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
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.
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.
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.
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.
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.
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?
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:
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.
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?
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.
That’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...
Books 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 ...
@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
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)
@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.
@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?
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".
@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 $\mathbb{R}^4$.
@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?
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?
@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
> "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 $\mathbb{M}$"
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)
@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.
There 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...
Books 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 ...
