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There's a nice answer to this question: Why is the scalar product of two four-vectors Lorentz-invariant? - that explains that a Lorentz transformation is one under which the inner product of two 4-vectors is invariant.

I know that the norm of the difference between two 4-vectors can be interpreted as the spacetime separation between corresponding events, and I understand the reason for invariance of the spacetime interval. But the inner product is more general than norm of difference, so what's the physical reason for saying that the LT should preserve the inner product of 4-vectors?

Alternatively I could phrase my question as follows: is there an explanation why a Lorentz transformation is equivalent to a change of basis? (because the inner product is invariant under a change of basis - the vectors essentially stay the same - but isn't generally invariant under a linear transformation)

Shirish Kulhari
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  • Not sure what you're look for in regards to a proof. Are you looking for a 'proof' of $\vec e_\alpha = \Lambda_\alpha^\beta\vec e_\beta$? – Alfred Centauri May 24 '20 at 13:24
  • @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. – Shirish Kulhari May 24 '20 at 13:27
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    Related reading: A few questions on passive vs active Lorentz transformations – Alfred Centauri May 24 '20 at 13:48
  • @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) – Shirish Kulhari May 24 '20 at 13:50
  • @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) – Shirish Kulhari May 24 '20 at 14:05
  • @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. – Vivek May 24 '20 at 15:04
  • @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) – Shirish Kulhari May 24 '20 at 15:39
  • @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 $v^T\eta w=v'^T\eta' w'$ – Shirish Kulhari May 24 '20 at 15:39
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    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". – Vivek May 24 '20 at 15:56
  • @Vivek: Perfectly clear. Thanks a lot! – Shirish Kulhari May 24 '20 at 15:58
  • 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. – Andreas Blass May 25 '20 at 00:28
  • @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). – Vivek May 25 '20 at 14:09

2 Answers2

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But the inner product is more general than norm of difference,

Actually it's not. If you have a norm, then you automatically get an inner product for free, through linearity, because $(u+v)\cdot(u+v)=u\cdot u+v\cdot v+2u\cdot v$.

Alternatively I could phrase my question as follows: is there a proof that a Lorentz transformation is equivalent to a change of basis? (because the inner product is invariant under a change of basis - the vectors essentially stay the same - but isn't generally invariant under a linear transformation)

Every Lorentz transformation is a change of basis, but not every change of basis is a Lorentz transformation. For example, you could do a change of basis in which the x and y axes become no longer perpendicular.

user265412
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    "If you have a norm, then you automatically get an inner product for free" Not necessarily. There are norms that don't induce inner products. – eyeballfrog May 25 '20 at 04:40
  • @eyeballfrog : That's interesting. Do such norms arise in the mathematical formulation of any of the Physics fields, or are they more of an edge case? – Shirish Kulhari May 25 '20 at 14:15
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    @ShirishKulhari The $L^1$ norm in function spaces is probably the most familiar norm to physicists that does not induce an inner product. Of course, physicists tend to be a little fast and loose with mathematics and will use the $L^2$ inner product on $L^1$ because it usually works out anyways. – eyeballfrog May 26 '20 at 05:50
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If a map preserves the length $\|{\bf a}\|$ of vectors, then it also preserves the inner product because $$ {\bf a}\cdot {\bf b}= \frac 12 (\|{\bf a}+{\bf b}\|^2- \|{\bf a}\|^2- \|{|\bf b}\|^2). $$ This identity is true for both the usual Euclidean length and inner product, and also the spacetime interval and Lorentz inner product.

mike stone
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