# Are the vectors and tensors of SR really so?

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## Main Question or Discussion Point

By the question in the title, I mean, do the so-called 4-vectors and tensors of SR transform as tangent vectors and tensors (in the sense of differential geometry) with respect to any transformation (local diffeomorphism) of the space-time coordinates or only with respect to Lorentz transformations? Thanks!

## Answers and Replies

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Fredrik
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Slightly longer: If we define the spacetime of SR as a smoth manifold, then yes, vectors like 4-velocity belong to the tangent space of the manifold, and everything you know from differential geometry can be applied to these vectors. This includes the stuff I did in this post. If the coordinate systems x and y in that calculation are global inertial coordinate systems that send the same point to 0, then the matrix with components ##\Lambda^\nu{}_\mu## is a Lorentz transformation.

Note that it also makes sense to define the spacetime of SR as a vector space. (This gives us a slightly different version of the theory in the sense that the statements that define the theory are slightly different, but the predictions the theory makes about results of experiments are the same). Then spacetime is ##\mathbb R^4## and the vectors are members of ##\mathbb R^4## too.

But, if you define SR in Minkowskian spacetime, then Lorentz transformations (Poincare if you include translations) are the ones that leave the manifold invariant.

Fredrik
Staff Emeritus
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Yes, they are the isometries of the metric. But the whole "transforms according to the tensor transformation law" thing doesn't really have anything to do with the metric. Unless of course you change the tensor transformation law to require that the transformation matrix is the Jacobian matrix of an isometry.

WannabeNewton