Question about a derivation: velocity and position

[Cathr]

I started studying Lagrangian mechanics, and the movement equation is like this:
d/dt (d/dz') L - d/dz L = 0 if the movement is on the z axis.

Now the problem is, let's say L = M(z')²/2 - Mgz. How do we derivate an expression depending of z with respect to z' and also , an expression depending of z' with respect to z? I know they're interdependent, but I am a bit confused.

Many thanks!

 

[BvU]

I started studying Lagrangian mechanics

Kudos ! In order to benefit even more from PF, a small investment in studying $\TeX$ would also be very beneficial
$${d\over dt}\left ( d\mathcal L \over d\dot z \right ) - {d\mathcal L \over dz} = 0$$ $$ {\mathcal L} = {\scriptstyle {1\over 2}} m\dot z^2 - mgz$$looks an awful lot better, doesn't it ?

The thing is to treat $z$ and $\dot z$ as independent variables in the Euler-Lagrange equations, even though you know $\dot z = {dz\over dt}$. In your case you substitute the lower one ($\mathcal L$) in the top one and take the derivatives: $$ {d\over dt}\Bigl ( m\dot z\Bigr ) - mg = 0 \quad \Rightarrow \quad \ddot z = g $$which is the expected equation of motion.

 

[Cathr]

Kudos ! In order to benefit even more from PF, a small investment in studying $\TeX$ would also be very beneficial
$${d\over dt}\left ( d\mathcal L \over d\dot z \right ) - {d\mathcal L \over dz} = 0$$ $$ {\mathcal L} = {\scriptstyle {1\over 2}} m\dot z^2 - mgz$$looks an awful lot better, doesn't it ?

The thing is to treat $z$ and $\dot z$ as independent variables in the Euler-Lagrange equations, even though you know $\dot z = {dz\over dt}$. In your case you substitute the lower one ($\mathcal L$) in the top one and take the derivatives: $$ {d\over dt}\Bigl ( m\dot z\Bigr ) - mg = 0 \quad \Rightarrow \quad \ddot z = g $$which is the expected equation of motion.

Thank you! Why should we treat them differently, is there an explanation for that?
And yes, I will definitely stydy LaTeX once I pass my exams :)

 

[BvU]

Why should we treat them differently, is there an explanation for that?

We don't treat them differently, on the contrary. Only as 'independent' varables
The Euler-Lagrange equation is established using calculus of variations. The path of least action leads to the EL equations. Deviations from the path can be in $z$ and also in $\dot z$

 

[Dale]

Technically the equation should be written
$${d\over dt}\left ( \partial\mathcal L \over \partial\dot z \right ) - {\partial\mathcal L \over \partial z} = 0$$
The difference between the total derivative, $d$, and the partial derivative, $\partial$, is precisely this. The partial derivative considers all other variables to be constants while the total derivative needs to compute how much any other variable changes with respect to any changes in the differentiation variable.

 

[lightarrow]

How do we derivate an expression depending of z with respect to z' and also , an expression depending of z' with respect to z? I

No, you *derive partially* an expression (the Lagrangian, in this case) depending on *both* z and z' (and depending on time, in general):
L = L(z, z', t).
The difference from the "total derivative" with respect to a variable u:
dL/du
and the partial derivative with respect to a variable q:
∂L/∂q
where in this case q can be z, or z' or t (because L, as I wrote, depends "explicitly" to those variables) and where u can be a variable from which the others can depend (z and z' in this case depends on t), has already explained to you.
Anyway, you should have some basic knowledge of calculus in more variables to understand all this, it's impossible to write an entire course on the subject here.

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