- #1
observer1
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I know this has been asked before: "Why is there a negative in the Lagrangian: L = T - V"
I have read the answers and am not happy with them so I tried to formulate my own justification and now ask if anyone could comment on it?
First, I am not happy with those who say "Because it works and F=ma comes out of it." I am just not happy with that, but I understand it and I suppose I can live with it.
Second, I am not happy with those who write that it is a special case of higher physics (I forget where I read this, but I did not understand the higher physics involving gravitational potentials... it was too advanced for me.
Finally, I am not happy with the documents -- many referenced here in previous posts -- that discuss how a thrown ball "wants" to maximize height and minimize its speed. Before I continue, let me add that I DO understand those discussions... sort of... But I am left with asking: Why doesn't the ball want to stop and get a cheeseburger? (Basically, I am unhappy with the anthropomorphic answers on that level... )
So...
Could people advise me on my answer? And I am going to "frame" this answer under the umbrella of a simple falling particle.
If I drop a ball, it will lose potential energy and gain kinetic energy. And there must be a continuous flow, with no loss of kinetic into potential. Now over the beginning and ending of the path, V (assuming the datum is the ground) becomes T. And these are large values and one should not expect THAT difference to be minimized.
However, during the "incremental motion" along the path (inside the integral), V must be continuously converted into T. Thus, it makes sense, in the integrand, to minimize the difference between T and V: it is essential that along the path, loss in V continuously becomes gain in T.
Does this make sense?
If not... can I take it from the top and could someone tell me why the "ball WANTS to maximize its height and minimize its speed? What is the ball "thinking?" Or give me another reason for the form of the Lagrangian other than the trivially obvious "because it produces F=ma."
I have read the answers and am not happy with them so I tried to formulate my own justification and now ask if anyone could comment on it?
First, I am not happy with those who say "Because it works and F=ma comes out of it." I am just not happy with that, but I understand it and I suppose I can live with it.
Second, I am not happy with those who write that it is a special case of higher physics (I forget where I read this, but I did not understand the higher physics involving gravitational potentials... it was too advanced for me.
Finally, I am not happy with the documents -- many referenced here in previous posts -- that discuss how a thrown ball "wants" to maximize height and minimize its speed. Before I continue, let me add that I DO understand those discussions... sort of... But I am left with asking: Why doesn't the ball want to stop and get a cheeseburger? (Basically, I am unhappy with the anthropomorphic answers on that level... )
So...
Could people advise me on my answer? And I am going to "frame" this answer under the umbrella of a simple falling particle.
If I drop a ball, it will lose potential energy and gain kinetic energy. And there must be a continuous flow, with no loss of kinetic into potential. Now over the beginning and ending of the path, V (assuming the datum is the ground) becomes T. And these are large values and one should not expect THAT difference to be minimized.
However, during the "incremental motion" along the path (inside the integral), V must be continuously converted into T. Thus, it makes sense, in the integrand, to minimize the difference between T and V: it is essential that along the path, loss in V continuously becomes gain in T.
Does this make sense?
If not... can I take it from the top and could someone tell me why the "ball WANTS to maximize its height and minimize its speed? What is the ball "thinking?" Or give me another reason for the form of the Lagrangian other than the trivially obvious "because it produces F=ma."
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