# Are the vectors and tensors of SR really so?

1. Mar 20, 2013

### quasar987

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!

2. Mar 20, 2013

### Fredrik

Staff Emeritus

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.

3. Mar 20, 2013

### TrickyDicky

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

4. Mar 20, 2013

### Fredrik

Staff Emeritus
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.

5. Mar 20, 2013

### WannabeNewton

All the same mechanisms of pullbacks and pushforwards still apply under diffeomorphisms. Remember that minkowski space is nothing more than $\mathbb{R}^{4}$ equipped with a riemannian metric but pullbacks and pushforwards don't care if there is a riemannian metric or not. The smooth structure is all we care about and in that sense we are just dealing with euclidean 4 - space.

6. Mar 20, 2013

### TrickyDicky

Just a small correction, that I know you are aware of, but just to be more precise, Minkowski is nothing more than $\mathbb{R}^{4}$ equipped with a pseudo-Riemannian (Lorentzian) metric. I know this distinction is not very important in flat spacetime, but it is when we go to GR's curved spacetime: hole argument, passive versus active transf., gauge vs not gauge...)
Also it is true that pullbacks and pushforwards each by themselves don't care if there is a Riemannian metric but it is also true that metrics give natural (musical) isomorphisms, that is, the raising and lowering of indices of tensors and vectors, IOW going back and forth from pullbacks to pushforwards.