Lorentz Covariant Mechanics: the case of the photon (#300)

Author: kloimhardt

src/mentat_collective/emmy/silcm_ch01.clj

@@ -16,6 +16,8 @@
             [mentat-collective.emmy.scheme :refer [define-1 let-scheme lambda]]
             [civitas.repl :as repl]))
 
+;; ##!!Draft Version!!
+
 ;; I investigate a Lorentz covariant Lagrangian that is not widely known. It is discussed in the textbooks of Greiner, "Systems of Particles and Hamiltonian Mechanics" (chapter 21), and also [on-line in: Cline, "Variational Principles in Classical Mechanics"](https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/17%3A_Relativistic_Mechanics/17.06%3A_Lorentz-Invariant_Formulation_of_Lagrangian_Mechanics).
 
 ;; Then I derive, along deBroglie's argument, the momentum-wavelength relation of a free particle.
@@ -174,12 +176,18 @@
 
 ;; The extended Lagrangian:
 
+(define (fourtuple->t v)
+  (ref v 0))
+
+(define (fourtuple->space v)
+  (apply up (rest v)))
+
 (define ((Lc-free-particle mass c) local)
   (let ((v (velocity local)))
     (* 1/2 mass (square c)
        (- (* (/ 1 (square c))
-             (square (apply up (rest v))))
-          (square (ref v 0))
+             (square (fourtuple->space v)))
+          (square (fourtuple->t v))
           1))))
 
 ;; According to chapter 4.3.2 of Struckmeier "Extended Lagrange and Hamilton Formalism for Point Mechanics and Covariant Hamilton Field Theory", this "case is largely disregarded in the literature covering the extended Hamilton- Lagrange formalism (cf., for instance, Lanczos, 1949; Dirac, 1950; Synge, 1960; Goldstein et al., 2002; Johns, 2005), namely there exist extended Lagrangians that are related to a given conventional Lagrangian without being homogeneous forms in the n + 1 velocities".
@@ -207,14 +215,19 @@
   ((Gamma Lc-test-path) 's))
 
 ^:kindly/hide-code
-(define path-dimension (atom true)) #_"true = 4D , false = 2D"
+(define path-dimension (atom 2))
 
 ^:kindly/hide-code
 (define three-path
-  (if @path-dimension
+  (case @path-dimension
+    4
     (up (literal-function 'x)
         (literal-function 'y)
         (literal-function 'z))
+    3
+    (up (literal-function 'x)
+        (literal-function 'y))
+    2
     (up (literal-function 'x))))
 
 (show-tex-expression
@@ -225,11 +238,17 @@
 
 ^:kindly/hide-code
 (define four-path
-  (if @path-dimension
+  (case @path-dimension
+    4
     (up identity
         (literal-function 'x)
         (literal-function 'y)
         (literal-function 'z))
+    3
+    (up identity
+        (literal-function 'x)
+        (literal-function 'y))
+    2
     (up identity
         (literal-function 'x))))
 
@@ -243,8 +262,14 @@
 
 ;; In calculating the Lagrangian action above, $m_0$ was $3$, $c$ set to $1$ and time was $10$. Thus the $-30$ stems from the $-m_0 c^2$.
 
+(define (Lc-momentum Lagrangian)
+  ((partial 2) Lagrangian))
+
+(define ((Lagrange-momentum Lagrangian) w)
+  (compose (Lc-momentum Lagrangian) (Gamma w)))
+
 (define ((Lagrange-equations Lagrangian) w)
-  (- (D (compose ((partial 2) Lagrangian) (Gamma w)))
+  (- (D ((Lagrange-momentum Lagrangian) w))
      (compose ((partial 1) Lagrangian) (Gamma w))))
 
 (show-tex-expression
@@ -257,20 +282,26 @@
 
 ^:kindly/hide-code
 (define Lc-path
-  (if @path-dimension
+  (case @path-dimension
+    4
     (up (literal-function 't)
         (literal-function 'x)
         (literal-function 'y)
         (literal-function 'z))
+    3
+    (up (literal-function 't)
+        (literal-function 'x)
+        (literal-function 'y))
+    2
     (up (literal-function 't)
         (literal-function 'x))))
 
-#_(define Lc-path (up (literal-function 't)
-                    (literal-function 'x)))
-
 (show-tex-expression
   (((Lagrange-equations (Lc-free-particle 'm_0 'c)) Lc-path) 's))
 
+(show-tex-expression
+  (((Lagrange-momentum (Lc-free-particle 'm_0 'c)) Lc-path) 's))
+
 ;;State the Lagrangian.
 
 (show-tex-expression
@@ -289,21 +320,41 @@
 ;; Non-homogeneous but symmetric Lagrangians of this type require a general implicit constraint: $L - pv = 0$ (Cline Eq. 17.6.12, Struckmeier Eq. 21.9)
 
 (define (Lc-Lagrange-constraint Lagrangian)
-  (- Lagrangian (* ((partial 2) Lagrangian) velocity)))
+  (- Lagrangian (* (Lc-momentum Lagrangian) velocity)))
 
 ;; Applied to the free particle (and multiplication with a constant factor), leads to (Struckmeier Eq. 21.12):
 
 (define Lc-constraint-s
-  (down (constantly 0) (compose (* (/ 2 (square 'c) 'm_0)
-                      (Lc-Lagrange-constraint (Lc-free-particle 'm_0 'c)))
-                   (Gamma Lc-path))))
+  (down (constantly 0)
+        (compose (* (/ 2 (square 'c) 'm_0)
+                    (Lc-Lagrange-constraint (Lc-free-particle 'm_0 'c)))
+                 (Gamma Lc-path))))
 
 ^:kindly/hide-code
 (show-eq
   (Lc-constraint-s 's))
 
 ;; The above is a constraint, i.e. needs to be zero, needs to vanish for all physical paths.
 
+;; For the free particle, we should think of it as a constraint on p, a constraint known as the mass shell or on-shell condition, usually written (thereby omitting the c) as $-p^2 = m^2$. This on-shell constraint is the correct version of the popular $E = mc^2$. Let's show that the above constraint indeed is the same as $-p^2 = m^2c^2$:
+
+(define p
+  ((Lagrange-momentum (Lc-free-particle 'm_0 'c)) Lc-path))
+
+(show-tex-expression
+  (p 's))
+
+;; (for defining the next "square", notice that for a 4-vector this involves a minus sign)
+
+(define ((foursquare c) v)
+  (- (square (fourtuple->space v))
+     (/ (square (fourtuple->t v)) (square c))))
+
+(show-eq
+  (down
+    (* (square 'm_0) (square 'c))
+    (- ((foursquare 'c) (p 's)))))
+
 ;; Contrary to constraints that depend on position, this constraint (being about velocity) cannot be easily treated via Lagrange multipliers.
 
 ;; We proceed to explicitely calculate $(Dt(s))^2$
@@ -319,9 +370,7 @@
 (show-eq
   (Dt-squared 's))
 
-;; The above is meant as an equality.
-
-;; We now upgrade the parameter $s$ to a function $s(t)$ which is the inverse of $t(s)$, $s = t^{-1}$. 
+;; We now upgrade the parameter $s$ to a function $s(t)$ which is the inverse of $t(s)$, $s = t^{-1}$.
 
 (define s (literal-function 's))
 
@@ -389,6 +438,152 @@
 
 ;; Thus, the function $s(t)$ is nothing but the proper time.
 
+;; ## The dynamic mass
+
+(define ((Lc-dynamic mass c) local)
+  (let ((q (coordinate local))
+        (v (velocity local)))
+    (* 1/2
+       (- (* (ref q 2)
+             (square (ref v 1)))
+          (* (ref q 2) (square c)
+             (square (ref v 0)))
+          (* (/ (square mass) (ref q 2)) (square c))))))
+
+(define dynamic-path
+  (up (literal-function 't)
+      (literal-function 'x)
+      (literal-function 'sigma)))
+
+;; For the record, I'd like to state the momentum of this particle, it will reveal an aha later.
+
+(show-tex-expression
+  ((foursquare 'c)
+   (((Lagrange-momentum (Lc-dynamic 'm 'c)) dynamic-path) 's)))
+
+;; That said, we move on to the equations of motion are
+
+(define dynamic-equations
+  ((Lagrange-equations (Lc-dynamic 'm 'c)) dynamic-path))
+
+(show-tex-expression
+  (dynamic-equations 's))
+
+;; We do not know yet what the first two equations means, but the last equation will tell.
+
+(define sigma
+  (let ((sigma-sq (square (literal-function 'sigma)))
+        (half-m2-c2 (* 1/2 (square 'c) (square 'm))))
+    ((eq-transformation
+       (fn [eq]
+         (sqrt
+           (* half-m2-c2
+              (/ 1
+                 (/ (+ (* sigma-sq eq)
+                       half-m2-c2)
+                    sigma-sq))))))
+     (down 0 (ref dynamic-equations 2)))))
+
+(show-eq
+  (sigma 's))
+
+;; Taking the $-p^2 = m^2$ on-shell constraint from above, we see that $\sigma(s) = m$
+
+(show-eq
+  (((eq-transformation
+       #(sqrt (+ (square 'c) (* (square 'c) %))))
+    Lc-constraint-s)
+   's))
+
+;; If you look at the equations of motion above, when you let $\sigma(s) \rightarrow m$, you get the motions of the free particle. All this can be achived much easier by directly setting $\sigma(s) = m(s) = m$ in the calculation of the equations of motion.
+
+(define (on-shell-path m)
+  (up (literal-function 't)
+      (literal-function 'x)
+      (constantly m)))
+
+
+(show-tex-expression
+  (((Lagrange-equations (Lc-dynamic 'm 'c))
+    (on-shell-path 'm))
+   's))
+
+;; Indeed these are the euqations of motion of the free farticle plus the on-shell constraint.
+
+
+;; ## The case of the Photon
+
+;; As stated above, the momentum for any mass in general is
+
+(show-tex-expression
+  ((foursquare 'c)
+   (((Lagrange-momentum (Lc-dynamic 'm 'c)) dynamic-path) 's)))
+
+;; We can set $m$ to zero in the Lagrangian
+
+(show-tex-expression
+  (((Lagrange-equations (Lc-dynamic 0 'c))
+    dynamic-path)
+   's))
+
+;; We see from the last equation that the square of the momentum of the massless particle is zero, $p^2 = 0$. It also means that $dx/dt = c$
+
+;; The first two equations are the same, as $ct' = x'$ also means $ct'' = x''$.
+;; The function $\sigma(s)$ remains completely arbitrary, and $s$ remains
+;; a parameter. Trying to get $s$ as a function of $t$ is in vain.
+
+;; Also the original constraint does not reveal anything new:
+
+(show-eq
+  (down 0
+        ((compose (Lc-Lagrange-constraint (Lc-dynamic 0 'c))
+                  (Gamma dynamic-path)) 's)))
+
+;;### The preferred parameter choice for massless particles
+;; But something still can be done. By making $s$ a functions (not of $t$ but of)
+;; some parameter $r$ and setting $\sigma(r)$ to be $\sigma(r) = ds/dr$,
+;; we can simplify the first two Lagrangian equations of motion to:
+
+^:kindly/hide-code
+(show-tex-expression
+((down (* -1 (square 'c) (D (D (literal-function 't))))
+         (D (D (literal-function 'x)))) 'r))
+
+;; Let's start the proof.
+;; The momentum is
+
+(show-tex-expression
+  (((Lagrange-momentum (Lc-dynamic 'm 'c)) dynamic-path) 's))
+
+;; Take the momentum for $x$.
+;; The key insight is recognizing the following:
+;; Since the free Lagrangian depends on $dx/ds$ only,
+;; the Lagrange equation is the derivative of this momentum. So:
+;; $d/ds(\sigma · dx/ds) = 0$,
+;; this means $\sigma · dx/ds = C$ (some constant).
+
+;; Now we introduce the variable $r$ and make s a function of $r$, $s(r)$.
+;; And we choose $\sigma$ to be $\sigma = ds/dr$.
+;; This is a common gauge choice or parametrization condition.
+
+;; Now: $dx/dr = dx/ds · ds/dr = dx/ds · \sigma$.
+;; But we know $\sigma · dx/ds = C$, so:
+;; $dx/dr = C$ (constant!). Therefore:
+
+;; $$\frac{d²x}{dr²} = 0$$
+
+;; To quote Zee, Gravitiy in a nutshell, p215:
+;; > A parameter that makes the equation of motion take on this simple form
+;; > is known as an affine parameter. We don’t have to solve any
+;; > equation of motion to know that (t, x, y, z) is proportional to (1, 1, 0, 0).
+;; > End of story. No need to parametrize this worldline.
+
+;; > But if you insist, you could write X=f(s)(1,1,0,0) with some monotonic
+;; > function $f$. Then $d^2X/ds^2 = (f''/f) · X$
+
+;; > To me, the pages some texts spend on the affine parametrization
+;; > of massless particles literally amounts to much ado about nothing.
+
 ;; ## The deBroglie wavelength
 
 ^:kindly/hide-code