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1

Your first line is actually wrong. The corrected one should be $$ \ dU = - F\cos\theta ds $$ . Since you're not adding/removing any energy, the force (whatever your potential energy is coming from) should increase the kinetic energy and decrease the potential energy.


1

A physical scenario will help you visualise some of the other answers, particularly Muphrid's and nonagon's. Imagine an aeroplane coming in to land, as a wheel of its landing gear touches the ground. If the tyre isn't spinning before contact, the point of contact on the tyre is moving at the same speed as the aeroplane, approximately $70{\rm m\,s^{-1}}$, ...


1

The above answers are all good but i want to give another example which really helped me with understanding what does it mean that the point of contact has a velocity of zero. Think of the 'spinning circular object' not as a ball, but instead think of it as a star polygon with an infinite amount of edges, for the sake of the example a very large number will ...


4

(1) Wikipedia (2) The equation $$\frac{d^2 x^\alpha}{d\tau^2}=0$$ is the equation of motion of a free particle in special relativity. It is in a sense a generalization of Newton's first law. It says that free particles move in straight lines (recall that second derivatives annihilate lines). Since in special relativity the Christoffel symbols are trivially ...


1

This is what A.Zee says about a tensor from his book Einstein Gravity in a Nutshell (Hardcover) A tensor is something that transforms like a tensor Long ago, an undergrad who later became a distinguished condensed matter physicist came to me after a class on group theory and asked me, 'What exactly is a tensor?' I told him that a tensor is something ...


1

The choice is exactly the same as for Euclidean space, which can be represented by the vector space $\mathbb R^3$, or by the associated affine space. Either works fine, and the only real difference is the existence, in the vector space, of a zero vector. In general, it is more formally correct to use the affine version, which has in-built formal support for ...


5

The confusion arises as in many cases no distinction is made between the manifold of space-time and its tangent bundle, or even a characteristic fibre of it. Flat space-time is assumed to be a pseudo-Riemannian manifold $M$ with a pseudo-metric tensor $\eta\in T^*M^{\otimes 2}$ defined everywhere. The vector Minkowski space is any fibre $T_pM$, which is ...


5

If I understood you correctly, you want to know the difference between the two. Take a look at this question (and answer): What are differences between affine space and vector space? Short answer: the only difference is that affine spaces don't have a special $\vec{0}$ element. But there is always an isomorphism between an affine space with an origin and ...


10

Physically speaking, there is no preferred origin in the spacetime of special relativity, therefore an affine space (equipped with a Lorentzian scalar product) is a better model than a vector space. The Lorentz group acts on the tangent space at each event, this space being isomorphic to the space of four-displacements. The whole invariance group is the ...


3

This answer won't be very long, because there's not all too much to say: It's a typo. This is clear from the context: The sentence describes graphene, as witnessed by the words "single layer", which is the characteristic property of graphene. The sentence occurs in a paper on graphene. The 'n' is found next to the 'm' on most keyboards.


0

Pressure is an external force, when applied on another body, the effect is easily seen on the outer part of body and it first affected the outer area of the body. In the case of stress, the molecular deformation is developed internal of the body and stress generate slowly slowly in the internal part of and object due to load. And simply pressure affected the ...


5

Why are states rays? (Answer to OP's 1. and 2.) One of the fundamental tenets of quantum mechanics is that states of a physical system correspond (not necessarily uniquely - this is what projective spaces in QM are all about!) to vectors in a Hilbert space $\mathcal{H}$, and that the Born rule gives the probability for a system in state $\lvert \psi ...


3

The notion of projective space is sensible for any vector space V (finite dimensional or not). As a set the projective space $P(V)$ is simply the set of one-dimensional sub vector spaces of the vector space. If you think briefly about it, any one-dimensional sub vector space (or ray as physicists call them) is given by a non-zero vector $v$ in $V$ and all ...


-2

Phoenix87 nicely summarizes the general situation. But in a pedestrian situation, with a given separable Hilbert space, switching to density operators to describe the states the whole issue disappears since they are invariant under changes of phase factors. Note that density operators are positive trace class operators with trace 1, pure states being ...


2

In addition ot Phoenix87's answer, which succintly summarizes the Hilbert space as it is now understood to emerge from the operator space structure of quantum mechanics, let me try to answer your questions more directly: A Hilbert space is nothing but a (complete) vector space with an inner product. As you said, this provides (a version of) the state space ...


6

The current mathematical formulation of Quantum Mechanics is based on the theory of Operator Algebras. The fundamental Axiom is that a mechanical system is described by a C*-algebra and the set of states is given by (a restriction of) the state space of the said C*-algebra. Hilbert spaces come into play from the representation theory of C*-algebras. Given a ...


0

I believe there might be a slight confusion between the whole "cause and effect" of the particle's position and resultant force throughout the motion. If you are able to define the position of a particle as a function of time, $\vec r(t)$, then, you can use that information on its own to determine the function of the resultant force as a function of time ...


0

Entropy is simply a useful thermodynamic function that defies our efforts at physical explanation. For details visit my website "All About Entropy".


0

In brief: Actually$ h$ is the height of body obtained by subtracting it absolute distance from centre of earth by Radius of earth. You should look for the derivation of this relation which first uses the original gravitational potential formula and then by neglecting and binomial expansion etc derives this approximate relation.


3

It's the first one. This is a really excellent observation! It's a fascinating fact of physics. Absolute potential energy is a silly idea. If you take a bunch of different objects, list their potential energies, and then add $100$ to each one, nothing will change about how the system behaves. We only talk about relative potential energy. The kinetic energy ...


1

You can always say that $\Delta S \ge \frac {Q}{T}$ irrespective of whether $Q$ is transferred reversibly or not, but for equality to hold you need a reversible process. If the process is irreversible then $\Delta S \gt \frac {Q}{T}$. In fact, the difference $\Delta S - \frac {Q}{T}$ is characteristic of the dissipation taking place in the system. These ...



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