Talk:Four-velocity

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unnamed thread

Simplified presentation a little while making notation more consistent with other relativity articles. ---Mpatel 19:06, 14 Jun 2005 (UTC)

Alternative approach using complex numbers

A work in progress

First Harmonic 22:48, 28 December 2006 (UTC)[reply]

Removed contentious phrase

I have removed the following sentence:

This observation, though trivial (as we will see from the formulas below), has a rather nice interpretation: in spacetime, an object is always in motion (at the speed of light!); it's just that in a rest frame, this motion is all in the time direction.

In my view this is misleading, although statements to this effect do appear in some "coffee-table" popularisations of relativity. An object moves in space. In spacetime it is a worldline and does not move at all. The four-velocity is just the normalised tangent vector to the worldline.--Dr Greg (talk) 16:58, 3 October 2008 (UTC)[reply]

Interpretation

Therefore light, and anything else traveling at light speed, do not experience the "flow" of time.

Well, if we, at rest, or uniformly moving (our four-velocity pointing in direction of our proper time axis), experience the flow of (our proper) time, couldn't we conjecture that anything traveling at speed of light along arbitrary spatial (spatial to us) axis would probably experience the flow of its own proper time just as well, like we do? OTOH, we know that speed of light remains the same in all frames of reference, which means that all observers, regardless of where their respective four-velocity vectors point will experience the light as moving perpendicular to their respective four-velocity vectors. What that tells us about the path or shape of photons and other massless particles? Or, more bluntly, do we need more dimensions then four to explain it? --147.91.1.43 (talk) 13:59, 9 October 2009 (UTC)[reply]

Does the "Theory of Relativity" section assume constant velocity?

It seems that everything in the "Theory of Relativity" section assumes constant velocity, without saying so. Otherwise, would we have $t=\gamma \tau$ ? Also, I find writing things like $x^{1}(\tau )=x^{1}(t)$ confusing. Isn't it bad form? I don't feel I know enough about this area to edit anything. — Preceding unsigned comment added by 66.188.89.180 (talk) 18:26, 8 January 2014 (UTC)[reply]

The Special Theory of Relativity is formulated with a definition of Four-velocity such that, for any object, it is always of constant magnitude, although its direction can change. It is a unit tangent vector. With this definition the Theory can be interpreted as a (multi-dimensional) geometry. For objects in free-fall the integral of the four-velocity is a geodesic (think "straight line" in flat space) in space-time. Cloudswrest (talk) 23:13, 17 October 2014 (UTC)[reply]

This was an error. I replaced the 3 ts with taus. 200.83.115.207 (talk) 19:51, 17 October 2014 (UTC)[reply]

Four-velocity is not a four vector under time inversion.

The proper time is not an invariant. It is a pseudo-invariant, with time-reversed observers disagreeing with each other on the sign of the proper time. You can see this because if t'=-t, dτ'=-dτ. The 4-velocity is the derivative of the 4 position (a 4-vector) with the proper time (a pseudo scalar under time reversal). This means that the 4-velocity is a pseudo 4-vector under time reversal. — Preceding unsigned comment added by 124.170.85.44 (talk) 07:46, 7 January 2015 (UTC)[reply]

The whole concept is a little murky; in particular, your statement is not valid until proper time has been suitably defined, and then only under a strange convention. I have yet to see proper time defined so as to be dependent on the reference frame of some arbitrary observer, as you have implicitly assumed. If you assume that the direction of proper time is assigned by some observer-independent convention to the world line (conceptually as the direction of time experienced by the entity that the world line), a time-reversed observer still sees the same proper time as a function of position along the worldline. A sensible convention would assign a direction of proper time to every (time-like) worldline so that the directions are all consistent with continuous deformations of any such worldline into another (all the while keeping the tangents time-like). Without having explicitly addressed which convention is being used, your statement above does not make sense. And as I indicated in my edit comment, this kind of thing does not belong in the lead in any event. As to it not being a four-vector, I agree that its interpretation as one must be highly constrained, but that is already clarified further down in the lead. —Quondum 14:47, 7 January 2015 (UTC)[reply]

I agree with Quondom that proper time has to be defined to be coordinate-independent; a change of coordinates cannot change the direction in which proper time increases. So, either you impose a restriction on allowable coordinates that prevents coordinate time being reversed (the article seems to be mostly written under an assumption of Minkowski coordinates rather than general coordinates), or else you allow the possibility that coordinate time could go backward relative to proper time (so

t=\pm \gamma \tau

, depending on choice of coordinates, and the temporal component of 4-velocity can be negative).

User:124.170.85.44's rewrite is inappropriate for several reasons:

the issues raised don't belong in the lead;
the terminology "vector with respect to boosts, ..." is technically incorrect; something either is or isn't a vector, without qualification.

-- Dr Greg talk 19:59, 7 January 2015 (UTC)[reply]

"I agree with Quondom that proper time has to be defined to be coordinate-independent; a change of coordinates cannot change the direction in which proper time increases." The problem is that I don't think the direction of proper time can be defined in a co-ordinate-independent way, because special relativity exhibits T-symmetry. If one observer sees a subluminal particle travelling along a wordline, passing first through four-position A, and then through B, there's nothing to stop another observer (with a clock running backward compared with the first), to see the same particle going first from B, and then to A. Who is right? It's not decideable under the postulates of special relativity. It can be proved that the proper time squared is an invariant for all inertial observers (the infinitesimal proper time squared is proportional to the infinitesimal space-time interval between two infinitesimally separated events), but not the proper time itself.

To look at the problem another way, think about how the four-velocity is calculated in co-ordinate time. Consider how 4-position transforms under time reversal T: $T:X=(t,{\vec {x}})=>X^{\prime }=(-t,{\vec {x}})=diag(-1,1,1,1)X$ .

The 4-velocity under time reversal transforms like this:

$T:U=\gamma (v)(1,{\vec {v}})=>U^{\prime }=\gamma (v)(1,-{\vec {v}})=diag(1,-1,-1,-1)U$ .

So if you choose to define a four-vector as a quantity that transforms the same way as the four-position under all transformations in $O(1,3)$ , the four-velocity is not a four vector, because it doesn't transform the same way under time reversal, it acquires an extra negative sign. It is however, a pseudo four-vector (it transforms the same way up to a sign under all transformations). I'm happy not to go into detail in the lead, so long as we don't call it a four-vector. — Preceding unsigned comment added by 124.170.85.44 (talk) 04:18, 8 January 2015 (UTC)[reply]

IMO, you are violating multiple WP guidelines, such as WP:OR, WP:RS and WP:CON, no matter how convinced you are of your own argument. Please start acting within the guidelines of the community, rather than pushing your own convictions. And start signing your posts. —Quondum 04:41, 8 January 2015 (UTC)[reply]

I'm new to editing (but not to reading!) wikipedia, so I apologize if I'm violating guidelines. However, I don't think this is original research; it follows quite immediately from definitions. Surely you don't want incorrect statements in the lead? If you can find anything actually wrong with my argument above, I'd like to hear it, but I think I've laid out the problem quite well. Either special relativity doesn't exhibit T-symmetry, or four-velocity is not a four-vector as that term is currently defined in wikipedia. - Sam — Preceding unsigned comment added by 124.170.85.44 (talk) 05:04, 8 January 2015 (UTC)[reply]

Can you provide any reputable sources which explain how and why 4-velocity is a pseudo-4-vector rather than a 4-vector? Zueignung (talk) 05:36, 8 January 2015 (UTC)[reply]