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Force on high moving object

  1. Mar 14, 2015 #1
    Lets assume the object is traveling in the x-direction.

    Ive been told that any force on the object in the x direction is given by F = maϒ3. And that any force in the y or z direction is given by F = F = maϒ.

    Why is there a difference between the x direction and the y, z direction?
     
  2. jcsd
  3. Mar 14, 2015 #2

    Orodruin

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    Because force is the change in momentum with time. An acceleration ##a## perpendicular to the direction of motion gives a different rate of change in the momentum than the same acceleration in the direction of motion.
     
  4. Mar 14, 2015 #3
    Is a different because of length contraction?
     
  5. Mar 14, 2015 #4

    Orodruin

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    No. It is different because the same change in momentum results in a different change in velocity depending on the direction.
     
  6. Mar 14, 2015 #5
    Why would the same change in momentum results in a different change in velocity?
     
  7. Mar 14, 2015 #6
    If you add some mass to a moving object, you must accelerate the mass to the same speed that the object has. A force is required to do that.

    In the case of accelerating an object to the direction that is perpendicular to its velocity, you are not adding any mass to the object.

    In the case of accelerating an object to the direction that is parallel to its velocity, you are adding mass to the object.



    (Note to physicists: That object there is one half of a system of two objects moving away from each other. The rest mass of that system increases when the objects are pushed in the direction of their motion. As all mass in that system is in motion the extra rest mass must be put to motion by a force)
     
    Last edited: Mar 14, 2015
  8. Mar 14, 2015 #7

    Ibix

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    I don't think jartsa is quite right there. Rest mass is invariant, short of a nuclear reaction.

    It's worth noting that even in Newtonian mechanics, forces parallel and perpendicular to a motion have different effects on the total momentum. If a body of mass m is travelling at velocity vx and then gains momentum mU, the final momentum could be m(vx+U) or [itex]m\sqrt{v_x^2+U^2}[/itex] depending on whether the force is parallel or perpendicular to the original motion.

    In relativity, one has to worry about not exceeding the speed of light. Without actually having worked through the maths, my feeling (by analogy with the above) is that a perpendicular force isn't increasing the object's total velocity as much, so has an easier time of it.
     
  9. Mar 14, 2015 #8

    Orodruin

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    This is a simple matter of taking the relativistic expression for momentum ##m\vec v/\sqrt{1+\vec v^2/c^2}## and differentiating with respect to time. You will notice that you get an expression that depends on ##\vec v \cdot \dot{\vec v} = \vec v \cdot \vec a##. Assuming ##\vec v \cdot \vec a = va## (acceleration parallel to velocity) will give you a different result than assuming ##\vec v \cdot \vec a = 0##. Also, and here is the real "killer", force is not parallel to the acceleration unless applied parallel or orthogonal to the direction of motion.
     
  10. Mar 14, 2015 #9

    PeterDonis

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    Only if the system is not interacting with anything else. In this case, since an external force is being applied to the system, there is an interaction, so you can't assume that the rest mass of the system is constant. Only if you included whatever is applying the force in the total system (so the total system has no external interactions) would the rest mass of the total system be constant--but in that case, the "total system" is more than what jartsa is including in the "system".
     
  11. Mar 14, 2015 #10

    PAllen

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    True, but the simple particle formulas referenced in OP, do assume invariant rest mass. Otherwise, as you know, you would have additional terms.
     
  12. Mar 14, 2015 #11

    Ibix

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    I read you as giving the same exceptions as the Newtonian case. Mass can change in the "bullet lodges in the target" kind of sense. In other words - did I just over-simplify, or did I miss some relativistic subtlety?

    My problem with jartsa's post (which I did not express well) was the second and third paragraphs. Surely if your method of applying a force adds mass (or relativistic mass, if that's a concept jartsa subscries to) to the accelerated object it does so whatever the direction of the force. Or at least, it cannot be assumed not to without some explanation.
     
  13. Mar 14, 2015 #12

    Steering an object does not change its relativistic mass, because steering does not require any energy, just force.

    Is acceleration perpendicular to velocity steering, you may ask. Yes it is.
     
  14. Mar 15, 2015 #13

    Ibix

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    Ah - you are thinking of a force like a uniform magnetic field on a charged particle. I was thinking of something like hitting an object with a bullet travelling perpendicular to the body's motion. I think the latter does change the kinetic energy of the body.
     
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