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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

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    $\begingroup$ 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. $\endgroup$
    – lpz
    Aug 13 at 13:37
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    $\begingroup$ 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. $\endgroup$ Aug 13 at 14:07
  • $\begingroup$ StephenG I don't know how to write equation so I need to post image. $\endgroup$ Aug 13 at 14:54
  • $\begingroup$ @SabiShrestha StephenG included a hyperlink to a good MathJax tutorial in their comment. $\endgroup$
    – J. Murray
    Aug 13 at 15:49
  • $\begingroup$ Related. $\endgroup$
    – Feynman_00
    Aug 13 at 20:08

1 Answer 1

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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.

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  • $\begingroup$ 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 ? $\endgroup$ Aug 13 at 14:09
  • $\begingroup$ It seems you got the main point. $\endgroup$
    – Qmechanic
    Aug 13 at 14:12
  • $\begingroup$ 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 ? $\endgroup$ Aug 13 at 15:06
  • $\begingroup$ 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 ? $\endgroup$ Aug 13 at 15:25
  • $\begingroup$ Sorry but I don't get your point on updated answer. $\endgroup$ Aug 13 at 15:49

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