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Formula for the acceleration felt by an accelerating object?

  1. Sep 23, 2008 #1
    what is the formula for the acceleration felt by an accelerating object? I've spent days googling and cant get a straight answer. some say its the derivative of proper velocity with respect to proper time but others say its the derivative of proper velocity with respect to coordinate time.

    its obviously the instantaneous acceleration of an object as perceived by an second object traveling at contant velocity equal to the velocity of the first object at that instantaneous moment. I've tried drawing a spacetime diagram and solving equations of lines and doing coordinate transformations but my formulas are a mess of 1/v's and vdeltav's and garbage that doesnt simplify to alpha=a*gamma^3 that I read about.
     
  2. jcsd
  3. Sep 23, 2008 #2
    http://en.wikipedia.org/wiki/Four-acceleration
    In special relativity, four-acceleration is a four-vector and is defined as the change in four-velocity over the particle's proper time:
    http://upload.wikimedia.org/math/b/6/d/b6de33bcaad9039d28577b89b0a60ae3.png
    where
    http://upload.wikimedia.org/math/f/d/d/fdd106b836057e1ca3479a6ffb79596d.png

    and γu is the Lorentz factor for the speed u. It should be noted that a dot above a variable indicates a derivative with respect to the time in a given reference frame, not the proper time τ.
    In an instantaneously co-moving inertial reference frame u = 0, γu = 1 and \dot\γu = 0, i.e. in such a reference frame
    http://upload.wikimedia.org/math/7/5/1/7515c263351fe55c99643cb76a9e3acd.png
    Therefore, the four-acceleration is equal to the proper acceleration that a moving particle "feels" moving along a world line.


    http://en.wikipedia.org/wiki/Four-vector
    the four-velocity of an \mathbf{x}(\tau) world line is defined by:
    http://upload.wikimedia.org/math/b/6/b/b6b9bfa43887f036a42076eaa3355fdd.png
    http://upload.wikimedia.org/math/0/5/e/05e21017a2762709240b89ee4314505f.png
     
    Last edited: Sep 23, 2008
  4. Sep 23, 2008 #3

    Fredrik

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    You're right about the definition. The formulas get easier if you use the "rapidity" [itex]\phi[/itex], defined by [itex]v=\tanh \phi[/itex]. Example:

    [tex]a=\frac{dv}{dt}=\frac{d\tau}{dt}\frac{d}{d\tau}\tanh\phi=\frac 1{\gamma^3}\frac{d\phi}{d\tau}=\frac\alpha{\gamma^3}[/tex]
     
  5. Sep 23, 2008 #4
    thanks.

    ok. I did a calculation. not for acceleration but for relative velocity. if a stationary observer measures the coordinate velocity of object A to be v and the coordinate velocity of object B to be v+dv then A will measure the velocity of B to be dv*gamma^2.

    so the derivative of that with respect to proper time should be a*gamma^3. which is what a read earlier. now I have to look up where I read that.
     
  6. Sep 23, 2008 #5
    ok. heres what I did:

    coordinate velocity=v=v(t)=t
    coordinate acceleration=a=dv/dt=1
    gamma=g=(1-v^2)^-1/2
    dg/dt=tg^3
    proper velocity=b=v*g

    db/dt=d(vg)/dt=vdg/dt+gdv/dt=vtg^3+g

    proper acceleration=a*gamma^3=gamma^3

    a*g^3=vtg^3+g
    g^3=g(vtg^2+1)
    g^2=vtg^2+1
    vt=(g^2-1)/g^2

    this appears to be correct. whew.


    using db/dtau one gets
    vt=(g-1)/g^2
    which as obviously wrong.
     
    Last edited: Sep 24, 2008
  7. Sep 24, 2008 #6
    if thats the case then I dont understand this quote:
    ' The proper acceleration 3-vector, combined with a null time-component, yields the object's four-acceleration ' (http://en.wikipedia.org/wiki/Proper_acceleration). below that proper acceleration is clearly and unambiguously defined as rate of change of proper velocity with respect to coordinate time.

    yet, In special relativity, four-acceleration is defined as the change in four-velocity over the particle's proper time.

    the three spacelike components of 4-velocity define a traveling object's proper velocity \gamma \vec{u} = d\vec{x}/d\tau i.e. the rate at which distance is covered in the reference map-frame per unit proper time elapsed on clocks traveling with the object.

    nor do I understand this:
    A=(0,a)
    Therefore, the four-acceleration is equal to the proper acceleration that a moving particle "feels" moving along a world line.
    (http://en.wikipedia.org/wiki/Four-acceleration)
     
    Last edited: Sep 24, 2008
  8. Sep 24, 2008 #7
    heres a simpler solution.

    go to http://calc101.com/webMathematica/derivatives.jsp#topdoit
    enter 'v[t]/sqrt[1-v[t]^2]' into the 'take the derivative of' box.
    enter 't' into the 'with respect to' box
    enter 'y' into the 'and again with respect to' box. (doesnt do anything but you must enter something)
    push the 'do it' button

    the result is a*gamma^3
     
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