# I Derivation of Euler-Lagrange equation with change of coordinates

1. Jul 28, 2016

### 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?

Last edited: Jul 28, 2016
2. Jul 28, 2016

### Orodruin

Staff Emeritus
$t$ does not depend on the coordinates.

3. Jul 28, 2016

### vanhees71

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

4. Jul 28, 2016

### Happiness

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

5. Jul 28, 2016

### ShayanJ

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

6. Jul 28, 2016

### Happiness

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.

7. Jul 28, 2016

### Orodruin

Staff Emeritus
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.

8. Jul 28, 2016

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

9. Jul 28, 2016

### 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)?

10. Jul 28, 2016

### Orodruin

Staff Emeritus
No, t still does not depend on the coordinates.

11. Jul 28, 2016

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

12. Jul 28, 2016

### Happiness

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

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

13. Jul 28, 2016

### Happiness

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).

14. Jul 28, 2016

### Orodruin

Staff Emeritus
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.

15. Jul 28, 2016

### vanhees71

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