2
$\begingroup$

I'm confused about the physical interpretation of the four-velocity $U^\mu=\frac{dx^\mu}{d\tau}$ in General Relativity. I know that it is a tangent vector to a particle's "worldline", but what does this mean more physically?

For example, I am comfortable with what $U^\mu$ means in Special Relativity. In your inertial frame, you cover a distance $\Delta x^\mu$ and your clock says time $\Delta \tau$ has passed, and by taking the limit as $\Delta \tau \to 0$ this defines your $U^\mu$.

But I'm unsure about what $U^\mu$ means in curved space, or even in an accelerated reference frame. In either case the frame is no longer an inertial frame, which makes it confusing to interpret $\tau$, because it's no longer the "proper time in a frame", there is no one frame we are working in.

$\endgroup$
  • 1
    $\begingroup$ Re your middle paragraph, in your inertial frame $\vec{U}$ is always just $(1, 0, 0, 0)$. Since in GR spacetime is always locally Lorentzian exactly the same is true in GR. $\endgroup$ – John Rennie Jun 8 '14 at 15:36
  • $\begingroup$ So the process of measuring $\Delta \tau$ and $\Delta x^\mu$ from within your spaceship is no longer meaningful in general relativity? Because the local inertial frame just follows you around? I'm actually more concerned about what $\vec{U}$ means outside of a momentarily co-moving frame, i.e. when it's not just $(1,0,0,0)$. Does anyone actually "see" or measure that $\vec{U}$ from their frame or is it just some kind of abstract mathematical quantity? $\endgroup$ – user82235 Jun 8 '14 at 16:06
3
$\begingroup$

There is no difference between the interpretation of the four-velocity in special relativity and general relativity. One way of stating the equivalence principle is that locally, GR reduces to SR.

Although it's true that expressions like $\Delta x^\mu$ don't make sense in GR for finite coordinate differences (displacement isn't a vector), infinitesimal differences do make sense. For a treatment in this style, see Nowik and Katz, Differential geometry via infinitesimal displacements, http://arxiv.org/abs/1405.0984 .

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Accelerated particles do not have a "fixed" inertial reference frame. However, we can define infinitely many momentarily comoving reference frames (MCRFs) at each event. These are all related to each other through simple rotations. The four velocity at each event therefore is the time component of that event's MCRF's basis. As the particle is accelerated, so its MCRF will keep changing and so will its four-velocity.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

The 4-velocity is defined to be normalized, i.e.:

$g_{\mu\nu}u^{\mu}u^{\nu}=-1 \; ,$

so if you choose comoving observers, for which $u_i=0$, then $u_{0} = \frac{1}{\sqrt{|g_{00}|}}$.

From the above formula you can read the difference between the time felt by the comoving observer and the proper time, indeed $\frac{dx^0}{d\tau} = \frac{1}{\sqrt{g_{00}}} \rightarrow d\tau^2 =g_{00}dx^0$.

this should clarify which kind of frame you should consider in your sentence: "..proper time in a frame", there is no one frame we are working in."

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The "tangent to a curve on a manifold" is by definition vector with components given in coordinates $x^\mu$ by: $$v^\mu=\frac{dx^\mu}{d\lambda},$$ in which the coordinate presentation of the curve is given by functions $x^\mu(\lambda).$ So the components of tangent vector in whatever coordinates and whatever manifold have the same interpretation - they tell you how quickly the coordinates of point $P$ "moving" alongside the curve change as the parameter of the curve changes. That is if I wish to move point $P$ alongside the curve by some very small amount $\epsilon$ of the parameter of the curve, the moved point will now have coordinates $\left.x^\mu\right|_{\text{of }P_\text{moved}}=\epsilon v^\mu+\left.x^\mu\right|_{\text{of }P_\text{original}}$.

STR and GR only gives physical meaning to the ingredients. The first ingredient is giving meaning to the curve - now the curve represents movement of real object and we will call it woldline.

The second ingredient is giving meaningful parameterization to worldline by using proper time. That is, parameterize the movement of the object by how much time passed on the objects own clock.

In noninertial frame this is as good parametrization as in objects rest frame. You just look at the objects clocks instead of your owns. So if you are observing for example astronaut with watch on his hand and you are using some (random) coordinates, then you get the components of his 4-velocity in your coordinates by writing down at what coordinates the astronaut is at the tick (of astronauts watch) in question (lets call it tick 1) ($x^\mu_1$) and writing down the coordinates at the next tick of astronauts watch (lets call it tick 2) ($x^\mu_2$) and using the high school velocity formula: $$v^\mu=\frac{x^\mu_2-x^\mu_1}{\text{tick}_2-\text{tick}_1},$$ assuming the tick is small enaugh for us to use finite formula with good enough precision.

In your inertial frame, you cover a distance

In your own frame you will never cover any (spatial) distance. The only nontrivial movement that you can observer is given by other objects with different rest frames.

which makes it confusing to interpret $\tau$, because it's no longer the "proper time in a frame", there is no one frame we are working in

The proper time of accelerated observer at each moment is defined by inertial frame in which the observer is at rest in the given time. This is how STR works. The question might remain, wheter such time is measurable so that we do not end up building a theory that is formulated w.r.t to some time that drops out of every measured quantity (like it was with absolute newtonian time in Lorentz theory). To the best of our knowledge, proper time of accelerated observer can indeed be measured.

So the only thing we need, is to ask manufacturer of the watch wether he can guarantee correct operation in the range of accelerations we are considering. If he can, we still check astronauts watch to determine the 4-velocity. In this regard, nothing changes.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.