Derivation of Euler-Lagrange equation with change of coordinates

[Happiness]

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?

 

[Orodruin]

$t$ does not depend on the coordinates.

 

[vanhees71]

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

 

[Happiness]

$t$ does not depend on the coordinates.

Suppose $(\frac{\partial x}{\partial t})_{y,z}\neq0$. Doesn't that imply $(\frac{\partial t}{\partial x})_{y,z}=((\frac{\partial x}{\partial t})_{y,z})^{-1}\neq0$?

 

[ShayanJ]

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" $\Leftrightarrow (\frac{\partial x}{\partial t})_{y,z}=0$!

 

[Happiness]

"$t$ does not depend on the coordinates" $\Leftrightarrow (\frac{\partial x}{\partial t})_{y,z}=0$!

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.

 

[Orodruin]

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.

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.

 

[vanhees71]

Well, of course you can have transformations from one set of generalized coordinates to another that depends explicitly on time. That's, e.g., useful if you want to describe the motion in non-inertial frames starting from the physics in an inertial frame. Still this has nothing to do with the variation, because by definition in Hamilton's principle time is not varied.

It turns out immidiately that the Euler-Lagrange equations are forminvariant under arbitrary diffeomorphisms
$$q^{\prime k}=Q^k(q,t).$$

 

[Happiness]

Indeed, the new coordinates may depend explicitly on time by (5.38) below.

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

 

[Orodruin]

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)?

No, t still does not depend on the coordinates.

 

[vanhees71]

I don't know what Eq. (5.41) is, but you should get the form-invariant Euler-Lagrange equations also in the new coordinates, i.e., you have given
$$L'(q,\dot{q},t)=L[x(q,t);\dot{x}(q,t),t].$$
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
$$\frac{\partial L'}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L'}{\partial \dot{q}^k}=0.$$

 

[Happiness]

No, t still does not depend on the coordinates.

Does that mean $\frac{\partial t}{\partial\dot{q_m}}=0$ and $\frac{\partial t}{\partial q_m}=0$? Why?

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

 

[Happiness]

I don't know what Eq. (5.41) is, but you should get the form-invariant Euler-Lagrange equations also in the new coordinates, i.e., you have given
$$L'(q,\dot{q},t)=L[x(q,t);\dot{x}(q,t),t].$$
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
$$\frac{\partial L'}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L'}{\partial \dot{q}^k}=0.$$

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).

 

[Orodruin]

Does that mean $\frac{\partial t}{\partial\dot{q_m}}=0$ and $\frac{\partial t}{\partial q_m}=0$? Why?

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

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.

 

[vanhees71]

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