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Relativistic Mechanics

  1. Jan 13, 2014 #1
    I want to know how force works in relatvistivc mechanics.I searched it and I know the formula but I didn't know how it works.Can you give some examples about force in Relativistic Mechanics.

    Thanks
     
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  3. Jan 13, 2014 #2

    ZapperZ

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    Let's start with this "formula" that you found. Please write it down and show it to us so that we know exactly what you already know and your starting point.

    Zz.
     
  4. Jan 13, 2014 #3
    Derivation
    Starting from

    [itex] \frac{d(\gamma(\mathbf{v}) \, \mathbf{v})}{dt} = m_0 \left( \frac{d \gamma(\mathbf{v})}{dt} \, \mathbf{v} + \gamma(\mathbf{v}) \frac{d\mathbf{v}}{dt} \right).[/itex]
    Carrying out the derivatives gives

    [itex] \mathbf{F} = \frac{\gamma(\mathbf{v})^3 m_0}{c^2} \left( \mathbf{v} \cdot \mathbf{a} \right) \, \mathbf{v} + \gamma(\mathbf{v}) m_0\, \mathbf{a}.[/itex]
    If the acceleration is separated into the part parallel to the velocity (a∥) and the part perpendicular to it (a⊥), so that:

    [itex] \mathbf{a} = \mathbf{a}_\parallel + \mathbf{a}_\perp\,,\quad \mathbf{v}\cdot\mathbf{a}_\perp = 0 \,,\quad \mathbf{v}\cdot\mathbf{a} = \mathbf{v}\cdot\mathbf{a}_\parallel \,[/itex],
    one gets

    [itex] \mathbf{F} = \frac{\gamma(\mathbf{v})^3 m_0}{c^2} \left( \mathbf{v} \cdot \mathbf{a}_\parallel \right) \, \mathbf{v} + \gamma(\mathbf{v}) m_0\, (\mathbf{a}_\perp + \mathbf{a}_\parallel ) \,[/itex] .
    By construction a∥ and v are parallel, so (v·a∥)v is a vector with magnitude v2a∥ in the direction of v (and hence a∥) which allows the replacement:

    [itex] (\mathbf{v}\cdot\mathbf{a}) \mathbf{v} = v^2 \mathbf{a}_\parallel [/itex]
    then

    [itex]\begin{align}
    \mathbf{F} & = \frac{\gamma(\mathbf{v})^3 m_0 v^{2}}{c^2} \, \mathbf{a}_{\parallel} + \gamma(\mathbf{v}) m_0 \, (\mathbf{a}_{\parallel} + \mathbf{a}_{\perp})\\
    & = \gamma(\mathbf{v})^3 m_0 \left( \frac{v^2}{c^2} + \frac{1}{\gamma(\mathbf{v})^2} \right) \mathbf{a}_{\parallel} + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_{\perp} \\
    & = \gamma(\mathbf{v})^3 m_0 \left( \frac{v^{2}}{c^2} + 1 - \frac{v^{2}}{c^2} \right) \mathbf{a}_{\parallel} + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_{\perp} \\
    & = \gamma(\mathbf{v})^3 m_0 \, \mathbf{a}_{\parallel} + \gamma(\mathbf{v}) m_0 \, \mathbf{a}_{\perp}
    \end{align}\,[/itex]
     
    Last edited by a moderator: Feb 26, 2014
  5. Jan 13, 2014 #4
  6. Jan 13, 2014 #5

    PAllen

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    That's all correct for force as measured in an inertial frame (there is a different and, in some ways, simpler formula for force as measured by an accelerating object; in SR these two forces are not the same, unlike Newtonian mechanics).

    So what, exactly, is your question. You wish it were simpler? I can't help you with that. Is there a specific question about the meaning or use? Then we can help.

    To clarify my first comment: imagine rocket propelled car on frictionless track. Now also imagine that you have a stationary rocket with thrust provided by super intense laser. The formula you quote measures how much thrust (measured, e.g. as reaction force to hold the stationary light rocket in place) the stationary rocket needs to achieve a given result. This will be different from the thrust required by the car's rocket; the latter is also the one that would correlate with acceleration measured by an accelerometer in the rocket (the gee force felt in the car by an occupant).
     
  7. Jan 13, 2014 #6

    Ken G

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    But to be clear, the two won't differ in the instant the rockets first turn on, correct?
     
  8. Jan 13, 2014 #7

    PAllen

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    Correct. The faster the rocket car is going, the more they differ.
     
  9. Jan 13, 2014 #8

    pervect

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    I couldn't make this out at all, so let me try to fix it a bit. I just put tex and /tex tags where they needed to be, the OP might want to quote the source.

     
  10. Jan 13, 2014 #9

    pervect

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    When you use four-velocities, four-accelerations, and four-forces, the laws become much simpler:

    If you have a point mass m, with cartesian coordinates t,x,y,z and ##\tau## representing proper time, and you parameterize all of [t,x,y,z] as functions of ##\tau##, i.e you write ##t(\tau), x(\tau), y(\tau), z(\tau)##

    Then the 4-velocity is a vector with 4 componenets: ##[u^0, u^1, u^2, u^3]##

    where ##u^0 = \frac{dt}{d\tau}##, ##u^1 = \frac{dx}{d\tau}##, ##u^2 = \frac{dy}{d\tau}##, ##u^3 = \frac{dz}{d\tau}##

    if you let ##\chi^0 = t##, ##\chi^1 = x##, ##\chi^2 = y##, ##\chi^3 = z## this can be compactly written as

    [tex]u^\mu = \frac{d\chi^\mu}{d\tau}[/tex]

    Continuing with the compact notation, then

    The 4-acceleration is ##a^\mu = \frac{d}{d\tau} u^\mu = \frac{d^2 \chi^\mu}{d \tau^2}##

    Then the 4-momentum is ##p^\mu = m \, u^\mu##
    the four-force is ##f^\mu = \frac{d p^\mu}{d\tau} = m a^u##

    Knowing the time dilation factor, you can find the components of the three-fource from the four-force as needed, knowing that the 3-force is ##\frac{dp}{dt}## while the 4-fource is ##\frac{dp}{d\tau}##
     
  11. Jan 13, 2014 #10

    PAllen

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    And to relate this to what I said, the 3 force is as measured in the inertial frame. The spatial components of the 4-force are thrust as measured in the accelerating body.
     
  12. Jan 14, 2014 #11
    Ok let me clear I want to say the difference between relative force and clasical force is depend observer speed and theres no another difference between them.
     
  13. Jan 14, 2014 #12
    Because relative force look complicate
     
  14. Jan 14, 2014 #13

    PAllen

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    If I understand what you're asking, what you could say is that Newtonian mechanics is correct to very high precision as long as all the objects concerned are moving slow compared to light in the frame you are using for the analysis. Note that if you have two high speed particles approaching each other, there is no such frame, so there is no way to avoid relativistic dynamics for such case.

    If you think relativistic force is complicated, what do you think about quantum fields? You just have to accept that speed is involved non-linearly when you have high energy particles (but then, again, one is rarely interested in expressions involving speed for such applications; you use formulas involving energy and momentum directly, usually without worrying about derivatives of gamma). Classical mechanics exercises involving springs and torque may be amusing to model in relativistic situation, but they don't exist in our current world. Even computing collision by the highest speed observed asteroid or meteorite requires no relativistic corrections - errors from lack of knowledge dwarf relativistic corrections.
     
  15. Jan 14, 2014 #14
    One typical example is of course when you use the Lorentz force as in accelerators that accelerates particles electromagnetically. Then you can use the formula precisely as written by pervect above:

    ##\frac{q}{m_0}(\mathbf{E}+\mathbf{v}\times\mathbf{B})=\gamma(\mathbf{v})^3 \, \mathbf{a}_{\parallel} + \gamma(\mathbf{v}) \, \mathbf{a}_{\perp}##

    The second typical example is if you want to incorporate gravity in your relativistic mechanics, unfortunately it does not work the same way. Naively replacing ##m## with ##m_0 \,\gamma{(\mathbf{v})}## results in:

    ##-\frac{GM}{r^3}\mathbf{r}=\gamma(\mathbf{v})^2 \, \mathbf{a}_{\parallel} + \, \mathbf{a}_{\perp}##

    This expression will result in only one third of the experimentally found anomalous perihelion shift, for instance. Nasa instead uses another formula that works adequately in determining the acceleration due to gravity (including relativity) in reasonably weak gravitational fields (such as found in the solar system) that looks like:

    ##\mathbf{a}=-\frac{GM}{r^3}(1-\frac{4GM}{rc^2}+\frac{v^2}{c^2})\mathbf{r}+\frac{4GM}{r^3}(\mathbf{r} \cdot \mathbf{v})\frac{v}{c^2}\mathbf{v}##

    http://descanso.jpl.nasa.gov/Monograph/series2/Descanso2_all.pdf [Broken]

    Expression (4-61)

    So if you want to to relativistic mechanics with gravity, stuff get complicated.
     
    Last edited by a moderator: May 6, 2017
  16. Jan 14, 2014 #15

    Bill_K

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    One-sixth, I believe. At least, that's what this paper says:

     
  17. Jan 14, 2014 #16
    It is easy to run (rewriting sligthly)

    ##\mathbf{a}=-\frac{GM}{r^3}{\mathbf{r} \cdot \mathbf{v}}\frac{1}{v}(1-\frac{v^2}{c^2}) + \frac{GM}{r^3}\frac{1}{v^2}\mathbf{r}\times\mathbf{v}\times{\mathbf{v}}##

    in a numerical integrator and see that you get precisely one third, I do not now how to do it analytically. I think the problem with the paper might be in the Lagrangian approach. If you replace the classical energy:

    ##E=m\frac{v^2}{2}- \frac{mgM}{r}##

    with the more relativistically looking:

    ##E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-\frac{mGM}{r}##

    you get the Lagrangian they are using. But the expression for the energy should really look like (I think):

    ##E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}(1-\frac{GM}{rc^2})##

    which is slightly different. The kinetic energy depends upon the potential energy...
     
    Last edited: Jan 14, 2014
  18. Jan 15, 2014 #17

    Bill_K

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    Note that he cites 11 previous references [7-11, 17-22] for this Lagrangian, and the factor of 1/6 that results from it, including an exercise in Goldstein! :wink:

    Of course this leaves open the question of whether the weak field limit of Schwarzschild is correctly represented by a simple 1/r potential. And actually the potential should include some velocity dependence. In isotropic coordinates,

    ds2 = [(1 - M/2r)/(1 + M/2r)]2 dt2 - (1 + M/2r)4 dx·dx ≈ (1 - 2M/r)dt2 - (1 + 2M/r)dx·dx

    As the Lagrangian for a geodesic we may therefore take

    L1 = (ds/dt)2 = (1 - 2M/r) - (1 + 2M/r)v2 = (1 - v2) - (2M/r)(1 + v2)

    or the square root of this expression,

    L2 = L11/2 ≈ (1 - v2)1/2 - (M/r)(1 + v2)/(1 - v2)1/2
     
    Last edited: Jan 15, 2014
  19. Jan 15, 2014 #18
    Note that if you take the classical

    ##\frac{d}{dt}{(m \mathbf{v})}=-\frac{GMm}{r^3} \mathbf{r}##

    and replace ##m## with ##\gamma(\mathbf{v})m##

    on both sides of the equation you end up with:

    ##-\frac{GM}{r^3}\mathbf{r}=\gamma(\mathbf{v})^2 \, \mathbf{a}_{\parallel} + \, \mathbf{a}_{\perp}##

    while if you take the first expression in this post and replace ##m## with ##\gamma(\mathbf{v})m## only on the left side of the equation you will instead end up with:

    ##-\frac{GM}{r^3}\mathbf{r}=\gamma(\mathbf{v})^3 \, \mathbf{a}_{\parallel} + \, \gamma(\mathbf{v})\mathbf{a}_{\perp}##

    The last expression is consistent with the Lagrangian in the paper you are referring to. I think replacing ##m## with ##\gamma(\mathbf{v})m## on both sides of the equation makes most sense, at least it is more in line with the equivalence principle. I have not checked it by I assume that the last expression results in an anomalous perihelion shift of one sixth of what is measured.
     
  20. Jan 15, 2014 #19

    Bill_K

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    I think that trying to derive the geodesic equations by guessing what replacement to make in Newton's law of motion makes no sense at all.

    What you've forgotten is the extra term that is present in the equation of motion that when the potential is velocity dependent. There are two possible ways to handle this. One is to let it modify the force, which will no longer be just - ∂V/∂x. The alternative is to let it modify the momentum, which will no longer be just γm.
     
    Last edited: Jan 15, 2014
  21. Jan 16, 2014 #20
    Okey then. Do note that the paper you have been referring to does just that. Replacing ##m## with ##\gamma{(\mathbf{v})}m## on the left side of ##\frac{d}{dt}(m\mathbf{v})=-\frac{GMm}{r^3}\mathbf{r}## and see where it leads is what the paper is all about.

    Actually I do not no if it is possible, or if it is just beyond me, to construct a Lagrangian that is consistent with using ##\gamma({\mathbf{v}})m## also on the force side of the equation.
     
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