- #1

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Why isn't ##\frac{\partial L}{\partial t}\frac{\partial t}{\partial \dot{q_m}}## included in (5.41), given that ##L## could depend on ##t## explicitly?

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- #1

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- #2

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##t## does not depend on the coordinates.

- #3

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And also ##t## is not varied in Hamilton's principle of least action.

- #4

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Suppose ##(\frac{\partial x}{\partial t})_{y,z}\neq0##. Doesn't that imply ##(\frac{\partial t}{\partial x})_{y,z}=((\frac{\partial x}{\partial t})_{y,z})^{-1}\neq0##?##t## does not depend on the coordinates.

- #5

ShayanJ

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"##t## does not depend on the coordinates" ## \Leftrightarrow (\frac{\partial x}{\partial t})_{y,z}=0##!Suppose ##(\frac{\partial x}{\partial t})_{y,z}\neq0##. Doesn't that imply ##(\frac{\partial t}{\partial x})_{y,z}=((\frac{\partial x}{\partial t})_{y,z})^{-1}\neq0##?

- #6

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Suppose a particle travels at constant velocity, i.e., ##x=vt##. Then ##\frac{\partial x}{\partial t}=v\neq0##. So does ##t## depends on ##x##? I'm confused."##t## does not depend on the coordinates" ## \Leftrightarrow (\frac{\partial x}{\partial t})_{y,z}=0##!

- #7

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You are considering a transformation taking x to q and t to t. Regardless of what transformation you have done, t only depends on t.Suppose a particle travels at constant velocity, i.e., ##x=vt##. Then ##\frac{\partial x}{\partial t}=v\neq0##. So does ##t## depends on ##x##? I'm confused.

- #8

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It turns out immidiately that the Euler-Lagrange equations are forminvariant under arbitrary diffeomorphisms

$$q^{\prime k}=Q^k(q,t).$$

- #9

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In that case, should ##\frac{\partial L}{\partial t}\frac{\partial t}{\partial\dot{q_m}}## be included in (5.41)?

- #10

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No, t still does not depend on the coordinates.In that case, should ∂L∂t∂t∂˙qm∂L∂t∂t∂qm˙\frac{\partial L}{\partial t}\frac{\partial t}{\partial\dot{q_m}} be included in (5.41)?

- #11

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$$L'(q,\dot{q},t)=L[x(q,t);\dot{x}(q,t),t].$$

where

$$\dot{x}^i=\dot{q}^k \frac{\partial x^i}{\partial q^k}+\partial_t x^i.$$

In this equation the ##\partial_t x^i## means the derivative of the explicit time dependence of ##x^i(q,t)##.

Then the equations of motion in the new coordinates read

$$\frac{\partial L'}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L'}{\partial \dot{q}^k}=0.$$

- #12

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Does that mean ##\frac{\partial t}{\partial\dot{q_m}}=0## and ##\frac{\partial t}{\partial q_m}=0##? Why?No, t still does not depend on the coordinates.

Clearly, ##\frac{\partial q_m}{\partial t}\neq0##.

- #13

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What you wrote is claim 5.2, where (5.41) is found. But the proof seems to omit the term ##\frac{\partial L}{\partial t}\frac{\partial t}{\partial\dot{q_m}}## in (5.41).

$$L'(q,\dot{q},t)=L[x(q,t);\dot{x}(q,t),t].$$

where

$$\dot{x}^i=\dot{q}^k \frac{\partial x^i}{\partial q^k}+\partial_t x^i.$$

In this equation the ##\partial_t x^i## means the derivative of the explicit time dependence of ##x^i(q,t)##.

Then the equations of motion in the new coordinates read

$$\frac{\partial L'}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L'}{\partial \dot{q}^k}=0.$$

- #14

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In general, for partial derivatives, ##\partial x^i/\partial y^j## is not equal to the reciprocal of ##\partial y^j/\partial x^i##. You need to consider what these functions actually are functions of and what variable change you are considering. In this case you are keeping t as an independent variable and its parial derivative wrt anything else is zero.Does that mean ##\frac{\partial t}{\partial\dot{q_m}}=0## and ##\frac{\partial t}{\partial q_m}=0##? Why?

Clearly, ##\frac{\partial q_m}{\partial t}\neq0##.

- #15

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There is no such term! It doesn't even make any sense, or how do you define it?

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