Minkowski Space Algebra Lets go farther with 4-vector algebra! For example, define the SCALAR PRODUCT of 2 4-vectors. A B ABSuppose A = (A 0,A 1,A 2,A 3 )  (A 0,A) is a 4-vector and B = (B 0,B 1,B 2,B 3 )  (B 0,B) is another 4-vector. The SCALAR PRODUCT of A & B is defined as: A  B  A 0 B 0 - A  B = A 0 B 0 - A 1 B 1 - A 2 B 2 - A 3 B 3 We can show that this scalar product is also a Lorentz invariant. That is, it does not change on a Lorentz Transformation between 2 inertial frames: AB ´´´A´B´ A  B = A 0 B 0 - A  B = A 0 ´ B 0 ´ - A´  B´ = A´  B´ Obviously, the invariant length or the magnitude of a 4-vector is a special case of this, where A = B.

To properly do much else with Minkowski space algebra, we need to distinguish 2 kinds (or classes) of 4-vectors: Contravariant & Covariant. This is done in detail in both Jackson & Goldstein. In order for you to understand this if you are reading these books, we need a brief discussion of these. ADEFINE a CONTRAVARIANT 4-VECTOR: A set of 4 numbers A = (A 0,A 1,A 2,A 3 )  (A 0,A) which transform with a Lorentz Transformation the same way that the 4 position coordinates (x 0,x 1,x 2,x 3 ) transform. By this definition, each 4- vector we’ve discussed is a contravariant 4-vector! But, note the change in notation from subscripts on the components to superscripts! Standard notation for contravariant vectors! DEFINE a COVARIANT 4-VECTOR: Each contravariant 4-vector has a covariant “partner” defined as the set of 4 numbers A c A c = (A 0,-A 1,-A 2,-A 3 )  (A 0,-A). Note the subscripts on the components! Standard notation for covariant vectors!

Now, we can again define the SCALAR PRODUCT. A BIf A =(A 0,A 1,A 2,A 3 )  (A 0,A) & B = (B 0,B 1,B 2,B 3 )  (B 0,B) are 2 contravariant 4-vectors, their SCALAR PRODUCT is obtained as the sum of the products of the components of A with the components of the covariant partner of B: AB A  B  A 0 B 0 - A  B = A 0 B 0 + A 1 B 1 + A 2 B 2 + A 3 B 3 = ∑ μ A μ B μ We can also define vector operators in 4d Minkowski spacetime. For example: the 4d generalization of the  operator: Recall that in 3d (Cartesian coordinates) we have:   (∂/∂x 1 )i + (∂/∂x 2 )j + (∂/∂x 3 )k (Vector operator!) In 4d spacetime, define the generalizations of this which have both contravariant & covariant forms: ❒  ([∂/∂x 0 ],  ) (Contravariant) ❒ c  ([∂/∂x 0 ],-  ) (Covariant)

AUsing this, along with the definition of the scalar product, if A is a contravariant 4-vector: A = (A 0,A 1,A 2,A 3 )  (A 0,A) we can define its 4-divergence as: ❒  A  (∂A 0 /∂x 0 ) +  A Going further, can define the 4d generalization of the Laplacian operator as: ❒ 2  ❒  ❒ c = (∂ 2 /∂x 0 ∂x 0 ) -  2  c -2 (∂ 2 /∂t 2 ) -  2 Mathematicians sometimes call this ❒ 2 operator the “D’Alembertian”. Obviously, it is the operator related to a wave equation. Jackson & Goldstein both use this type of notation to write Maxwell’s equations of E&M.

In 3d space tensors are defined in terms of their transformation properties under rotations. In 4d Minkowski spacetime, tensors are defined in terms of their transformation properties under Lorentz transformations. Just as for 4 Vectors (tensors of rank 1), there are contravariant & covariant tensors of all ranks. Also there are “mixed” tensors with one contravariant (superscript) index & one covariant (subscript). See Goldstein & Jackson for details. Its worth noting that, as discussed in Goldstein & Jackson, Maxwell’s equations for the electric field E & the magnetic field B in Minkowski space, can be written elegantly in terms of a 2 nd rank contravariant (or covariant) Minkowski tensor called F, with elements F αβ which are ordinary vector components of the B field (spacelike parts) & the E field (timelike parts). See Goldstein & Jackson! This takes us too far away from mechanics! Take E&M II (Phys. 6306!).

Sect. 7.7: Relativistic Kinematics (with lots of input from Marion!) Consider 2 colliding particles m 1, m 2. If the 2 velocities are significant fractions of c, we must use relativity to analyze. Recall: In the standard (Newtonian) treatment of the 2 body problem, we change coordinates from r 1,r 2 to Center of Mass coordinate R & relative coordinate r. As we’ve seen, mass & energy are intimately related in Relativity, so it’s no longer (rigorously) meaningful to speak of a “Center of Mass” system. Instead, a “Center of Momentum” system is used. The Center of Momentum (COM) system is DEFINED as the system in which the total (3d) linear momentum is zero. To analyze a collision: Choose the COM system as inertial frame S´ & the lab system as inertial frame S. To go from one to another just apply a Lorentz transformation!

Procedure: Solve the problem in the COM system & then transform back to the lab system with a Lorentz transformation. Assume an elastic collision. Figure: 2 colliding particles m 1, m 2. Views of the collision in the COM system and in the lab system.

2 particles m 1,m 2. In the lab system S, choose coordinates so that m 1 moves initially along the x axis with velocity u 1. Assume initially m 2 is at rest, u 2 = 0. In the COM system S´, m 2 initially moves with velocity u 2 ´.  S´ is moving with respect to S with velocity u 2 ´ in the opposite direction as the initial motion of m 1. Figure. Goal: Find the scattering angle Θ in the COM frame, the scattering angle θ in the lab frame & the final momenta p 3 & p 4 in the lab frame. Recall definition of relativistic 4-Vector momentum: P = (p 0,p) = γ u m(c,u) = [(E/c),p]

COM frame definition:  p 1 ´ = p 2 ´ (1) Along the x direction. The 3d momentum is: p = γ u mu γ u  [1 - (β u ) 2 ] -½, β u  (u/c), Identity: γ u β u  [(γ u ) 2 -1] ½ Solve for γ u ‘s & then get the u’s.  (1) becomes: γ 1 ´m 1 u 1 ´= γ 2 ´m 2 u 2 ´ (2) Consider: γ i ´u i ´= γ i ´c(u i ´/c) = cγ i ´β i ´ = c[(γ i ´) 2 -1] ½  (2) is: m 1 c[(γ 1 ´) 2 - 1] ½ = m 2 c[(γ 2 ´) 2 - 1] ½ (3) The Lorentz transformation from the lab frame S to the COM frame S´ (moving with u 2 ´) gives: p 1 ´ = γ 2 ´[p 1 - ( β 2 ´/c)E 1 ] (4) Also have the (definitions): p 1 = γ 1 m 1 u 1, E 1 = γ 1 m 1 c 2 (5) Solve by combining (2)-(5) to get γ 1 ´& γ 2 ´ in terms of γ 1.

Results are (student exercise!), defining q  (m 1 /m 2 ) γ 1 ´ = [γ 1 + q][1 + 2γ 1 q +q 2 ] -½ (6a) γ 2 ´ = [γ 1 + q -1 ][1 + 2γ 1 q -1 + q -2 ] -½ (6b) Note that the final momenta of both m 1 & m 2 have both x & y components in both frames! Now, use conservation of momentum again in the S´ frame & get p 3x ´ = p 1 ´cosΘ, p 3y ´= p 1 ´sinΘ. Similarly for p 4x ´, p 4y ´.

Finally, write equations for the inverse Lorentz transformation from the COM system S´ back to The lab system S after scattering. Final Lab frame Momentum of m 1 (see figure): p 3x = γ 2 ´[p 3x ´ + (β 2 ´/c)E 1 ´] Algebra (combining a number of previous equations) gives: p 3x = m 1 cγ 1 ´γ 2 ´[β 1 ´cosΘ + β 2 ´] (a) Similarly, get p 3y = m 1 cγ 1 ´β 1 ´sinΘ (b) From (a) & (b) get lab frame scattering angle θ: tanθ = (p 3y /p 3x ) = [(γ 2 ´) -1 sinΘ][cosΘ + (β 2 ´/β 1 ´)] -1

Similarly for m 2 : Final Lab frame Momentum of m 2 (see figure): p 4x = γ 2 ´[p 4x ´ + (β 2 ´/c)E 2 ´] Algebra (combining a number of previous eqtns) gives: p 4x = m 2 c(γ 2 ´) 2 β 2 ´(1 - cosΘ) (c) Similarly, we get p 4y = -m 2 cγ 2 ´β 2 ´sinΘ (d) From (c) & (d), get the lab frame recoil angle ζ: tanζ = (p 4y /p 4x ) = [(γ 2 ´) -1 sinΘ][1- cosΘ] -1

A special case of interest is m 1 = m 2.  γ 1 ´ = γ 2 ´ = [(½) + (½)γ 1 ] ½  tanθ = [(½) + (½)γ 1 ] -½ (sinΘ)(1 + cosΘ) -1 and tanζ = -[(½) + (½)γ 1 ] -½ (sinΘ)(1 - cosΘ) -1