It is well known that the Lagrangian of a classical free particle equal to kinetic energy. This statement can be derived from some basic assumptions about the symmetries of the space-time. Is there any similar reasoning (eg. symmetry based or geometrical) why the Lagrangian of a classical system is equal kinetic energy minus the potential energy? Or it is just because we can compare the Newton's equations with the Euler-Lagrange equation and realize how they can match?
Subscribe to:
Post Comments (Atom)
Understanding Stagnation point in pitot fluid
What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...
-
I have an hydrogenic atom, knowing that its ground-state wavefunction has the standard form $$ \psi = A e^{-\beta r} $$ with $A = \frac{\bet...
-
I stand up and I look at two parallel railroad tracks. I find that converge away from me. Why? Can someone explain me why parallel lines s...
-
At room temperature, play-dough is solid(ish). But if you make a thin strip it cannot just stand up on it's own, so is it still solid? O...
-
This image from NASA illustrates drag coefficients for several shapes: It is generally accepted that some variation of the teardrop/airfoil...
-
Sorry if this question is a bit broad but I can't find any info on this by just searching. The equation q = neAL where L is the length o...
-
Sometimes I am born in silence, Other times, no. I am unseen, But I make my presence known. In time, I fade without a trace. I harm no one, ...
-
I want to know what happens to the space a black hole crosses over as our galaxy travels through space.
No comments:
Post a Comment