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B Ordinary and proper velocity

  1. Jan 25, 2017 #1
    good morning,

    I read that are defined two type of 'velocities' in Special relativity:

    ordinary v1= dl/dt where l and t are measured in ground-based frame

    proper v2= dl/d(tau), a hybrid quantity where l is the same as ordinary v1 but tau is the proper, rest time

    I can't understand the physical meaning of this last definition.... Is it possible to get an extensive explanation?
    thank you in advance
    mark
     
  2. jcsd
  3. Jan 25, 2017 #2

    Vanadium 50

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    Where did you find this "proper velocity"? A proper velocity is the velocity in one's rest frame, and it's pretty hard to make that anything other than zero.
     
  4. Jan 25, 2017 #3

    Ibix

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    Are you thinking of the coordinate velocity, a three-vector whose components are ##dx/dt## etc, versus the four velocity, a four vector whose components are ##dx^i/d\tau##?
     
  5. Jan 25, 2017 #4

    FactChecker

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    Can you be more specific about what you find confusing? A moving object has a different elapsed time measurement from a stationary observer. Proper velocity of a moving object measured relative to an observer would be the distance traveled measured by a stationary observer divided by the elapsed time measured by the moving object. So the distance traveled and the elapsed time are measured in two different reference frames.

    A discussion of the physical implications of proper velocity can be found in https://en.wikipedia.org/wiki/Proper_velocity
     
  6. Jan 25, 2017 #5

    PAllen

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    The OP is describing celerity, which is sometimes referred to as proper velocity (at least per wikipedia; I have always called it celerity):

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

    The physical description is simply how fast I (in a rocket, say) am traveling per some reference frame's distance measure (different from my own rest frame). The idea is to accept some standard distance measure not related to my own frame (e.g. a solar system frame), but find the rate I am traversing that distance per my own clock. Extending this to GR, and abstracting it, one can talk about travel rate per a reference foliation (abstraction of the key feature of some reference coordinates; we don't need the whole coordinates to be specified; a given foliation plus a reference world line uniquely determine a basis for celerity computations) by my own clock. In this generalization, cosmological recession rate may be considered to be an example of celerity (using the standard foliation, and picking an arbitrary comoving observer as a reference world line). Celerity has no upper bound in either SR or GR.
     
    Last edited: Jan 25, 2017
  7. Jan 25, 2017 #6
    There seem to be three schools of thought about what "proper" should mean in relativity.

    The first is what you say, Vanadium 50: proper [whatever] is a traveler's [whatever] as measured in the traveler's own instantaneous inertial frame. I prefer this convention because it's consistent, although "proper distance" doesn't quite fit (since there we're dealing not with a traveler's rest frame, but rather with the rest frame in which two spatially separated events occur simultaneously). Close enough, though. Under this convention, "proper velocity" isn't a useful term because, as you say, it's just the zero vector.

    The second convention is inconsistent in that it uses the first convention for some quantities (e.g., proper time, proper acceleration, proper [rest] energy), but also uses "proper" for proper-time derivatives of certain vector quantities (notably velocity and force). So "proper velocity" is ##d \vec r / d \tau##, and "proper force" is ##d \vec p / d \tau##. I don't like this convention. Aside from the fact that it's inconsistent, it also claims for itself the term "proper force," which would be a perfectly useful concept under the first convention (unlike proper velocity).

    The third convention, like the second convention, borrows from the first convention (proper time, proper energy, etc.), but it uses "proper velocity" and "proper force" as synonyms for the velocity and force four-vectors. This is arguably more objectionable than the second convention, since we already have terms like "four-velocity" and "four-force" that fit the bill.

    I take it you agree with me that the first convention is best.

    Unfortunately, "proper velocity" as defined in the second convention is extremely popular in the literature, and the alternative term "celerity" (which would be preferable IMO) is quite rare.
     
  8. Jan 25, 2017 #7
  9. Jan 25, 2017 #8
    (By the way, if you're ever making notes for other people to read, the PDF I linked in the above post is a great example of how not to do it.)
     
  10. Jan 25, 2017 #9

    PAllen

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    Note that proper force under the second convention you describe is actually the same as proper force under the first convention. So really, the only outlier is proper velocity, for which I obviously prefer the term celerity.
     
  11. Jan 25, 2017 #10
    Not so.

    Under the second convention, proper force is ##d \vec p / d \tau##.

    Under the first convention, proper force is the force acting on a traveler as measured in the traveler's own instantaneous inertial frame. Namely, it's the traveler's mass times the traveler's proper acceleration:

    ##\vec f_0 = m \vec a_0 = m \gamma^3 \vec a##.
     
  12. Jan 25, 2017 #11
    (And really it's only the magnitude of ##\vec f_0## that's interesting, because it's an invariant.)
     
  13. Jan 25, 2017 #12

    PeroK

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    It seems to me that celerity, if anything else, should be improper velocity.
     
  14. Jan 25, 2017 #13

    DrGreg

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    There's one more reason why I don't like to call ##d\vec{x}/d\tau## proper velocity, and that's because it may tempt some people to think that ##d^2\vec{x}/d\tau^2## is proper acceleration, which it isn't.
     
  15. Jan 25, 2017 #14
    And to correct myself, that last equation should be:

    ##\vec f_0 = m \vec a_0 = \dfrac{m \gamma^3 \vec a}{\gamma_\perp}##,

    where ##\gamma_\perp = \dfrac{1}{\sqrt{1 - \left[ \left( \frac{v}{c} \right) \sin{\theta} \right] ^2}}##, and ##\theta## is the angle between ##\vec v## and ##\vec a##.

    (What I gave before was only for the special case that ##\vec v \parallel \vec a##.)
     
  16. Jan 25, 2017 #15

    PAllen

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    Yes, I was thinking of the first above as a 4-force, in which case its norm is, indeed, the same as proper force by the rest frame definition. However, as 3 vector, as you intended, it is the spatial part of the 4-vector, and its 3 norm has no useful meaning, in the general case.
     
  17. Jan 25, 2017 #16

    PAllen

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    I had never seen this formulation before you posted it in another thread. This time I wanted to derive it. It can be derived (looking for it) pretty easily from the general expression for norm of the 4-acceleration:

    γ2 √(a2 + γ2v2v'2) or

    γ2 √(a2 + γ2(v⋅a)2)
     
    Last edited: Jan 25, 2017
  18. Jan 25, 2017 #17
    Have you left out the time component?

    Setting ##c=1##, four-acceleration is:

    ##\big( \gamma^4 (\vec v \cdot \vec a), \, \gamma^2 \vec a + \gamma^4 (\vec v \cdot \vec a) \vec v \big)##.

    Its norm is then the square root of the inner product:

    ##\sqrt{\Big[ \gamma^4 (\vec v \cdot \vec a) \Big]^2 - \Big[ \gamma^2 \vec a + \gamma^4 (\vec v \cdot \vec a) \vec v \Big]^2}##

    Do some careful and tedious algebra (remembering that ##1 - v^2 = \gamma^{-2}##), and you can reduce that to:

    ##\gamma^3 a i \sqrt{ \dfrac{(\vec v \cdot \vec a)^2}{a^2} + 1 - v^2 }##.

    From there, invoke the cosine definition of the dot product (where ##\theta## is the angle between ##\vec v## and ##\vec a##):

    ##\gamma^3 a i \sqrt{ v^2 \cos^2{\theta} + 1 - v^2 }##.

    Simplify, using the fact that ##(v \sin{\theta})## is the component of ##\vec v## that's perpendicular to ##\vec a##. (And of course the ##i## is just an artifact of my chosen sign convention.)
     
  19. Jan 25, 2017 #18

    PAllen

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    I didn't leave out anything. The expresssions I gave were for the 4 norm itself. It is easy to get from the second to the 3 norm of your 3 vector.
     
  20. Jan 25, 2017 #19
    I don't quite follow, but it looks like you do, so cheers!
     
  21. Jan 25, 2017 #20

    PAllen

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    It is really easy to rearrange my second expression into your next to last expression.
     
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