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Unformatted text preview: Scott Hughes 17 March 2005 Massachusetts Institute of Technology Department of Physics 8.022 Spring 2005 Lecture 12: Forces and Fields in Special Relativity 12.1 Transformation of the electric field It is very easy to see how the electric field transforms by considering a special case. Examine a capacitor in its rest frame. Orient the capacitor such that its plates are parallel to the x y plane. We take the capacitor to have charge density σ on the lower plate, σ on the upper plate. We take the plates to be square, with each side having length L . The total amount of charge on the plates is thus Q = σL 2 ; the electric field between the plates is ~ E = 4 πσ ˆ z : L The field inside the capacitor is represented by the field lines sketched here. Consider now this capacitor “boosted” to a velocity v in the x direction. Charge is a “Lorentz invariant”: all observers agree that the total charge on the plates is Q . Because of the Lorentz contraction, the plate’s dimension along the direction of motion is seen to be L/γ . v γ L/ The charge density is thus augmented: σ = Q/ ( L × L/γ ) = γQ/L 2 = γσ . The electric field is therefore augmented as well: ~ E = 4 πγσ ˆ z . Notice that the field lines are denser in this picture: keep in mind that density of field lines tells us how strong the field is. 110 How about if the capacitor is oriented differently? Let’s rotate it so that the plates lie in the y z plane. The electric field in the rest frame is thus ~ E = 4 πσ ˆ x . When this capacitor is boosted in the x direction, the plates appear closer together, but the charge density does not change : σ = σ with this orientation: ~ E = ~ E = 4 πσ . v L Notice that the density of field lines are not changed in this picture, reflecting the fact that the field is not changed by the boost. Even though we have only looked at the electric fields for a rather special circumstance, the rule we deduce from this is in fact quite general: Components of the electric field perpendicular to the velocity are augmented by the Lorentz transformation; components that are parallel to the velocity are unaffected . Mathematically, E ⊥ = γE ⊥ E  = E  . 12.2 Transformation of energy and momentum Our next goal will be to understand how forces transform between different frames of refer ence. To do this, we first must understand how to describe energy and momentum in special relativity. Consider an object whose rest mass — the mass that we would measure when this object is at rest with respect to us — is m . If this object moves with velocity ~u , what is its energy and its momentum? In the “oldfashioned” mechanics you came to know and love in 8.012, we said that a moving body has a kinetic energy E kin = 1 2 m  ~u  2 111 and a momentum ~ p = m~u ....
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 Spring '10
 ScottHughes
 Special Relativity, Frame, Electric charge

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