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

[Fredrik]

Yes, a compressed spring has slightly less mass than an uncompressed one.

This sounds very wrong to me. You have to perform work to compress a spring, so the compressed spring should "contain more energy" and therefore weigh more. This isn't meant to be a rigorous argument. It's just my intuition talking, but what I'm saying also agrees with what snoopies actually concluded. What he found was that the uncompressed spring has mass $(3-2\gamma)$ times the mass of the compressed spring, and since $\gamma>1$, this is <1, so the compressed spring weighs more.

I'm still confused by this though. Shouldn't the 0 component of the 4-momentum of the spring in its center of mass frame be the same before and after the acceleration? Before the acceleration, the spring carries the energy required to compress it, and after, it's oscillating at first (kinetic energy), and comes to rest after a while (when the kinetic energy of the oscillation has been converted to heat), so it shouldn't lose energy (relativistic mass) until it has radiated away a significant part of the heat.

 

[Dale]

And quantitatively? Is my operational premise correct -- the sums of each component of four-momentum must be the same both before and after acceleration?

Yes, that is the conservation of four-momentum.

This sounds very wrong to me. You have to perform work to compress a spring, so the compressed spring should "contain more energy" and therefore weigh more. This isn't meant to be a rigorous argument. It's just my intuition talking, but what I'm saying also agrees with what snoopies actually concluded. What he found was that the uncompressed spring has mass $(3-2\gamma)$ times the mass of the compressed spring, and since $\gamma>1$, this is <1, so the compressed spring weighs more.

Woops! How embarassing. You are exactly correct. I simply meant to agree with snoopies662's math and re-phrase the math into english, but I got the math to english translation exactly backwards. Sorry for the confusion.

I'm still confused by this though. Shouldn't the 0 component of the 4-momentum of the spring in its center of mass frame be the same before and after the acceleration? Before the acceleration, the spring carries the energy required to compress it, and after, it's oscillating at first (kinetic energy), and comes to rest after a while (when the kinetic energy of the oscillation has been converted to heat), so it shouldn't lose energy (relativistic mass) until it has radiated away a significant part of the heat.

I was assuming no oscillation, that all of the energy of compression went into the KE of the other masses. If there is remaining KE in the form of oscillation then eventually that will convert to heat and radiated away and the mass of the cool uncompressed non-oscillating spring will be less than the mass of the cool uncompressed oscillating spring which was less than the mass of the cool compressed non-oscillating spring. (Assuming I didn't make another english error).

 

[Dale]

I have had another look at rapidity and it appears that rapidity (r) can be expressed as the instantaneous terminal velocity that is attained by a particle with constant proper acceleration (a) after a proper time interval (tau) relative to an inertial frame that was comoving with the particle at the time tau=0.

Oh, that is very interesting. So rapidity is the speed measured by an inertial guidance unit.

The more I think about that the more I like it.

I wonder if that works for arbitrary accelerations, or only 1 dimensional ones

 

[granpa]

here is another article on rapidity. I can't make much sense out of it

http://www.geocities.com/Area51/Nebula/3735/sr.html

on page 2 it reads:

Let's write comprehensible expressions for different types of acceleration.
dv/dt=d2r/dt2.
Let's note in advance, that dv/dt = 1·dv/dt = g0dv/dt.
dv/dt=d(dr/dt)/dt = gd2r/dt2.
dv/dt = g1dv/dt.

db/dt = d(dr/dt)/dt = g3v(vdv/dt)/c2 + gdv/dt.
If v || dv/dt, then db/dt = g3dv/dt.
If v is perpendicular to dv/dt, then db/dt = gdv/dt.

db/dt = d(dr/dt)/dt = g4v(vdv/dt)/c2 + g2dv/dt.
If v || dv/dt, then db/dt = g4dv/dt.
If v is perpendicular to dv/dt, then db/dt = g2dv/dt.

dr/dt = (c·arth(v/c))' = g2dv/dt.
dr/dt = g3dv/dt,
Comparing the upper expressions, we can conclude that dr/dt = db/dt, at least if v || dv/dt. (I think this is an error. it should read dr/dtau=db/dt)r is of course rapidity
b is evidently proper velocityhere is another article:
http://Newton.umsl.edu/philf//a1dwuzzl.html

at the bottom it reads

alpha(which is evidently acceleration felt by an accelerated object)=c^2delta gamma/delta x=delta proper velocity/delta t=cdelta rapidity/delta tau

 

[Dale]

By the way granpa, if "proper velocity" really is the velocity given by an inertial guidance system (as suggested by kev above), then I can accept the term "proper" better. I didn't like the word "proper velocity" previously because I would have always assumed that would be 0, but now there is a reasonable way that someone could measure their own velocity without reference to an external measurement. That is starting to sound more "proper" to me.

 

[granpa]

no that's not proper velocity. I got that wrong. I've been trying to make that clear.

proper acceleration would presumably be the acceleration felt by an accelerating object. according to what I've been able to dig up on google the integral of proper acceleration with respect to coordinate time is proper velocity. the intergral with respect to tau is rapidity. (but I'm not even sure about that)

proper acceleration is also the derivative of gamma with respect to distance(?).

http://Newton.umsl.edu/philf//a1dwuzzl.html

 

[granpa]

but that can't be right either.
if delta proper velocity/delta t=cdelta rapidity/delta tau then I get v=r.

heres an interesting link:
http://physics.nmt.edu/~raymond/classes/ph13xbook/node59.html

and especially here:
http://Newton.umsl.edu/run//traveler.html

http://Newton.umsl.edu/run//trveqn05.gif
http://Newton.umsl.edu/run//trveqn06.gif

 

[granpa]

according to this:
http://Newton.umsl.edu/run//trveqn05.gif

the invariant acceleration is regular acceleration * gamma^3

time dilation accounts for one gamma
length contraction accounts for a second gamma.
I suppose that loss of simultaneity must account for the third gammaaccording to JesseM:
if two clocks are a distance L apart in their rest frame, and in your frame they are moving at some speed v in a direction parallel to the axis between them, and the clocks are synchronized in their own rest frame, then in your frame they will always be out-of-sync by vL/c^2, with the leading clock showing a time that's behind the trailing clock.

v must be proper velocity.

 

[granpa]

http://Newton.umsl.edu/run//traveler.html

In this section, we cover less familiar territory, namely the equations of relativistic acceleration. Forces if defined simply as rates of momentum change in special relativity have no frame-invariant formulation, and hence Newton's 2nd Law retains it's elegance only if written in coordinate-independent 4-vector form. It is less commonly taught, however, that a frame-invariant 3-vector acceleration can be defined (again also in context of a single inertial frame). We show that, in terms of proper velocity and proper time, this acceleration has three simple integrals when held constant. Moreover, it bears a familiar relationship to the special frame-independent rate of momentum change felt by an accelerated traveler.

`proper acceleration'' for a given object, which is the same to all inertial observers:
http://Newton.umsl.edu/run//trveqn05.gif
http://Newton.umsl.edu/run//trveqn06.gif
Note that both vperp and the ``transverse time-speed'' gammaperp are constants, and hence both proper velocity, and longitudinal momentum p|| = mw||, change at a uniform rate (with respect to coordinate time) when proper acceleration is held constant. If motion is only in the direction of acceleration, gammaperp is 1, and Delta p/Delta t = m alpha in the classical tradition

http://Newton.umsl.edu/run//trveqn04.gif
In classical kinematics, the rate at which traveler energy E increases with time is not frame-independent, but the rate at which momentum p increases is invariant. In special relativity, these rates (when figured with respect to proper time) relate to each other as time and space components, respectively, of the acceleration 4-vector (four-acceleration is the change in four-velocity over the particle's proper time). Both are frame-dependent at high speed. However, we can define proper force separately as the force felt by an accelerated object. We show in the Appendix that this is simply Fo = m alpha. That is, all accelerated objects feel a frame-invariant 3-vector force Fo in the direction of their acceleration. The magnitude of this force can be calculated from any inertial frame, by multiplying the rate of momentum change in the acceleration direction (with respect to coordinate time) times gammaperp, or by multiplying mass times the proper acceleration alpha. The classical relation F = dp/dt = mdv/dt = md^2x/dt^2 = ma then becomes:

http://Newton.umsl.edu/run//trveqn21.gif

Even though the rate of momentum change joins the rate of energy change in becoming frame-dependent at high speed, Newton's 2nd Law for 3-vectors thus retains a frame-invariant form.
Although they depend on the observer's inertial frame, it is instructive to write out the rates of momentum and energy change in terms of the proper force magnitude Fo. The classical equation relating rates of momentum change to force is dp/dt = F = ma i||, where i|| is the unit vector in the direction of acceleration. This becomes

http://Newton.umsl.edu/run//trveqn08.gif

 

[yuiop]

according to this:
http://Newton.umsl.edu/run//trveqn05.gif

the invariant acceleration is regular acceleration * gamma^3

time dilation accounts for one gamma
length contraction accounts for a second gamma.
I suppose that loss of simultaneity must account for the third gamma
...

Acceleration is distance/(time^2)

distance' = distance*gamma

time' = time/gamma

distance'/(time'^2) = (distance*gamma)/(time^2/gamma^2) = distance/(time)^2 *gamma^3


That is where gamma^3 comes from.

 

[yuiop]

Oh, that is very interesting. So rapidity is the speed measured by an inertial guidance unit.

The more I think about that the more I like it.

I wonder if that works for arbitrary accelerations, or only 1 dimensional ones

By the way granpa, if "proper velocity" really is the velocity given by an inertial guidance system (as suggested by kev above), then I can accept the term "proper" better. I didn't like the word "proper velocity" previously because I would have always assumed that would be 0, but now there is a reasonable way that someone could measure their own velocity without reference to an external measurement. That is starting to sound more "proper" to me.

Hi Dalespam, the inertial guidance system velocity (I like that description by the way ;) is the rapidity as you stated in post #28 but you appear to have switched to calling it the "proper velocity" in your last post.

I agree with you that proper velocity appears to be misnamed in the texts and references, but we appear to be stuck with that situation and in the end it is just about semantics. The non intuitive definitions caused me to make some erronous statements earlier in this thread and this post is an attempt to try and clear things up.

Rapidity is defined as atanh(v) and is additive. It is also the velocity that would be calculated by an onboard accelerometer and onboard clock in an inertial guidance system and as Dalespam has pointed out, can be determined without any reference to any coordinate system. wow!

Proper velocity could be described as the spatial component of the four velocity. That is a description given by granpa earlier except he stated proper velocity is the "magnitude of the spatial component of the four velocity." As far as I know, proper velocity is still a vector quantity and has three vectors. (I might be wrong on this aspect, I am not quite sure - clarification welcome ;)

...

http://en.wikipedia.org/wiki/Proper_velocity

Proper-velocity, the distance traveled per unit time elapsed on the clocks of a traveling object, equals coordinate velocity at low speeds. At any speed it equals momentum per unit mass, and it therefore has no upper limit. It is one of three related derivatives in special relativity (coordinate velocity v=dx/dt, proper-velocity w=dx/dτ, and Lorentz factor γ=dt/dτ) that describe an object's rate of travel. Each of these is also simply related to a traveling object's hyperbolic velocity angle or rapidity η.

THe Wikipedia description above seems to be merging the descriptions of proper velocity and rapidity. The part that describes proper velocity as the momentum per unit mass seems to be the common usage in most texts.

Momentum (p) in special relativity is defined as

p = (mv)*gamma

Old texts used to interpret that equation as

p = (m*gamma) * (v) = relativistic mass * velocity.

That interpretation is discouraged in modern texts and the "correct" interpretation is:

p= (m) * (v*gamma) = rest mass * proper velocity.

Proper velocity (V) is the "dilated velocity" where:

$$V = \frac{v}{\sqrt{1-v^2/c^2} }$$

According to the Wikipedia article, the proper velocity is the 3-vector you get if you take the four-velocity $dx^\mu/d\tau$ and throw away the 0 component.
...

The description above by Fredrik of proper velocity is also pretty good.

Unlike rapidities, proper velocities are not additive in the normal way.

To find the sum (W) of two proper veolcities (v1 and v2) use

W = (v1+v2)*(gamma1*gamma2)

Ref: http://Newton.umsl.edu/run//traveler.html

 

[Dale]

Hi Dalespam, the inertial guidance system velocity (I like that description by the way ;) is the rapidity as you stated in post #28 but you appear to have switched to calling it the "proper velocity" in your last post.
...
THe Wikipedia description above seems to be merging the descriptions of proper velocity and rapidity.

That is probably the source of my confusion. I guess I will have to go back to disliking "proper velocity" then.

 

[granpa]

heres how I think of proper velocity. remember that all velocities are relative to some other object. if we use the Earth to measure our velocity against then imagine that the universe is filled with clocks which are stationary from Earth's point of view and all the clocks are synchronized with Earth time. the clocks act as mile markers for the rocket to measure the distance it has traveled. divide the distance by the proper time of the rocket to get proper velocity.

actually they don't have to be clocks. I'm just used to thinking that way.

 

[Dale]

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.

 

[Jonathan Scott]

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.

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.

 

[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

$$\theta = arctan (0.5) \approx 26.6 ^ {\circ}$$

but the rapidity between us is

$$\phi = arctanh (0.5) \approx 31.5 ^ {\circ}$$.

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

$$\theta = arctan (0.5) \approx 26.6 ^ {\circ}$$

but the rapidity between us is

$$\phi = arctanh (0.5) \approx 31.5 ^ {\circ}$$.

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

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

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

$$\left(\begin{array}{cc} \cosh\phi & \sinh\phi \\ \sinh\phi & \cosh\phi \\ \end{array}\right)$$​

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

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

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.

 

[thljcl]

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