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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
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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.

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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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