# Tag Info

## Hot answers tagged classical-mechanics

4

"The number of degrees of freedom can be defined as the MINIMUM number of independent coordinates that can specify the position of the system completely" (wikipedia) In your case the number is ONE, because you only need to know the position of the particle along the curve. It doesn't matter if the curve is not a line, or even contained on a plane, because ...

3

You may know about it already, but you can find an excellent account of Lagrangian Mechanics on manifolds in the book Mathematical Methods of Classical Mechanics by V. Arnold. Also to specifically address your question: $L:TM \rightarrow \mathbb{R}$ so that $L$ is a 1-form; ie $L \in \Omega^1 M$ which is neccessary to integrate over a ...

3

The word "relativistic" means "compatible with principles of Special Relativity". This usually implies that we can no longer use the "classical" picture of universal stationary space and time. Instead we talk about 4-dimensional space-time. The word "quantum" means compatible with principles of quantum mechanics. You can look them up on wikipedia etc. But ...

1

The rule is simple: regardless of which forces are acting, if the motion is accelerated then there is a net force, otherwise, there not net force. The only kind of non-accelerated motion is motion in a straight line at uniform speed. In particular you options: a)wrong, at it is accelerated in the curved part. We do not know in the straight part. b)wrong, ...

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You should think of the definite integral operation as a function of two arguments: a region over which to integrate (here, $[x_0,x_1]$), and another function $f$ called the integrand (here, $f:\xi \mapsto (E-V(\xi))^{-\frac{1}{2}}$). So first of all, in my definition of $f$ above, we could have used (almost) any other symbol instead of $\xi$ and the ...

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As the counter example given by Herr_Mitesh shows it is not true and this is because the lagrangian is not uniquely determined. In physics sometimes you don't have to think like in mathematics and in this case you must content yourself thinking that if the lagrangian does not contain x as a variable that is enough for the condition of homogeneity to be ...

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