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I Acceleration without force is possible in relativity

  1. Jan 27, 2016 #1
    In prime frame, if Fz =0 & ratio Fx/Fy is equal to ( v/c2 . Uy)/(1-V .Ux/c2) then after transformation in S’ frame F’x becomes F’x = 0 because
    F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2) ----transformation equation
    Let, consider one situation
    In frame S :- Now, just consider that on magnetic substance on frictionless platform magnetic forces are acting in X-direction & in Y-direction. Magnetic force Fx is so adjusted by software program that ratio Fx/Fy is always equal to ( v/c2 . Uy)/(1-V .Ux/c2).

    Then, Forces Fx (very small) & Fy in this frame will create acceleration ax & ay in direction x & y.

    Observer frame S’ is moving with velocity V with relative to frame S then in frame S’ :-

    There is acceleration in X' direction because ax’= ax/{r3. (1-ux. v/c2) 3 } where r =1/(1-v2/c2) 0.5 but there is no force in X'- direction because

    as F’x = Fx – ( v/c2 . Fy. Uy)/(1-V .Ux/c2) & as Fx/Fy=( v/c2 . Uy)/(1-V .Ux/c2)

    So, F’x =0

    Means, in this case in frame S’
    there is acceleration in X’-direction but no force is present in X’-direction.
    In relativity, is this possible that in some frame, there is acceleration but no force in X-direction.
     
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  3. Jan 27, 2016 #2

    Orodruin

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    I did not have time to read your entire setup, but yes, it is possible to have acceleration but no force in the x-direction. The force is given by the change in momentum, not the change in velocity. The momentum in the x-direction is given by ##p_x = m\gamma v_x##, where ##\gamma = 1/\sqrt{1-v^2}## generally depends on all the velocity components. It is therefore possible to change ##v_x## and keep the same momentum in the x-direction if you change the other velocity components so that ##\gamma## changes accordingly to compensate.
     
  4. Jan 27, 2016 #3

    Ibix

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    Acceleration is not necessarily parallel to the force producing it in relativity, no. This follows from the definition of force as $$\vec{F}=\frac{d\vec{p}}{dt}$$where ##\vec{p}=\gamma m\vec{v}##.
     
  5. Jan 27, 2016 #4
    Thanks, interesting
     
  6. Jan 27, 2016 #5

    Dale

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    This is one reason to use four-vectors instead of three-vectors.
     
  7. Jan 27, 2016 #6

    Orodruin

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    4-acceleration is also not necessarily proportional to 4-force. It only holds of the mass of the object is constant.
     
  8. Jan 27, 2016 #7
     
    Last edited by a moderator: Apr 18, 2017
  9. Jan 27, 2016 #8

    Orodruin

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    This is not the general definition, it is only true when the mass ##m## is constant. The general expression is ##F = dP/d\tau = mA + \dot m V##.
     
  10. Jan 27, 2016 #9
    Oh, I didn't realize that you were talking about objects which are gaining/losing invariant mass. (You'd only mentioned mass in your post.) But even in that case, the equation is identical to its non-relativistic version with just the 3-vectors replaced by 4-vectors, and this corroborates Dale's point.
     
  11. Jan 27, 2016 #10
    This is very interesting, but there are things that still are not clear to me. At which point would the acceleration no longer affect the object(or, more importantly, is the object gaining speed)? I've figured that momentum had to do with acceleration without force, but in what way?
     
  12. Jan 27, 2016 #11

    Ibix

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    The object can always consider itself to be instantaneously at rest in some inertial frame, so it can always accelerate. The velocity will be asymptotic to c.

    You'll never have an acceleration without a force. However, you may have an acceleration which has components perpendicular to the impressed force.
     
  13. Jan 27, 2016 #12
    Is there any possible way you can apply momentum in a way that you essentially "fall" forward(or the direction equal to forward)?
     
  14. Jan 27, 2016 #13

    Ibix

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    I have no idea what you are asking. What forces are you thinking about on what objects?
     
  15. Jan 27, 2016 #14
    In the event that the force(propulsion) acting upon the object(any object that has descent mass and density) no longer acts on the object, is it possible that momentum can be used to allow the object to "fall" forward? The object is in space, therefor the only forces would be inertia and the propulsion.
     
  16. Jan 27, 2016 #15

    Ibix

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    Inertia isn't a force, not in Newtonian nor relativistic physics. If there are no forces on an object (e.g. a rocket in space with the engines off far from any mass) then it will continue in a straight line at constant speed. This is true in Newtonian and relativistic physics.
     
  17. Jan 28, 2016 #16
    I am happy, many intelligent understand that
    When we transform forces for Fx =0 , F'x is not zero but when acceleration is transformed then for ax = 0 , a'x =0 (By transformation equation)
    this create above situation after transformation that there may be acceleration in X-direction but no force in X-direction.
    When I read some initial post, I think problem is solved, but now I think problem is not solved.

    In SR, in any frame:-
    Fx = d/dt { r. mo . Ux } where r = (1- U2/C2)-1/2
    Now, I differentiate above
    Fx = mo. r . dUx/dt + mo. Ux . dr/dt
    Fx = mo. r. ax + mo. Ux . r3. (U/C2) . a
    This equation clearly shows that
    i.e. if 'ax' is not zero then 'Fx' can not be zero.
    Mean' if there is acceleration in X-direction then there is Fx in X-direction.

    Is there any alternative mathematics is available which proves that even ax is present, Fx may not present in SR.
     
  18. Jan 28, 2016 #17

    A.T.

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    Please use LaTeX, or nobody will read your math. What did you assume about the change of U when Ux changes?
     
  19. Jan 28, 2016 #18

    Ibix

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    Hint: re-read Orodruin's reply.

    There is a guide to how to use LaTeX linked immediately below the box where you type your reply. Please read it and use it. Then we will not have to struggle to understand straightforward maths. You can write your final expression (quote my post to see how I did it) as $$F_x=\gamma m_0 a_x+\gamma^3 m_0u_x\frac {u}{c^2}a$$which is a lot clearer, and may help you to spot your error.
     
    Last edited: Jan 28, 2016
  20. Jan 30, 2016 #19
    This is interesting. (for every acceleration in x-direction, force Fx exist or not.)
    Point 1:- Orodruin is true.
    If we use transformation equation in relativity, then for every set up, there is one value of Fx for which F'x =0
    (as given in post 1)
    but for every acceleration in prime frame there is acceleration in non-prime frame.
    This happen due to nature of transformation equations in relativity for force & acceleration.
    This can create following situation in non-prime frame,
    there is acceleration in X-direction but no force in X-direction --------(1)

    Point 2:- Now, I do calculation in same frame only,

    In any frame, for getting force in X-direction, I have to differentiate Px by time t.
    ( I have not assumed anything. I am just plainly differentiating)
    [itex] p_x = m_0\gamma u_x [/itex]
    after differentiation. I find following equation
    [itex]F_x=\gamma m_0 a_x+\gamma^3 m_0u_x\frac {u}{c^2}a [/itex]
    Means, in any situation
    If there is acceleration [itex]a_x [/itex] then there must be Fx because 1St portion [itex]\gamma m_0 a_x [/itex] can not be zero.
    At every case, if there is acceleration in X-direction then there is force in X-direction---------(2)


    Now, (1) & (2) out puts are completely opposite. What is wrong.
     
  21. Jan 30, 2016 #20

    Ibix

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    Try writing ##u=\sqrt {u_x^2+u_y^2+u_z^2}## and doing the differentiation again.
     
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