have some invertible transformation $T$, it's equivalent to a change of basis. Let $v,w$ be representations of two 4-vectors in the standard Minkowski basis with the usual standard metric representation $\eta=\text{diag}(-1,1,1,1)$. $v'=Tv$ and $w'=Tw$ will then be representations of the same vectors but in the new basis. Also, the metric will have some new representation in the new basis (call it $\eta'$). Now since we've just changed the basis, the inner product of these two vectors should be the same regardless of basis representation (cont'd) | |
| May 24, 2020 at 15:04 | comment | added | Vivek | @ShirishKulhari Nope. Lorentz transformations are defined as a special class of invertible transformations obeying the $M\eta M^T = \eta$ condition, $\eta$ being the Minkowski metric tensor. It's assumed that you're transforming from one inertial frame using cartesian coordinates to another. | |
| May 24, 2020 at 14:05 | comment | added | Shirish | @AlfredCentauri : One potentially naive doubt: any invertible linear transformation on Minkowski space would preserve the spacetime interval, right? (because invertible transformations are like a change of basis and the norm is invariant under a change of basis) | |
| May 24, 2020 at 13:50 | comment | added | Shirish | @AlfredCentauri : Thanks! I will go through it. At first glance it seems to address the doubt. Another thing I've found is that the LT being invertible is enough for us to treat it like a change of basis (but that's just mathematically speaking and the link you gave should give additional physical insights) | |
| May 24, 2020 at 13:48 | comment | added | Alfred Centauri | Related reading: A few questions on passive vs active Lorentz transformations | |
| May 24, 2020 at 13:30 | vote | accept | Shirish | ||
| May 24, 2020 at 13:27 | comment | added | Shirish | @AlfredCentauri: I've edited the question. What I meant was, not every linear transformation is a change of basis. How do we show that a Lorentz transformation is indeed a change of basis? Sure, a Lorentz transformation preserves the norm and so does a change of basis, but that doesn't necessarily imply that an LT is just a change of basis. | |
| May 24, 2020 at 13:25 | history | edited | Shirish | CC BY-SA 4.0 |
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| May 24, 2020 at 13:24 | comment | added | Alfred Centauri | Not sure what you're look for in regards to a proof. Are you looking for a 'proof' of $\vec e_\alpha = \Lambda_\alpha^\beta\vec e_\beta$? | |
| May 24, 2020 at 13:23 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| May 24, 2020 at 13:16 | answer | added | mike stone | timeline score: 6 | |
| May 24, 2020 at 13:15 | answer | added | user265412 | timeline score: 6 | |
| May 24, 2020 at 12:56 | history | asked | Shirish | CC BY-SA 4.0 |