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Let's consider a particle in a rotating coordinate system, bearing in mind that in classical mechanics, the description in a rotating coordinate system is just a kind of mathematical sleights, so it does not mean anything special in physical substance.
First, we assume that the velocity of the particle is u. To have more discussion, first let's see what a rotating coordinate system really means in mathematics. As we know, the most convenient coordinate system to describe transitional motion is the Cartesian System. However, in some cases, it's not the most convenient. We must bear in mind that a group of new coordinates does not mean anything new in physical substance here in classical mechanics. They just make a difference in mathematics. We first construct a group of new coefficients to describe the motion of a particle.
We know, in a stable polar coordinate system, r=(x2+y2)^(1/2) ( in order to describe in a simple way, we just discuss the 2-dimensional cooradinate system here, its methods could be easily broadened into a 3-dimensional space if we know the Euler Angle. However, that's too conplicated in mathematical calculation.) θ= arctany/x. Here, in a rotating system fixed to a rotating rigid body, we now put the time t in the expression ofθ, for example:θ=arctany/x-Ωt. As we know, we could establish a new set of vectors at every point of the space as basic vectors. In the case of a stable polar system, we make:
(er,eθ)=(partial(x,y)/partial(r,θ))T(means the transposed matrix)(ex, ey). In the rotating general coordinate system, we also use this kind of vector sets, just ignoring the existence of t.
However, when we inspect the time differential of a vector, we could find that the exact full time derivative of a vector A is written as: DA/Dt=dA/dt+Ω×A, which we called the Boisson Formula of Absolute Derivative and Relative Derivative of time. Here Ω=Ωez and the d/dt here means the time derivative ignoring the existence of Ωt, and this is the instantaneous change of a vector which an observer rotating with the body could realize if he does not know his foothold is undergoing a rotation. The mathematical derivation is omitted here. It's not a difficult task for everyone.
So, the really derivative of the vector r=rer with respect to time is Dr/Dt=u=dr/dt+Ω×r=v+Ω×r. The v is the speed observed by a rotating observer. The total kinetic energy of the particle is T=1/2m(v+Ω×r)2, and the potential energy is U, so the Lagrangian of the particle is:
L=1/2m(v+Ω×r)2-U
We expand the first term as follows:
L=1/2m(v2+2v(Ω×r)+(Ω×r)2)-U
We must notice that the canonical momentum satisfies p=mv just in the case of the Cartesian Coordinate System. Here in the rotating case, the canonical momentum is (if we regard the v as the general velocity of the particle):
P=partialL/partialv=mv+mΩ×r, as we have seen, the real momentum in the fixed coordinate system.
If we want to work out the mechanical energy in a rotating system, we must go under the principles as follows:
First, the kinetic evergy does not vary while time elapses in a conserved system of mechanics.
Second, the kinetic energy is of course a scalar quantity.
Third, any absolute differential and relative differential with respect to the same physical quantity are the same when they are exerted on a scalar quantity.
So, the Euler-Lagrange Equation could also describe the real trajectory of the particle here as :
d/dt(mv+mΩ×r)=partialL/partialr=mv×Ω+mΩ×(Ω×r)-gradU. When we have completed the simple mathematical arrangements of this equation, we could find:
d/dt(mv)=-gradU+2mv×Ω+mΩ×(Ω×r). Haha, this is the familiar formula we have already known about the Corliolis Force and Centrifugal Force.
We know that the scalar quantity v(partialL/partialv)-L is a time independent scalar quantity. Because the the principle 3 I have listed, indeed, we could say that it's absolutely time independent. Any scalar quantity is time independent if its relative time derivative is zero. (So we do not need to calculate its absolute differential.)
After very simple calculation, we find: the energy E could be defined as v(partialL/partialv)-L=E=E0-Ω(mr×u)=E0-MΩ(where E0 is the mechanical energy calculated in the Cartesian Coordinates.)
The importance of this formula could be seen if we have read the Statistical Physics Part I written by Landau and Lifgarbagez, ξ34). In the quantised case of statistical physics in a rotating system, the probability density matrix of a system is exp(F'-H+ΩL), where F' is the free energy calculated in the rotating system, H is the Hamiltonian Operator written in the Cartesian System, and L is the angular momentum operator. Obviously, we could regard the operator H'=H-ΩL as the equivalent Hamiltonian Operator in a rotating system with the angular velocity Ω!
I am just a senior high school student, if there are any mistakes in this article, please point out and correct for me, thanks.
I'd like to discuss about what the E=E0-MΩ really means. When we calculate the Gibbs Distribution, we have Aexp(F'-E0+MΩ). Could anyone tell me the deepest meaning of the physical sense of the E0-MΩ?
First, we assume that the velocity of the particle is u. To have more discussion, first let's see what a rotating coordinate system really means in mathematics. As we know, the most convenient coordinate system to describe transitional motion is the Cartesian System. However, in some cases, it's not the most convenient. We must bear in mind that a group of new coordinates does not mean anything new in physical substance here in classical mechanics. They just make a difference in mathematics. We first construct a group of new coefficients to describe the motion of a particle.
We know, in a stable polar coordinate system, r=(x2+y2)^(1/2) ( in order to describe in a simple way, we just discuss the 2-dimensional cooradinate system here, its methods could be easily broadened into a 3-dimensional space if we know the Euler Angle. However, that's too conplicated in mathematical calculation.) θ= arctany/x. Here, in a rotating system fixed to a rotating rigid body, we now put the time t in the expression ofθ, for example:θ=arctany/x-Ωt. As we know, we could establish a new set of vectors at every point of the space as basic vectors. In the case of a stable polar system, we make:
(er,eθ)=(partial(x,y)/partial(r,θ))T(means the transposed matrix)(ex, ey). In the rotating general coordinate system, we also use this kind of vector sets, just ignoring the existence of t.
However, when we inspect the time differential of a vector, we could find that the exact full time derivative of a vector A is written as: DA/Dt=dA/dt+Ω×A, which we called the Boisson Formula of Absolute Derivative and Relative Derivative of time. Here Ω=Ωez and the d/dt here means the time derivative ignoring the existence of Ωt, and this is the instantaneous change of a vector which an observer rotating with the body could realize if he does not know his foothold is undergoing a rotation. The mathematical derivation is omitted here. It's not a difficult task for everyone.
So, the really derivative of the vector r=rer with respect to time is Dr/Dt=u=dr/dt+Ω×r=v+Ω×r. The v is the speed observed by a rotating observer. The total kinetic energy of the particle is T=1/2m(v+Ω×r)2, and the potential energy is U, so the Lagrangian of the particle is:
L=1/2m(v+Ω×r)2-U
We expand the first term as follows:
L=1/2m(v2+2v(Ω×r)+(Ω×r)2)-U
We must notice that the canonical momentum satisfies p=mv just in the case of the Cartesian Coordinate System. Here in the rotating case, the canonical momentum is (if we regard the v as the general velocity of the particle):
P=partialL/partialv=mv+mΩ×r, as we have seen, the real momentum in the fixed coordinate system.
If we want to work out the mechanical energy in a rotating system, we must go under the principles as follows:
First, the kinetic evergy does not vary while time elapses in a conserved system of mechanics.
Second, the kinetic energy is of course a scalar quantity.
Third, any absolute differential and relative differential with respect to the same physical quantity are the same when they are exerted on a scalar quantity.
So, the Euler-Lagrange Equation could also describe the real trajectory of the particle here as :
d/dt(mv+mΩ×r)=partialL/partialr=mv×Ω+mΩ×(Ω×r)-gradU. When we have completed the simple mathematical arrangements of this equation, we could find:
d/dt(mv)=-gradU+2mv×Ω+mΩ×(Ω×r). Haha, this is the familiar formula we have already known about the Corliolis Force and Centrifugal Force.
We know that the scalar quantity v(partialL/partialv)-L is a time independent scalar quantity. Because the the principle 3 I have listed, indeed, we could say that it's absolutely time independent. Any scalar quantity is time independent if its relative time derivative is zero. (So we do not need to calculate its absolute differential.)
After very simple calculation, we find: the energy E could be defined as v(partialL/partialv)-L=E=E0-Ω(mr×u)=E0-MΩ(where E0 is the mechanical energy calculated in the Cartesian Coordinates.)
The importance of this formula could be seen if we have read the Statistical Physics Part I written by Landau and Lifgarbagez, ξ34). In the quantised case of statistical physics in a rotating system, the probability density matrix of a system is exp(F'-H+ΩL), where F' is the free energy calculated in the rotating system, H is the Hamiltonian Operator written in the Cartesian System, and L is the angular momentum operator. Obviously, we could regard the operator H'=H-ΩL as the equivalent Hamiltonian Operator in a rotating system with the angular velocity Ω!
I am just a senior high school student, if there are any mistakes in this article, please point out and correct for me, thanks.
I'd like to discuss about what the E=E0-MΩ really means. When we calculate the Gibbs Distribution, we have Aexp(F'-E0+MΩ). Could anyone tell me the deepest meaning of the physical sense of the E0-MΩ?