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I The Electric field of a moving charge according to Purcell

  1. Mar 26, 2017 #1
    In Berkeley Physics Course (Volume II) on Electricity and Magnetism:

    https://www.scribd.com/doc/128728926/Electricity-and-Magnetism-Berkeley-Physics-Course-Purcell

    ....Purcell discusses the invariance of charge (Section 5.4), the electric field measured in different frames of reference (Section 5.5), and the field of point charge moving at constant velocity (Section 5.6).

    What I find peculiar is that toward the end of Section 5.5, he states that if one observer were to view a charge as stationary in one frame, and moving relative to second observer in another frame, the longitudinal component of the electric field is the same according to both observers.

    However, in section 5.6, he starts with one observer where the charge is already moving in one frame, while a relatively moving second observer sees a different longitudinal component of this charge's electric field.

    To see what I mean, look at figure 5.13. Also, upon looking at equation 12 in Section 5.6, it appears that the transverse electric field scales with the Lorentz factor, while the longitudinal electric field (where sin(theta) = 0) scales inversely to the square of the Lorentz factor. What's going on here?

    Anyway, it would appear to me that if one were to have increased transverse components to the electric field of a moving charge upon switching to a different observer, then invariance of charge would require that longitudinal components of the electric field would have to be less in this other frame so that the total surface integral of the electric field on a Gaussian surface enclosing the charge remains unaffected.
     
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  3. Mar 31, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Mar 31, 2017 #3

    pervect

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    Have you looked at the effects of the relativity of simultaneity? (I'm assuming you're familiar with that concept, if not, we'd need to digress). If you have an E field at some point at some time, the transverse component gets compressed and the longitudinal field doesn't as Purcell says.

    When you switch from a moving frame to a stationary frame, though, the simultaneity convention is different between the two frames.

    If you have some charge Q, the E field of that charge is proportiaonl to Q/r^2 in the frame in which the charge is at rest, but that relation is not correct in other frames. You can do the Lorentz transform of the E-field between frames, creating a magnetic field (not discussed), strengthening the transverse field, and leaving the longituidinal field unchanged. But after doing this, one needs to adjust time as well, in the moving frame the E-field isn't constant, and it has a different notion of time (a different simultaneity convention, and of course time dilation) than the non-moving frame.

    So if the time is t in the stationary frame, you use the Lorentz transform to find the time t' in the moving frame, and you transform the fields that exist at time t by the longituidnal/transverse rule, then you have the fields at time t' in the moving frame.
     
  5. Apr 6, 2017 #4
    Okay, so to paraphrase in short, after accounting for the relativity of simultaneity, then we should have a situation where the longitudinal components of the electric field are the same in any frame.

    After another look at Purcell's book, I see now that he explains it differently than you have. He has a length contraction of the distance r' between the charge and the source in the moving frame, contracted compared to r in the rest frame, that compensates for the otherwise reduced electric field in the longitudinal direction. He considers a scenario where two test charges meet at a point in space-time, where they see the same magnitude of longitudinal component of the electric field at this same space-time point, but a different angular dependence due to seeing this charge at different distances.

    However, I found this bizzare. If I had a electrically-charged rotating wheel, the parts of the wheel may alternate between longitudinal and transverse motion, which if applied to Purcell's explanation, would mean this effective contraction of distance r' relative to r would vary for each part. lf so, if this wheel had an arbitrarily small radius, the parts of this wheel would essentially be located at nearly the same point in space-time, but the contraction he decribes would result in some rather bizzare implications if the distance between a charge and the charged wheel were large relative to the radius of the wheel, especially moreso if the wheel's rotation were relativistic.

    I am also concerned about how the relativity of simultaneity could explain such a particular case as a relativistically-spinning, electrically-charged wheel of arbitrarily small radius, one which is essentially point-like according to a distant charge.

    Also, what would identical electric field measurement devices arranged on a surrounding sphere pick up? Surely the total flux extrapolated from these discrete detectors should remain unaffected by whatever rotation rate assumed by the electrically-charged wheel!

    However, if an array of sensors essentially sharing the same rest frame were to see only alterations to the transverse electric fields, then wouldn't the divergence of these components have a dependence on the spin-speed of the electrically-charged wheel that carries over to the combined (transverse + longitudinal) electric fields?


    Sincerely,

    Kevin M.
     
    Last edited: Apr 6, 2017
  6. Apr 7, 2017 #5

    pervect

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    I don't have Purcell, and I can't say I follow your concerns, except that the seem murky, and that getting sidetraced into analyzing a harder problem (rotating frames) isn't going to be the best solution.

    If you'd like to try an alternate approach, you could use the Lienard-Wiechert potentials to solve directly for the electric (and magnetic, if you're interested) fields of a moving charge.

    Some links:

    https://en.wikipedia.org/w/index.php?title=Liénard–Wiechert_potential&oldid=761833056
    http://farside.ph.utexas.edu/teaching/em/lectures/node124.html#e10.247
     
  7. Apr 7, 2017 #6

    vanhees71

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    Isn't this again the usual confusion caused by Purcell's overcomplicated treatment of relativistic electromagnetics. Instead of first introducing the appropriate math (tensor algebra on pseudo-Euclidean (here Minkowski) space), he tries to be "pedagogical" :rolleyes:. For a very clear treatment of covariant E&M, see Landau&Lifshitz, vol II. Admittedly this book is the opposite: It's giving exactly the math you need without any ambition to be "pedagogical" :biggrin:.

    If you have a situation, where a charge distribution is at rest in one inertial reference frame, then the most simple way is to calculate the four-potential and then boost it to any other inertial reference frame you like (or you evaluate the electric field and then apply the Lorentz boost to the electromagnetic field components directly; this is most elegantly done in the ##\mathrm{SO}(3,\mathbb{C})## representation which applies to the Riemann-Silberstein vector ##\vec{E}+\mathrm{i} \vec{B}##, using Gaussian or Heaviside-Lorentz units of course).

    As mentioned by @pervect above, you can also directly calculate the four-potential in any frame, using the retarded propagator (which is a manifestly covariant scalar distribution if the Lorenz gauge is chosen) or the Jefimenko equations (which is also the application of the retarded propagtor) to get the em. field components directly.
     
  8. Apr 7, 2017 #7
    I looked at the results for the electric fields, and see expressions taking the dot product of the velocity and the distance vector r. It appears the longitudinal components of the E field calculated this way is dependent on the velocity of the source as observed in that frame. The force caused by this does not appear to correspond to a "magnetic" force.
     
  9. Apr 7, 2017 #8
    The attraction and replusion between wires was explained by Purcell in a particular way. To simplify this, one can say if you have two identical current elements side by side, they would attract because opposite charges would "see" a length contraction that like charges would not. If you reversed one of the currents, the opposite would be the case.

    So what happens if two current elements were lined up or "co-linear"? It would seem that the electric field observed by a charge in one of the elements would depend on its own rest frame, but since other charges in that same element would move differently, wouldn't they see a different electric field via 1/gamma^2? Doesn't this likely mean that there are different forces on them which do not cancel, after applying transformations, resulting on a net force on said element acting along a postion vector connecting the two elements? If I am not mistaken, this would be longitudinal tension which cannot be described by a magnetic Lorentz force. Furthermore, positive and negative charges in each element would observe opposing forces, which implies that the would be magnetic induction forces dependent on how fast these elements are moving apart along the position vector r. I don't see how the Lorentz Force could explain this, since its E is derived by the independent observer.

    The above of course assumes that the longitudinal electric fields are frame dependent as you say.

    Purcell says the transformations compensate for this, but he is implying that the effective distance to the source is a function of its velocity in addition to its distance along space time. So, he uses longitudinal field weakening to impose charge invariance, but he apparently discards that somehow after applying Lorentz transformations to r.

    Strange.
     
  10. Apr 7, 2017 #9

    pervect

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    Let's try this again - Hopefully my deleted post got deleted before it caused any confusion. (add- I can see that it didn't. Sorry.)

    Let's suppose we have a stationary charge Q, in some frame of reference S, that has coordinates t,x,y,z.

    We can write down the electric field in frame S fairly easily, I won't go through the formulas, but it's pretty straightforwards, the field points radially outward, is indepenent of time, and dies off as ##1/r^2 = 1/(x^2 + y^2 + z^2)## The magnetic field in frame S is zero,

    Now we ask the question - what are the electric fields in some frame S', moving with a velocity ##\beta = v/c##? This should be the same as the field of a moving charge we calculate by other methods. We are applying a principle here, that says that if we know E and B in one frame, we can find E', and B' in another frame. We'll let the coordiantes for S' be t', x', y', and z'.

    We proceed in two steps. First we find E in terms of t', x', y', and z', the coordinates we use in frame S'. Then we transform E into E'.

    Additionally, in frame S', we have a magnetic field (unlike in frame S), but it doesn't seem to be of interest, so I won't discuss it.

    The equations we need are the following. There is a mapping from points in S to points in S', known as the Lorentz transform. Hopefully this is already familiar to you and is just a review, though from some of your other comments perhaps it is not a review. If it's not a review, you might have to ask about this in detail in another post, it's important to understand this first before proceeding with the more complex problem of how the electromagnetic field transforms.

    It's also important to understand that t' is not the same as t, due to time dilation and the relativity of simultaneity.

    If S' is moving with some velocity ##\beta = v/c## in the x direction, we can write:

    $$t' = \gamma (t - \beta x/c) \quad x' = \gamma (x - \beta c t) \quad y'=y \quad z'=z$$

    This is the Lorentz transform.

    This allows us to find E in terms of t' and x'. But what we desire to compute is not E, but E', the value of the field in S'. To do this we need to take the additional step that Purcell describes, in which ##E'_x = E_x## and ##E'_y = \gamma E_y## and ##E'_z = \gamma E_z## . This step is described in words by saying "longitudinal components don't change, transverse components get multipled by gamma.

    This is what Purcell was describing, you can also find it in wiki <<here>>., using the parallel and transverse symbolism instead of x,y, and z. Because x is the direction of motion I chose, the x components are the parallel components and the y and z components are the transverse components.

    Because the B field in frame S was zero, we don't have to worry about them in this particular case. In general, though we do, the wiki link has the equations that address what would have happened if the B field was non-zero.
     
    Last edited by a moderator: Apr 9, 2017
  11. Apr 7, 2017 #10
    I think I have a fairish understanding of the details, but what has bothered me is the question of the invariance of charge. The value of a charge is the result of the divergence of the electric field. In the case of a stationary point charge, in Cartesian coordinates, its electric field's x-components, y-components, and z-components contribute equally to its divergence. However, if indeed the transverse but not longitudinal components get compressed due to length contraction in frame S', then how in how in Earth can the divergence of the electric field of a charge be an invariant when you are essentially increasing the divergence of transverse electric field components by the Lorentz factor? How is it possible to have quantized electric charges if you have something like that going on?
     
    Last edited by a moderator: Apr 9, 2017
  12. Apr 7, 2017 #11

    PAllen

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    At some point you need to consult a more advanced source. Purcell was a valiant attempt to show how EM demands a relativistic treatment, while being presented such that freshman physics students can use it. The freshman goal presents enormous limits on the completeness of arguments. I cannot give a useful recommendation because the first good treatment of relativistic charge invariance I studied is 5 decades out of print.
     
  13. Apr 8, 2017 #12

    vanhees71

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    Sigh. Is there any E&M textbook that provides more confusion about the relativistic formulation than Purcell? That will be hard to achieve. Just learn tensor calculus on Minkowski space and THEN do the physics. You find an introduction in my SRT FAQ:

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

    If I find the time, I'll write the E&M part soon. For the time being, here are the transformation laws for the electric and magnetic components of the electromagnetic field for a rotation-free boost in arbitrary direction:

    http://theory.gsi.de/~vanhees/faq/edyn/node4.html#1.2.60

    The special case of electrostatic fields is here (in German; sorry):

    http://theory.gsi.de/~vanhees/faq/edyn/node13.html
     
  14. Apr 8, 2017 #13

    pervect

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    If you look at the equation for the electric field of a moving charge, you should get the same answer as from http://farside.ph.utexas.edu/teaching/em/lectures/node125.html

    The field points in the radial direction, at an angle ##\theta## the electric field from <<this link>> is:

    $$||E|| = \frac{e}{4 \pi \epsilon_0 r^2} \frac{\gamma}{(1 + \beta^2 \gamma^2 \cos^2 \theta)^\frac{3}{2}}$$

    I've taken some liberties with the notation, calculating the magnitude of E (since we know it points in the radial direction), and letting ##\beta = v/c##. ##\gamma## is used on the webpage, it's ##1/\sqrt{1-\beta^2}## as usual.

    The term
    $$\frac{e}{4 \pi \epsilon_0 r^2}$$
    is just the Coulomb field. So we see that at ##\theta=+/-90##, which is the transverse direction, the field is boosted by a factor of ##\gamma##. But when ##\theta=0## or ##\theta=180##, in the longitudinal direction, the field is:

    $$\frac{\gamma}{(1 + \beta^2 \gamma^2 )^\frac{3}{2}}$$

    If we take for example ##\beta = .995## so that ##\gamma = 10.01 \approx 10##, we see that the transverse field is about 10 times the columb field of a stationary charge, but the longitudinal field is about ##10 / 100^\frac{3}{2} \approx .01##. times the columb field of a stationary charge.

    If you're really ambitious, you can set up and do the integral. I think I may have done that once in the very distant past, but it's pretty messy looking. The E-field is normal to the surface of the sphere, but you'll have to do a 3d integral to get the total charge from Gauss' law.

    So I suspect from your comments is that you don't understand the steps I outlined as well as you think you do. I rather suspect you'll find you forgot to compensate for the Lorentz contraction between frame S and S', at a guess.

    So the next step is to go back to your text and make sure that you get the same answer (.01) for the longitudinal field under the same conditions using the method that you're studying, and compare the results to your text rather than a webpage of somewhat unknown provenance. (Though I don't think there's any problem with it, I don't have my E&M textbook handy to check it).

    Note that arguing that E' = E in the longitudinal direction is not sufficient to calculate the longitudinal field, one also needs to do the Lorentz transform.
     
  15. Apr 8, 2017 #14
    I have been under the impression that Lorentz transformation effectively takes the reduced longitudinal fields and ends up making them independent of the velocity, but here it seems that you are saying the reverse is true, that Lorentz transformation is essentially responsible for the Lorentz-invariance of the charge.
     
  16. Apr 8, 2017 #15
    To illustrate my concerns further, imagine a system of two charged particles moving in inertial frame S. Let's have a primary lab observer at rest in frame S. Let's have a secondary observer riding on one of the moving charges, and this observer is at rest in frame S'. Let's say these charges are backed by much mass, so acceleration is negligible. If charge is frame-invariant, then the longitudinal fields should be reduced to compensate for the increased transverse fields. So the longitudinal fields would be different in frame S than they are frame S'. Where did I mess up?
     
    Last edited: Apr 8, 2017
  17. Apr 8, 2017 #16
    http://www.physicspages.com/2015/03...mation-of-force-electric-and-magnetic-fields/

    ^ The above link (based on Griffiths) brings up a derivation, but it does not preserve the divergence of the electric field under Lorentz boosts. According to the page above, longitudinal components of the electric field remain unaffected by Lorentz boosts. Under the predication that charge is due to the divergence of electric fields, Gauss' law would imply that the value of a charge is non-invariant.

    Does the mean that Gauss' law is an approximation for moving charges?

    Or does it mean the above site link misrepresents the electric field of a moving charge?

    If the former and not the latter, wouldn't that mean that heated systems would acquire electric charge from thermal motion of electrons? Surely there is something wrong with that! Heating something up doesn't generally produce electrostatic charge, and velocities of electrons due to thermal excitations are orders of magnitude greater than drift velocities typical in electrical conductors. So if such variance of charge was real, the effects would be enormous.

    It appears clear the charge should be frame-invariant, but then what does this say about charge confined to a straight conductor? The electic field acting on a co-linear current element would be percieved differently by free electrons in metal than by the metal "lattice". Do these differences of longitudinal electric fields/forces viewed by conduction electrons vs. by the metal "lattice" vanish after Lorentz transformations are applied to each such that each longitudinal electric field/force is calculated from a common inertial frame? If not, it seems we have net longitudinal forces between electrically-neutral current elements.

    Sincerely,

    Kevin M.
     
    Last edited: Apr 8, 2017
  18. Apr 8, 2017 #17
    After a boost electric field is unchanged at every point in the boost direction. But every point in the boost direction is closer to the source of electric field, because the universe shrinks in a boost.

    Right?
     
  19. Apr 8, 2017 #18
    I can see, via the argument by Purcell that if we are comparing the field of a charge observed by two distant observers in relative motion who are located at the same distant point in space-time, that the length contraction would act parallel to the boost direction, compensating the otherwise different longitudinal electric field. However, if the source of electric field consists of two charges in relative motion at the same point in space-time, would you have to apply different Lorentz contraction factors to distance to each source? If not, it follows that the contraction of r is solely dependent on the velocity of the observer, though not "absolute" velocity of course. But what is this Lorentz factor then if not dependent on the velocities of the source? If, as indicated by Griffiths' the transverse electric flux is altered on Lorentz boosts of the observer, and if, as indicated by Purcell, the value of a charge is invariant with respect to its motions, it would seem that the relative velocity, position, and acceleration between charges and observers provides inadequate information about the electric fields of a charge as viewed by that observer. Some "total Lorentz boost" must exist for the observer which accounts for the apparent increase in electric flux from sources of electric fields (elementary charges) after vs. before Lorentz boosts of the observer, if we hold both Griffiths' derivation of electric fields through Lorentz transforms, while at the same time enforcing the constancy of electric charge with respect to motions of its source over time providing conservation, without invariance, of charge (the time component of the four-current is not invariant).
     
    Last edited: Apr 8, 2017
  20. Apr 8, 2017 #19
    I think these are true sentences:

    1: Observer's acceleration towards or away from a charge does not change the measured electric field.
    2: Observer's tangential acceleration relative to a charge changes the measured electric field.
    3: Charge's acceleration towards or away from an observer changes the measured electric field.
    4: Charge's tangential acceleration relative to an observer changes the measured electric field.

    From 1 we get: Relative radial motion between a charge and an observer has no effect on measured electric field.
    From 3 we get: Relative radial motion between a charge and an observer has an effect on measured electric field.

    There seems to be a contradiction. Is that contradiction a problem?

    Addition: Charge's acceleration towards or away from an observer changes the measured electric field, after enough time has passed. I have a feeling that time is important here, because of the relativity of simultaneity.
     
    Last edited: Apr 8, 2017
  21. Apr 9, 2017 #20

    vanhees71

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    I'm really very surprised that this simple thing can be so much confused. So let's start from scratch to derive the field for the situation that it is an electrostatic one in a reference frame ##\Sigma##, i.e., in this frame there's a electric point charge ##q## at rest in the origin. Let ##\Sigma'## the frame, where the charge moves with velocity ##v## in ##x^1## direction (i.e., ##\Sigma'## moves with velocity ##-v## in ##x^1## direction). I work in Heaviside Lorentz units with the vacuum-speed of light ##c=1##.

    The spacetime coordinates transform as ##x^{\prime \mu}={\Lambda^{\mu}}_{\nu} x^{\nu}##,
    $$x^{\prime 0}=\gamma (x^0+v x^1), \quad x^{\prime 1}=\gamma (v x^0+x^1), \quad x^{\prime 2}=x^2, \quad x^{\prime 3}=x^3.$$
    Here ##\gamma=1/\sqrt{1-v^2}##.

    We'll also need the inverse transformation
    $$x^{0}=\gamma (x^{\prime 0}-v x^{\prime 1}), \quad x^{1}=\gamma (-v x^{\prime 0}+x^{\prime 1}), \quad x^{2}=x^{\prime 2}, \quad x^{3}=x^{\prime 3}.$$

    If we only want to work with standard real quantities the most simple way to calculate the fields is to use the four-potential. In the original frame we have a Coulomb field with
    $$(A^{\mu})(x)=\begin{pmatrix} \phi(\vec{x}) \\ 0 \\0 \\0 \end{pmatrix}, \quad \phi(\vec{x})=\frac{q}{4 \pi |\vec{x}|}.$$
    The four-potential is a vector field and thus transforms as
    $$A^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} A^{\mu}(x)={\Lambda{\mu}}_{\nu} A^{\mu} (\hat{\Lambda}^{-1} x').$$
    This gives
    $$(A^{\prime \mu})=\begin{pmatrix} \gamma \phi(\vec{x}) \\ \gamma v \phi(\vec{x}) \\ 0 \\ 0 \end{pmatrix},$$
    We have
    $$|\vec{x}|=\sqrt{\gamma^2 (x^{\prime 1}-v x^{\prime 0})^2+ (x^{\prime 2})^2+ (x^{\prime 3})^2}.$$
    The electromagnetic field components are then given by
    $$\vec{E}'=-\partial_0' \vec{A}'-\vec{\nabla}' A^{\prime 0}=\frac{\gamma q}{4 \pi [\gamma^2(x^{\prime 1}-v x^{\prime 0})^2+x^{\prime 2}+x^{\prime 3}]^{3/2}} \begin{pmatrix} x^{\prime 1}-v x^{\prime 0} \\ x^{\prime 2} \\ x^{\prime 3} \end{pmatrix}$$
    and
    $$\vec{B}'=\vec{\nabla}' \times \vec{A}'=\vec{v} \times \vec{E}'. \qquad (*)$$
    The latter equation is immediately clear from
    $$\vec{A}'=\vec{v} A^{\prime 0} \; \Rightarrow \; \vec{B}'=-\vec{v} \times \vec{\nabla}' A^{\prime 0}=\vec{v} \times (\vec{E}'+\partial_0' \vec{A}').$$
    Since ##\vec{v} \times \partial_0' \vec{A'}=\vec{v} \times \vec{v} \partial_0' A^{\prime 0}=0##, we indeed get (*).

    It's easy to check that, of course, these are solutions of the Maxwell equations for the given charge-current distribution. In the original frame we have
    $$j^{\mu}=q \delta^{(3)}(\vec{x}) \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
    It's again a vector field, such that
    $$(j^{\prime \mu})=q \gamma \delta^{(3)}(\vec{x}) \begin{pmatrix} 1 \\ v \\ 0 \\ 0 \end{pmatrix}.$$
    Now
    $$\delta^{(3)}(\vec{x})=\delta[\gamma (x^{\prime 1}-v x^{\prime 0})] \delta(x^{\prime 2}) \delta(x^{\prime 3}) = \frac{1}{\gamma} \delta(x^{\prime 1}-v x^{\prime 0})\delta(x^{\prime 2}) \delta(x^{\prime 3}),$$
    and thus
    $$(j^{\prime \mu})=q \delta(x^{\prime 1}-v x^{\prime 0})\delta(x^{\prime 2}) \delta(x^{\prime 3}) \begin{pmatrix} 1 \\ v \\ 0 \\ 0 \end{pmatrix}.$$
     
    Last edited: Apr 9, 2017
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