Is electric current relative?











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Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?






electromagnetism electric-current relative-motion







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  • Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
    – The Photon
    3 hours ago















up vote
2
down vote

favorite












Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?






electromagnetism electric-current relative-motion







share|cite|improve this question






















  • Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
    – The Photon
    3 hours ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?






electromagnetism electric-current relative-motion







share|cite|improve this question













Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?






electromagnetism electric-current relative-motion




electromagnetism electric-current relative-motion






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share|cite|improve this question











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asked 3 hours ago









ado sar

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1488












  • Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
    – The Photon
    3 hours ago


















  • Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
    – The Photon
    3 hours ago
















Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago




Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago










2 Answers
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Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.



For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is



$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$



where $gamma=1/sqrt{1-v^2/c^2}$.



If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.



Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.






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    Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
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      down vote













      Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.



      For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is



      $$begin{align}
      crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
      J_x^prime &= J_x \
      J_y^prime &= J_y \
      J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
      end{align}$$



      where $gamma=1/sqrt{1-v^2/c^2}$.



      If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.



      Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.






      share|cite|improve this answer



























        up vote
        5
        down vote













        Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.



        For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is



        $$begin{align}
        crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
        J_x^prime &= J_x \
        J_y^prime &= J_y \
        J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
        end{align}$$



        where $gamma=1/sqrt{1-v^2/c^2}$.



        If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.



        Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.






        share|cite|improve this answer

























          up vote
          5
          down vote










          up vote
          5
          down vote









          Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.



          For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is



          $$begin{align}
          crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
          J_x^prime &= J_x \
          J_y^prime &= J_y \
          J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
          end{align}$$



          where $gamma=1/sqrt{1-v^2/c^2}$.



          If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.



          Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.






          share|cite|improve this answer














          Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.



          For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is



          $$begin{align}
          crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
          J_x^prime &= J_x \
          J_y^prime &= J_y \
          J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
          end{align}$$



          where $gamma=1/sqrt{1-v^2/c^2}$.



          If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.



          Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 2 hours ago









          G. Smith

          3,673917




          3,673917






















              up vote
              3
              down vote













              Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.






              share|cite|improve this answer

























                up vote
                3
                down vote













                Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.






                  share|cite|improve this answer












                  Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  akhmeteli

                  17.6k21740




                  17.6k21740






























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