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Gauss' Law for spherical shell vs Coulomb's law, regarding reativity

  1. Jun 25, 2013 #1
    Shalom

    We are used to hearing that Coulomb's law doesn't settle with the relativity principle that nothing moves faster than the speed of light, in the sence that it embeds 'Action in a Distance'. Meaning that if somthing changes in r1 at time t1, and we write the law for any t before t1+(|r1-r2|/c), (r2 is the where the test charge is) then the law doesn't represent reality, because the 'knowledge' about the change hasn't reached r2 yet.

    And it is often said that Gauss' law fixes that because of its local nature. But what I can't figure out is:

    (1) how does one settle that with one of the famous implementations of Gauss' law, the one for a spherical shell with a charge in its center. When we use the Divergance law to find the same form of Coulomb's law, resulting from Gauss' Law. How does this implementation not violate the relativity principle, violated by Coulomb's law (nothing travels faster than the speed of light)?

    (2) another related question is how does Gauss' law, or Maxwell's laws express that the information about the change in r1 travels at speed c? (other than the wave equation please, and other than being local). I just can't see how Gauss' law shows that principle, which Coulomb's law couldn't.

    And I would really love to hear from anyone who might have a better vision and be able to see what I seem to be missing here.

    Thank you in advance
     
  2. jcsd
  3. Jun 25, 2013 #2
    Gauss's law works with relativity because charge is conserved and charges can't move faster than light. Therefore, if you draw a Gaussian sphere around a charge with a radius of R, you know that the charge was still inside the sphere at R/c (quantity of time) in the past. So you can calculate the flux of E based on charge inside the sphere NOW, since it's the same as the charge inside the sphere in the past. Since Gauss's law is part of the Maxwell's equations which can be written in covariant form, it obeys relativity.

    If you imagine a Gaussian sphere, and a charge that is moving across the surface of the sphere, the flux of E suddenly changes when the charge crosses the sphere. This means that the "disturbance" travels at least as fast as the fastest charge.
     
  4. Jun 25, 2013 #3

    WannabeNewton

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    In order for EM to be compatible with SR you need all of Maxwell's equations so that you can write them down in a manifestly Lorentz covariant manner (so that they transform correctly under Lorentz boosts); Gauss's law on its own is not enough.

    Again, Gauss's law on its own is not enough to show that changes in the EM field travel at c. Also, you can derive all of Maxwell's equations using Coulomb's law and special relativity and deduce it in the usual way.
     
  5. Jun 26, 2013 #4
    the form of the law resulting from the sphere impementation

    yes, that's exactly what I'm reffering to. I totally agree. my problem is how to settle that with the resulting formula (from the spherical shell implemantaion). The resulting formula looks just like Coulomb's law, while at the same time it is not supposed to be limited to electrostatic cases as is Coulomb's law. Because it is generated using Gauss' law.

    How is it expressed in the final connection between the field on the sphere and the charge within it, derived from Gauss' law using the divergence theorem ,(which looks just like Coulomb's law), that it is only true for a charge in the middle at time no earlier than (t-R/c), if E(t) is the field?

    thank you
     
  6. Jun 26, 2013 #5
    can you elaborate on writing them down in a Lorentz covarient manner?

    Thank you.
    I would really appreciate it if you ellaborate a bit on writing Maxwell's equations in a Lorentz covariant manner:

    1. how is that done? what does it look like? any specific refference would be great too
    2. How does it show that any information about any change, let's say in the point particle in the origin of a system of coordinates at time t, doesn't really get to another point r, until (t+|r|/c) ? (as opposed to the formula of Coulomb's law

    thank you
     
  7. Jun 26, 2013 #6

    WannabeNewton

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    (1) In special relativity we can write down Maxwell's equations in the following form: ##\partial^{a}F_{ab} = 4\pi j_{b}## and ##\partial_{a}F_{bc} - \partial_{b}F_{ac} + \partial_{c}F_{ab} = 0## where ##F_{ab}## is the electromagnetic field strength tensor defined in terms of the 4-potential ##A_{a}## as ##F_{ab} = \partial_{a}A_{b} - \partial_{b}A_{a}## and ##j^{b}## is the 4-current density. These two sets of equations comprise all of the classical Maxwell equations. They are said to be manifestly Lorentz covariant because the equations are written only in terms of quantities that transform correctly under Lorentz boosts.

    (2) I'm not sure what you mean by this. The above form of Maxwell's equations are still Maxwell's equations so the fact that changes in the electromagnetic field only propagate at ##c## (in all inertial reference frames) still follows in the usual way from the wave equation that comes out of Maxwell's equations.
     
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