Primacy of conservation laws

DaleSpam

Relative to the reference frame. That is how velocities are always defined.

Studiot

So where in this empty infinite universe is the origin of this reference frame?

DaleSpam

Wherever you choose, that is one of the decisions you have to make when you define a reference frame.

Studiot

But I didn't, becuse there isn't one, or alternatively one can't.

If we let the velocity of the particle be v or 10¹⁰v, what difference would it make to this universe (or the particle)?

DaleSpam

Huh? What would make you think that? There are an infinite number of reference frames you can define.


It would change the momentum and any other frame variant quantities. It would not change any frame invariant quantities.

K^2

What is your thesis. State it clearly. You are not arguing a point right now, you are just bickering about stuff. That is not a way to have a conversation.


Classical Mechanics is non-Relativistic. There is an absolute velocity. Fact that you aren't specifying velocity just means that you don't know its velocity, because you are being intentionally shifty about posing the question.

In Classical Mechanics, a particle has a determinate position x,y,z for any given time t. I can take position between two times and obtain velocity. That's it. If you insist that velocities are relative, then you are the one who is not sticking to the constraints of Classical Mechanics.

Now, you might be tempted to bring up Galilean Relativity, but it's not true relativity in sense that it is indistinguishable from preferred system. Just because I can re-write equations doesn't mean that particles doesn't have an intrinsic absolute velocity. It is only when you start considering motion of light relative to other objects that the concept of velocity as relative quantity becomes irrefutable. But by now you've ventured into Special Relativity where force is a relative quantity as well.

Hetware

How did you get mgy?

Isn't it simpler to just write down mass times acceleration, and be done with it?

K^2

For a mechanical system, Lagrangian is equal to total kinetic energy of the system minus the total potential energy of the system. Later gives you the mgy term. There is also a term that goes with lambda. That's due to the constraint.

Mass times acceleration is easier when you have a trivial degree of freedom with a trivial constraint. The more general problem becomes, the more effort it becomes to write out the correct equations for each DoF. In contrast, you can always write down the Lagrangian with the same amount of effort, and get the equations of motion by differentiating.