Emmy, the Algebra System: Lorentz Covariant Mechanics
##!!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”.
Then I derive, along deBroglie’s argument, the momentum-wavelength relation of a free particle.
Extended Lagrangian
I start with the conventional case
(define ((L-free-particle mass) local)
(let ((v (velocity local)))
(* 1/2 mass (square v))))(define (test-path t)
(up (+ (* 4 t) 7)
(+ (* 3 t) 5)
(+ (* 2 t) 1)))(show-tex-expression
(test-path 't))\[\begin{pmatrix}\displaystyle{4\,t + 7} \cr \cr \displaystyle{3\,t + 5} \cr \cr \displaystyle{2\,t + 1}\end{pmatrix}\]
(Lagrangian-action (L-free-particle 3.0) test-path 0.0 10.0)435.0The 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 (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”.
(define (Lc-test-path s)
(up s
(+ (* 4 s) 7)
(+ (* 3 s) 5)
(+ (* 2 s) 1)))(show-tex-expression
(Lc-test-path 's))\[\begin{pmatrix}\displaystyle{s} \cr \cr \displaystyle{4\,s + 7} \cr \cr \displaystyle{3\,s + 5} \cr \cr \displaystyle{2\,s + 1}\end{pmatrix}\]
(Lagrangian-action (Lc-free-particle 3.0 1.0) Lc-test-path 0.0 10.0)405.0That’s 30 less, let’s see where we lost those 30
(show-tex-expression
((Gamma test-path) 't))\[\begin{pmatrix}\displaystyle{t} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{4\,t + 7} \cr \cr \displaystyle{3\,t + 5} \cr \cr \displaystyle{2\,t + 1}\end{pmatrix}} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{4} \cr \cr \displaystyle{3} \cr \cr \displaystyle{2}\end{pmatrix}}\end{pmatrix}\]
Can you guess the following? (if not, use the sidebar!)
^{:kindly/kind :kind/hidden}
(show-tex-expression
((Gamma Lc-test-path) 's))(show-tex-expression
((Gamma three-path) 't))\[\begin{pmatrix}\displaystyle{t} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{x\left(t\right)}\end{pmatrix}} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{Dx\left(t\right)}\end{pmatrix}}\end{pmatrix}\]
(show-tex-expression
((compose (L-free-particle 'm) (Gamma three-path)) 't))\[\frac{1}{2}\,m\,{\left(Dx\left(t\right)\right)}^{2}\]
(show-tex-expression
((Gamma four-path) 's))\[\begin{pmatrix}\displaystyle{s} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{s} \cr \cr \displaystyle{x\left(s\right)}\end{pmatrix}} \cr \cr \displaystyle{\begin{pmatrix}\displaystyle{1} \cr \cr \displaystyle{Dx\left(s\right)}\end{pmatrix}}\end{pmatrix}\]
Note the number 1 on top of the last 4-tuple. This will be relaxed later.
(show-tex-expression
((compose (Lc-free-particle 'm_0 'c) (Gamma four-path)) 's))\[- {c}^{2}\,m_0 + \frac{1}{2}\,m_0\,{\left(Dx\left(s\right)\right)}^{2}\]
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 ((Lagrange-momentum Lagrangian) w))
(compose ((partial 1) Lagrangian) (Gamma w))))(show-tex-expression
(((Lagrange-equations (L-free-particle 'm)) three-path) 't))\[\begin{bmatrix}\displaystyle{m\,{D}^{2}x\left(t\right)}\end{bmatrix}\]
(show-tex-expression
(((Lagrange-equations (Lc-free-particle 'm_0 'c)) four-path) 's))\[\begin{bmatrix}\displaystyle{0}&\displaystyle{m_0\,{D}^{2}x\left(s\right)}\end{bmatrix}\]
The first \(0\) is because we set \(t=s\) for the path. This reduces the extended Lagrangian to the “conventional” case. Reassuring. Now we can allow arbitrary functions \(t(s)\), making the path truely 4-dimensional.
(show-tex-expression
(((Lagrange-equations (Lc-free-particle 'm_0 'c)) Lc-path) 's))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,m_0\,{D}^{2}t\left(s\right)}&\displaystyle{m_0\,{D}^{2}x\left(s\right)}\end{bmatrix}\]
(show-tex-expression
(((Lagrange-momentum (Lc-free-particle 'm_0 'c)) Lc-path) 's))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,m_0\,Dt\left(s\right)}&\displaystyle{m_0\,Dx\left(s\right)}\end{bmatrix}\]
State the Lagrangian.
(show-tex-expression
((compose (Lc-free-particle 'm_0 'c) (Gamma Lc-path)) 's))\[\frac{-1}{2}\,{c}^{2}\,m_0\,{\left(Dt\left(s\right)\right)}^{2} + \frac{-1}{2}\,{c}^{2}\,m_0 + \frac{1}{2}\,m_0\,{\left(Dx\left(s\right)\right)}^{2}\]
According to Struckmeier p80, “In contrast to the non-relativistic description, the constant rest energy term \(− 1/2 mc^2\) in the extended Lagrangian is essential”. I guess this constant term makes the Lagrangian “not being a homogeneous form in the velocities”.
It is very interesting that a constant term can lead to a new formalism. Cline: “It provides a plausible manifestly-covariant Lagrangian for the one-body system, but serious problems exist extending this to the N-body system when N>1”.
Nobody possesses a Lorentz covariant N-body mechanics. Goldstein “Classical Mechanics”, chapter 7.9: “Hitherto, there does not exist a satisfying description of the relativistic many body system.”
The Constraint on the four dimensional path
The Lagranian above looks very nice because it is symmtreic in t, x, y, z. But there is a caveat which ultimately brings it in line with the orthodox formulation.
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 (* (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))))\[\begin{bmatrix}\displaystyle{0}&\displaystyle{=}&\displaystyle{\frac{{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2} - {c}^{2} - {\left(Dx\left(s\right)\right)}^{2}}{{c}^{2}}}\end{bmatrix}\]
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))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,m_0\,Dt\left(s\right)}&\displaystyle{m_0\,Dx\left(s\right)}\end{bmatrix}\]
(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)))))\[\begin{bmatrix}\displaystyle{{c}^{2}\,{m_0}^{2}}&\displaystyle{=}&\displaystyle{{c}^{2}\,{m_0}^{2}\,{\left(Dt\left(s\right)\right)}^{2} - {m_0}^{2}\,{\left(Dx\left(s\right)\right)}^{2}}\end{bmatrix}\]
Contrary to constraints that depend on position, this constraint (being about velocity) cannot be easily treated via Lagrange multipliers.
We proceed to explicitely calculate \((Dt(s))^2\)
(define t (literal-function 't))(define Dt-squared
((eq-transformation
#(- (square (D t)) %))
Lc-constraint-s))\[\begin{bmatrix}\displaystyle{{\left(Dt\left(s\right)\right)}^{2}}&\displaystyle{=}&\displaystyle{\frac{{c}^{2} + {\left(Dx\left(s\right)\right)}^{2}}{{c}^{2}}}\end{bmatrix}\]
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))Inverse means that \(t(s(y)) = y\). While in the math rendering there is a minor risk of confusion, in the underlying code we do not need the auxilary y, the function t and the symbol 't are clearly distinct:
(define t°s (down (comp t s) identity))(show-eq
(t°s 't))\[\begin{bmatrix}\displaystyle{t\left(s\left(t\right)\right)}&\displaystyle{=}&\displaystyle{t}\end{bmatrix}\]
The derivative results in the number one
(show-eq
((D t°s) 't))\[\begin{bmatrix}\displaystyle{Dt\left(s\left(t\right)\right)\,Ds\left(t\right)}&\displaystyle{=}&\displaystyle{1}\end{bmatrix}\]
With this interlude, we return to the constraint. Just by formally replacing \(s \rightarrow s(t)\), it reads as
(show-eq
(Dt-squared (s 't)))\[\begin{bmatrix}\displaystyle{{\left(Dt\left(s\left(t\right)\right)\right)}^{2}}&\displaystyle{=}&\displaystyle{\frac{{c}^{2} + {\left(Dx\left(s\left(t\right)\right)\right)}^{2}}{{c}^{2}}}\end{bmatrix}\]
We’d like to get rid of the function t and only keep the symbol/parameter 't. For this, we multiply the above equation by \((Ds(t))^2\)
(define eq1
((eq-transformation
#(* ((square (D s)) 't) %))
(* (Dt-squared (s 't)))))(show-eq eq1)\[\begin{bmatrix}\displaystyle{{\left(Dt\left(s\left(t\right)\right)\right)}^{2}\,{\left(Ds\left(t\right)\right)}^{2}}&\displaystyle{=}&\displaystyle{\frac{{c}^{2}\,{\left(Ds\left(t\right)\right)}^{2} + {\left(Ds\left(t\right)\right)}^{2}\,{\left(Dx\left(s\left(t\right)\right)\right)}^{2}}{{c}^{2}}}\end{bmatrix}\]
We introduce the following slighly sloppy notation
(show-eq
((down three-path (comp three-path s)) 't))\[\begin{bmatrix}\displaystyle{\begin{pmatrix}\displaystyle{x\left(t\right)}\end{pmatrix}}&\displaystyle{=}&\displaystyle{\begin{pmatrix}\displaystyle{x\left(s\left(t\right)\right)}\end{pmatrix}}\end{bmatrix}\]
The derivative of this definition is
(show-eq
((D (down three-path (comp three-path s))) 't))\[\begin{bmatrix}\displaystyle{\begin{pmatrix}\displaystyle{Dx\left(t\right)}\end{pmatrix}}&\displaystyle{=}&\displaystyle{\begin{pmatrix}\displaystyle{Ds\left(t\right)\,Dx\left(s\left(t\right)\right)}\end{pmatrix}}\end{bmatrix}\]
With this the constraint becomes
(show-eq
constraint-1)\[\begin{bmatrix}\displaystyle{1}&\displaystyle{=}&\displaystyle{\frac{{c}^{2}\,{\left(Ds\left(t\right)\right)}^{2} + {\left(Dx\left(t\right)\right)}^{2}}{{c}^{2}}}\end{bmatrix}\]
Remember that, as shown above, the number one appears because of \(s = t^{-1}\)
Isolating \(Ds(t)\) on the right hand side
(show-eq
((eq-transformation
#(sqrt (- % (/ ((square (D three-path)) 't) (square 'c)))))
constraint-1))2026-02-08T20:17:32.056Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (+ (expt c 2) (* -1 (expt ((D x) t) 2)))) in e3
2026-02-08T20:17:32.056Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (expt c 2)) in e3
2026-02-08T20:17:32.058Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (= (sqrt (expt c 2)) (expt c 1)) in simsqrt1
2026-02-08T20:17:32.062Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (= (sqrt (expt ((D s) t) 2)) (expt ((D s) t) 1)) in simsqrt1
\[\begin{bmatrix}\displaystyle{\frac{\sqrt {{c}^{2} - {\left(Dx\left(t\right)\right)}^{2}}}{c}}&\displaystyle{=}&\displaystyle{Ds\left(t\right)}\end{bmatrix}\]
The above equation, which is just another expression of the constraint \(L - pv = 0\), is conventionally written as below, introducing the famous \(\gamma\) factor of Special Relativity.
\[\frac{ds}{dt} = \sqrt{1- \frac{v^2}{c^2}} = \sqrt{1 - \beta ^ 2} = \frac{1}{\gamma}\]
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)))\[- {c}^{2}\,{\left(Dt\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2} + {\left(Dx\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2}\]
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))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,{D}^{2}t\left(s\right)\,\sigma\left(s\right) - {c}^{2}\,Dt\left(s\right)\,D\sigma\left(s\right)}&\displaystyle{Dx\left(s\right)\,D\sigma\left(s\right) + {D}^{2}x\left(s\right)\,\sigma\left(s\right)}&\displaystyle{\frac{\frac{1}{2}\,{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2} + \frac{-1}{2}\,{c}^{2}\,{m}^{2} + \frac{-1}{2}\,{\left(Dx\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2}}{{\left(\sigma\left(s\right)\right)}^{2}}}\end{bmatrix}\]
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))2026-02-08T20:17:32.088Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (* 2N (expt (sigma s) 2))) in e3
2026-02-08T20:17:32.088Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? 2N) in e3
2026-02-08T20:17:32.088Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? 2N) in e1
2026-02-08T20:17:32.089Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (expt (sigma s) 2)) in e1
2026-02-08T20:17:32.089Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (= (sqrt (expt (sigma s) 2)) (expt (sigma s) 1)) in simsqrt1
2026-02-08T20:17:32.094Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (* 2N (expt c 2) (expt m 2))) in e3
2026-02-08T20:17:32.094Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (+ (* 2N (expt c 2) (expt ((D t) s) 2)) (* -2N (expt ((D x) s) 2)))) in e3
2026-02-08T20:17:32.095Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? 2N) in e2
2026-02-08T20:17:32.095Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (expt c 2)) in e2
2026-02-08T20:17:32.095Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (expt c 2)) in e1
2026-02-08T20:17:32.096Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (expt m 2)) in e1
2026-02-08T20:17:32.096Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (= (sqrt (expt m 2)) (expt m 1)) in simsqrt1
2026-02-08T20:17:32.099Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? 2N) in c3
2026-02-08T20:17:32.099Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (+ (* 2N (expt c 2) (expt ((D t) s) 2)) (* -2N (expt ((D x) s) 2)))) in c3
2026-02-08T20:17:32.105Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? 1N) in e3
2026-02-08T20:17:32.105Z runnervmwffz4 WARN [emmy.util.logic:22] - Assuming (non-negative? (+ (* (expt c 2) (expt ((D t) s) 2)) (* -1N (expt ((D x) s) 2)))) in e3
\[\begin{bmatrix}\displaystyle{\sigma\left(s\right)}&\displaystyle{=}&\displaystyle{\frac{c\,m}{\sqrt {{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2} - {\left(Dx\left(s\right)\right)}^{2}}}}\end{bmatrix}\]
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))\[\begin{bmatrix}\displaystyle{c}&\displaystyle{=}&\displaystyle{\sqrt {{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2} - {\left(Dx\left(s\right)\right)}^{2}}}\end{bmatrix}\]
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))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,m\,{D}^{2}t\left(s\right)}&\displaystyle{m\,{D}^{2}x\left(s\right)}&\displaystyle{\frac{1}{2}\,{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2} + \frac{-1}{2}\,{c}^{2} + \frac{-1}{2}\,{\left(Dx\left(s\right)\right)}^{2}}\end{bmatrix}\]
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)))\[- {c}^{2}\,{\left(Dt\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2} + {\left(Dx\left(s\right)\right)}^{2}\,{\left(\sigma\left(s\right)\right)}^{2}\]
We can set \(m\) to zero in the Lagrangian
(show-tex-expression
(((Lagrange-equations (Lc-dynamic 0 'c))
dynamic-path)
's))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,{D}^{2}t\left(s\right)\,\sigma\left(s\right) - {c}^{2}\,Dt\left(s\right)\,D\sigma\left(s\right)}&\displaystyle{Dx\left(s\right)\,D\sigma\left(s\right) + {D}^{2}x\left(s\right)\,\sigma\left(s\right)}&\displaystyle{\frac{1}{2}\,{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2} + \frac{-1}{2}\,{\left(Dx\left(s\right)\right)}^{2}}\end{bmatrix}\]
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)))\[\begin{bmatrix}\displaystyle{0}&\displaystyle{=}&\displaystyle{\frac{1}{2}\,{c}^{2}\,{\left(Dt\left(s\right)\right)}^{2}\,\sigma\left(s\right) + \frac{-1}{2}\,{\left(Dx\left(s\right)\right)}^{2}\,\sigma\left(s\right)}\end{bmatrix}\]
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:
\[\begin{bmatrix}\displaystyle{- {c}^{2}\,{D}^{2}t\left(r\right)}&\displaystyle{{D}^{2}x\left(r\right)}\end{bmatrix}\]
Let’s start the proof. The momentum is
(show-tex-expression
(((Lagrange-momentum (Lc-dynamic 'm 'c)) dynamic-path) 's))\[\begin{bmatrix}\displaystyle{- {c}^{2}\,Dt\left(s\right)\,\sigma\left(s\right)}&\displaystyle{Dx\left(s\right)\,\sigma\left(s\right)}&\displaystyle{0}\end{bmatrix}\]
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.
(repl/scittle-sidebar)