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Relativistic Effects on the E-field

  1. Jul 15, 2010 #1

    mysearch

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    I was trying to read up on the effects of the Lorentz transforms on electromagnetic fields in the stationary and moving frames and came across the following equation for the E-field:

    [1] [tex] E= \frac {Q}{4 \pi \epsilon_0 r^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta) ^{3/2}} [/tex]

    The first part of the equation appears to be the standard electrostatic expression derived from Coulomb’s law, while the second expression encompasses a relativistic component based on velocity [tex]( \beta = v/c}) [/tex] and the angle [tex]( \theta )[/tex] formed between the position, charge and angle of motion along the x-axis. As such, I am assuming [1] returns the value of (E) as measured in the moving frame. If so, I would have thought you could determine the relative strength of [Ex] and [Ey] to compare against the value predicted by the Lorentz transforms in [3]. As the value of [Ex] aligns to the direction of charge motion, (sin0=0), while [Ey] is perpendicular to the motion (sin90=1). If so, the relativistic component of [1] would reduce to:

    [2] [tex]E_X \Rightarrow 1- \beta ^2; E_Y \Rightarrow \frac {1}{(1- \beta^2) ^{1/2}} [/tex]

    By way of orientation, the following Lorentz transforms define the prime (*) frame being stationary, while in the moving frame, the velocity (v) of the source charge [Q] is again along the x-axis:

    [3] [tex] E_X^* = E_X; E_Y^* = \frac {E_Y - vB_Z}{(1- \beta^2) ^{1/2}}; E_Z^* = \frac {E_Z + vB_Y}{(1- \beta^2)^{1/2}} [/tex]

    Now there appears to be some discrepancy in the implications of [2] and [3]. We know that the moving frame will have an associated magnetic (B) field, which will not exist in the stationary (prime*) frame. However, I am assuming that [3] can be reversed by simply changing the sign of [v], since the moving frame moves with a negative velocity with respect to the stationary frame:

    [4] [tex] E_X = E_X^*; E_Y = \frac {E_Y^* + vB_Z^*}{(1- \beta^2) ^{1/2}}; E_Z = \frac {E_Z^* - vB_Y^*}{(1- \beta^2) ^{1/2}} [/tex]

    However, we also know that there can be no magnetic fields in the stationary (prime*) frame; therefore [4] can presumably reduce to the form:

    [5] [tex] E_X = E_X^*; E_Y = \frac {E_Y^*}{(1- \beta^2) ^{1/2}}; E_Z = \frac {E_Z^*}{(1- \beta^2) ^{1/2}} [/tex]

    As such, I would have thought that the relativistic components in [2] and [5] should be directly comparable. However, this is only true for the E-fields perpendicular to the motion, i.e. Ey & Ez, but not (Ex).

    Any thoughts?
     
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  3. Jul 15, 2010 #2

    mysearch

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    Sorry, I figured out the discrepancy for myself:

    [1] [tex] E= \frac {q}{4 \pi \epsilon_0 r^2}* \frac {1- \beta^2}{(1- \beta^2 sin^2 \theta)^{3/2}} [/tex]

    For (Ex), I also had to update the value of [r] in the denominator of [1], such that [r=x] and [x] is subject to length contraction in the moving frame:

    [2] [tex] E_X= \frac {q}{4 \pi \epsilon_0 (x \sqrt{1- \beta ^2})^2}*(1- \beta^2) [/tex]

    As a footnote question, is relativistic electrodynamics discussed in the relativity forum or the classical forum?
     
  4. Jul 23, 2010 #3

    mysearch

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    I was wondering whether anybody knows enough about relativity and electrodynamics to help with an issue I am trying to resolve. The attached diagram is attempting to show the electric (E) and magnetic (B) forces in the stationary and moving frames. The model assumes (v=0.5c), where (Q) and (q) are both positive charges, which are conceptually tied together to resist the repulsive electrostatic force between them. At relativistic speeds, I am assuming that the inherent propagation speed [c] has to be taken into consideration, such that it takes (t) units of time for the E-field to propagate from (Q) and arrive at (q), where (r=ct).

    If the stationary frame is transposed into the moving frame, it must also take a finite time (t’) for the E-field to propagate between (Q) and (q), although this observed time is now subject to time dilation, which increased the path from [r] to [r’=ct’]. As such, the shortest path between (Q) and (P’) is now defined by [r’], which implies that [E’] is reduced at [P’], due to the increase in [r’], in the moving frame. However, there is an additional B-field in the moving frame due to the relative velocity [v] of (Q). The magnitude and direction of the B-field, as shown in the diagram, is defined by the polarity of [Q], its velocity (v) and the effective angle between (P’), (Q) and the axis of motion [x]. The B-field is subject to the same propagation delay as the E-field.

    At (t1’), in the moving frame, (q) arrives at (P’). However, while (Q) will have also moved by a corresponding distance, as they have the same velocity [v], both the strength and direction of the E and B field vectors sourced by [Q] still appear to be defined by its position at [t0’] at [P’] at time [t1’]. As far as I can determine, [q] approaches the B-field at [P’] at right-angles, such that the magnetic force on [q] would point vertical downwards, as shown and in accordance with the RH rule. So I am also assuming that the size of the magnetic force corresponds to the reduced E-force at [P’], when multiplied by the factor (v/c)^2, see equations below.

    The only problem with this model is that I can’t see how the vector addition of E and B forces at [P’], in the moving frame, can be reconciled with just the E force at [P] in the stationary frame. As such, I am assuming the logic is wrong somewhere and would appreciate any insights. Thanks

    Basic force equations

    [tex] F_E= \frac {qQ}{4 \pi \epsilon_0 r^2} [/tex]

    [tex] F_B= \frac {qQ}{4 \pi \epsilon_0 r^2} \frac {v^2}{c^2} [/tex]
     

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  5. Jul 24, 2010 #4
    Have a look at chapter 8 of Special Relativity by AP French that does exactly what you're trying to do.

    Essentially, you use the fact that the force experienced by a charge in its proper frame when moving in a static Coulomb field depends only on its positon there and not its velocity acceleration etc. You then transform the coordinates of this event to any other frame: In a frame where the particle is stationary, the force is E. In another where it's moving, the force is the sum of E and B there. The important point is that you can say with certainty that the force experienced by moving charges in their proper frame at the same space, time locations in another frame is identical.

    The tricky part is defining what is meant by the force on a particle in a frame other than its proper frame. One way is to define it as F = dp/dt as usual and then to see how the results of an experiment defining F this way in the proper frame of the particle, transforms to another frame using the LT equations. If in this other frame you still want to define F as d/dt something then something = gamma *m. i.e. F = d/dt gamma*m. From this, you can show that F'_x' = F_x, gamma*F'_y' = F_y where primed is the proper frame of the moving charge.

    Hope this helps.
     
  6. Jul 25, 2010 #5

    mysearch

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    Unfortunately, I don’t have access to this text, although I have come across the name in the following reference, which I found to be very useful in understanding the basic premise of the relativistic effects in electrodynamics. http://puhep1.princeton.edu/~mcdonald/examples/EM/vankampen_ejp_29_879_08.pdf [Broken]

    Just for reference, I believe this text provides a similar argument regarding relativistic effects as per the following links.
    http://physics.weber.edu/schroeder/mrr/MRRnotes.pdf
    http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf
    http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html

    What I didn’t originally understand, which none of the references above explicitly explains, is how a current–carrying wire is electrical neutral while still supporting a magnetic field. While this is possibly not a topic for this forum, I believe the following link might be helpful as it seems to be critical to the relativistic solution.
    http://cnx.org/content/m32246/latest/
    Your last sentence seem to be a critical point, which I would like to understand better because I am confused as to whether force is considered to be invariant under a Lorentz transform to another equivalent frame of reference. At a basic level, the electric field E is defined as the force per change [E=F/q]. If charge q is invariant, but E is not, it would suggest that the electric force F associated with E cannot be invariant. Of course, in the context of this discussion, the net force might be be compensated by an additional magnetic force.
    I have quickly looked at a few derivations of force, linked to frame transforms, based on the idea of the rate of change of momentum and I think I understand the inference that [F=ma=m*dv/dt=dp/dt]. However, the electrostatic equation of E has no explicit reference to velocity, i.e, it is essentially the inverse square law of distance [R].

    [tex] F_E= \frac {qQ}{4 \pi \epsilon_0 r^2} [/tex]

    [tex] F_B= \frac {qQ}{4 \pi \epsilon_0 r^2} \frac {v^2}{c^2} [/tex]

    Of course, the equation of force associated with the B-field does make reference to velocity [v], but it is unclear how mass [m] aligns to this definition and therefore the idea of momentum in this context is also unclear. However, the key issue for me in respect to diagram attached to post #3 is the inference that the electric force Fe’ is smaller in the moving frame due to [ct’>ct], which the magnetic force Fb’ would have to compensate to match the electric force Fe in the stationary frame. Unfortunately, the diagram suggests the Fb’ does not align to Fe’ and would actually reduce the net force. So would still like to understand where this diagram is wrong.
    Yes, many thanks. I also appreciated your comments in another thread to which I have also replied. See cross-reference for other electrodynamic equations:
    https://www.physicsforums.com/showthread.php?t=409967
     
    Last edited by a moderator: May 4, 2017
  7. Jul 29, 2010 #6

    Dale

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