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Equilibrium in Multiple Inertial Frames

  1. Jan 11, 2010 #1
    Two massive lead balls are sprayed with a thin veneer of positive charge. They are isolated and at rest in inertial frame K. The positive charge is such that the electrostatic force of repulsion exactly cancels the gravitational force of attraction. Viewed from frame K’, the Lorentz force on either ball will not equal the electrostatic force of repulsion in K. Yet each ball moves with a constant velocity in K’. Comments?
     
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  3. Jan 11, 2010 #2

    Mentz114

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    I don't believe you. Please, show your working.
     
  4. Jan 11, 2010 #3

    Dale

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    In any other frame there is a magnetic field and a current also.
     
  5. Jan 11, 2010 #4
    LOL. Actually I'm retired and strictly an amateur physicist. If I offended your GRT sensibilities, please accept my apologies. What I had in mind was a magnetic field/magnetic force entity in the domain of mass/mass interactions.
     
  6. Jan 11, 2010 #5

    Mentz114

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    No offence taken, I just think you made a mistake in your calculation. As DaleSpam has said, one would expect the forces to re-balanced by the appearance of a magnetic field.
     
  7. Jan 11, 2010 #6
    Actually, if the charges are at rest on the y-axis of IFR K, and if the repulsive electrostatic forces between them is F, then in K' the total electric and magnetic force on either charge will be F' < F. Since the charges are in equilibrium in K' as they are in K (they move with constant, identical velocities in K'), then presumably the net force on either charge is zero in K', as it is in K. The implication is that the gravitational forces of attraction are also less in K'. Note that in general the orientation of the charges could be anything in K. Each would still presumably have zero total force acting on it in K'.
    A variation on this theme is by having the 2 positive charges held at rest in K by a spring. The question then becomes, what effect does motion have on the spring constant? Here again, the charge orientation in K could be anything, for example they could both lie on the x-axis.
     
  8. Jan 11, 2010 #7

    Mentz114

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    All IFR's will see the same overall force.

    In order to convince me otherwise, you'll have to show a calculation. I'm not going to work this out because I already know the answer.

    You might find this helpful

    http://en.wikipedia.org/wiki/Electromagnetic_tensor

    You're trying to change the subject. Let's deal with your misunderstanding of the original problem.
     
    Last edited: Jan 11, 2010
  9. Jan 11, 2010 #8

    pervect

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    All inertial observers will see the same 4-force, dp/dtau, where tau is proper time. However, the force, dp/dt, where t is the time coordinate, will vary depending on the frame of reference that defintes t. dp/dt = (dp/tau) / (dt/dtau) and time dilation will make dt/dtau change depending on the frame of reference.

    In the weak field limit, one will find that there is a gravitomagnetic force that needs to be included so that the gravitational force transforms correctly. To really do the problem with full GR requires abandoning the idea that gravity is a force however.

    See for instance http://en.wikipedia.org/w/index.php?title=Gravitomagnetism&oldid=337231566 to find out more about gravitomagnetism and the weak-field GEM formalism.

    The full GR treatment would be too confusing , take too much time, and is not really needed at this point.
     
  10. Jan 12, 2010 #9

    Mentz114

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    Pervect has explained the gravitational side. I suppose Newton's gravity is not relativistic. My calculation may be inapprorpriate in curved spacetime but I tried working out the case where the boost is perpendicular to the charge axis. The force vector is calculated for K and K'. The boost is in the x-direction, charge axis in the y-direction.

    [tex]
    \begin{align*}
    F^{mn}\eta_{an}J^a=\left[ \begin{array}{cccc}
    0 & 0 & E_y & 0 \\\
    0 & 0 & 0 & 0 \\\
    -E_y & 0 & 0 & 0 \\\
    0 & 0 & 0 & 0 \end{array} \right]
    \left[ \begin{array}{cccc}
    -1 & 0 & 0 & 0 \\\
    0 & 1 & 0 & 0 \\\
    0 & 0 & 1 & 0 \\\
    0 & 0 & 0 & 1 \end{array} \right]
    \left[ \begin{array}{c}
    -cq \\\
    0 \\\
    0 \\\
    0 \end{array} \right]
    =
    \left[ \begin{array}{c}
    0 \\\
    0 \\\
    -cqE_y \\\
    0 \end{array} \right]
    \end{align*}
    [/tex]

    Now boost the field tensor

    [tex]
    \begin{align*}
    F'^{pq} = \lambda^{p}_{n}\lambda^{q}_{m}F^{mn}=
    \left[ \begin{array}{cccc}
    0 & 0 & \gamma E_y & 0 \\\
    0 & 0 & -\beta\gamma Ey & 0 \\\
    -\gamma E_y & \beta\gamma Ey & 0 & 0 \\\
    0 & 0 & 0 & 0 \end{array} \right]
    \end{align*}
    [/tex]

    and the current,

    [tex]
    \begin{align*}
    J'^n &= \lambda^{n}_{p}J^p
    &=
    cq\left[ \begin{array}{c}
    -\gamma \\\
    \gamma\beta \\\
    0 \\\
    0 \end{array} \right]
    \end{align*}
    [/tex]

    Multiply them to get the force vector

    [tex]
    \begin{align*}
    F'^{mn}\eta_{an}J'^a=
    \left[ \begin{array}{cccc}
    0 & 0 & \gamma E_y & 0 \\\
    0 & 0 & \-beta\gamma Ey & 0 \\\
    -\gamma E_y & \beta\gamma Ey & 0 & 0 \\\
    0 & 0 & 0 & 0 \end{array} \right]
    \left[ \begin{array}{cccc}
    -1 & 0 & 0 & 0 \\\
    0 & 1 & 0 & 0 \\\
    0 & 0 & 1 & 0 \\\
    0 & 0 & 0 & 1 \end{array} \right]
    \left[ \begin{array}{c}
    -\gamma cq \\\
    \gamma\beta cq \\\
    0 \\\
    0 \end{array} \right]
    =
    cq\left[ \begin{array}{c}
    0 \\\
    0 \\\
    -\gamma^2(1-\beta^2)E_y \\\
    0 \end{array} \right]
    \end{align*}
    [/tex]

    So, from this it looks as if the force is invariant. However, I'm well known for getting calculations wrong ...
     
    Last edited: Jan 12, 2010
  11. Jan 12, 2010 #10

    atyy

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  12. Jan 12, 2010 #11

    pervect

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    I didn't look at the calculation closely, but I"m pretty sure you calculated the four-force, which is a geometric invariant.

    I.e. you calculated the rate of change of momentum p with respect to proper time tau, which is a geometric object, and not the rate of change of momentum p with respect to coordinate time t, which is not a geometric object, but gives the traditional "3-force".
     
  13. Jan 12, 2010 #12

    Physics Monkey

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    There are even bizarre gravitational objects called extremal charged black holes which have exactly the property that the gravitational attraction between them is compensated by electrostatic repulsion. Of course, they remain in equilibrium no matter what frame you view them from.

    Edit: I see atyy already provided a link.
     
  14. Jan 12, 2010 #13
    Fair enough. In what follows, let “G” stand for the Greek letter “gamma” (NOT the gravitational constant).
    Let q1 = q2 = q > 0 be at rest (1) at the origin of IRF K, and (2) on the positive y-axis of K respectively. B = 0 everywhere. If E is the field of q1 at q2, then the Lorentz force experienced by q2 is electrostatic; it points toward positive y and has the magnitude
    F = qE.
    Let K’ move in the positive x-direction of K at speed v. Then q1 and q2 both move in the negative x’-direction of K’ at speed v. The Lorentz force in K’ has magnitude
    F’ = q(E’ + vB’),
    where B’ points in the negative z’-direction and vB’ points in the negative y’-direction. qE’ points in the positive y’-direction. The values of E’ and B’ can be obtained from the standard EM field transformations (See Griffiths, 10.119). According to these transformations,
    E’ = GE, (where G = 1 / ((1-v^2/c^2)^(1/2))
    B’ = -GvE / c^2.
    Thus
    F = q (GE – vGvE/c^2)
    = GqE(1-v^2/c^2)
    =qE / G
    =F (1-v^2/c^2)^(1/2)
    In brief, the Lorentz force in K’ still points toward positive y’, but its magnitude is less than the Lorentz force in K.
    Since each particle moves with a common, constant velocity in K’, the net force on each particle must be zero in K’, quite as it is in K. Evidently the gravitational, attractive force in K’ is only 1/G what it is in K. This equilibrium condition must generally hold for all orientations of q1 and q2 in K. It suggests (a) that there is a counterpart to EM’s B field when masses move, and (b) that the total force between masses is specified by a counterpart to the Lorentz force law of EM.
     
  15. Jan 12, 2010 #14

    Dale

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  16. Jan 12, 2010 #15

    Mentz114

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    GRDixon,

    thanks for showing your calculation, which looks correct. I think the difference in our results is that I've done a fully relativistic version with 4-vectors, and yours is in 3-D with the space-contraction grafted in. They have different interpretations. My calculation is dp/dtau, and yours is dp/dt.
     
  17. Jan 12, 2010 #16

    George Jones

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    The paper at atty's link is about magnetically charged objects that have opposite magnetic charges, and that are held in equilibrium by an external magnetic field. In 1966, Bonner considered a similar situation, but with a "strut" instead of an external magnetic field. In 1947, Majumdar and Papapetrou (independently) found equilibrium solutions for an arbitrary number of extremal electrically charged objects that are distributed randomly. The Majumdar-Papapetrou solutions require neither struts nor an external field, and so represent a balance between gravity and electrostatic repulsion.

    All of these, the paper to which atyy links, Bonner's work, the Majumdar-Papapetrou solutions, and other equilibrium solutions, are referenced and discussed briefly in an interesting book that I picked up a few weeks ago, Exact Space-Times in General Relativity by Jerry B, Griffiths and Jury Podolsky. Particularly relevant are subsection 10.8.1 Equilibrium configurations with distinct sources of section 10.8 Axially symmetric electrovacuum space-times and section 22.5 Majumdar-Papapetrou solutions.
     
  18. Jan 12, 2010 #17

    Physics Monkey

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    Thanks George, I glanced too briefly at atyy's link to realize it wasn't talking about what I was talking about. Thanks also for the book mention, I'll see if I can find it.
     
  19. Jan 12, 2010 #18

    Mentz114

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    There's a transcription error in my post #9. The signs on the magnetic bits of F' are transposed. I can't edit that post now.
     
  20. Jan 15, 2010 #19
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