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Minkowski Force

  1. Dec 12, 2013 #1
    I am studying relativity by myself. There is one problem in the book which says that the 4-dot product of the Minkowski force and proper velocity is zero. But again it say that qE.u = change in energy over time. Is there a contradiction? If not, Am I missing something important.

    here q is the charge
    E is the field
    and u is a velocity vector.

    Thanks
     
  2. jcsd
  3. Dec 12, 2013 #2

    WannabeNewton

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    There's no contradiction there whatsoever if by "time" you mean the proper time ##\tau## along the worldline of the interacting charged particle.

    In flat space-time the equation of motion for a charged particle with 4-velocity ##u^{\mu}## and 4-momentum ##p^{\mu}## interacting with an electromagnetic field ##F_{\mu\nu}## is given by ##u^{\nu}\partial_{\nu}p^{\mu} = \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} = qF^{\mu}{}{}_{\nu}u^{\nu}##. A background observer with 4-velocity ##\xi^{\mu}## sees an electric field given by ##E^{\mu} = F^{\mu}{}{}_{\nu}\xi^{\nu}##. Boosting to the rest frame ##(t,\vec{x})## of this background observer, we have ##E^{\mu} = F^{t \mu} ## hence ##\frac{\mathrm{d} E}{\mathrm{d} \tau} = qF^{t \mu}u_{\mu} = qE^{\mu}u_{\mu}## as desired, where ##E = p^t## is the energy of the charged particle as measured by this background observer.

    This bears no relation to the fact that ##a^{\mu}u_{\mu} = u^{\nu}u_{\mu}\partial_{\nu}u^{\mu} = \frac{1}{2}u^{\nu}\partial_{\nu}(u^{\mu}u_{\mu}) = 0##.
     
  4. Dec 12, 2013 #3

    PeterDonis

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    By "proper velocity", I think the book means what is more often called 4-velocity:

    http://en.wikipedia.org/wiki/Four-velocity

    Since 4-velocity is a unit vector (in 4-dimensional spacetime), its length can't change; the only thing that can change is its direction. So any force (more precisely, any 4-acceleration) that changes the 4-velocity must be perpendicular to it (in 4-dimensional spacetime), since otherwise the length of the 4-velocity would change, not just its direction.

    Changing the direction (in spacetime) of the 4-velocity corresponds to changing the ordinary 3-velocity of the object relative to a fixed reference frame, which changes the object's energy in that frame. So applying a force can indeed change the object's energy.
     
  5. Dec 13, 2013 #4

    vanhees71

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    Perhaps, it's good to give a brief derivation of this in principle not too difficult concept of covariant equations of motion for classical point particles. It's often screwed up in textbooks, because it's mixed with non-covariant descriptions (which in practice can have their value too, of course, but when deriving equations of motion, a manifest covariant description is always of advantage!).

    The idea is to formulate the equation of motion of a point particle at presence of external fields (e.g., an electromagnetic field, created by some charge-current distributions) in terms of four vectors. The equation of motion also must use time derivatives, but time is not a scalar but the temporal component of the space-time-position four-vector, [itex]x^{\mu}[/itex]. Thus the first step is to think about a scalar paramater that can be used instead of time.

    Now it is good to use parameters that are somehow "natural" for the given problem. In our case, we always have a preferred frame of reference, which is the instantaneous rest frame of the particle, which is defined as that inertial frame, where the particle is momentarily at rest, i.e., at an instant of time you Lorentz boost to the momentary rest frame of the particle. Then you can use the coordinate time of the observer in this reference frame, the "proper time", [itex]\tau[/itex].

    It's of course cumbersome to formulate an equation of motion in these instantaneous rest frames along the particle's trajectory. Thus, it's much better to stay in the fixed (inertial) "lab frame" and look for a covariant definition of the proper time. To that end let [itex]x^{\mu}(\tau)[/itex] the world line of the particle in Minkowski space and consider a small proper-time increment, during which [itex]x^{\mu}[/itex] changes by [itex]\mathrm{d} x^{\mu}[/itex]. Then the only scalar we can build from this four-vector increment is
    [tex]\mathrm{d} \tau^2 = \mathrm{d} x_{\mu} \mathrm{d} x^{\mu},[/tex]
    where I use units with [itex]c=1[/itex] and the west-coast metric [itex]\eta_{\mu \nu}=\mathrm{diag}(1,-1,-,1-1)[/itex].

    Obviously, for the instantaneous rest frame of the, [itex]\mathrm{d} \tau[/itex] is the increment of the observer's time in this frame. Here, we have tacitly assumed that [itex]\mathrm{d} x^{\mu}[/itex] is always time like, i.e., [itex]\mathrm{d} x^{\mu} \mathrm{d} x_{\mu}>0[/itex]. This is no constraint as long as we do not consider massless particles or, even worse, tachyons. Here, we'll restrict ourselves to the case of massive particles.

    The next idea is to find a covariant definition of momentum. Using [itex]m[/itex] as the invariant mass (rest mass) of the particle, we can define the four-momentum,
    [tex]p^{\mu}=m\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},[/tex]
    which obviously is a four-vector. It turns out that the temporal component is the kinetic energy of the particle (including its "rest energy", [itex]E_0=m[/itex]).

    Then the covariant equation of motion reads
    [tex]\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu},[/tex]
    where [itex]K^{\mu}[/itex] is obviously a four vector, called the Minkowski Force.

    At the first glance this seem to be four equations of motion, instead of three as for Newtonian dynamics. This is, however, not true, because there is a "hidden" constraint. Using the definition of proper time, we find
    [tex]p_{\mu} p^{\mu}=m^2=\text{const}[/tex]
    as a constraint. This is the "on-shell" condition for classical particles, and it must be identically fulfilled along the trajectory (world line) of the particle. Taking the proper-time derivative of this constraint gives
    [tex]p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=p_{\mu} K^{\mu}=0.[/tex]
    Since [itex]p^{\mu}[/itex] is always time like, [itex]K^{\mu}[/itex] must be space like, and only three of the four components are independent, and the Minkowski force must be given such that this constraint is always fulfilled. Thus, in relativistic equations of motion, usually the force depends not only on the momentary position of the particle but also on its momentum.

    An example is the electromagnetic force on a particle in a given external electromagnetic field of the particle. In this case the force reads
    [tex]K^{\mu}=\frac{q}{m} F^{\mu \nu} p_{\mu} = q F^{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.[/tex]
    The "on-shell" constraint for the Minkowski Force is fulfilled in this case, because [itex]F_{\mu \nu}[/itex] is an antisymmetric tensor and thus [itex]K_{\mu} p^{\mu}=F_{\mu \nu} p^{\mu} p^{\nu}=0[/itex].

    Writing this out in terms of the non-covariant conventional field components [itex](\vec{E},\vec{B})[/itex], you'll see that the spatial components give the Lorentz force with the proper velocity instead of the usual one:
    [tex]\vec{K}=q \left (\frac{\mathrm{d}t}{\mathrm{d} \tau} \vec{E}+\frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \times \vec{B} \right).[/tex]
    Now you can rewrite the spatial part of the covariant equation, very easily into the non-covariant three-dimensional form by using
    [tex]\mathrm{d} \tau=\sqrt{1-\vec{v}^2} \mathrm{\mathrm{d} t},[/tex]
    where
    [tex]\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t},[/tex]
    which are NOT the spatial components of a four-vector, because we take the derivative wrt. coordinate time instead of proper time. However, then we can write
    [tex]\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\frac{\mathrm{d} \tau}{\mathrm{d} t} \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau}=\sqrt{1-\vec{v}^2} \vec{K}.[/tex]
    Thus, the non-covariant conventional force is
    [tex]\vec{F}=\sqrt{1-\vec{v}^2} \vec{K},[/tex]
    and for the electromagnetic force you get the usual Lorentz-force expression in terms of the usual three-velocity:
    [tex]\vec{F}=q (\vec{E}+\vec{v} \times \vec{B}).[/tex]

    One must, of course, keep in mind that the relation of momentum with the three-velocity is
    [tex]\vec{p}=m \frac{\mathrm{d} \vec{x}}{\mathrm{d}\tau} = m \frac{\vec{v}}{\sqrt{1-\vec{v}^2}}.[/tex]
     
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