@TedShifrin When your own students refuse to come to office hours because you suck at teaching, you probably have a lot of time to browse stackexchange.
It seems like with spaces that are 'twisted' rather than have holes of a certain dimension the forms fail to pick up on it. After thinking about it for a few hours my best guess at an explanation is that when you have a "hole" of dimension n, it fails to affect all the n-k forms and then suddenly there is an n form that fails to be exact, whereas the twist constrains forms regardless of their dimension allowing d to have full rank all the way through. Dunno if that's right or can be formalized.
Oh, I see what you mean about full rank. You mean it on the boundary level. Yes, right. Or there might not be any nonzero closed forms at all. I guess the previous $d$ still has full rank.
@TedShifrin What an immature, self absorbed, and condescending thing to say to somebody who is on the face of things putting in an effort to learn the discipline. It is well acknowledged that Spivak is or at least close to the most challenging introductory text for Calculus........some people forget that there was a time that they struggled with things too......... anyways no more bad energy on that.
@TedShifrin Just wanted to say, I appreciate your comment below this question. I flagged two comments; one has been deleted after being handled by a mod; the first still remains, and my flag is pending. I just wanted to say thanks for your response there.
Not sure I have earned the credit you gave me there, @TedShifrin . I wasn't thinking about the "boundary" explicitly (although maybe I was implicitly without realizing it).
What I was thinking about what that the sphere is the universal cover of RP2. And I was trying to come up with 1-forms and 2-forms that would or wouldn't induce well-defined 1-forms or 2-forms on RP2. My reasoning was that if it were possible for forms to tell me something about the 'twist' in the space, then either (1) there needed to be some forms on RP2 that cannot work on the sphere OR (2) there need to be some form…
And I think I convinced myself that (1) is impossible. Every well-defined form on a space seems to induce a well-defined form on its universal cover.
(2) is possible, you just have to have a form that acts at a point and its antipode in an asymmetric way, but seems to restrict things in a way where if $\omega$ is an n-form on the universal cover that is not well-defined on the covered space, then a non-zero $d \omega$ is similarly ill-defined. So you don't run into the situation that would generate non-trivial cohomolgy where the twist disallows $\omega$ but preserves what was $ d \omega$,…
123 I don't understand the question. Are you asking about the fact that we can compute torque about any point, and we get different answers depending on the point we choose?
@123 This reflects a fundamental difference in moving in a straight line (translating) vs. moving in a circle (rotating).
Suppose you move along some path in space. There is a unique answer to the question "how far away is the final point from the starting point?" It's some distance, and that distance doesn't depend on how I choose to overlay coordinates onto space.
@123 However, if instead I ask "what angle does the final point make with the initial point?" Then that does NOT have a unique answer. For example, if I choose the origin of my coordinates to be exactly in the middle between where you started and where you ended, then the answer will be 180 degrees (assuming that you did not start and end in the same spot). However, suppose I choose my origin to be very far away from where you were moving around.
In that case, teh angle that you moved through will be very small with respect to this very far away origin
The fact that torque gives a different answer depending on what point we calculate the torque with respect to follows exactly from this fact that the angle that you move through depends on where we choose our origin to be
@KevinDriscoll Pls share equation because F = ma i don't see use of angle. In M = r x F , i can see r distance can be change by me and cross product said it need angle between r and F
Indeed, so once you recognize that torque is all about angles, well then we better NOT get a unique answer. Because the angle that you move through depends on what point you take to be the origin.
@123 https://imgur.com/GazBhsJ
Consider the path shown here in blue. If you take the origin to be at the black point, then the angle that the path moves through is 180 degrees. However, if you take the red point to be the origin, the angle that the path moves through is 0 degrees (it starts and ends at the same angle with respect to the origin).
So when we ask questions like "how fast is the angle changing over time as you move along this path?" (AKA "what is the angular velocity?") and "how quickly is the angular velocity changing?" (AKA "what is the torque?"), you should expect to get different answers depending on which point you take to be the origin
I am reading principle of mechanics synge griffith. I am still confused with translation and rotation of a rigid body.
rotation is more complex in which 3 cases arises (1) when there is no fixed point (free body) (2) When fixed point inside the body (3) When fixed point outside the body.
Pls explain simple case. If we have isolated rigid body and we apply only single force on it. What conditions need to fulfill for a body have pure linear motion?
But if the object is moving along any line that does not connect the center of mass of the object to the origin, then formally speaking it is rotating about that origin
So to be as general as possible, if a single force is acting on a body, it must act along the line connecting the center of mass to the origin of the coordinate system in order for the motion to be purely linear
Multiple forces is slightly more complicated. If you add all the forces together to find the net force, then the net force must act along a line connecting the center of mass and the origin.
AND the sum of all the torques about the center of mass of the object must also sum to zero
@KevinDriscoll If this is the case for body purely linear motion under multiple forces. I can conclude all the forces must be concurrent and the point of concurrency is the body COM. Is it correct?
The first condition precludes the possibility that the object is rotating about the origin. And the second condition precludes the possibility that the object is rotating about its own center of mass.
Ummm, no I don't think the forces need to be concurrent.
The origin is the point in your coordinate system that you label (0,0,0). In order to describe the motion of the object using cartesian coordinates, you must first choose which point to call the origin of your coordinate system.
Suppose in this image that the origin is the green point, the black square is some rigid body. Each of the red arrows represents a force of equal magnitude F. Suppose that the upward and downward areas are all an equal distance from the COM.
@KevinDriscoll In this example all upward forces balanced downward forces. Yes upward and downward forces are not pass through COM, the only unbalanced force is at left side of the body which pass through the COM of the body which reside between the rectangular box. Yes also forces are not concurrent but is has pure transation.
@KevinDriscoll Great this example help me to better understand the topic. how to think this upward and downward force because they cancel each other.
No, there is no such case. The geometry is impossible. If there are >2 forces, than either they all act along the same line, in which case the "point of concurrency" is the entire line of action OR they act along different lines, but in Euclidian geometry non-parallel lines always intersect at exactly 1 point. So if all the lines pass through the COM, then it is the unique point where they all intersect.
@KevinDriscoll One more important question. Is it possible for free rigid body can have torque nonzero if line of action of all the forces pass through COM. Is it possible? If it is not possible then it is the only condition for pure translation.
@KevinDriscoll You said in this comment. If all the forces line of action pass through COM body can have nonzero torque and can rotate about some external point. What does it means pls share example.
The torque about the green point is (0, F, 0) x (-d, 0 , 0) = (0, 0, -F d)
So this object will move upward in a straight line (assuming it is at rest to start), but that counts as "rotating" about the green point (the torque is non-zero)
Most people would not consider an object moving in a straight line as "rotating," but if you take "rotating" to mean there is a non-zero torque about SOME point, then such motion is rotating
Right, so F is the force. And r is the position of the point where the force acts on the body relative to the point that you're calculating the torque about
So to calculate the torque about any point, you just change what you take 'r' to be
I don't think there is a benefit in this case to using the torque. It gives you exactly the same information as using the forces and F = ma in this case. But it is important that we understand this case and that our system is consistent, ie we do not run into any contradictions by considering the torque in this unusual way.
Torque relates only to rotations by definition. By defining the torque to be r x F, it is automatically 0 when F acts along r and always non-zero when F does not act along r.
@KevinDriscoll Force is not confusing topic due to its single value, we can understand magnitude of force intuitively and its effect later for other bodies . But torque does not have single answer it change depend on reference point.
@KevinDriscoll I think in this your example if we fixed green point or green point is the force position object it shows pure circular motion.
The benefit is that it lets us correctly describe how a single object can be rotating at different rates about many different points at the same time. For example, it can be rotating about its center of mass and also its center of mass can be moving in a circle about some external point at the same time
Like how the Earth both revolves on its axis and rotates around the Sun at the same time
The disadvantage, I guess, is that this can be kinda confusing because people sometimes expect torque to have a unique answer like force. But such a construction would be impossible. There isn't a useful definition of torque that is independent of what point you are considering rotations around like there is with force.
@KevinDriscoll Do i consider this comment as, If i considered me as reference point, when i look rotating object at short distance it look slow rotating , then the same object i see from far it looks faster?
Yes, but be careful here about what you are holding fixed when you make this comparison
For example, given an object that is moving at a fixed speed, it rotates much faster about you if you are close to it than it does when you are far away
@KevinDriscoll If object is rotating at fixed speed , say spinning about it COM. I can calculate torque about me as reference point and i can distance from body.
Sure, yes. Although if the COM is at rest and the object is spinning about its COM, you will get 0 calculating the torque about you (or any external point)
Fun fact I found : One known fact in general topology 'If $q:X\to Y$ is a quotient map and $Z$ is a locally compact T_2 space, then $q\times 1:X\times Z\to Y\times Z$ is a quotient map' is also called Whitehead theorem.
I saw an exercise where they were asking the extremas of $\displaystyle f(x,y)= \frac{xy+x\sin(y)}{(x^2+y^2)^{\alpha} }$
Not sure how the author pretends to compute the partials in y.
The max i can say is that for $(x,y)\neq 0$ we have $\dfrac{\partial f}{\partial x} = (y+\sin (y))\left(-\left((2 {\alpha}-1) x^{2}-y^{2}\right)\right)\left(x^{2}+y^{2}\right)^{-{\alpha}-1} = 0 \iff (2{\alpha}-1) x^{2}-y^{2} = 0$
Which yields $y = \pm \sqrt{2\alpha-1} x$ and $\alpha > \dfrac{1}{2}$
But $\dfrac{\partial f}{\partial y}$ is really ugly. Even for $\alpha = 1$
@Odestheory12 sorry, i don't have time at the moment to help you. it was a suggestion, note that if you extremise with the constraint $x^2+y^2 = r$ you need to do more than just differentiate with respect to $x$.
The expression that I got for $Q(x)$ was $x^2-(b+c)x+b^2-bc+c^2$ through long division. Completing the square on $2Q(x)$ gives $2(x-\frac{b+c}{2})^2+\frac{3(b-c)^2}{2}$. The question states that $2Q(x)$ can be rewritten as the sum of three expressions, each of which is a perfect square. But as you can see there is only two perfect squares in my expression. Not sure what they want me to do
It is well known that $\int 1/(1+x^2) = arctan(x) $ but why when i do the following substitute i get rubbish results? Let $ sinh(u) = x , cosh(u) * du = dx $ then $\int 1/(1+x^2) = \int \frac{cosh(u)*du}{1+sinh^2(u)} = \int \frac{du}{coshu}$ mathematica gives after resubstuting $ u = arcsinh(x) $ the final function to be $ gd (Sinh^-1(x)$
I really have no idea about that subject i just threw a suggestion, sometimes i get inspired by monkeys doing wierd stuff so maybe this inspires you :D
Suppose you have an element in the compliment with no epsilon such that it is completly in the compliment. then it means it has to for all epsilon none trivially intersect with your graph
Well ... as i said in the integration theory, your set has jordan volume of zero, it has thus finite number of squares covering it, you can take hte interior of these squares and you have a finite cover. thus your set is compact
You try it be contradiction. If some point is not in the interior of compliment, thus for all epsilon it must cut the graph none trivially. If such is the case, then you show the point must be in the graph because (draw it you will see what im ean) it only is the case for such point to itersect always with the graph because the graph is a line.. this is a contradiction to the point being frm the compliment & oyu sowed all points in compliment can be fitted in balls inside compiment thus its open
but i have to go now, good luck trying that) it should not be hard
As a tip remember that any point you take with radius of epsilon will have the coordinates $(a_x+\epsilon, a_y+\epsilon) $ remember oyur graph consists of points of the construction $(c_x,0) $ it is now easy to construct a contradiction
Why is it when i integrate functions such as $ \sqrt(1-x^2) $ and $\frac{1}{\sqrt(1-x^2}$ using the substution $cos(u) = x $ i get the same result but with a minus sign difference as when i integrate with $sin(u) = x$?
The first minus sign results in from the derivation of $cos(u)$ but then i do not recieve a second one anywhere to cancel it out!
For example: $cos(u) = x , -sin(u)*du=dx$ then we get $\int \sqrt (1-x^2) = -1*\int sin^2(u)du = - * (1/2u-1/2 sin(u)cos(u))$ resubstuting u yields the result... The answer is given to be as i written but both signs positiv... ?
hi, why is the following true? Let $f : \mathbb{C} \rightarrow \mathbb{C} \cup \{\infty\}$ be meromorphic, and suppose $\mathcal{F}$ is a family of meromorphic functions define on the complex plane, such that for every $g \in \mathcal{F}$, $g^{-1}(1) = f^{-1}(1) \neq \emptyset$ and $g^{-1}(2) = f^{-1}(2) \neq \emptyset$. Then for any $a \in f^{-1}(2)$, there is some $B_{\epsilon}(a)$ for which $g(B_{\epsilon}(a))$ does not contain $1$ for any $g$
we have no additional assumptions on the family $\mathcal{F}$ by the way
The Gudermannian function is a way of connecting the hyperbolic & circular functions for people who're allergic to complex numbers. ;) But it also shows you some stuff that may not be immediately obvious when you just think of the hyperbolics as the circulars with imaginary args.
@love_sodam I don't think it's possible. I don't know how to prove it, but it wouldn't surprise me if there's a clever argument involving interlocking rings of 8 squares. It's easy enough to ensure all 7 numbers are in any 3×3 block, eg the following Latin square.
@Jam: So you're ignoring my first comment. Assume $n=2$ for starters. Do NOT use the same notation $dy$ for both the double integral and the line integral over the boundary. You will make mistakes like crazy that way.
and i think we should use some lemmas to prove that the graph $\mathcal{G}(g)$ is closed. to recall, we are trying to prove that $\{(x,y,z)\in\mathbb{R}^3|z=0,(x,y)\in[0,1]^2\}$.
i guess if it harmonic you can show that the integral at the boundary is the value at the centre and then show it is also equal to the integral over the whole sphere
but now starting with the integral over the sphere i need to somehow differentiate and get the integral at the boundary at least what my intutition says but literally i dont know what im allowed to do or how
@sevdaicmis No, the $g(x,y)=0$ doesn't belong there. You're talking about the graph of a continuous function $g$ on $[0,1]^2$. You want to prove the graph is a compact subset of $\Bbb R^3$. Easiest to use sequences.
@sevdaicmis What set is closed? This is getting frustrating.
Write $\int_0^r\int_0^{2\pi} u(r,\theta)r\,d\theta\,dr$ and you want to differentiate with respect to $r$. As I said, constants need to be taken care of, as well. You need to use the Leibniz rule — differentiating with respect to $r$ will have a term from the Fundamental Theorem of Calculus and a term differentiating the integrand with respect to $r$.
well, yes. the question was to find a two variable function of which the graph is compact. i've chosen $g:[0,1]^2\to\mathbb{R},g(x,y)=0$. isn't $\{(x,y,z)\in\mathbb{R}^3|z=0,(x,y)\in[0,1]^2\}$ the graph of $g$?
because, i don't see a general way to select an $\epsilon$ for radius of the ball. that's why i divide the set into four parts, to more easily construct balls.
@TedShifrin Yep. Engineers use it. It's a bit like the emptyset symbol, except it has to be circular, whereas the emptyset symbol often looks like a slashed zero.
Write out the product rule carefully. I think there's an error. And we want the average over the sphere, too, so we expect to divide the surface integral by $r^{n-1}$.
