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Acceleration measured in another inertial frame?

  1. Sep 25, 2011 #1
    Suppose, in the first case, that an object (initially at rest in frame S) accelerates, for whatever reason upward, with a force F. Let M be the relativistic mass of the object.
    The force measured in frame S is relativistically given by:

    F_s=γ^2 M〖a_par〗^S+M〖a_ort〗^S

    where 〖a_par〗^S is the accelration that is parrelel to the velocity and 〖a_ort〗^S is the accelration that is orthogonal to the velocity. IN this case, since the intial velocity is 0, the γ=0.

    F_s= M(〖a_par〗^S +〖a_ort〗^S)
    F_s= M (a^S)

    Let frame S' exists as measured by an obesrver that travels say a velocity v downard.
    The force measured in this frame must obey the same law, except that fact that all these measurements are in the S' reference frame.

    F_s'=γ'^2 M〖a_par〗^S'+M〖a_ort〗^S'

    the intial velocity is no longer 0 as this is in a nother frame of referce where the intial velocity is v. the force direction shouldnt be affected so it still points upward and the velocity measured in frame s' is upward. thus a_ort=0

    F_s'=γ^2 M(a^S')

    i can seperately proove that the realtivsitc mass does not vary in seeperate inertial frames of reference. taking the above statement as a matter of fact, even if it may not be true, just how can i figure out what (a^S') is in terms of (a^S) if the inertial motion is parrelel to the direction of accelration. i have an idea to take out such a calcluation and that is the follwoing


    the observer in frame s measures the final velocity u^S after a time say t'. his meaurement of time is not the proper time since the locations at which he measures the velocities are different. the observer in frame S' should at that moment when the velocity of the particle is u^S meaure the velocity given by the realtivsitic addition of this vleocity as his realtive motion.

    u^S'= (u^S + v)/ (1+(u^S*v)/c^2)

    the time it took to reach that velocity in frame S' is not that same as in frame S. Notince how observer in frame s' meausres the proper time of the event as the location at which he examines the vlekocity occur at the same point. so frame s' meaures a time t.

    the final velocity in frame S is beacuase of a_s
    the final velocity in frame S' is because of a_s'

    i cannot take out the mathematics of it.
    can some1 tell me the solution

    i think (if my work is coorect) i whould be getting something of this sort:

    a_s'=a_s/ (1-v^2/c^2)

    for the parrel case.


    if the frame S' travels say to the right realtive to frame S (or the relative velocity is perp to the aaccelration, then i also want to know what the accelration across rframes is given the above formulations.)

    thanks for some1 just tellin gme. i really need it
     
  2. jcsd
  3. Sep 25, 2011 #2

    pervect

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    Would you be interested in an attempt to work out the problem with four-vectors? The current notation and set-up seems rather cumbersome.
     
  4. Sep 25, 2011 #3
    i understand it must be very cumber some :(. and so i dont mind. But just post the answer. obviousl ythis is not a very science thing to aks but in the matter of umergency, i need the solution immideatley. ty
     
  5. Sep 25, 2011 #4

    pervect

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    There's a wiki write-up at http://en.wikipedia.org/w/index.php?title=Four-acceleration&oldid=445796931

    If you have an object with a 3-acceleration of a in some frame S, with components a_x, a_y, a_z, the 4-acceleration in frame S will be

    (0,a_x,a_y,a_z)

    The magnitude of the 4-vector hence gives the magnitude of the proper acceleration.

    The 4-acceleration in an arbitrary frame will be given by a Lorentz boost of the 4-acceleration in the rest frame. If we boost with a velocity [itex]\beta = v/c[/itex] in the z direction,


    [itex](-\beta\,\gamma a_z / c, a_x, a_y,\gamma\,a_z)[/itex]

    Conceptually, a is dU/dtau, where U is the four-velocity and tau is proper time
    In case you aren't famliar with U, U is the derivative of the 4-position (t,x,y,z) with respect to proper time tau, i.e. (dt/dtau, dx/dtau, dy/dtau/, dz/dtau)

    The wiki article has an expression (which I haven't checked) that writes the 4-acceleration in terms of the 3-acceleration a and the 3-velocity u

    I wrote my post to follow the wiki conventions, though you'll usually see a lower-case u used to write the 4-velocity.

    I haven't double-checked wiki's writeup, I will do a bit more when I get some time.
     
  6. Sep 26, 2011 #5
    thank you so much! iam not familiar with this notation but ill learn!

    sorry this must be such a stupid question, but can u pls explain teh first component ((0 in the rest frame but −βγa in the other frame). ty in advance

    Edit: i unsretnad the first coordinate refers to its time measurements, and since its steadily chagning its velocty at some rate, its clock steadily gets slower at some rate. but ok im not ready, i havent studied relativit beyond the scope of lorentz transformations in complete depth. i will get it eventually. But i have an idea in physics which may al lb wrong together but better to explore, and this piece is needed so for now please i just need the answer. my question will be restated:

    if i have a parrticle and it is begin accelerated upwards as measured by an observer A at rest, then another observer B going v downwards will measure another accelration. What is the accleration B measures wiht respect to the accelration A measures.

    If i have a particle and it is being acclerated upwards as measured by an observer A at rest, then another observer B going to the left v will measure another accleration. What is the acceleration B measures with respect to accleration A measures.
     
    Last edited: Sep 26, 2011
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