OK, thanks. I was confused by the remark "This trivial problem emphasizes that the forces of constraint--here the tension in the rope--appear nowhere in the Lagrangian formulation."
Let's say I want to be extremely formal. How would I proceed? The constraint is ##x_1+x_2=l##, where ##x_i## is...
The first chapter in Goldstein's Classical Mechanics ends with 3 examples about how to apply Lagrange's eqs. to simple problems. The second example is about the Atwood's machine. The book says that the tension of the rope can be ignored, but I don't understand why. The two masses can move...
You're probably thinking about the eigendecomposition of the inertia matrix. This is something unrelated to that.
Here's the lecture:
It turns out we're assuming that ##\boldsymbol\omega## is parallel to the principal axis ##\hat{\boldsymbol e}## so, by the transport theorem, the inertial...
When we solve Euler's differential equations for rigid bodies we find the angular acceleration ##\dot{\boldsymbol\omega}## and then the angular velocity ##\boldsymbol\omega##. Integrating ##\boldsymbol\omega## is less straightforward, so we start from a representation of the attitude, take its...
1. Homework Statement
Suppose that a smooth differential ##n-1##-form ##\omega## on ##\mathbb{R}^n## is ##0## outside of a ball of radius ##R##. Show that $$
\int_{\mathbb{R}^n} d\omega = 0.
$$
2. Homework Equations
$$\oint_{\partial K} \omega = \int_K d\omega$$
3. The Attempt at a...
@fresh_42 I see it now. Thank you so much for your very detailed post! The book I'm reading does define the pullback of maps on manifolds. I got confused because it doesn't give an explicit formula for the pullback of forms. Instead, it says that the pullback can be extended to differential...
It seems to me fresh_42 gave the same exact definition I'm using: ##(\phi^* \nu)(p) = \nu(\phi(p)) = (\nu\circ\phi)(p)##.
His expression for differential forms is just a property of the ##d## operator, according to my book.
In ##(f^*(w))(X_p) := w(f_* X_p)## you do the pullback on ##w## by...
Speaking of push-forward, one book says that it's also called differential, but another book defines the differential differently: ##df(X_p) = X_p(f)##. Which is it?
I also noticed that your definition of pull-back is somewhat different from mine. Your definition is ##(f^\star w)(X_p) :=...
Basically, you have defined the tangent space and the cotangent space by the push-forward and pull-back induced by a map ##F:M\to N.## One can also note that the matrix associated with ##F_*## is just the Jacobian matrix of ##F## (which is more or less equivalent to your remark about the Chain...
I understand that a covector is just a vector, but can we say that a cotangent space is just a tangent space? They're both vector spaces but are they both tangent to the manifold at a point? To me "tangent" means that it has to do with derivations, whereas cotangent means it's related to...
In the exercises on differential forms I often find expressions such as $$
\omega = 3xz\;dx - 7y^2z\;dy + 2x^2y\;dz
$$ but this is only correct if we're in "flat" space, right?
In general, a differential ##1##-form associates a covector with each point of ##M##. If we use some coordinates...
In my head I was identifying ##x^i## with ##\phi^i## and so the ##x^i## were local coordinates directly on ##U##.
In the meantime I'd like to thank you all for your patience!
I'm still not completely sure. We assumed that ##(U, \phi)## is a coordinate chart from the start. Let's say ##\phi## goes from the manifold ##M## to ##\mathbb{R}^n## and we also have some coordinates ##x^i:\mathbb{R}^n\to\mathbb{R}##. Then let's define ##\phi^i = x^i\circ\phi##. Would ##(U...
Maybe I'm starting to see the problem with my definition: the coordinates would just be a local parametrization of the curve but maybe I'd lack enough structure to do regular calculus over them. The cleanest way is to define ##\phi## from ##M## to ##\mathbb{R}^n## and then the coordinates on...
It says:
--- starts ---
Let ##(U,\phi)## be a coordinate chart with ##p \in U## and suppose that ##\phi(p)=q##. If ##x^1,\ldots,x^n## are the standard coordinate functions on ##\mathbb{R}^n## then ##q## has coordinates ##(x^1(q),\ldots,x^n(q))##. Thus we can write $$
\phi(p) =...
Whether something is an abuse or not depends on the definitions you choose. You decided to define the ##x_i## the way you did, which makes my notation an abuse. But the point is this: what do you gain by using your definition? Or in other words, what do I lose by not using it?
We're not talking about tangent spaces here. The manifold could be non-smooth. We're talking about the charts ##(U,\phi)## where ##\phi:U\to\mathbb{R}^n## is an homeomorphism. The point is whether we need to introduce coordinates ##x_1,\ldots,x_n## explicitly or if ##x_i## is just ##\phi_i##...
An element of ##\mathbb{R}^n## has the form ##(x_1,\ldots,x_n)## by definition, so we must have ##\phi(p) = (\phi_1(p),\ldots,\phi_n(p))##. Why should we define a coordinate system for ##\mathbb{R}^n## when ##\mathbb{R}^n## is already a cartesian product? I would agree with you if we were...
Let ##M## be an ##n##-dimensional (smooth) manifold and ##(U,\phi)## a chart for it. Then ##\phi## is a function from an open of ##M## to an open of ##\mathbb{R}^n##. The book I'm reading claims that coordinates, say, ##x^1,\ldots,x^n## are not really functions from ##U## to ##\mathbb{R}##, but...
Thank you all for your answers. As for QM I don't think I'll ever learn about it as I'm learning Mechanics to better understand underactuated robotics and locomotion in particular.
I'm reading Scheck's book about Mechanics and it says that Newton's first law is not redundant as it defines what an inertial system is. My problem is that we could say the same about Newton's second law. Indeed, Newton's second law is only valid, in general, for inertial systems, so it also...
Why are there (at least) two definitions of a tensor? For some people a tensor is a product of vectors and covectors, but for others it's a functional. While it's true that the two points of view are equivalent (there's an isomorphism) I find having to switch between them confusing, as a...