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Minkowski space with the signature $(+---)$ can be described by $\Bbb{C}^{1,3}$ whilst with the signature $(-+++)$ by $\Bbb{C}^{3,1}$ (I am using $\Bbb{C}$ instead of $\Bbb{R}$ to allow for Wick rotations). Given a standard vector (or tensor) in one of these spaces: $$\vec v=v_i\hat e^i, \quad T=T^{ij} \hat e_i \otimes \hat e_j$$ what is the standard isomorphism(?) used to form the equivalent vector/tensor in the other?

My guess is that for vectors the isomorphism is defined to keep the $v^i$ components the same in both spaces - is this correct?.

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  • $\begingroup$ I don't understand why you imagine there needs to be an isomorphism. It's the same manifold, same tangent space. Components depend on the choice of coordinate system, but we're not discussing changing the coordinate system. $\endgroup$
    – user4552
    Aug 30, 2017 at 23:15
  • $\begingroup$ @BenCrowell Please can you explain why we are working on the same manifold? Manifolds depend on topological spaces; which depend on open sets; which depends on the metric defined on the metric space; which is different for the two spaces. $\endgroup$ Aug 31, 2017 at 3:33

2 Answers 2

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  1. If a quantity $Q(\eta)$ depends on the metric tensor $\eta_{\mu\nu}$, then the quantity is $Q(-\eta)$ in the opposite signature convention.

  2. Whether a quantity $Q(\eta)$ depends on the metric tensor $\eta_{\mu\nu}$ or not depends on conventions. Different authors have different conventions. The key is to choose a consistent convention.

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  • $\begingroup$ *Whether a quantity Q(η) depends on the metric tensor ημν or not depends on conventions. * Really? What's an example of someone defining a convention in which Q depends on the metric? $\endgroup$
    – user4552
    Aug 30, 2017 at 23:10
  • $\begingroup$ E.g. the 4-gauge potential $A_{\mu}$ in E&M often depends on the Minkowski signature convention, cf. this Phys.SE post. $\endgroup$
    – Qmechanic
    Aug 31, 2017 at 8:21
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Just to expand a little on Qmechanic's answer on the difficulties that are involved, consider if we did indeed make the substitution $v^\mu\mapsto v^\mu, u_\mu\mapsto-u_\mu$ for all vectors and covectors in the space. This indeed inverts all scalar products $u_\mu v^\mu$. But would this be an isomorphism?

Maybe -- but consider that $\eta_{\mu\nu} = a_\mu b_\nu + c_\mu d_\nu + \dots$ for some finite set of covectors $(a, b, c, d, \dots).$ Under this mapping we find that $\eta_{\mu\nu}\mapsto\eta_{\mu\nu}$ and what's really happened is that the $[0, 2]$-tensor which we called the "metric tensor" before, now no longer describes properly the inner product. Is this really "structure preserving" in the relevant way?

We can perhaps make instead the substitution $v^\mu\mapsto i~v^\mu, u_\mu\mapsto i~u_\mu$ to do something similar, now it is more clear that $\eta_{\mu\nu}\mapsto -\eta_{\mu\nu}$ under this mapping while simultaneously $u_\mu v^\mu$ inverts, so we can keep the same object as metric tensor. Similarly the inverse metric $\eta^{\mu\nu}\mapsto -\eta^{\mu\nu}$ and this seems appropriate.

There is one more nasty place where I can see that this might not be an isomorphism. I am talking above in abstract index notation. There are no components, properly speaking. The "components" of vectors are based on the interactions of vectors with other vectors; they tell you how to build this one up out of those ones. That is, you have specified up-front some unit vectors $\hat e^\mu_{0,1,2,3}$ and found some covectors $\hat e_\mu^{0,1,2,3}$ such that $$\hat e_\mu^{m}~\hat e^\mu_{n} = \delta^m_n.$$ Let us ignore the fact that our "unit vectors" have all reversed the sign of their norm, and still regard them as proper "unit vectors" (since I think that that's probably the proper thing -- probably the definition of "unit" flips its sign for all of these when we go from $-+++$ to $+---.$)

Since this is an inner product, and we inverted all of the inner products, we must find that this is no longer an appropriate expression; under the above mapping we find that this $\mapsto -\delta^m_n$ instead and therefore actually (if we hold the unit vectors constant) we must find $\hat e_\mu^{0,1,2,3} \mapsto -\hat e_\mu^{0,1,2,3}$ just because we need to choose these new unit covectors. So, even after this rather mysterious isomorphism we find that we have to modify any theorems which depended on any sort of algebra with the actual vector components, since these touched the unit covectors. Not great.

Well, this mapping does indeed induce $v^m \mapsto v^m$ for our vector components, so that's not too bad. Lacking this oddity, our covector components simply map to $v_m \mapsto -v_m$ as the covectors themselves to. Maybe that's all of the difficulties involved and this is suitable. But I wouldn't stake my life on it -- in particular I am very happy that our orientation tensor has not changed $\epsilon_{\lambda\mu\nu\omicron}\mapsto \epsilon_{\lambda\mu\nu\omicron}$ so our unit vectors still have the proper "handedness", but I would worry that somewhere down in the "plumbing" I would find that secretly all of our magnetic fields switched sign or whatever because somehow this application of a PT-operator has swapped out the right-hand-rule for the left-hand-rule on the 3D subspace or so.

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