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As someone who as a student was good at math and bad at physics, I'd rather disagree. Take for example the gravity formula F_{A->B} = G * M_A * M_B/d(A,B)^3. It
by S4M 10y ago
As someone who as a student was good at math and bad at physics, I'd rather disagree. Take for example the gravity formula F_{A->B} = G * M_A * M_B/d(A,B)^3. It's very easy to interpret that, but where does that come from? I can see why the gravity gets lower when the distance between A and B is higher, but why does it depend of the inverse of the cube of the distance? The formula is very easy to read, but since I didn't know where it came from, I would forget easily.
On the other hand, it would have been intellectually pleasing(at least for me) to know how Physicists came with it, what's the motivation behind 1/d(A,B)^3, how did people come up with G, how it was measured, etc.
- 4ad 10y ago> F_{A->B} = G * M_A * M_B/d(A,B)^3 Your formula is wrong. It's either: F_{a} = F_{b} = G * m_a * m_b/r^2 or \vec{F} = G * m_a * m_b/|r|^3 * {\vec{r}} > but where does that come from? You find that by studying physics, not history of physics. > it would have been intellectually pleasing(at least for me) to know how Physicists came with it, what's the motivation behind 1/d(A,B)^3, how did people come up with G, how it was measured, etc. Yes, understanding a specific change in the state of physics at time T is most certainly helped by understanding it at time T-1. But make no mistake here, this is still physics and not history of physics (whatever that means).
- S4M 10y agoThanks for correcting me. The fact that I remembered the formula wrong proves that I didn't understand that formula well enough to remember it. It was been about 13 years since I didn't do any physics, and I forgot most of it, however, I remember well the math I learnt because I understood them much better.
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- blahblah3 10y agoit's r^2 in the denominator. one idea is to think of the geometry: the surface area of a sphere is proportional to r^2. therefore if you think of a constant "influence" at a given distance, you might divide it by the surface area (to get r^2 in the denominator)
- 4ad 10y agoYes, it's trivial to deduce Newton's law of universal gravitation from the divergence theorem. But to deduce that the divergence theorem applies, you need the lagrangian and the least action principle. Conceptually the least action principle is the fundamental concept in classical mechanics, everything is derived from it, plus symmetry. Historically this has all been done backwards. Once you have Newton's law of universal gravitation, it's trivial to prove that the gravitational flux must be constant for every enclosing surface. It's easy, in terms of mathematics required, to come up with a least action principle from that (albeit not as easy as the inverse deduction). However, if you think of Newton's laws as fundamental, it is not a very natural thing to do. Why would you do it? Lagrange was a very deep thinker to see why the least action principle is the truly fundamental thing. He was doing this way before Noether's theorem. He was doing this before the concept of energy was formalized! They knew that a quantity with what we now know as units of energy was conserved, and they knew momentum was conserved, but they didn't know what these things were, especially as no other form of energy except kinetic and potential energy was known back then. Truly a visionary thinker.
- selimthegrim 10y agohttp://math.ucr.edu/home/baez/classical/gravitational2.pdf http://math.ucr.edu/home/baez/classical/gravitational2.pdf