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I am confused from the equation 6, why we get Euler-Lagrange equation from equation 8 but not from equation 6?

Why we need to use $\zeta$ as invariant parameter in equation 8 even we already have invariant parameter $s$ in equation 6, in order to get relativistic euler-lagrange equation?

Reference: Relativistic mechanics satya prakash page no.402

Qmechanic
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Sabi Shrestha
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    Actually, you can get the Euler Lagrange equations (ELE) from (6) if you use $x^0$ as your parameter and vary $\vec x$, at the cost of losing sight of the Lorentz invariance (still present but less obvious). As indicated, an arbitrary variation of (6) wouldn’t satisfy the condition $u^2=1$. Another way to remedy that would be to add a Lagrange multiplier. Note that in (8), while you do get the correct equations, you have a gauge invariance since the $\zeta$ parameter is arbitrary, and you only have $3$ physical degrees of freedom, rather than the $4$ used in the variational approach. – LPZ Aug 13 '22 at 13:37
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    Posting images of text and math is very strongly discouraged here. Please use a combination of text and Mathjax instead. It's the site standard and images cannot be usefully searched by the site search engine. – StephenG - Help Ukraine Aug 13 '22 at 14:07
  • StephenG I don't know how to write equation so I need to post image. – Sabi Shrestha Aug 13 '22 at 14:54
  • @SabiShrestha StephenG included a hyperlink to a good MathJax tutorial in their comment. – J. Murray Aug 13 '22 at 15:49
  • Related. – Mr. Feynman Aug 13 '22 at 20:08
  • In the book Analytical Mechanics for Relativity and Quantum Mechanics, Oliver Johns, chapter 5 calculus of variations and the general parametric method are discussed and this is the only book I could find that addresses this question in detail. This is in fact a general problem in calculus of variations. – Mr. Feynman Aug 13 '22 at 20:15

1 Answers1

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  1. Assuming that we are talking about a massive point particle, we know that the arclength $$s~=~c\tau\tag{A}$$ is the speed of light $c$ times the proper time $\tau$ (up to an additive constant), and the 4-velocity $$u^{\mu}~:=~\frac{dx^{\mu}}{d\tau}\tag{B}$$ satisfies $$u^{\mu}u_{\mu}~\stackrel{(A)+(B)+(3)}{=}~c^2\tag{7}.$$ [For the overall sign, compare with the Minkowski sign convention (3).]

  2. The most important point (which Prakash doesn't seems to explain) is now that in the stationary action/Hamilton's principle (in contrast to e.g. Maupertuis' principle) the integration region $[\zeta_1,\zeta_2]$ for the world-line parameter $\zeta$ is kept fixed and the same for all paths/trajectories.

    Also note that the 4 position coordinates $x^{\mu}$ are to varied independently (say, within timelike curves), and that the quantity $$ \dot{x}^{\mu}\dot{x}_{\mu}, \qquad \dot{x}^{\mu}~:=~\frac{dx^{\mu}}{d\zeta}, \tag{C}$$ is not fixed (but say, positive).

  3. The main reason that we cannot pick the arclength $s$ (or equivalently the proper time $\tau$) as the world-line parameter $\zeta$ is that the integration region $[s_1,s_2]$ should then be fixed, but this contradicts the fact that neighboring paths/trajectories clearly generically have different arclengths.

    Moreover, Prakash points out that if $\zeta=\tau$ then the 4 position coordinates $x^{\mu}$ cannot be varied independently because of the constraint (7), cf. eqs. (A)+(B)+(3), i.e. there are only 3 independent position variables, so the variational principle (in its current form) does also not work for this reason.

Qmechanic
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  • Make me correct if I am wrong, What I understood from your answer is that each trajectories have it's own proper time or arclength so that each trajectories have its own region of integration so it doesn't make sense to compare arclength of those trajectories within different region of integration with each other so in order to compare the arc length of those trajectories within the same region of integration [$/zeta_1,/zeta_2$] we have to introduce another invariant parameter $/zeta$ such that we can compare and pick up the minimum trajectory. Right ? – Sabi Shrestha Aug 13 '22 at 14:09
  • It seems you got the main point. – Qmechanic Aug 13 '22 at 14:12
  • But what's the point of using the normalization of four velocity $u^{\mu} u_{\mu}$=$c^{2}$ ? Is it all that we only want to consider the four velocity of massive particle has to satisfy that normalization property ? – Sabi Shrestha Aug 13 '22 at 15:06
  • From your point no.3 That means when $u^{\mu} u_{\mu}=1$ (in geometric units) then $s$=$\zeta$ which again make difficulty to compare neighbouring trajectories due to different region of integration is that right ? – Sabi Shrestha Aug 13 '22 at 15:25
  • Sorry but I don't get your point on updated answer. – Sabi Shrestha Aug 13 '22 at 15:49
  • But I still don't get the point of using $u^{\mu} u_{\mu}=c^{2}$ will you please make me clear ? – Sabi Shrestha Aug 13 '22 at 16:08
  • I updated the answer. – Qmechanic Aug 13 '22 at 16:24
  • All we want is to vary $x^{\mu}$ w.r.t only one invariant parameter i.e $\zeta$ which do not affect by the constraint (7) but if we choose invariant parameter to be $\tau$ then by constraint (7) we cannot vary $x^{\mu}$ w.r.t $\tau$ because for every trajectories there is a unique Interval for every $\tau$, is that right understanding? – Sabi Shrestha Aug 13 '22 at 16:33