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Imagine a four-dimensional affine space $\mathcal{M}$ with the standard metric $\eta = \text{diag} (1,-1,-1,-1)$. Let $\mathcal{C}$ be a worldline of a point particle parametrized by an affine parameter $\tau$ (proper time). If we introduce some (Cartesian) coordinates $\{x^\mu\}$, the trajectory of the particle is given by the relations $x^\mu(\tau)$. The four-velocity of the particle is then defined to be a tangent to the worldline, $u^\mu = d x^\mu/d\tau$ and is an element of a tangent bundle $T\mathcal{M}$. In a different set of coordinates $\{x'^\mu\}$ the four-velocity changes components according to the standard Lorentz transformation, but geometrically does not change. OK, so far, so good.

Imagine now that we are trying to describe the same motion in a $3+1$ fashion. If we stick to the $\{x^\mu\}$ coordinates, our evolution parameter becomes $t=x^0$, and we describe the motion of the particle with a 3-velocity $v^i = d x^i / dt$.

My question is, from the point of view of a $4D$ Minkowski space $\mathcal{M}$, where does the 3-velocity $v^i$ live? Is it an element of a subspace of a tangent bundle $T\mathcal{M}$?

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  • $\begingroup$ I'm not sure what your question is. The 3-velocity is just the spatial component of the 4-velocity in an appropriate frame, it doesn't live somewhere special, it's just part of the 4-velocity. $\endgroup$ – ACuriousMind Sep 15 '15 at 16:53
  • $\begingroup$ If I am not mistaken, $u^\mu = (\gamma c,\gamma v^i)$, so the spatial component of 4-velocity has an extra $\gamma$ factor. $\endgroup$ – user17116 Sep 15 '15 at 17:12
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When asking where the 3-velocities live you might be looking for a geometrical explanation in a different sense than the more algebraic given above, one more related to metric geometry, even though the previous answer is the most correct one. Well just in case you might have been thinking of the claim about the space of relativistic velocities having hyperbolic geometry in 3 dimensions, the reason for that is that the restricted group of Lorentz transformations are projective instead of affine in 3 dimensions, and this projective group (PSL2,C) is isomorphic to the group of orientation-preserving isometries of hyperbolic space.

Then of course Minkowski space is a 4-dimensional real affine space, not a geometry in the sense of metric geometry and has nothing to do with hyperbolic geometry, nor hyperbolic space foliates Minkowski space.

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  • $\begingroup$ This is very interesting. Could you recommend some relevant literature? $\endgroup$ – user17116 Sep 20 '15 at 17:20
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    $\begingroup$ For instance Naber's "The geometry of Minkowski spacetime", for a more relaxed account Penrose "The road to reality" has nice sections about the space of velocities and hyperbolic geometry in a couple of chapters. $\endgroup$ – Walterscott Sep 20 '15 at 17:52
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There is a very geometrical answer. If you assume that the basis 4-vectors (hereafter called vectors) are not just vectors but are multivectors then they are a subspace of dimension four inside an associative, distributive non commutative algebra of dimension $2^4$ when $vv=\eta_{\mu\nu}v^\mu v^\nu.$

Such an algebra has scalars, vectors, bivectors trivectors and quadvectors such as $1,$ $e_1,$ $e_1e_2,$ $e_1e_2e_3,$ and $e_1e_2e_3e_4.$ And if you had a vector such as $\alpha^\mu e_\mu$ then you can multiply it by any unit timelike vector such as $e_0$ and you get $$e_0\alpha^\mu e_\mu=\alpha^01+\alpha^ie_0e_i.$$

Which has a scalar part and a bivector part. The bivectors that share a common, fixed, timelike factor (such as $e_0$) form a 3 dimensional subspace, the space of relative vectors. If you take the scalar $\alpha^0$ and the bivector $\alpha^ie_0e_i$ and add them together you get a multivector that only needs to be multiplied by $e_0$ (on the left) to get the original vector. So the scalar and the bivector are the two pieces of information you need. And you can use any unit timelike vector to break a vector into a scalar and a bivector and you don't ever need a basis either if you just work with the algebra.

So instead of having just a four dimensional vector space at each point you can have the whole $2^4$ dimensional multivector algebra at each point.

And you will have to learn to work with this spacetime algebra eventually if you do relativistic quantum mechanics. So your only choices are when to learn it and whether to learn it geometrically as well as algebraically or just algebraically. And if you do it just algebraically you'll have tons and tons and tons of formulas, none of which mean anything to you. But you can make predictions just fine that way if you like random algebra formulas.

But 3-velocity is more than the spatial part of the 4-velocity.

You can multiply any vector by a timelike vector to get a scalar and a bivector. The four velocity or a multiple of the four velocity. Any vector. You asked where they live. And they live naturally and basis independently in the union of the multivector algebra generated by the tangent vectors at an event.

Is [the 3-velocity] an element of a subspace of a tangent bundle $T\mathcal{M}$?

Absolutely not. But no matter which frame someone wants to use at a point, the bivector they need is there in the algebra and the algebra doesn't care about your basis or your frame, it contains any plane. It contains the electromagnetic field (which can be decomposed in the same way into a vector and a trivector, i.e. get your electric and magnetic fields, which are a frame dependent breaking apart of one geometric object).

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  • $\begingroup$ But 3-velocity is more than the spatial part of the 4-velocity. $\endgroup$ – user17116 Sep 15 '15 at 17:16
  • $\begingroup$ @LBO You can multiply any vector by a timelike vector to get a scalar and a bivector. The four velocity or a multiple of the four velocity. Any vector. You asked where they live. And they live naturally and basis independently in the union of the multivector algebra generated by the tangent vectors at an event. $\endgroup$ – Timaeus Sep 15 '15 at 17:21

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