Partial derivative confusion in lagrangian mechanics

Question

An excerpt from an answer:

Consider the Lagrangian:

$L = (0.5)m(\dot{x}^2+\dot{y}^2 +\dot{z}^2)-U$.

The lagrange equation :

$\dfrac{\partial{L}}{\partial{x}}=\dfrac{d}{dt}\dfrac{\partial{L}}{\partial{\dot{x}}}$.

Evaluating this, we get:

$ -\frac{\partial{U}}{\partial{x}} = F_{x} $.

My issue with the evaluation:

The evaluation suggests $\frac{\partial{\dot x}}{\partial{x}}=0$ and $\frac{d}{dt}\left(\frac{\partial{U}}{\partial{\dot x}}\right)=0$, which unfortunately doesn't make sense to me.

How can we say in general that $\dot{x}$ is always not a function of $x$? We might have a situation where, say,$\dot{x} =ax+b$. How do we justify $\frac{\partial{\dot x}}{\partial{x}}=0$ for such a scenario?

If the answer is somehow "treat everything except $x$ as constant", then I will need some justification on what comes under "everything",especially $\dot{x}$, which might have a dependence on $x$ as I point out.

Answer 1

The whole point of lagrangian mechanics is to start rather more abstractly from a configuration space of two variables, without any attached trajectory. At the moment, we can vary $x(t)$ and $x’(t)$ independently, and this independent variation of these quantities just corresponds to different paths between two events. Now Lagrange discovered (and this is where the physics comes in), that if you draw all the possible paths between two events in this configuration space, the one that extremizes the action is the one that nature follows.

So to answer your question: the $x(t)$ and $x’(t)$ are unrelated because they are the configuration space variables first and foremost, and are completely independent. Only when you fix a path do they have any correlation.

Answer 2

It is a question of what you are given and what do you want to find in the problem. For example, you are given that the object is moving only due to a field. You need to find the position and velocity. Then the Lagrange equations will give you the relationship between position and velocity. There is no knowledge of the link before you solve the equation. So before you solve the equation, $x$ and $\dot x$ are independent variables.

What you are asking is a different question: what if there is an explicit relationship between position and velocity, such as the one you give in the example? Then there must be some unknown in the Lagrange equation. And that unknown is the force: what's the force that can move the object according to this constraint?

To show this by example, say a particle is moving under the action of the gravitational filed at the surface of the Earth. Then the force is $\vec F=-mg\hat y$, and you can write the potential as $U=mgy$. It's easy to calculate that $$v_y=\sqrt{v_y(0)^2-2gy}$$ Now what if $v_y=ay+b$? Then the answer is that you have the wrong potential. There is an extra force that will make the particle move that way. So using $U=mgy$ is incompatible with your problem.