Items tagged with physics

I need yours hepl.  I work with the physics paсkage and I set:

with(Physics)

Setup(mathematicalnotation = true)

 Coordinates(X)

Setup(Dgammarepresentation = standard)

Setup(spaceindices = uppercaselatin)

Define(m, m5, y, p, mm, pp)

I try to square the next value: 

W := Dgamma[mu]*d_[mu]+M+Psigma[A]*aa[A]-mm*Dgamma[0]-m5*Dgamma[0]*Dgamma[5]+I*Dgamma[5]*Psigma[B]*pp[B]+I*Dgamma[5]*y

("*" is multiplication)

W*W

And after that I want to simplify it:

Simplify(W*W)

I guess that matter is owing to d_[`~mu`]. If I remove this term:

E:=Psigma[A]*aa[A]-mm*Dgamma[0]-m5*Dgamma[0]*Dgamma[5]+I*Dgamma[5]*Psigma[B]*pp[B]+I*Dgamma[5]*y

And if i do:

E*E

Then next error emerges:

What is it?

 

I have to prove the following:

So I do not need the explicit derivative of the function Psi(r,t) . The metric is:

ds^2=(1-rg/r)*dt^2-(1-rg/r)^(-1)*dr^2

I am in the case of a collapsing star that emit radiation during the collapsing.  And I do not need to have a rotating black hole so that the reason I dont have dt*dr term in the metric, and I fix theta and phi.  So if you look in the Maple file attach to this post, I don't manage to obtain what I need to prove the equality between the two aspect of the same calculation.

Plese, take into account that I am sort of novice with the Physcis package and that the question is not part of an exam.

Thank you in advance for your help. 

Mario Lemelin

dAlembertian.mw

 

 

Hello! 

For the last couple of days I've been trying really hard to solve the linear PDE 

dR/dt = -dRdH/dqdp + dRdH/(dpdq) . Where R is a function R(t,q(t),p(t)) and H is the hamiltonian H=  p^2/2 +q^2 +2*q .

(dH/dp= p and dH/dq= -2q-2), q and p depends on the time t, and I'm supposed to solve the PDE and then plot the gaussian distribution (2D). 

I tried doing this:

pde := diff(R(t, q1(t), p1(t)), t) = -(diff(R(t, q(t), p(t)), q(t)))*p(t)+(diff(R(t, q(t), p(t)), p(t)))*(-2*q(t)-2)

But pdsolve(pde) gives me:  "Error, (in pdsolve/info) the name of the indeterminate function must be given". 

When I change q(t) to q and p(t) to p I get:

R(t, q, p) = _F1(p^2-2*q^2-4*q, -(1/2)*ln(sqrt(2)*q+p+sqrt(2))*sqrt(2)+t)

And then I'm lost. How do I solve this PDE in maple? 

Thankful for any help 

 

 

I need yours hepl.  I work with the physics paсkage and I set:

with(Physics)

Setup(mathematicalnotation = true)

 Coordinates(X)

Setup(Dgammarepresentation = standard)

Setup(spaceindices = uppercaselatin)

Define(M, aa, mu, mu5, Pi, eta)

M_[mu, mu5] := Dgamma[mu]*d_[mu]+M+Psigma[A]*aa[A]-mu*Dgamma[0]-mu5*Dgamma[0]*Dgamma[5]+i*Dgamma[5]*Psigma[B]*Pi[B]+i*Dgamma[5]*eta

And next:

Dagger(M_[mu, mu5])

How is Maple explained that  

Dagger(d_[mu])=d_[mu]

conjugate(M)=M

conjugate(aa[A])=aa[A]

conjugate(i)=i

and so on?

Dear Friends, I work with physics paсkage. I have a quation. I don't understend how one works with metrics. For example, let:

Nice!
Very good!

1) It doesn't work. Why? (I want exactly gamma_[A,B], rather than g_[A,B], because as i guess gamma_[A,B] has a signature [1,1,1] but g_[A,B] has a signature [-1,-1,-1])

 2) And how may I see what is matrices g_[A, B], gamma_[A, B] explicitly? That is I know how to see what is g_[mu, nu], for this one needs write "g_[];".  But how may I see g_[A, B] and gamma_[A, B] in explicitly forms?

3) Why command Trace(g_[mu, nu]))  does not work?"

Dear Friends, I work with physics paсkage. And I don't know how to simplify the next expression: Dgamma[mu]*a[mu]*Dgamma[nu]*a[nu]

(I want to obtain  the well-known result a2 )

The command "Simplify" doesn't work in this case.

Dear Friends, I work with physics paсkage. One allows to calculate in four dimensional space-time. But in addition, I need to calculate in three dimensional space. For example, I need t0 use the next scalar products: xAy and zawa  where A=0,1,2,3 and a=1,2,3. How may I do it?

Could you tell about manual (book) at maple which tells how to make calculations in quantum field theory (Grassmann algebra, a Lie algebra, producing functional, fermionic determinants) and high-temperature quantum field theory (partition functions, thermodynamic potentials)?

Hi everybody;

I have a problem with Physics[diff] command. When I run the following code, error messages appear where the Physics[diff] command exist. What is the source of error? How can I fix it?

Thanks in advance

Q1.mw

Hi there,

I have a big polynomial expression involving powers of x and y, that comes from expanding a function in powers of x and y in polynomial form (I use series(convert(series(a,x=0,10),polynom),y=0,10) ). I want to multiply each of the terms by the factorial of the power of x and y it has. How can I do this?
I tried using Physics[Coefficient](a,x) but I get the error: it cannot compute the degree of the expression.
I tried using a double for with a double coeff to get each of the coefficients and the maybe be able to multiply them but I get the error "unable to compute coeff".

Is it because as expanding the series I have the term +O(y^11) that it cannot compute it?


[Edit]
I managed to substitute the x terms using subs(x^3=3!*x^3,x^5=5!*x^5,a). Obviously this is not very efficient since I need to write the substitution for each term, and since the ploynom is grouped in powers of y, this does not work for y (neither does algusbs).
 

[Edit 2]:

an example of it would be:
 

restart; z:=1/2*log((1+y+x)/(1+y-x)): a:=diff(z,x)*h: i:=int(series(convert(series(a,x=0,12),polynom),y=0,12),x);
with result 
i := -(1/6)*x^3-(1/8)*x^5-(11/112)*x^7-(31/384)*x^9-(193/2816)*x^11+(x+(2/3)*x^3+(7/10)*x^5+(41/56)*x^7+(109/144)*x^9+(1093/1408)*x^11)*y

And I want the coefficients for each x and y power to be multiplied by the factorial of those powers.

 

Thank you!

I am trying to solve a system of equations of motion of gravitational field and in this way, I deal with a second order differential equation containing Dirac delta function as follows:

Eq9 := -(l[f]^2/l[g]^2+2)*(diff(G(r, t), r, r)-2*(diff(G(r, t), r))/r)+l[f]^2*(diff(F(r, t), r, r)-2*(diff(F(r, t), r))/r)/l[g]^2+2*l[f]^2*(m^2*l[f]^2-3)*(G(r, t)/r^2-F(r, t)/r^2)/l[g]^2 = l[g]*E*exp(Pi*t*l[f]^2*(1-2*kappa)/l[g]^2)*Dirac(t)*Dirac(r)*Dirac(z)

and

Eq10 := diff(G(r, t), r, r)-2*(diff(G(r, t), r))/r-2*kappa*(diff(F(r, t), r, r)-2*(diff(F(r, t), r))/r)-(2*(m^2*l[f]^2-3))*(G(r, t)/r^2-F(r, t)/r^2) = l[g]*E*exp(Pi*t*l[f]^2*(1-2*kappa)/l[g]^2)*Dirac(t)*Dirac(r)*Dirac(z)

 want to solve these equations with MAPLE software symbolically.
Can anyone guide me in this way, please?

The material below was presented in the "Semantic Representation of Mathematical Knowledge Workshop", February 3-5, 2016 at the Fields Institute, University of Toronto. It shows the approach I used for “digitizing mathematical knowledge" regarding Differential Equations, Special Functions and Solutions to Einstein's equations. While for these areas using databases of information helps (for example textbooks frequently contain these sort of databases), these are areas that, at the same time, are very suitable for using algorithmic mathematical approaches, that result in much richer mathematics than what can be hard-coded into a database. The material also focuses on an interesting cherry-picked collection of Maple functionality, that I think is beautiful, not well know, and seldom focused inter-related as here.

 

 

Digitizing of special functions,

differential equations,

and solutions to Einstein’s equations

within a computer algebra system

 

Edgardo S. Cheb-Terrab

Physics, Differential Equations and Mathematical Functions, Maplesoft

Editor, Computer Physics Communications

 

 

Digitizing (old paradigm)

 

• 

Big amounts of knowledge available to everybody in local machines or through the internet

• 

Take advantage of basic computer functionality, like searching and editing

 

 

Digitizing (new paradigm)

• 

By digitizing mathematical knowledge inside appropriate computational contexts that understand about the topics, one can use the digitized knowledge to automatically generate more and higher level knowledge

 

 

Challenges


1) how to identify, test and organize the key blocks of information,

 

2) how to access it: the interface,

 

3) how to mathematically process it to automatically obtain more information on demand

 

 

 

 

                                           Three examples


Mathematical Functions

 

"Mathematical functions, are defined by algebraic expressions. So consider algebraic expressions in general ..."

The FunctionAdvisor (basic)

 

"Supporting information on definitions, identities, possible simplifications, integral forms, different types of series expansions, and mathematical properties in general"

Examples

   

General description

   

References

   

 

Differential equation representation for generic nonlinear algebraic expressions - their use

 

"Compute differential polynomial forms for arbitrary systems of non-polynomial equations ..."

The Differential Equations representing arbitrary algebraic expresssions

   

Deriving knowledge: ODE solving methods

   

Extending the mathematical language to include the inverse functions

   

Solving non-polynomial algebraic equations by solving polynomial differential equations

   

References

   

 

Branch Cuts of algebraic expressions

 

"Algebraically compute, and visualize, the branch cuts of arbitrary mathematical expressions"

Examples

   

References

   

 

Algebraic expresssions in terms of specified functions

 

"A conversion network for arbitrary mathematical expressions, to rewrite them in terms of different functions in flexible ways"

Examples

   

General description

   

References

   

 

Symbolic differentiation of algebraic expressions

 

"Perform symbolic differentiation by combining different algebraic techniques, including functions of symbolic sequences and Faà di Bruno's formula"

Examples

   

References

   

 

Ordinary Differential Equations

 

"Beyond the concept of a database, classify an arbitrary ODE and suggest solution methods for it"

General description

   

Examples

   

References

   

 

Exact Solutions to Einstein's equations

 

 

Lambda*g[mu, nu]+G[mu, nu] = 8*Pi*T[mu, nu]

 

"The authors of "Exact solutions toEinstein's equations" reviewed more than 4,000 papers containing solutions to Einstein’s equations in the general relativity literature, organized the whole material into chapters according to the physical properties of these solutions. These solutions are key in the area of general relativity, are now all digitized and become alive in a worksheet"


The ability to search the database according to the physical properties of the solutions, their classification, or just by parts of keywords (old paradigm) changes the game.

More important, within a computer algebra system this knowledge becomes alive (new paradigm).

• 

The solutions are turned active by a simple call to one commend, called the g_  spacetime metric.

• 

Everything else gets automatically derived and set on the fly ( Christoffel symbols  , Ricci  and Riemann  tensors orthonormal and null tetrads , etc.)

• 

Almost all of the mathematical operations one can perform on these solutions are implemented as commands in the Physics  and DifferentialGeometry  packages.

• 

All the mathematics within the Maple library are instantly ready to work with these solutions and derived mathematical objects.

 

Finally, in the Maple PDEtools package , we have all the mathematical tools to tackle the equivalence problem around these solutions.

Examples

   

References

   

 

Download:  Digitizing_Mathematical_Information.mw,    Digitizing_Mathematical_Information.pdf

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

This presentation is on an undergrad intermediate Quantum Mechanics topic. Tackling the problem within a computer algebra worksheet in the way shown below is actually the novelty, using the Physics package to formulate the problem with quantum operators and related algebra rules in tensor notation.

 

Quantization of the Lorentz Force

 

Pascal Szriftgiser1 and Edgardo S. Cheb-Terrab2 

(1) Laboratoire PhLAM, UMR CNRS 8523, Université Lille 1, F-59655, France

(2) Maplesoft

 

We consider the case of a quantum, non-relativistic, particle with mass m and charge q evolving under the action of an arbitrary time-independent magnetic field "B=Curl(A(x,y,z)), "where `#mover(mi("A",mathcolor = "olive"),mo("→"))` is the vector potential. The Hamiltonian for this system is

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

where `#mover(mi("p",mathcolor = "olive"),mo("→"))` is the momentum of the particle, and the force acting in this particle, also called the Lorentz force, is given by

 

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t))

 

where `#mover(mi("v",mathcolor = "olive"),mo("→"))` is the quantized velocity of the particle, and all of  H, `#mover(mi("p",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("A",mathcolor = "olive"),mo("→"))` and `#mover(mi("F",mathcolor = "olive"),mo("→"))` are Hermitian quantum operators representing observable quantities.

 

In the classic (non-quantum) case, `#mover(mi("F"),mo("→"))` for such a particle in the absence of electrical field is given by

 

`#mover(mi("F"),mo("→"))` = `&x`(q*`#mover(mi("v"),mo("→"))`, `#mover(mi("B"),mo("→"))`) ,

 

Problem: Departing from the Hamiltonian, show that in the quantum case the Lorentz force is given by [1]

 

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`))

 

[1] Photons et atomes, Introduction à l'électrodynamique quantique, p. 179, Claude Cohen-Tannoudji, Jacques Dupont-Roc et Gilbert Grynberg - EDP Sciences janvier 1987.

 

Solution

 

We choose to tackle the problem in Heisenberg's picture of quantum mechanices, where the state of a system is static and only the quantum operators evolve in time according to

``

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`ℏ`

 

Also, the algebraic manipulations are simpler using tensor abstract notation instead of the standard 3D vector notation. We then start setting the framework for the problem, a system of coordinates X, indicating the dimension of the tensor space to be 3 and the metric Euclidean, and that we will use lowercaselatin letters to represent tensor indices. In addition, not necessary but for convenience, we set the lowercase latin i to represent the imaginary unit and we request automaticsimplification so that the output of everything comes automatically simplified in size.

 

restart; with(Physics); interface(imaginaryunit = i)

Setup(mathematicalnotation = true, automaticsimplification = true, coordinates = X, dimension = 3, metric = Euclidean, spacetimeindices = lowercaselatin, quiet)

[automaticsimplification = true, coordinatesystems = {X}, dimension = 3, mathematicalnotation = true, metric = {(1, 1) = 1, (2, 2) = 1, (3, 3) = 1}, spacetimeindices = lowercaselatin]

(1)

 

Next we indicate the letters we will use to represent the quantum operators with which we will work, and also the standard commutation rules between position and momentum, always the starting point when dealing with quantum mechanics problems

 

Setup(quantumoperators = {F}, hermitianoperators = {A, B, H, p, r, v, x}, realobjects = {`ℏ`, m, q}, algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0}, hermitianoperators = {A, B, H, p, r, v, x}, quantumoperators = {A, B, F, H, p, r, v, x}, realobjects = {`ℏ`, m, q, x1, x2, x3, %dAlembertian, Physics:-dAlembertian}]

(2)

 

Note that we start not indicating F as Hermitian, in order to arrive at that result. The quantum operators A, B, and F are explicit functions of X, so to avoid redundant display of this functionality on the screen we use

 

CompactDisplay((A, B, F)(X))

A(x1, x2, x3)*`will now be displayed as`*A

 

B(x1, x2, x3)*`will now be displayed as`*B

 

F(x1, x2, x3)*`will now be displayed as`*F

(3)

Define now as tensors the quantum operators that we will use with tensorial notation (recalling: for these, Einstein's sum rule for repeated indices will be automatically applied when simplifying)

 

Define(x, p, v, A, B, F, quiet)

{A, B, F, p, v, x, Physics:-Dgamma[a], Physics:-Psigma[a], Physics:-d_[a], Physics:-g_[a, b], Physics:-KroneckerDelta[a, b], Physics:-LeviCivita[a, b, c], Physics:-SpaceTimeVector[a](X)}

(4)

The Hamiltonian,

H = (`#mover(mi("p",mathcolor = "olive"),mo("→"))`-q*`#mover(mi("A",mathcolor = "olive"),mo("→"))`(X))^2/(2*m)

in tensorial notation, is given by

H = (p[n]-q*A[n](X))^2/(2*m)

H = (1/2)*Physics:-`^`(p[n]-q*A[n](X), 2)/m

(5)

Generally speaking to arrive at  ```#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`)) what we now need to do is

1) Express this Hamiltonian (5) in terms of the velocity

 

And, recalling that, in Heisenberg's picture, quantum operators evolve in time according to

diff(O(t), t) = I*Physics:-Commutator(H, O(t))/`ℏ`

 

2) Take the commutator of H with the velocity itself to obtain its time derivative and, from `#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t)) , that commutator is already the force up to some constant factors.

 

To get in contact with the basic commutation rules between position and momentum behind quantum phenomena, the quantized velocity itself can be computed as the time derivative of the position operator, i.e as the commutator of x[k] with H

I*Commutator(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, x[k])/`ℏ`

I*Physics:-Commutator(H, x[k])/`ℏ` = (1/2)*(I*q^2*Physics:-AntiCommutator(A[n](X), Physics:-Commutator(A[n](X), x[k]))-I*q*Physics:-AntiCommutator(p[n], Physics:-Commutator(A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics:-KroneckerDelta[k, n]*`ℏ`)/(`ℏ`*m)

(6)

This expression for the velocity, that involves commutators between the potential A[n](X), the position x[k] and the momentum p[n], can be simplified taking into account the basic quantum algebra rules between position and momentum. We assume that A[n](X)(X) can be decomposed into a formal power series (possibly infinite) of the x[k], hence all the A[n](X) commute between themselves as well as with all the x[k]

 

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

{%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}

(7)

(Note: in some cases, this is not true, but those cases are beyond the scope of this worksheet.)

 

Add these rules to the algebra rules already set so that they are all taken into account when simplifying things

 

Setup(algebrarules = {%Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0})

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(8)

Simplify(I*Physics[Commutator](H, x[k])/`ℏ` = (1/2)*(I*q^2*Physics[AntiCommutator](A[n](X), Physics[Commutator](A[n](X), x[k]))-I*q*Physics[AntiCommutator](p[n], Physics[Commutator](A[n](X), x[k]))-2*(q*A[n](X)-p[n])*Physics[KroneckerDelta][k, n]*`ℏ`)/(`ℏ`*m))

I*Physics:-Commutator(H, x[k])/`ℏ` = (-A[k](X)*q+p[k])/m

(9)

The right-hand side of (9) is then the kth component of the velocity tensor quantum operator, the relationship is the same as in the classical case

v[k] = rhs(I*Physics[Commutator](H, x[k])/`ℏ` = (-A[k](X)*q+p[k])/m)

v[k] = (-A[k](X)*q+p[k])/m

(10)

and with this the Hamiltonian (5) can now be rewritten in term of the velocity completing step 1)

simplify(H = (1/2)*Physics[`^`](p[n]-q*A[n](X), 2)/m, {SubstituteTensorIndices(k = n, (rhs = lhs)(v[k] = (-A[k](X)*q+p[k])/m))})

H = (1/2)*m*Physics:-`^`(v[n], 2)

(11)

For step 2), to compute

 `#mover(mi("F",mathcolor = "olive"),mo("→"))` = m*(diff(v(t), t)) and m*(diff(v(t), t)) = I*m*Physics:-Commutator(H, v(t)[k])/`ℏ` 

 

we need the commutator between the different components of the quantized velocity which, contrary to what happens in the classical case, do not commute. For this purpose, take the commutator between (10) with itself after replacing the free index

Commutator(v[k] = (-A[k](X)*q+p[k])/m, SubstituteTensorIndices(k = n, v[k] = (-A[k](X)*q+p[k])/m))

Physics:-Commutator(v[k], v[n]) = -q*(Physics:-Commutator(A[k](X), p[n])+Physics:-Commutator(p[k], A[n](X)))/m^2

(12)

To simplify (12), we use the fact that if f  is a commutative mapping that can be decomposed into a formal power series in all the complex plan (which is assumed to be the case for all A[n](X)(X)), then

Physics:-Commutator(p[k], f(x, y, z)) = -I*`ℏ`*`∂`[k](f(x, y, z))

where p[k]"=-i `ℏ` `∂`[k] " is the momentum operator along the x[k] axis. This relation reads in tensor notation:

Commutator(p[k], A[n](X)) = -I*`ℏ`*d_[k](A[n](X))

Physics:-Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X])

(13)

Add this rule to the rules previously set in order to automatically take it into account in (12)

Setup(Physics[Commutator](p[k], A[n](X)) = -I*`ℏ`*Physics[d_][k](A[n](X), [X]))

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X]), %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(14)

Physics[Commutator](v[k], v[n]) = -q*(Physics[Commutator](A[k](X), p[n])+Physics[Commutator](p[k], A[n](X)))/m^2

Physics:-Commutator(v[k], v[n]) = -I*q*`ℏ`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2

(15)

Also add this other rule so that it is taken into account automatically

Setup(Physics[Commutator](v[k], v[n]) = -I*q*`ℏ`*(Physics[d_][n](A[k](X), [X])-Physics[d_][k](A[n](X), [X]))/m^2)

[algebrarules = {%Commutator(p[k], p[n]) = 0, %Commutator(p[k], A[n](X)) = -I*`ℏ`*Physics:-d_[k](A[n](X), [X]), %Commutator(v[k], v[n]) = -I*q*`ℏ`*(Physics:-d_[n](A[k](X), [X])-Physics:-d_[k](A[n](X), [X]))/m^2, %Commutator(x[k], p[l]) = I*`ℏ`*Physics:-KroneckerDelta[k, l], %Commutator(x[k], x[l]) = 0, %Commutator(A[k](X), x[l]) = 0, %Commutator(A[k](X), A[l](X)) = 0}]

(16)

Recalling now the expression of the Hamiltonian (11) as a function of the velocity, one can compute the components of the force operator  "()Component(v*B,k)=m (v[k])=(i m [H,v[k]][-])/`ℏ`"

F[k](X) = I*m*%Commutator(rhs(H = (1/2)*m*Physics[`^`](v[n], 2)), v[k])/`ℏ`

F[k](X) = I*m*%Commutator((1/2)*m*Physics:-`^`(v[n], 2), v[k])/`ℏ`

(17)

Simplify this expression for the quantized force taking the quantum algebra rules (16) into account

Simplify(F[k](X) = I*m*%Commutator((1/2)*m*Physics[`^`](v[n], 2), v[k])/`ℏ`)

F[k](X) = (1/2)*q*(-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])+Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))+Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X])))

(18)

It is not difficult to verify that this is the antisymmetrized vector product `&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`). Departing from `#mover(mi("B",mathcolor = "olive"),mo("→"))` = `&x`(VectorCalculus[Nabla], `#mover(mi("A",mathcolor = "olive"),mo("→"))`) expressed using tensor notation,

B[c](X) = LeviCivita[c, n, m]*d_[n](A[m](X))

B[c](X) = -Physics:-LeviCivita[c, m, n]*Physics:-d_[n](A[m](X), [X])

(19)

and taking into acount that

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`), k) = `ε`[b, c, k]*v[b]*B[c](X) 

multiply both sides of (19) by `ε`[b, c, k]*v[b], getting

LeviCivita[k, b, c]*v[b]*(B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = -Physics:-LeviCivita[b, c, k]*Physics:-LeviCivita[c, m, n]*Physics:-`*`(v[b], Physics:-d_[n](A[m](X), [X]))

(20)

Simplify(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = -Physics[LeviCivita][b, c, k]*Physics[LeviCivita][c, m, n]*Physics[`*`](v[b], Physics[d_][n](A[m](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[m], Physics:-d_[k](A[m](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(21)

Finally, replacing the repeated index m by n 

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[m], Physics[d_][k](A[m](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X])))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(v[b], B[c](X)) = Physics:-`*`(v[n], Physics:-d_[k](A[n](X), [X]))-Physics:-`*`(v[n], Physics:-d_[n](A[k](X), [X]))

(22)

Likewise, for

 Component(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`), k) = `ε`[b, c, k]*B[b]*B[c](X) 

multiplying (19), this time from the right instead of from the left, we get

Simplify(((B[c](X) = -Physics[LeviCivita][c, m, n]*Physics[d_][n](A[m](X), [X]))*LeviCivita[k, b, c])*v[b])

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[m](X), [X]), v[m])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(23)

SubstituteTensorIndices(m = n, Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[m](X), [X]), v[m])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))

Physics:-LeviCivita[b, c, k]*Physics:-`*`(B[c](X), v[b]) = Physics:-`*`(Physics:-d_[k](A[n](X), [X]), v[n])-Physics:-`*`(Physics:-d_[n](A[k](X), [X]), v[n])

(24)

Simplifying now the expression (18) for the quantized force taking into account (22) and (24) we get

simplify(F[k](X) = (1/2)*q*(-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n])+Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))+Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))), {(rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](v[b], B[c](X)) = Physics[`*`](v[n], Physics[d_][k](A[n](X), [X]))-Physics[`*`](v[n], Physics[d_][n](A[k](X), [X]))), (rhs = lhs)(Physics[LeviCivita][b, c, k]*Physics[`*`](B[c](X), v[b]) = Physics[`*`](Physics[d_][k](A[n](X), [X]), v[n])-Physics[`*`](Physics[d_][n](A[k](X), [X]), v[n]))})

F[k](X) = (1/2)*q*Physics:-LeviCivita[b, c, k]*(Physics:-`*`(v[b], B[c](X))+Physics:-`*`(B[c](X), v[b]))

(25)

i.e.

`#mover(mi("F",mathcolor = "olive"),mo("→"))` = (1/2)*q*(`&x`(`#mover(mi("v",mathcolor = "olive"),mo("→"))`, `#mover(mi("B",mathcolor = "olive"),mo("→"))`)-`&x`(`#mover(mi("B",mathcolor = "olive"),mo("→"))`, `#mover(mi("v",mathcolor = "olive"),mo("→"))`))

in tensor notation. Finally, we note that this operator is Hermitian as expected

(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))-Dagger(F[k](X) = (1/2)*q*Physics[LeviCivita][b, c, k]*(Physics[`*`](v[b], B[c](X))+Physics[`*`](B[c](X), v[b])))

F[k](X)-Physics:-Dagger(F[k](X)) = 0

(26)



Download:  Quantization_of_the_Lorentz_force.mw,   Quantization_of_the_Lorentz_force.pdf


Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Editor, Computer Physics Communications

hi every one, i want to get derivative with sacalar function psi, first time ordinary derivative and second time covariant derivative , how should i write that ?!

hi every one ! i want to use Assume option to simplify some expression ! but it is not working ! what should i do !?

i have assume that ( a+b+c=0) and i want maple returns me exp(a+b+c) =1 ! but it does not ! what should i do !?


restart:with(Physics):

Assume(a+b+c=0):

about(a+b+c)

a+b+c:

  is assumed to be: 0

 

simplify(exp(a)*exp(b)*exp(c))

exp(a+b+c)

(1)

simplify(exp(a+b+c))

exp(a+b+c)

(2)

 


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