Lecture Notes on Electric and Magnetic Fields in Relativity - Prof. Darin E. Acosta, Study notes of Physics

Download Lecture Notes on Electric and Magnetic Fields in Relativity - Prof. Darin E. Acosta and more Study notes Physics in PDF only on Docsity! PHY2061 Enriched Physics 2 Lecture Notes Relativity 6 Relativity 6 Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor. Transformation of the Electric Field Electric charge is invariant under motion. In other words, electric charge, like mass, is a property of a particle and is invariant under transformations. It is a Lorentz Invariant quantity. If this were not so, physics would look different in different reference frames. The light emitted by hydrogen would have different colors beyond just the usual Doppler shift. In fact, hydrogen might not even bind if the positive and electric charges did not transform the same way. Thus, Gauss’ Law holds in all reference frames: ( ) '( )S t S t d d′ ′⋅ = ⋅∫ ∫E a E a Two reference frames with a relative velocity between them would agree on the charge enclosed by a surface, even if the surface integral is computed using the coordinates of S or S’. Consider 2 parallel sheets of charge infinite in extent with opposite charge density: σ σ σ+ −= − = . By Gauss’ Law in the rest frame of the two sheets, we can solve for the electric field arising from each sheet. Take as a Gaussian surface a box enclosing just one sheet or the other, with area on the top and bottom faces parallel to the sheet: A x= ∆ ∆ z 0 0 0 0 2 2 A AE A E A E A E A E E σ σ ε ε σ σ ε ε + − + + − − + − + − + = + = = = You can neglect the field arising from one sheet affecting the flux through the box enclosing the other sheet because the net flux will be zero through the surface (no charge enclosed from the other sheet). By the superposition principle, the two fields add, and we see that outside the two sheets, E = 0. Between the 2 sheets,: 0 E σ ε = D. Acosta Page 1 10/11/2005 PHY2061 Enriched Physics 2 Lecture Notes Relativity 6 Now consider a Lorentz Transformation to a frame where the 2 sheets are moving at a velocity v with respect to an observer. How do the previously considered Gaussian surfaces change? First of all, if we do not change the definition of the surface, the charge enclosed must remain invariant as stated before. However, the geometry of the box has changed. The dimension parallel to the direction of motion has contracted, so the area of the top and bottom surfaces has contracted. In the S’ frame where the sheets are in motion, Gauss’ Law stated in that frame becomes: 0 0 encq AE A σ ε ε ′ ′ = = But since the dimension parallel to the motion is length contracted, the area in the S’ frame is 2 2 1 where 1 / AA v c γ γ ′ = = − So ( ) 0 0 A AE E σ γ ε γσ ε ′ = ′⇒ = In other words, the charge density has increased: σ γσ′ = And the electric field perpendicular to the plates (and to the direction of motion) has increased: E Eγ⊥ ⊥′ = If we had instead considered a Lorentz Transformation in a direction perpendicular to the plates (i.e. parallel to the electric field direction), there would have been no length contraction of the plate dimensions since they are perpendicular to the direction of motion. So the charge density would not change in the S’ frame, nor would the electric field: E E′ = D. Acosta Page 2 10/11/2005 PHY2061 Enriched Physics 2 Lecture Notes Relativity 6 ( ) ( )( ) ( ) net 0 2 20 2 2 0 0 0 0 0 0 0 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 02 2 0 00 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1 2 λλ λ λ γ γ γ λ γ β β λ γ β β β β β β β β β β β βλ γ β β β β β β β β β βλ λ β β γ γ γ γβ β + + − + − + + − + + + + ′ ′ ′ ′ ′= + = − ⎛ ⎞ ⎜ ⎟= − ⎜ ⎟′ ′− −⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ = −⎜ ⎟ ⎜ ⎟⎛ ⎞ ⎛ ⎞− + − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟− +⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎛ ⎞− +⎜ ⎟= − ⎜ ⎟− − + − − +⎝ ⎠ − = = − − − = − 0 2 1 where as usual 1 λ γβ β γ β + = − which is indeed not zero. We can use this net charge density to calculate the electric field at the position of the test charge (which is at rest in this frame): Appling Gauss’ Law to an infinitely long line charge, we consider a cylindrical surface: net 0 net 0 0 0 2 2 2 2 r r LE r L E r r λπ ε λ λ γβ β πε πε + ′ ′ ′Φ = = ′ ′ = = − ′ ′ The force on a test charge of charge q is toward the wire (since the wire has a net negative charge in this frame) if q>0: 0 0 2 2r r qF qE r λ γβ β πε + ′ ′= = ′ Thus, the particle which was originally at rest, will accelerate toward the wire (in the direction perpendicular to the Lorentz Transformation). D. Acosta Page 5 10/11/2005 PHY2061 Enriched Physics 2 Lecture Notes Relativity 6 Now, let’s transform this force back to the original frame where the test charge was in motion and the wire had no net charge density. In this frame, the force perpendicular to the direction of the Lorentz Transformation direction is: 1F F γ⊥ ′= ⊥ , so 0 0 2 since 2r qF r r λ β β πε + ′= = r i Now is just the current carried in the wire. Also, note that 02 vλ + = 0 1 4 K πε = . So, 0 2 2 0 2 2 2 2r q v v Ki kiF qv qv c r c r r λ πε + ⎛ ⎞ ⎛= = =⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎞ ⎟ ⎠ where we have defined a new constant: 2 Kk c = The term in parentheses can be defined to be the “magnetic field” arising from a current in a wire: 2 where for this exampler kiF qvB B r = = We have just derived the force acting on a charged particle when it moves at a velocity v with respect to a current i. Note that the force is in a direction perpendicular to the velocity direction. This force is called the magnetic force, with an associated magnetic field. However, it really is nothing new. In the rest frame of the test charge, only an electric field existed. But when transformed to the lab frame, the electric field changed into a magnetic field that only operates on charges with non-zero velocities. The magnetic force equation can be written more generally in vector form as: q=F v×B The force is the cross-product of the velocity and magnetic field vectors, multiplied by the charge of the particle: ( ) ( ) ( )ˆ ˆy z z y x z z x x y y xq v B v B q v B v B q v B v B= − − − + −F x ˆy z When combined with the electric force law, we have the full Lorentz Force equation: ( )q= +F E v ×B D. Acosta Page 6 10/11/2005