From Special Relativity to Feynman Diagrams.pdf

We readily obtain y β v y β v β v y β 2 v y β 2

Info icon This preview shows pages 30–33. Sign up to view the full content.

View Full Document Right Arrow Icon
) we readily obtain: y = β( V ) y = β( V )β( V ) y = β 2 ( V ) y β 2 ( V ) = 1 , which implies β( V ) = ± 1. On the other hand, since we have orientated y and y in the same direction, we must have β( V ) 1. By the same token we also find z = z . Thus the first three equations of the transformations ( 1.33 ) take the simple form: x = α( V )( x V t ), (1.40) y = y , (1.41) z = z . (1.42) Let us now consider the fourth equation involving the time variable t . Solving the first of ( 1.40 ) with respect to t we find: t = 1 V x x α( V ) , (1.43) Using the same argument which led to ( 1.37 ), if we consider S in motion with velocity V with respect to S , the equation obtained from ( 1.43 ) by replacing t with t , x with x and V with V must also be true: t = − 1 V x x α( V ) = − 1 V α( V )( x V t ) x α( V ) = α( V ) t + 1 V 1 α( V ) α( V ) x . (1.44)
Image of page 30

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
1.3 Lorentz Transformations 19 Fig. 1.8 Light signal as seen by S and S We may then rewrite the transformation ( 1.44 ) as follows: t = α( V ) t + δ( V ) x , (1.45) where we have set δ( V ) = 1 V 1 α( V ) α( V ) . (1.46) By simple considerations we have reduced the problem of determining all the coef- ficients in ( 1.33 ), to that of computing a single function α( V ) . This coefficient will be now determined by implementing the principle of con- stancy and isotropy of the speed of light. Let us suppose that at t = t = 0, when O O , a light (or electromagnetic wave) source emits a signal isotropically, see Fig. 1.8 . According to this principle, the signal propagates isotropically with the same constant speed c for both observers S and S . Thus with respect to the two frames the wave front of the electromagnetic signal will be described by spheres of radii r = ct and r = ct respectively. The equations for the wave front of the spherical wave are thus given by: x 2 + y 2 + z 2 c 2 t 2 = 0 , (1.47) for the observer S , and x 2 + y 2 + z 2 c 2 t 2 = 0 , (1.48) for the observer S . Since the four coordinates ( x , y , z , t ) and ( x , y , z , t ) refer to the same physical events, that is the locus of points reached by the signal at a fixed time, they must hold simultaneously. We must then have: x 2 + y 2 + z 2 c 2 t 2 = κ x 2 + y 2 + z 2 c 2 t 2 , (1.49) where κ is a constant. If we now substitute the expression of x , y , z , t in terms of x, y, z, t as given by ( 1.40 ), ( 1.41 ), ( 1.42 ) and ( 1.45 ), in ( 1.49 ), we obtain:
Image of page 31
20 1 Special Relativity x 2 + y 2 + z 2 c 2 t 2 = κ α 2 ( x V t ) 2 + y 2 + z 2 c 2 t + δ x ) 2 , (1.50) and this relation must be an identity in ( x , y , z , t ) . Conparing the coefficients of z and y on both sides, we immediately find κ = 1 . Next, equating the coefficients of t 2 , one finds: α 2 ( V ) V 2 = c 2 1 α 2 ( V ) α( V ) = ± 1 1 V 2 c 2 . Since at t = t = 0, x and x have the same orientation, we conclude that: α( V ) = α( V ) = 1 1 V 2 c 2 . (1.51) One can easily verify that, with the above value of α( V ) , also the coefficients of x 2 and xt are equal. The transformation laws ( 1.40 ), ( 1.41 ), ( 1.42 ) and ( 1.44 ) now take the following final form: x = γ ( V )( x V t ), (1.52) y = y , (1.53) z = z , (1.54) t = γ ( V ) t V c 2 x , (1.55) where γ ( V ) 1 1 V 2 c 2 > 1 . (1.56) Equations ( 1.52 ) are the Lorentz transformations . They represent the correct trans- formation laws connecting two inertial frame, which allow to extend the principle of relativity to electromagnetism.
Image of page 32

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 33
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern