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Operational definition of four velocity

  1. Feb 16, 2009 #1
    Dear Friends

    Bearing in mind the special relativity definition of "observer" (a system of rods and clock disposed everywhere in the space), what's the
    operational definition of four velocity (and therefore of four acceleration, four force, etc)?

    Obviously, no problems in detecting the displacement's components of the
    traveling particle (in the observer's reference frame) but how about the proper time tau? I'm mulling about two possibilities:

    (1)
    The observer can measure the time using its clocks and then use
    dt = gamma dtau
    (but I don't like it, because it repose on too theoretical premises and it's no really "experimental fashioned" )

    (2)
    Time can be measured in the respect of the reference frame comoving with the traveling particle, but this is not an inertial one so more dilemmas
    rise up.

    I would love to hear your thoughts about that.

    Thanks for unscrambling my English.
    Have a great day !!

    Barbara Da Vinci
    Rome
     
  2. jcsd
  3. Feb 16, 2009 #2

    DrGreg

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    What you are really asking for is an operational definition of proper time [itex]\tau[/itex]. The proper time of an object is the time recorded by a clock carried by the object. The clock hypothesis postulates that an accelerating clock ticks at the same rate as a co-moving inertial clock. I.e. [itex] dT/d\tau = 1 [/itex] where T is the time recorded by an inertial observer who, at that moment, is at rest relative to the accelerating object. Note that proper time is defined only for the object itself, you can't extend it to events that are not on the object's worldline.

    That's the operational definition. The mathematical definition is

    [tex]d\tau^2 = dt^2 - \frac{dx^2 + dy^2 + dz^2}{c^2}[/tex]​

    where (t, x, y, z) are any inertial coordinates (whether co-moving or not), which is equivalent to your (1).

    Either way, the 4-velocity is then

    [tex]\textbf{V} = (dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau)[/tex]​
     
  4. Feb 16, 2009 #3
    I agree, you got fully the main point. Using a metaphor, you are essentially confirming the clock hypothesis to be a sort of relativistic Euclid's fifth postulate, an assumption no way coming out from the standard hypotheses behind the Lorentz transformations. Do I understand it right ?

    > an inertial observer who, at that moment, is at rest relative to the accelerating object

    Well, it's a widely spread but somewhat paradoxical concept: "at rest" entail an observation over a finite time interval. Thinking the word line approximated by means of a sequence of infinitesimal extended straight lines doesn't sound entirely good to me: measure and integration theory is a subtle and tricky business ...

    Thanks for your reply.
    Best regards!

    Barbara Da Vinci
    Rome
     
  5. Feb 16, 2009 #4
    While having an heavy Google session about the topic we are here talking about, I found
    (http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html)
    what in the following I quote (may be someone is interested ...). Bye.

    "The clock postulate is not meant to be obvious, and it can't be proved. It's not merely some kind of trivial result obtained by writing special relativity using non-cartesian coordinates. Rather, it's a statement about the physical world. But we don't know if it's true; it's just a postulate. For instance, we can't magically verify it by noting that the Lorentz transform is only a function of speed, because the Lorentz transform is something that's built before the clock postulate enters the picture. Also, we cannot simply wave our arms and maintain that an acceleration can be treated as a sequence of constant velocities that each exist only for an infinitesimal time interval, for the simple reason that an accelerating body (away from gravity) feels a force, while a constant-velocity body does not. Although the clock postulate does speak in terms of constant velocities and infinitesimal time intervals, there's no a priori reason why that should be meaningful or correct. It's just a postulate! This is just like the fact that even though a 1000-sided polygon looks pretty much like a circle, a small piece of a circle can't always be treated as an infinitesimal straight line: after all, no matter how small the circular arc is, it will always have the same radius of curvature, whereas a straight line has an infinite radius of curvature. It also won't do to simply define a clock to be a device whose timing is unaffected by its acceleration, because it's not clear what such a device has got to do with the real world: that is, how well it approximates the thing we wear on our wrist.
     
  6. Feb 17, 2009 #5

    DrGreg

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    Perhaps it would have been better for me to say "an inertial frame in which the object's velocity ([itex]d\textbf{x}/dt[/itex]) is zero at that moment".

    Or, to put it another way, the clock hypothesis postulates that whenever, in an inertial frame [itex](t,x,y,z)[/itex],

    [tex]\frac{dx}{dt} = \frac{dy}{dt} = \frac{dz}{dt} = 0[/tex]..........(1)​

    then

    [tex]\frac{dt}{d\tau} = 1[/tex]...........(2)​

    At any event along the object's path you will be able to find an inertial frame satisfying (1) so [itex]d\tau[/itex] can be defined, at that event only, via (2). To calculate [itex]\tau[/itex] along the whole path you have to integrate [itex]d\tau[/itex].

    I'm assuming you understand integration and differentiation in calculus.
     
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