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I was reading the part that Euler-Lagrange equation holds even on changing the coordinates. In the book by David Morin, the author talks of geometrical picture of the change of coordinates. He makes the following point,

"If you plot a function and then stretch the horizontal axis in an arbitrary manner, a stationery value will be a stationary value after stretching."

My question is, how is change of coordinates equivalent to stretching the horizontal axis?

Qmechanic
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Ankur Singh
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    That is true of stationary value, stretching the function horizontally merely changes it's location on the x-axis; but this is true not only of stationary values, its true of any value; so I'm not sure what the relevance of this is - perhaps if you quote more of Morins text ... – Mozibur Ullah Dec 19 '17 at 16:05
  • -1 A stationary value changes eg under a rotation of the co-ordinate system. What is the wider context in which the quote is made? Probably you are taking what the author says out of context. – sammy gerbil Dec 19 '17 at 17:03
  • While reading the principal of stationary action, I encountered the statement that E-L equation is coordinate independent. In order to clarify that this statement holds even on changing the coordinates, the author says that changing the coordinates is same as stretching the horizontal axis. Since on stretching the horizontal axis, the position of stationary points (f'=0) doesn't change, one can easily see that E-L equation holds even when one changes the coordinate (new coordinate system must be function of original coordinates only, not of generalized velocities). – Ankur Singh Dec 20 '17 at 07:49
  • My question was how we can say that stretching the axis is analogous to switching the coordinates. – Ankur Singh Dec 20 '17 at 07:50

2 Answers2

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David Morin's pictorial argument of horizontal stretching is already pretty convincing. Alternatively, covariance follows from the fact that the functional derivative/EL-expression $$ \frac{\delta S[q]}{\delta q^i(t)}~=~\frac{\partial q^{\prime j}}{\partial q^i}\frac{\delta S[q]}{\delta q^{\prime j}(t)}\tag{1}$$ transforms as a co-vector [i.e. a $(0,1)$ tensor] under coordinate transformations $$q^{\prime j}~=~f^j(q,t), \tag{2}$$ cf. e.g. this Phys.SE post.

Qmechanic
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If one wants to rescale the horizontal axis, then is simply means that one scale the independent variable of a function $$ f(x) \rightarrow f(ax) , $$ where $a$ is an arbitrary constant. For example, if $$ f(x) = \cos(kx) , $$ and $a=2$ then $$ f(2x) = \cos(2kx) . $$ The latter will oscillate twice as fast as the former when plotted as a function of $x$.

flippiefanus
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