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Force in Special Relativity

  1. Feb 14, 2007 #1
    Hi...

    Please can anyone explain me why we can talk about force in special relativity? I thought that special relativity is valid only in inertial reference frames and the force has to product acceleration. And so the reference frame is not inertial and the principle of relativity doesn't have to be valid. And so the formula F=ma doesn't have to be true in this refernce frame. What's wrong with this thinking? My understanding of inertial frames? Thank you very much for explanation.
     
  2. jcsd
  3. Feb 14, 2007 #2

    robphy

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    Note that we don't have problems talking about force in Newtonian kinematics, which also has a notion of a mechanical principle of relativity and a notion of inertial frames.

    In the modern presentations, special relativity is physics in a non-curved ("flat") spacetime on R^4. In this view, "special relativity is valid only in inertial reference frames" is not strictly correct. Rather, "some methods of special relativity are valid only in inertial reference frames"... for example, one can only perform a global Lorentz boost to transform from one inertial frame to another inertial frame.

    Accelerated frames of reference are akin to using curvilinear coordinates (e.g. polar coordinates) in Euclidean space. Its use may require calculus, in addition to algebra and trigonometry which are often sufficient for many discussions with inertial frames. (Calculation of curvature in these non-rectangular coordinates will still yield zero curvature.) Admittedly, some of these calculus-based methods were developed with General Relativity (where spacetime is not a flat R^4)... but were later seen to be useful in Special Relativity.
     
  4. Feb 14, 2007 #3
    In an inertial frame you can still measure velocities, observe collisions, and talk about conservation of momentum. From there, it isn't difficult to discuss forces (rates of change of momentum). If two bodies interact with some force, the usual idea would be to describe this from a third reference frame (which isn't interacting or accelerating) hence, what difficulty is there?
     
    Last edited: Feb 14, 2007
  5. Feb 14, 2007 #4
    The correct formula in BOTH Newtonian and SR dynamics is [tex]F=\frac {dp}{dt}[/tex]


    In SR , the above becomes:

    [tex]F=m \frac {d (\gamma v)}{dt}[/tex]

    since
    [tex]p=m \gamma v [/tex]

    and

    [tex]\gamma=\frac {1}{\sqrt(1-v^2/c^2)}[/tex]
     
  6. Feb 15, 2007 #5
    So did I understand correctly that we can talk in special relativity about small change in momentum as well as we can talk about classic mechanics only in small velocities? In larger change in momentum we would have to use general relativity right?
     
  7. Feb 15, 2007 #6

    pervect

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    No. You might try re-reading robphy's response, or perhaps try

    http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

    (emphasis mine)

    One can deal with accelerating objects without using accelerating frames. This is as true in classical mechanics as it is with relativity.

    There are other points to be made, but in the interest of keeping things simple I'll only make this one - that one can deal with accelerating objects without using accelerating frames.
     
  8. Feb 15, 2007 #7

    rbj

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    nakurusil, just a note about LaTeX (since this appears to be a repeating pattern with your posts here), you need to group stuff together with { .. } for an operator like \sqrt to apply to the entire contents.

    such as
    [tex]\gamma=\frac {1}{ \sqrt{1-v^2/c^2} }[/tex]
     
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