What is the physical interpretation of four-velocity in special relativity?

[yuiop]

I suspect that the word "proper" in this context was originally introduced in the older meaning of "its own" as in the French "propre" and the English word "property". That is, "proper" quantities are quantities expressed relative to aspects (in particular the time rate) of the object itself.

I think the part that Dalespam objects to (and I do to a certain extent) is that the spatial component of proper velocity is not a proper measurement. It is distance as measured in another reference frame and not distance as measured by the traveller with the clock. In other words he using his own clock but someone else's rulers.

For example a rocket traveling from here to Mars would measure the distance as shorter (as per length contraction) than the distance measured by an observer on Earth or Mars. The proper velocity is measured using the clock of the rocket (which is a proper measurement) and the distance as measured by an observer on Earth on Mars (which is therefore not a proper measurement). So "proper velocity" is a misnomer. It should be called something like the dilated three velocity.

[EDIT] I suppose you could justify the term "proper velocity" on the grounds that the distance is measured by one observer at rest with respect to the markers that define the distance and the time is measured by another observer at rest with respect to the moving object which is having its velocity determined. A funny way of doing things, but I guess it works.

 

[DrGreg]

Yeah, I understand the definition, I just don't think it is very "proper". I wouldn't mind the concept if it were called something else.

Some authors use the term "celerity" instead of "proper velocity". I prefer this because it doesn't cause confusion over what's "proper".

It should also be noted that a few authors have described proper velocity as "rapidity", but that term is now generally taken to be something else (\tanh^{-1}(|\textbf{v}|/c)).

Also to respond to a point raised earlier in this thread, whereas both velocity and celerity are 3-vectors, rapidity is a scalar, so isn't much use for non-linear motion. The rapidity of A relative to B is, in terms of 4D geometry, the (hyperbolic) angle between A and B's 4-velocities.

 

[snoopies622]

The rapidity of A relative to B is, in terms of 4D geometry, the (hyperbolic) angle between A and B's 4-velocities.

A long time ago I was taught that an angle is two rays with a common origin that is measured with a protractor, so I am having trouble with 'hyperbolic' angles. If a rocket ship moves away from me in the x direction at v=0.5c and I decide to draw this on a spacetime diagram (x vs. ct in my frame of reference) the angle the ship's world line makes with mine is

<br /> \theta = arctan (0.5) \approx 26.6 ^ {\circ}<br />

but the rapidity between us is

<br /> \phi = arctanh (0.5) \approx 31.5 ^ {\circ}<br />.

Is there a way to draw spacetime diagrams such that the angles that appear are the rapidities? Is a 'hyperbolic angle' really an angle?

 

[granpa]

its the area swept out between the origin and the hyperbola.

 

[DrGreg]

Rapidity and 4-velocity

A long time ago I was taught that an angle is two rays with a common origin that is measured with a protractor, so I am having trouble with 'hyperbolic' angles. If a rocket ship moves away from me in the x direction at v=0.5c and I decide to draw this on a spacetime diagram (x vs. ct in my frame of reference) the angle the ship's world line makes with mine is

<br /> \theta = arctan (0.5) \approx 26.6 ^ {\circ}<br />

but the rapidity between us is

<br /> \phi = arctanh (0.5) \approx 31.5 ^ {\circ}<br />.

Is there a way to draw spacetime diagrams such that the angles that appear are the rapidities? Is a 'hyperbolic angle' really an angle?

"Hyperbolic angle" is not literally an angle in the same way that "spacetime interval" ds is not literally a distance, but, nevertheless, we can mentally picture it as a distance and in many (but not all) ways it resembles a distance. You can't measure a spacetime interval directly using a ruler on a spacetime diagram, and similarly you can't measure a rapidity using a protractor either.

In 3D Euclidean geometry, distance, angle and scalar product are related by

\textbf{u} \cdot \textbf{v} = ||\textbf{u}|| \, ||\textbf{v}|| \, \cos\theta​

In 4D Minkowski geometry, the spacetime interval, rapidity and metric are related by

\textbf{U} \cdot \textbf{V} = ||\textbf{U}|| \, ||\textbf{V}|| \, \cosh\phi​

where \textbf{U} = (U_t, U_x, U_y, U_z) and V are both timelike 4-vectors (e.g. 4-velocities),

\textbf{U} \cdot \textbf{V} = g_{\mu\nu}U^{\mu}V^{\nu} = U_tV_t - U_xV_x - U_yV_y - U_zV_z​

and ||U|| denotes the spacetime interval

||\textbf{U}|| = \sqrt{\textbf{U} \cdot \textbf{U}}​

When a velocity is along the x-axis (i.e. U lies in the tx-plane and the observer's V lies along the t-axis) we can also write

\textbf{U} = (c\cosh\phi, c\sinh\phi, 0, 0)​

In 2D Euclidean geometry, rotation through an angle \theta is represented by the matrix

<br /> \left(\begin{array}{cc}<br /> \cos\theta &amp; -\sin\theta \\<br /> \sin\theta &amp; \cos\theta \\<br /> \end{array}\right) <br />​

In 2D (1+1) Minkowski geometry, the Lorentz transform for velocity c\tanh\phi is represented by the matrix

<br /> \left(\begin{array}{cc}<br /> \cosh\phi &amp; \sinh\phi \\<br /> \sinh\phi &amp; \cosh\phi \\<br /> \end{array}\right) <br />​

You can see the resemblance between angles in Euclidean geometry (with trig functions) and rapidities in Minkowski geometry (with hyperbolic functions), which is why the term "hyperbolic angle" is used.

 

[robphy]

Geometrically...
rapidity can be defined as
- (adding to granpa's description) the area enclosed by the two unit-radii (i.e. 4-velocities) and the unit-hyperbola (effectively a sector of a Minkowski-unit-circle),
- the Minkowski arc-length of the portion of the unit-hyperbola intercepted by the two radii
Both are analogues of the Euclidean definitions.

You can't measure a spacetime interval directly using a ruler on a spacetime diagram, and similarly you can't measure a rapidity using a protractor either.

You can if you use the right kind of ruler and protractor (i.e. the analogous tools for Minkowski geometry).

 

[snoopies622]

...you can't measure a rapidity using a protractor...

OK, I just wanted to be certain. Recently I noticed that an ordinary rotation of axes transformation

\left(\begin{array}{cc}\cos\theta &amp; -\sin\theta \\\sin\theta &amp; \cos\theta \\\end{array}\right)

using ct on the vertical axis and xi on the horizontal (where i^2 = -1) produces the Lorentz transformation almost immediately, but even in this case the angle of rotation as measured with a protractor isn't the rapidity. Alas.

This relationship

\textbf{U} \cdot \textbf{V} = ||\textbf{U}|| \, ||\textbf{V}|| \, \cosh\phi

however, does provide me with another use for the four-velocity vector, which helps to answer my original question from the beginning of this thread, so thank you for that.

 

[DrGreg]

Proper acceleration and rapidity

I came to this thread rather late. Having now read some of the earlier posts, I can point out, for a particle moving in straight line, with proper acceleration \alpha(\tau), rapidity \phi(\tau) (relative to some inertial observer) and proper time \tau, we have

\alpha = c \frac {d\phi} {d\tau}​

or, equivalently,

\phi(\tau) = \int \frac{\alpha(\tau)}{c} \, d\tau \, + \, \phi_0​

(The value of the "constant of integration" \phi_0 depends on which inertial observer is measuring the rapidity.)

For explanation, see my posts #13, #14 & #15 (and a correction in post #28) in this thread[/color].

 

[thljcl]

Here is my answer for four-velocity. http://word-view.officeapps.live.co...mb6VB+4dzY9I9HrYgXquJAFnA=2&ad=en-US&popout=1