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12:12 AM
What's a pullback
2 hours later…
1:45 AM
@ACuriousMind essentially all the interesting/recent stuff is non-perturbative
1:56 AM
that's a very biased statement and not really true, there are people who do interesting stuff e.g. approximation methods that sum infinite diagrams (a bit like large-N expansions) etc, but I think it's fair to say that "simple" perturbation theory where you have a small coupling constant and compute the first few orders of the expansion is less interesting in condmat applications than in HEP
6 hours later…
7:29 AM
Hi @JohnRennie sir
@123 hi :-)
Pls explain free vector, fixed vector and sliding vector.
Can fixed vector move along the line of action
I don't know to be honest. I don't think those are standard terms so their exact meaning would be dependent on the context.
In rotation, can we slide vector along the line of action.
Suppose you are sliding on ice then your weight is a vector pointing vertically down, and this vector moves as you slide on the ice. I guess this could be an example of a vector moving.
@123 I'm not sure what that means.
7:34 AM
Coplanar couples can be moved along the plane?
I am having problem to understand moment. If couple of forces act near the corner plane. How it rotate, also what is reference axis about plane rotate in couple case. Is that center of mass or anything else.
I think that's too vague a question to be usefully answered. We would have to consider a specific example.
I am reading mechanics book. They don't clear this point. That's why I am having trouble. We can take uniform circular disk as an example.
What book is it?
How can I share a picture using mobile?
If you tell me the name of the book I may be able to find a copy.
7:42 AM
Pls see the link
Oops let me copy the link again
Ah. It worked on the second attempt for some reason.
In my textbook it is written that the activity of the solute follows Henry's law which is stated as follows:ai= γ i ° Xi .I see this to be no different from the way Raoult's law was defined apparently.Except for the specific symbol of gamma nought.What exactly is gamma nought?Please tell me how Raoult's law and Henry's law are different from each other in the context.An in depth answer will be highly appreciated.
@JohnRennie I am adding chapter in the same link. Pls see.
In any system that is at equilibrium you can choose any point and take moments about that point. If the system is at equilibrium the total moment about any point must be zero.
@JohnRennie Did you get my query?
7:49 AM
They've chosen to take moments about point B as it produces a nice simple equation for the forces.
@JohnRennie pls see the chapter pages are uploading.
@JohnRennie sir do you use what'sapp. Is it possible for you to contact and chat this way? If possible and also time permit you.
No, I don't use social media.
Now chapter is completely uploaded.
Ookay my luck.. Pls see the content of chapter.
I need to break off now for a bit. I'll be back in about half an hour.
Ookay thanks.
8:37 AM
9:13 AM
Hello @john Sir I have uploaded few chapters of this book. Pls see chapter-2
You need to be clearer what you are asking.
They used sentence. "to avoid the reaction at point we take moment". What is meant by that?
@123 The moment of a force about a point is force x perpendicular distance. Yes?
So if the force passes through the point then the perpendicular distance is zero and that force can be ignored when calculating moments. OK so far?
9:18 AM
Oookay.. Yes
So when we are analysing a mechanics problem we often choose our point for taking moments so that one of more forces passes through that point. Then we can ignore that force because its moment about our chosen point is zero.
This is what the book means by "to avoid the reaction"
It means that the force(s) passing through the point can be ignored.
Aaah I see. It means whatever forces passes through point we take it zero.
@JohnRennie what is point of rotation for couple of forces?
@123 there isn't a simple answer to that because it depends on the system.
What if we take simple uniform circular disk
How it is possible if we apply couple of force acting at the near corner of uniform circular disk object rotate about center of mass?
Suppose you have a disk pivoted at its centre, and you apply a force tangential to its edge. Like this:
9:28 AM
If we apply single Force on a plane and point O is fixed point does fixed point also act as a reaction force in opposite to applied force to make a couple?
Now, the disk cannot have any linear momentum because it is fixed at the pivot point so it cannot translate - only rotate. Yes?
@JohnRennie thanks for sharing a picture. It means where ever we apply force in circular disk we take it as tangential force.
@123 You can apply the force at any angle you want. I've drawn it tangential to keep the example simple.
@123 if the linear momentum is constant at zero then the net force must be zero, so there must be some other force that opposes the force F.
In couple there is no fixed pivot point. What happened in this case. Also I am having problem the role of fixed point by single Force. Does fixed point feel equal and opposite force to behave as couple.
And the only place that force can act is at the pivot, because that is the only place the disk is connected to anything. So we are going to have a force acting at the pivot:
9:34 AM
@JohnRennie how can I share direct picture in this page.
Aaaaaaaah.. I see thats what I meant to say
And these two forces form a couple. The couple is F x r
In this case one of the forces acts at the centre of rotation, so if we take moments about the centre the blue force has a zero moment. That means the total moment is just F x r
Thank you so much for clear me very important point. What if there is fixed point on circular disk and we apply couple of force at the near corner of disk.
Where is the fixed point? At the edge of the disk?
Yes you are right but I am looking this phenomenon as couple of force where center of rotation is fixed. And we apply only red force the fixed point feel equal and opposite blue force at center. Does it correct?
Fixed point is at center
@123 Yes
@123 I don't know what you mean. What does couple of force at the near corner of disk mean?
9:41 AM
I meant to say that we have one fixed point at center. And we are applying couple of forces not single Force at near corner of disk. How disk behave.
Where are the two forces being applied?
Near corner of disk. How can I share a picture here directly.
Like that?
@123 disks don't have a corner ...
No no. Both forces at same side above the fixed point
But equal and opposite
@123 like that?
9:49 AM
No take red force little lower distance d apart. But above fixed point
I considered it as disk not circle
I made picture how can I share here.
@123 upload it to Imgur and post the link here.
Let me try or I will share you on one drive
But you made last picture okay
Pls see this link. Red and blue are couple. Point A is fixed on uniform circular disk
Consider my picture same as you shared last picture
OK ... ?
10:01 AM
Yes this one. But your picture is correct explaination
So what are you asking?
Also there are more possible cases what if red force stays the same position. And we apply purple force little away from its line of action.
I am asking what happened in this situation how to calculate this situation
You just take moments about A. That gives you the total torque, and then use torque = moment of inertia x angular acceleration.
Ookay.. Thanks @JohnRennie sir. What if in same situation there is no fixed point.
If you have a free body then the following two principles apply:
10:09 AM
1. calculate the net linear force F. Then the centre of mass accelerates in the direction of this force with acceleration given by F = ma. This always applies whether the force is acting through the COM or not.
2. the object will rotate about its centre of mass. Calculate the net torque about the centre of mass, then use τ = I α to calculate the angular acceleration.
Thank you so much @JohnRennie sir. Google and books not giving me this level of answer and satisfaction to understand phenomenon.
Give me your brain.
1 hour later…
11:19 AM
Q: What is the meaning of “bc” in physics stack exchange?

Web Development ProThat’s all. This might sound stupid, but that’s it. I got a comment on one of my questions. It read: Is your problem with the result or the Mathematical formalism? Bc you can easily show that the number of images is positive and arbitrarily large, which is the same as 'infinity', or in the, in t...

2 hours later…
1:01 PM
Q: Colsed question

A plus bBooks on general properties of matter and fluids. Can anyone please tell why this was closed and down voted . The topic comes under properties of bulk matter which includes theory of elasticity and Fluid flow, pressure, statics, surface tension . And a good physics ...

5 hours later…
6:00 PM
How do you prove the equation $d\textbf{k}= \partial_\nu k^{[\mu\nu]}...$ ? I am not so expert with differential forms
this is the definition of the volume element
with epsilon a pseudotensor
@newUser What is there to prove? That's just the definition of the exterior derivative of $k$.
@ACuriousMind isn't the definitionof the exterior derivative $d\textbf{k} = \partial_\alpha k^{[\mu\nu]} dx^\alpha \wedge (d^{n-2}x)_{\mu\nu}$?
@ACuriousMind I don't understand why you say that's in the picture is the definition of exterior derivative
6:22 PM
@newUser sorry, you're right, it's not
but then what do you want to prove that equation from?
you haven't given any additional information about $k$ one could derive that expression from
given that $\partial_\mu k^{[\mu\nu]}=0$, I want to prove that the form is closed
@ACuriousMind do you need this information to express $d\textbf{k}$ as shown in the picture?
I'm very confused about what the givens are here and what exactly you want to prove
@ACuriousMind consider the Electromagnetic field strength $F_{\mu\nu}$. In vacuum, it satisfies $\partial_\mu F^{\mu\nu} = 0$. Now, consider the 2-form in d=4, $*F= F^{\mu\nu}(d^2x)_{\mu\nu}$. I want to show that, in vacuum, this form is closed $d*F=0$
@ACuriousMind the form $\textbf{k}$ would be the dual of the field strength
Is that clear now?
I read "there is no consensus on the topology of spacetime. This means that
frequently used notions like “neighborhood”, “coordinate” and “continuity” are actually not well defined from the mathematical point of view. The Lorentz metric, being non-positive definite, does not define any topology: its role is actually to introduce causality. Many proposals have been made to fix that topology, but none has obtained general acceptance."
Why does defining “neighborhood”, “coordinate” and “continuity” need topology?
@newUser Alright, I've got it- it is just the definition of the derivative: $\mathrm{d}x^\alpha \wedge (\mathrm{d}^{n-2}x)_{\mu\nu} = \frac{1}{2}\delta^\alpha_\mu (\mathrm{d}^{n-1}x)_\nu - \frac{1}{2}\delta^\alpha_\nu (\mathrm{d}^{n-1}x)_\mu$ - you have to think about what $\mathrm{d}x^\alpha \wedge \mathrm{d}^{n-2}x$ is in terms of $\mathrm{d}^{n-1}x$ - note it is non-zero only when $\alpha = \mu$ or $\alpha = \nu$, hence the $\delta$s
(no guarantee for the prefactors, but if you work it out carefully, then contract with the $\delta$s with the $\partial_\alpha$ in front of it you should get the expression your source has)
6:39 PM
@Bohemianrelativist what do you mean? How would you define a neighbourhood without a topology?
@ACuriousMind oh yeah.. thank you. I was only trying to use the properties of the Levi-Civita tensor
@Bohemianrelativist 1. In order to even have a manifold on which you can define a Lorentz metric, you need to know the topology. The topology of the manifold is fixed through its coordinate charts, you cannot have a manifold and not know its topology. Whatever you're quoting is nonsense. 2. If you don't understand why you need a topology for the notion of continuity, I suggest you look up the definition of continuity.
@Bohemianrelativist Continuous maps, neighbourhoods and coordinates are all defined on topological spaces which by definition carry a topology
@ACuriousMind but I often read in literature the term specetime manifold is used without a caveat like the above. If a spacetime cannot have a well-defined topology, why use the term specatime manifold?
can you provide a source?
6:47 PM
@Bohemianrelativist who claims "spacetime cannot have a well-defined topology"?
That's just nonsense
@ACuriousMind I quoted the above from the book Teleparallel Gravity-An introduction by Ruben Aldrovandi and José Geraldo Pereira. Do they nonsense?
I don't know the book or the authors, and I don't know if they "do nonsense", but the passage you quoted definitely is incorrect as written.
I suspect there's some confusion here because people sometimes say "we don't know the topology of the universe" when they mean that we don't know from local observations whether it is e.g. compact (and just very large) or non-compact. It's not that it has no topology, it's that spacetime is a manifold, but we don't know which one. A more correct statement would be "We don't know which manifold spacetime is" or "We don't know the homotopy type of spacetime".
7:05 PM
@ACuriousMind you mean the universe/spacetime has a topology but we don't know what kind of topology it has?
@Bohemianrelativist I'm saying we know it's a manifold (and hence a topological space with a well-defined topology) but we don't know which manifold it is
Locally in coordinates it's always just $\mathbb{R}^{4}$ topologically due to how coordinates work
Globally - who knows? All sort of things like wormholes and such might exist somewhere, altering the topology from the naive $\mathbb{R}^4$.
7:17 PM
@ACuriousMind But when we get a specific spacetime by solving Einstein's equation, say, the Schwarzchild spacetime, we know the topology of this spacetime, don't we?
@Bohemianrelativist you don't get a spacetime from solving the equations - the equations are for the metric on a manifold you must already have
solving the EFE just gives you a metric that turns the manifold into a pseudo-Riemannian manifold, i.e. a spacetime
7:32 PM
If we calculate electrostatic equilibrium points of n charges placed at vertices of the polygon, they tend to move towards the side of that polygon with increasing n. Taking limit for a ring; the ring should be at equilibrium??? What am I missing?
1 hour later…
8:52 PM
@Bohemianrelativist I just read that comment on page 4. 'The Lorentz metric, being non-positive definite, does not define any topology'
@ACuriousMind they say doing that is purely operational and hasn't gained consensus, I remember thinking about the non-positive-definiteness before and the fact a metric space doesn't satisfy this but hmm
@Slereah where is your math now
This is actually pretty interesting, it's basically saying that GR people are even more hand-wavey than QFT people...
@RyanUnger all this time you've apparently been doing topology in a more hand-wavey way than a particle physicist...
what is true is that the Lorentz metric does not define a "metric" in the sense of metric space, unlike in the case of a Riemannian manifold where you get an actual metric that induces a topology. But that topology is just the same topology as that of the underlying manifold, cf. math.stackexchange.com/a/981303/143136
there's nothing hand-wavey here - in order to even talk about a metric tensor, you need a manifold, but a manifold is already a topological space
9:06 PM
> "A caveat: there is no consensus on the topology of spacetime. This means that frequently used notions like "neighborhood", "coordinate" and "continuity" are actually not well-defined from the mathematical point of view. The Lorentz metric, being non-positive definite, does not define any topology: its role is actually to introduce causality. Many proposals have been made to fix that topology, but none has obtained general acceptance.
> In practice, physicists make implicitly a purely operational option: they use an underlying Euclidean $\mathbb{E}^4$ when eventually using global coordinates, or when talking about "continuous" fields. In order to have causality, they then superpose an additional Lorentz metric, making of $\mathbb{E}^4$ a Minkowski $\mathbb{M} = \mathbb{E}^{3,1}$.
> This is the causal space. They finally deform $\mathbb{E}^{3,1}$ into a Riemannian space $\mathbb{R}^{3,1}$, of the same signature, so as to locally preserve causality. This $\mathbb{R}^3$ has, at each point, a tangent space which is identical to the causal Minkowski $\mathbb{M}$"
Honestly, it sounds as if these people just don't understand how differential geometry works.
Well this teleparallel stuff is all about questioning the basic tenets of gr, I'm not sure what to think, it's a pretty strong claim
in order to do that properly it would be even more important to understand the math correctly :P
@ACuriousMind what do you mean by saying "you get an actual metric that induces a topology"? Don't you say topology is given before having a metric?
If you start from the definition of a metric on a set of points, it has to induce a topology on the underlying set of points (as long as the axioms are satisfied)
9:15 PM
@Bohemianrelativist When you have a metric space, you can define a topology on it solely via the metric. For a Riemannian manifold, the topology defined in this way from the Riemannian metric coincides with the topology already given by the nature of the manifold as a topological space.
A: How are the pseudo-Riemannian metric tensor properties restricted by the manifold topology in pseudo-Riemannian manifolds?

WalterscottThere is a basic problem with the premise of the question as stated in the first paragraph. A pseudo-Riemannian(Lorentzian) metric tensor cannot produce a metric in the mathematical sense of a distance function(in semi-Riemannian manifolds the concept of length does not have a meaning), so it doe...


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