The derivative of velocity with respect to a coordinate

[olgerm]

Why $\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial x}=-mg$? why $\frac{\partial \dot{x}}{\partial x}=0$?


Why $\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial \dot{x}}=m*\dot{x}$ ? why $\frac{\partial x}{\partial \dot{x}}=0$?

Does it assume that speed is same at every location?
I know that $\dot{x}=\frac{\partial x}{\partial t}$.

 

[PeroK]

Why $\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial x}=-mg$? why $\frac{\partial \dot{x}}{\partial x}=0$?


Why $\frac{\partial (0.5*m*\dot{x}^2-m*g*x)}{\partial \dot{x}}=m*\dot{x}$ ? why $\frac{\partial x}{\partial \dot{x}}=0$?

Does it assume that speed is same at every location?
I know that $\dot{x}=\frac{\partial x}{\partial t}$.

The Lagrangian is treated as a function of independent variables. In this context $x$ and $\dot{x}$ are treated as independent.

These are dependent along the trajectory of a particle, but the Lagrangian is a function on the whole of your phase space, not just one trajectory.

In phase space a particle may have position $x$ and any velocity $\dot{x}$.

 

[Periwinkle]

The Lagrangian is treated as a function of independent variables. In this context $x$ and $\dot{x}$ are treated as independent.

These are dependent along the trajectory of a particle, but the Lagrangian is a function on the whole of your phase space, not just one trajectory.

In phase space a particle may have position $x$ and any velocity $\dot{x}$.

$x (t)$ is a function depends on $t$ and describes the movement of a mass point. Its Lagrange function $$ \frac 1 2 m {\left[\dot{x}(t)\right]}^2 -mgx(t) $$ which contains the unknown function $x (t)$. But if the function $x (t)$ is known, it depends only on the independent variable $t$.

The $$ \frac {d} {dt} \frac {\partial L} {\partial \dot x } - \frac {\partial L} {\partial x} = 0$$ Lagrange equation is written using the following substitution (use the left-hand element to replace the right) $$ t \to t $$ $$ x(t) \to u $$ $$ \dot x(t) \to v. $$ Of course, the partial derivative of the Lagrange function cannot be taken by $x(t)$ or $\dot{x} (t)$, because they are not independent variables. But the partial derivatives of $u$ and $v$ can be calculated for the function $L (u, v, t)$ and then replace it with $x(t)$ and $\dot{x} (t)$ as above. The usual way to write Lagrange equations is the shorthand version of the above procedure.

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Lagrangian mechanics do not use the phase space at all but instead use the configuration space.

The general expression of the momentum is $$ p= \frac {\partial L} {\partial \dot q } $$ which is not the product of the speed and the mass.

 

[olgerm]

In this context $x$ and $\dot{x}$ are treated as independent.

I believe that. But generally mathematical formulas should be unambiguously understandable without any comments or context needed.

 

[olgerm]

And if the system consists of many moving objects then L is still same for all objects and langrange 2. kind equation is true for every degree of freedom and corresponding momentum?

For example 2 pointbodies of which one is stuck on 2D area I get 5 equations:

$L=\frac{m_1*(v_{x1}^2+v_{y1}^2+v_{z1}^2)+m_2*(v_{x2}^2+v_{y2}^2)}{2}+\frac{m_1*m_2*G-q_1*q_2*k_E}{\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}}$
$\frac {d} {dt} \frac {\partial L} {\partial v_{x1} } - \frac {\partial L} {\partial x_1} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{x2} } - \frac {\partial L} {\partial x_2} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{y1} } - \frac {\partial L} {\partial y_1} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{y2} } - \frac {\partial L} {\partial y_2} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{z1} } - \frac {\partial L} {\partial z_1} = 0$
?

 

[HomogenousCow]

I believe that. But generally mathematical formulas should be unambiguously understandable without any comments or context needed.

If you’ve followed through the derivation of the Euler-Lagrange equation it’s all very clear and apparent.

And if the system consists of many moving objects then L is still same for all objects and langrange 2. kind equation is true for every degree of freedom and corresponding momentum?

For example 2 pointbodies of which one is stuck on 2D area I get 5 equations:

$L=\frac{m_1*(v_{x1}^2+v_{y1}^2+v_{z1}^2)+m_2*(v_{x2}^2+v_{y2}^2)}{2}+\frac{m_1*m_2*G-q_1*q_2*k_E}{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}$
$\frac {d} {dt} \frac {\partial L} {\partial v_{x1} } - \frac {\partial L} {\partial x_1} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{x2} } - \frac {\partial L} {\partial x_2} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{y1} } - \frac {\partial L} {\partial y_1} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{y2} } - \frac {\partial L} {\partial y_2} = 0$
$\frac {d} {dt} \frac {\partial L} {\partial v_{z1} } - \frac {\partial L} {\partial z_1} = 0$
?

Yes that’s true

 

[olgerm]

Is same true about hamiltonian, that I can write both Hamilton's equations about all degrees of freedom and corresponding momentums, but Hamiltonian itself is same?

 

[HomogenousCow]

Yes, there is one hamiltonian for your entire system the same way there is one Lagrangian. The Lagrangian/Hamiltonian describes the collective time evolution of your entire system, including all the particles and all the fields.