1. Relativistic Invariance (Lorentz invariance) The laws of physics are invariant under a transformation between two coordinate frames moving at constant velocity w.r.t. each other. (The world is not invariant, but the laws of physics are!)

2. Review: Special Relativity Speed of light = C = |r 2 – r 1 | / (t 2 –t 1 ) = |r 2 ’ – r 1 ’ | / (t 2 ’ –t 1 ‘ ) = |dr/dt| = |dr’/dt’| Einstein’s assumption: the speed of light is independent of the (constant ) velocity, v, of the observer. It forms the basis for special relativity.

3. This can be rewritten: d( C t) 2 - |dr| 2 = d( C t’) 2 - |dr’| 2 = 0 d( C t) 2 - dx 2 - dy 2 - dz 2 = d( C t’) 2 – dx’ 2 – dy’ 2 – dz’ 2 d( C t) 2 - dx 2 - dy 2 - dz 2 is an invariant! It has the same value in all frames ( = 0 ). |dr| is the distance light moves in dt w.r.t the fixed frame. C 2 = |dr| 2 /dt 2 = |dr’| 2 /dt’ 2 Both measure the same speed!

4. http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2 A Lorentz transformation relates position and time in the two frames. Sometimes it is called a “boost”.

5. How does one “derive” the transformation? Only need two special cases. Recall the picture of the two frames measuring the speed of the same light signal. Next step: calculate right hand side of Eq. 1 using matrix result for cdt’ and dx’. Eq. 1 Transformation matrix relates differentials a b fh a + b f + h

6. = 0 = 1 = -1

8.  c[-  + v/c] dt =0  = v/c

10. But, we are not going to need the transformation matrix! We only need to form quantities which are invariant under the (Lorentz) transformation matrix. Recall that (cdt) 2 – (dx) 2 – (dy) 2 – (dz) 2 is an invariant. It has the same value in all frames ( = 0 ). This is a special invariant, however.

11. Suppose we consider the four-vector: ( E/c, p x, p y, p z ) (E/c) 2 – ( p x ) 2 – ( p y ) 2 – ( p z ) 2 is also invariant. In the center of mass of a particle this is equal to (mc 2 / c) 2 – (0) 2 – (0) 2 – (0) 2 = m 2 c 2 So, for a particle (in any frame) (E/c) 2 – ( p x ) 2 – ( p y ) 2 – ( p z ) 2 = m 2 c 2

12. covariant and contravariant components* *For more details about contravariant and covariant components see http://web.mst.edu/~hale/courses/M402/M402_notes/M402-Chapter2/M402-Chapter2.pdf

13. The metric tensor, g, relates covariant and contravariant components covariant components contravariant components

14. Using indices instead of x, y, z covariant components contravariant components

15. 4-dimensional dot product You can think of the 4-vector dot product as follows: covariant components contravariant components

16. Why all these minus signs ? Einstein’s assumption (all frames measure the same speed of light) gives : d( C t) 2 - dx 2 - dy 2 – dz 2 = 0 From this one obtains the speed of light. It must be positive! C = [dx 2 + dy 2 + dz 2 ] 1/2 /dt

17. Four dimensional gradient operator covariant components contravariant components

18. 4-dimensional vector component notation x µ  ( x 0, x 1, x 2, x 3 ) µ=0,1,2,3 = ( ct, x, y, z ) = (ct, r) x µ  ( x 0, x 1, x 2, x 3 ) µ=0,1,2,3 = ( ct, -x, -y, -z ) = (ct, -r) contravariant components covariant components

19. partial derivatives  /  x µ   µ = (  /  (ct),  /  x,  /  y,  /  z) = (  /  (ct),  ) 3-dimensional gradient operator 4-dimensional gradient operator

20. partial derivatives  /  x µ   µ = (  /  (ct), -  /  x, -  /  y, -  /  z) = (  /  (ct), -  ) Note this is not equal to   They differ by a minus sign.

21. Invariant dot products using 4-component notation x µ x µ =  µ=0,1,2,3 x µ x µ (repeated index one up, one down)  summation) x µ x µ = (ct) 2 -x 2 -y 2 -z 2 Einstein summation notation covariant contravariant

22. Invariant dot products using 4-component notation  µ  µ =  µ=0,1,2,3  µ  µ (repeated index  summation ) =  2 /  (ct) 2 -  2  2 =  2 /  x 2 +  2 /  y 2 +  2 /  z 2 Einstein summation notation

23. Any four vector dot product has the same value in all frames moving with constant velocity w.r.t. each other. Examples: x µ x µ p µ x µ p µ p µ  µ  µ p µ  µ  µ A µ

24. For the graduate students: Consider ct = f(ct,x,y,z) Using the chain rule: d(ct) = [  f/  (ct)]d(ct) + [  f/  (x)]dx + [  f/  (y)]dy + [  f/  (z)]dz d(ct) = [  (ct)/  x ] dx = L 0 dx dx  = [  x  /  x ] dx = L  dx First row of Lorentz transformation. Summation over implied 4x4 Lorentz transformation.

25. For the graduate students: dx  = [  x  /  x ] dx = L  dx  /  x  = [  x /  x  ]  /  x = L   /  x Invariance: dx  dx  = L   dx  L  dx = (  x  /  x  )(  x  /  x )dx  dx = [  x  /  x ] dx  dx =   dx  dx = dx dx

26. Lorentz Invariance Lorentz invariance of the laws of physics is satisfied if the laws are cast in terms of four-vector dot products! Four vector dot products are said to be “Lorentz scalars”. In the relativistic field theories, we must use “Lorentz scalars” to express the interactions.