Maple Questions and Posts

Maple code for finding optimum maximum value of fu...

Asked by: MANUTTM 30

August 06 2022

2 16

Maple code for finding optimum maximum value of function TOTP given below when D is kept in the range of 300 to 2000?Also give maple code to find 3Dplot to determine the convexity of function TOTP with respect to y and Sr?

Hi currently i do my master thesis and i need hel...

Asked by: Mohamed Mohsin 5

August 06 2022

0 3

D^γ S(t)=(b-ev)S-α_1 (1-ev)SI_1-α_2 (1-ev)SI_2-dS,
D^γ I_1 (t)=α_1 (1-ev)SI_1-(β_1+d_1+d)I_1,                                           
D^γ I_2 (t)=α_2 (1-ev)SI_2-(β_2+d_2+d) I_2                  , 
D^γ R(t)=β_1 I_1+β_2 I_2+evS-dR

why PDEtools:-Solve gives exception but not solve?...

Asked by: nm 8552

August 05 2022

2 5

Found case where PDEtools:-Solve gives exception but not solve.

Is this known issue?

>	interface(version);

>

restart;

>	result:=1/2(Dirac(1,-t+4+k)+Dirac(1,t-4+k)-Dirac(1,-t+4+k)cos(-t+4+k)+2Dirac(-t+4+k)sin(-t+4+k)+2sin(t-4+k)Dirac(t-4+k)-Dirac(1,t-4+k)*cos(t-4+k))/k; result:=simplify(result) ; PDEtools:-Solve(result = 0,t) assuming t::real

(1/2)*(Dirac(1, -t+4+k)+Dirac(1, t-4+k)-Dirac(1, -t+4+k)*cos(-t+4+k)+2*Dirac(-t+4+k)*sin(-t+4+k)+2*sin(t-4+k)*Dirac(t-4+k)-Dirac(1, t-4+k)*cos(t-4+k))/k

Error, (in assuming) when calling 'RootOf'. Received: 'expression independent of, _Z, RootOf'

>	solve(result = 0,t) assuming t::real

>

Download exception_from_Solve_august_8_2022.mw

Maple in Social Science

by: Lenin Araujo Castillo 1790 Product: Maple 2022

August 05 2022

2 0

UNIVERSIDAD AUTÓNOMA METROPOLITANA
UNIDAD XOCHIMILCO

15º Foro de Investigación de las Matemáticas Aplicadas a las Ciencias Sociales "Reflexiones sobre Educación y Matemáticas”,

"Learning of mathematical functions using applications with Maple, for students of Social Sciences"

The students of the first cycles have a low level of learning in the subject of functions, and the arrival of the pandemic worsened the understanding of this content. Increasing the use of ICT in great magnitude improving learning.
To determine the relationship between learning and mathematical functions using applications with Maple, for students of Social Sciences. The experimental method was used using the scientific software Maple applied to students of Social Sciences.

In Spanish.

Lenin Araujo Casillo

Ambassador of Maple

how to read the "originalview" or the range of x a...

Asked by: nm 8552

August 04 2022

2 6

I create a plot, then later on, I want to use the same view for a second plot or using the same starting x, ending x and starting y and ending y. For reasons no need to get into into, I need to do this and I have no access to this information at this time. I only have access to the p variable used to store the plot into as in this example below.

I am not sure what is the correct way to do read this information from the PLOTS structure as I never used it before.

Help describes the PLOTS structure under "PLOT and PLOT3D Data Structures" , but do not see how is one supposed to read different parts of it. I could have missed it. Maple help pages are very hard for me to read in order to obtain the information needed. From help it says:

The Maple plotting functions, plot, plot3d, and others, produce PLOT and PLOT3D data structures describing the images to be displayed.

By trial and error, for standard plot this worked

p:=plot(sin(x),x=-1..10):
op([4,1,2,2],p)

returns

"originalview" = [-0.9862500000 .. 9.9862500000, -1.0000000000 .. 1.0000000000]

I see also that command plottools:-getdata(p): returns the data. So I could also do (it is stored in matrix, in third entry):

p0:=plottools:-getdata(p);
startingx:=p0[3][1,1];
startingy:=p0[3][1,2];
endingx:=p0[3][-1,1];
endingy:=p0[3][-1,2];

Will the above work all the time? should this command be used to obtain this information or is there a better way?

When I lprint(p) I see these fields have actual names. For example , here is screen shot of lprint(p)

it looks like a RECORD data structure. But when I tried p:-originalview Error, `p` does not evaluate to a module

The question is: What is the correct way to find what x range and y range was used for an earlier plot, given one has access only to p variable used to store the output of the plot command? This is for 2D plot only.

Has anyone produced a MATLAB command read-across -...

Asked by: wswain 230

August 04 2022

0 3

Hello,

Has anyone produced a MATLAB command read across? I was reviewing some MATLAB code for the approach to the proc being written for similar functionality. They used a command "deal" as a way to distribute a Matrix sizing across multiple Matrix variables at once.

-> I.e., EOM matrices of MASS, DAMPENING, And STIFFNESS are all equal size set once initial size is declared with mass being first by number of elements and then the deal function to propgate the sizing to the X, X_Dot, Etc. and other matrices as one definition line assigned to many matrix variables.

I am not a Matlab coder to know why this command "deal" even exists as it seems quite possible to do this in a couple lines or "I think" with the XX_New := copy(Matrix_basic) maple lines.

I seem to always have to find the read across when updating or using Matlab function codes with a new Maple project. I loath retaining the matlab language. Sorry MathWorks folks.

P.S., Is the maplet the only menthod to get input from a user at time of running the worksheet. the "Input" command in other languages has the built in window for user entry. I find the GetInput, InputDialog, or other via maplet a bit cumbersome and code intensive (esp. tracking the instantiation handle for the interaction. I also found the return of "list" and "symbol" to be problematic for a simple "get integer" request.

Regards,
Bill

how to set DynamicSystems continuoustimevar to so...

Asked by: nm 8552

August 04 2022

2 1

I have differential equation with x as independent variable that I want to convert to state space representation.

But DynamicSystems will not accept y(x) as the output variable, since it defaults to y(t).

Help under "Description of the Model of a Linear System Object" says it is possible to change the independent variable by changing continuoustimevar setting. But help does not give an example how to do this and all my tries failed.

How to do this? Attached is worksheet showing the problem

Download DS_problem.mw

Deriving 4D relativistic Lorentz transformations

by: ecterrab 13431 Product: Maple

August 04 2022

7 0

Following the previous post on The Electromagnetic Field of Moving Charges, this is another non-trivial exercise, the derivation of 4-dimensional relativistic Lorentz transformations, a problem of a 3rd-year undergraduate course on Special Relativity whose solution requires "tensor & matrix" manipulation techniques. At the end, there is a link to the Maple document, so that the computation below can be reproduced, and a link to a corresponding PDF file with all the sections open.

Deriving 4D relativistic Lorentz transformations

Freddy Baudine⁽¹⁾, Edgardo S. Cheb-Terrab⁽²⁾

(1) Retired, passionate about Mathematics and Physics

(2) Physics, Differential Equations and Mathematical Functions, Maplesoft

Lorentz transformations are a six-parameter family of linear transformations that relate the values of the coordinates of an event in one inertial reference system to the coordinates of the same event in another inertial system that moves at a constant velocity relative to the former. An explicit form of can be derived from physics principles, or in a purely algebraic mathematical manner. A derivation from physics principles is done in an upcoming post about relativistic dynamics, while in this post we derive the form of mathematically, as rotations in a (pseudo) Euclidean 4 dimensional space. Most of the presentation below follows the one found in Jackson's book on Classical Electrodynamics [1].

The computations below in Maple 2022 make use of the Maplesoft Physics Updates v.1283 or newer.

Formulation of the problem and ansatz

The problem is to find a group of linear transformations,

that represent rotations in a 4D (pseudo) Euclidean spacetime, and so they leave invariant the norm of the 4D position vector ; that is,

For the purpose of deriving the form of , a relevant property for it can be inferred by rewriting the invariance of the norm in terms of . In steps, from the above,

from where,

or in matrix (4 x 4) form, ,

where is the transpose of . Taking the determinant of both sides of this equation, and recalling that , we get

The determination of is analogous to the determination of the matrix (3D tensor ) representing rotations in the 3D space, where the same line of reasoning leads to . To exclude reflection transformations, that have and cannot be obtained through any sequence of rotations, because they do not preserve the relative orientation of the axes, the sign that represents our problem is +. To explicitly construct the transformation matrix , Jackson proposes the ansatz

Summarizing: the determination of consists of determining entering such that followed by computing the exponential of the matrix .

Determination of

In order to compare results with Jackson's book, we use the same signature he uses, , and lowercase Latin letters to represent space tensor indices, while spacetime indices are represented using Greek letters, which is already Physics' default.

(1)

Start by defining the tensor whose components are to be determined. For practical purposes, define a macro to represent the tensor and use to represent its components

`&Lopf;`[`~mu`, nu] = Matrix(%id = 36893488153289603060)

(2)

${Lambda, `&Lopf;`[`~mu`, nu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], Physics:-LeviCivita[alpha, beta, mu, nu]}$

(3)

Next, from (see above in Formulation of the problem) one can derive the form of . To work algebraically with representing matrices, set these symbols as noncommutative

(4)

From

(5)

it follows that

(6)

(7)

Expanding the exponential using exp(`&Lopf;`) = Sum(`&Lopf;`^k/factorial(k), k = 0 .. infinity) , and taking into account that the matrix product can be rewritten as, the left-hand side of (7) can be written as

(8)

Multiplying by

(9)

Recalling that , and that for any matrix , ,

(10)

Physics:-`*`(exp(Physics:-g_[`~alpha`, `~mu`]*Physics:-g_[beta, nu]*`&Lopf;`[`~beta`, alpha]), exp(`&Lopf;`[`~mu`, nu])) = 1

(11)

To allow for the combination of the exponentials, now that everything is in tensor notation, remove the noncommutative character of

(12)

exp(`&Lopf;`[`~beta`, alpha]*Physics:-g_[beta, nu]*Physics:-g_[`~alpha`, `~mu`]+`&Lopf;`[`~mu`, nu]) = 1

(13)

Since every tensor component of this expression is real, taking the logarithm at both sides and simplifying tensor indices

(14)

(15)

So the components of

`&Lopf;`[`~μ`, nu] = Matrix(%id = 36893488151939882148)

(16)

satisfy (15). Using TensorArray the components of that tensorial equation are

{2*L[1, 1] = 0, 2*L[2, 2] = 0, 2*L[3, 3] = 0, 2*L[4, 4] = 0, -L[1, 2]+L[2, 1] = 0, L[1, 2]-L[2, 1] = 0, -L[1, 3]+L[3, 1] = 0, L[1, 3]-L[3, 1] = 0, -L[1, 4]+L[4, 1] = 0, L[1, 4]-L[4, 1] = 0, L[3, 2]+L[2, 3] = 0, L[4, 2]+L[2, 4] = 0, L[4, 3]+L[3, 4] = 0}

(17)

Simplifying taking these equations into account results in the form of we were looking for

`&Lopf;`[`~μ`, nu] = Matrix(%id = 36893488153606736460)

(18)

This is equation (11.90) in Jackson's book [1]. By eye we see there are only six independent parameters in , or via

(19)

(20)

This number is expected: a rotation in 3D space can always be represented as the composition of three rotations, and so, characterized by 3 parameters: the rotation angles measured on each of the space planes . Likewise, a rotation in 4D space is characterized by 6 parameters: rotations on each of the three space planes, parameters and , and rotations on the spacetime planes, parameters . Define now using (18) for further computing with it in the next section

${Lambda, `&Lopf;`[`~mu`, nu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], Physics:-LeviCivita[alpha, beta, mu, nu]}$

(21)

Determination of

From the components of in (18), the components of can be computed directly using the command. Then, following Jackson's book, in what follows we also derive a general formula for in terms of and gamma = 1/sqrt(-beta^2+1) shown in [1] as equation (11.98), finally showing the form of as a function of the relative velocity of the two inertial systems of references.

An explicit form of in the case of a rotation on the plane can be computed by taking equal to zero all the parameters in (19) but for and substituting in

(22)

`&Lopf;`[`~μ`, nu] = Matrix(%id = 36893488153606695500)

(23)

Computing the matrix exponential,

Lambda[`~μ`, nu] = Matrix(%id = 36893488151918824492)

(24)

Lambda[`~μ`, nu] = Matrix(%id = 36893488151918852684)

(25)

This is formula (4.2) in Landau & Lifshitz book [2]. An explicit form of in the case of a rotation on the plane can be computed by taking equal to zero all the parameters in (19) but for

(26)

`&Lopf;`[`~μ`, nu] = Matrix(%id = 36893488151918868828)

(27)

Lambda[`~μ`, nu] = Matrix(%id = 36893488153289306948)

(28)

Rewriting `%&Lopf;`[`~mu`, nu] = K[`~i`]*Zeta[i]+S[`~i`]*omega[i]

Following Jackson's notation, for readability, redefine the 6 parameters entering as

$'{LM[1, 2] = `ζ__1`, LM[1, 3] = `ζ__2`, LM[1, 4] = `ζ__3`, LM[2, 3] = `ω__3`, LM[2, 4] = -`ω__2`, LM[3, 4] = `ω__1`}'$

${`&Lopf;`[1, 2] = zeta__1, `&Lopf;`[1, 3] = zeta__2, `&Lopf;`[1, 4] = zeta__3, `&Lopf;`[2, 3] = omega__3, `&Lopf;`[2, 4] = -omega__2, `&Lopf;`[3, 4] = omega__1}$

(29)

(Note in the above the surrounding backquotes '...' to prevent a premature evaluation of the left-hand sides; that is necessary when using the command.) With this redefinition, becomes

$Library:-RedefineTensorComponent({`&Lopf;`[1, 2] = zeta__1, `&Lopf;`[1, 3] = zeta__2, `&Lopf;`[1, 4] = zeta__3, `&Lopf;`[2, 3] = omega__3, `&Lopf;`[2, 4] = -omega__2, `&Lopf;`[3, 4] = omega__1})$

`&Lopf;`[`~μ`, nu] = Matrix(%id = 36893488151939901668)

(30)

where each parameter is related to a rotation angle on one plane. Any Lorentz transformation (rotation in 4D pseudo-Euclidean space) can be represented as the composition of these six rotations, and to each rotation, corresponds the matrix that results from taking equal to zero all of the six parameters but one.

The set of six parameters can be split into two sets of three parameters each, one representing rotations on the planes, parameters , and the other representing rotations on the planes, parameters . With that, following [1], (30) can be rewritten in terms of four 3D tensors, two of them with the parameters as components, the other two with matrix as components, as follows:

(31)

${Lambda, `&Lopf;`[mu, nu], Physics:-Dgamma[mu], K[i], Physics:-Psigma[mu], S[i], Zeta[i], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], omega[i], Physics:-LeviCivita[alpha, beta, mu, nu]}$

(32)

The 3D tensors and satisfy the commutation relations

(33)

(34)

(35)

(36)

The matrix components of the 3D tensor , related to rotations on the planes, are

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (0), ( 4, 2 ) = (0), ( 1, 2 ) = (1), ( 3, 2 ) = (0), ( 1, 3 ) = (0), ( 4, 3 ) = (0), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (1), ( 3, 3 ) = (0), ( 2, 4 ) = (0), ( 1, 4 ) = (0), ( 2, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 1 ) = (0), ( 3, 4 ) = (0) ] )

(37)

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (1), ( 4, 2 ) = (0), ( 1, 2 ) = (0), ( 3, 2 ) = (0), ( 1, 3 ) = (1), ( 4, 3 ) = (0), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (0), ( 3, 3 ) = (0), ( 2, 4 ) = (0), ( 1, 4 ) = (0), ( 2, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 1 ) = (0), ( 3, 4 ) = (0) ] )

(38)

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (0), ( 4, 2 ) = (0), ( 1, 2 ) = (0), ( 3, 2 ) = (0), ( 1, 3 ) = (0), ( 4, 3 ) = (0), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (0), ( 3, 3 ) = (0), ( 2, 4 ) = (0), ( 1, 4 ) = (1), ( 2, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 1 ) = (1), ( 3, 4 ) = (0) ] )

(39)

The matrix components of the 3D tensor , related to rotations on the 3D space planes, are

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (0), ( 4, 2 ) = (0), ( 1, 2 ) = (0), ( 3, 2 ) = (0), ( 1, 3 ) = (0), ( 4, 3 ) = (1), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (0), ( 3, 3 ) = (0), ( 2, 4 ) = (0), ( 1, 4 ) = (0), ( 2, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 1 ) = (0), ( 3, 4 ) = (-1) ] )

(40)

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (0), ( 4, 2 ) = (-1), ( 1, 2 ) = (0), ( 3, 2 ) = (0), ( 1, 3 ) = (0), ( 4, 3 ) = (0), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (0), ( 3, 3 ) = (0), ( 2, 4 ) = (1), ( 1, 4 ) = (0), ( 2, 2 ) = (0), ( 2, 3 ) = (0), ( 4, 1 ) = (0), ( 3, 4 ) = (0) ] )

(41)

array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (0), ( 4, 2 ) = (0), ( 1, 2 ) = (0), ( 3, 2 ) = (1), ( 1, 3 ) = (0), ( 4, 3 ) = (0), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (0), ( 3, 3 ) = (0), ( 2, 4 ) = (0), ( 1, 4 ) = (0), ( 2, 2 ) = (0), ( 2, 3 ) = (-1), ( 4, 1 ) = (0), ( 3, 4 ) = (0) ] )

(42)

	Verifying the commutation relations between and

The tensor is now expressed in terms of these objects as

(50)

where the right-hand side, without free indices, represents the matrix form of . This notation makes explicit the fact that any Lorentz transformation can always be written as the composition of six rotations

(51)

(52)

`%&Lopf;`[`~μ`, nu] = (array( 1 .. 4, 1 .. 4, [( 3, 1 ) = (zeta__2), ( 4, 2 ) = (-omega__2), ( 1, 2 ) = (zeta__1), ( 3, 2 ) = (omega__3), ( 1, 3 ) = (zeta__2), ( 4, 3 ) = (omega__1), ( 4, 4 ) = (0), ( 1, 1 ) = (0), ( 2, 1 ) = (zeta__1), ( 3, 3 ) = (0), ( 2, 4 ) = (omega__2), ( 1, 4 ) = (zeta__3), ( 2, 2 ) = (0), ( 2, 3 ) = (-omega__3), ( 4, 1 ) = (zeta__3), ( 3, 4 ) = (-omega__1) ] ))

(53)

which is the same as the starting point (30)

The transformation Lambda[`~mu`, nu] = exp(`%&Lopf;`[`~mu`, nu]) , where `%&Lopf;`[`~mu`, nu] = K[`~i`]*Zeta[i] , as a function of the relative velocity of two inertial systems

As seen in the previous subsection, in , the second term, , corresponds to 3D rotations embedded in the general form of 4D Lorentz transformations, and is the term that relates the coordinates of two inertial systems of reference that move with respect to each other at constant velocity . In this section, is rewritten in terms of that velocity, arriving at equation (11.98) of Jackson's book [1]. The key observation is that the 3D vector , can be rewritten in terms of , where and c is the velocity of light (for the rationale of that relation, see [2], sec 4, discussion before formula (4.3)).

Use a macro - say ub - to represent the atomic variable (this variable can be entered as `#mover(mi("β"),mo("&circ;")`. In general, to create atomic variables, see the section on Atomic Variables of the page 2DMathDetails ).

(54)

${Lambda, `&Lopf;`[mu, nu], Physics:-Dgamma[mu], K[i], Physics:-Psigma[mu], S[i], Zeta[i], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], omega[i], Physics:-LeviCivita[alpha, beta, mu, nu], `#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[j]}$

(55)

With these two definitions, and excluding the rotation term we have

(56)

(57)

(58)

From this expression, the form of can be obtained as in (24) using and simplifying the result taking into account that is a unit vector

(59)

$exp(lhs(`%&Lopf;`[`~μ`, nu] = (array( 1 .. 4, 1 .. 4, [( 3, 3 ) = (0), ( 2, 3 ) = (0), ( 4, 2 ) = (0), ( 1, 1 ) = (0), ( 1, 2 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]*arctanh(beta)), ( 4, 4 ) = (0), ( 4, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]*arctanh(beta)), ( 3, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]*arctanh(beta)), ( 3, 4 ) = (0), ( 4, 3 ) = (0), ( 1, 4 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]*arctanh(beta)), ( 3, 2 ) = (0), ( 1, 3 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]*arctanh(beta)), ( 2, 4 ) = (0), ( 2, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]*arctanh(beta)), ( 2, 2 ) = (0) ] )))) = simplify(LinearAlgebra:-MatrixExponential(rhs(`%&Lopf;`[`~μ`, nu] = (array( 1 .. 4, 1 .. 4, [( 3, 3 ) = (0), ( 2, 3 ) = (0), ( 4, 2 ) = (0), ( 1, 1 ) = (0), ( 1, 2 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]*arctanh(beta)), ( 4, 4 ) = (0), ( 4, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]*arctanh(beta)), ( 3, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]*arctanh(beta)), ( 3, 4 ) = (0), ( 4, 3 ) = (0), ( 1, 4 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]*arctanh(beta)), ( 3, 2 ) = (0), ( 1, 3 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]*arctanh(beta)), ( 2, 4 ) = (0), ( 2, 1 ) = (`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]*arctanh(beta)), ( 2, 2 ) = (0) ] )))), {`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]^2 = 1})$

exp(`%&Lopf;`[`~μ`, nu]) = Matrix(%id = 36893488153234621252)

(60)

It is useful at this point to analyze the dependency on the components of of this matrix

Matrix(%id = 36893488151918822812)

(61)

We see that the diagonal element depends on two instead of only one component of . That is due to the simplification with respect to side relations , performed in (60), that constructs an elimination Groebner Basis that cannot reduce at once, using the single equation (59), the dependency of all of the elements and to a single component of . So, to reduce further the dependency of the element, this component of (60) requires one more simplification step, using a different elimination strategy, explicitly requesting the elimination of

((`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2)*(-beta^2+1)^(1/2)-`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2-`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2+1)/(-beta^2+1)^(1/2)

(62)

$simplify(((`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2)*(-beta^2+1)^(1/2)-`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2-`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2+1)/(-beta^2+1)^(1/2), {`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]^2 = 1}, {`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1], `#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2]})$

(-(-beta^2+1)^(1/2)*`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]^2+`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3]^2+(-beta^2+1)^(1/2))/(-beta^2+1)^(1/2)

(63)

This result involves only , and with it the form of becomes

exp(`%&Lopf;`[`~μ`, nu]) = Matrix(%id = 36893488151918876660)

(64)

Replacing now the components of the unit vector by the components of the vector divided by its modulus

`#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[1] = beta[1]/beta, `#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[2] = beta[2]/beta, `#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[3] = beta[3]/beta

(65)

and recalling that

exp(`%&Lopf;`[`~μ`, nu]) = Lambda[`~μ`, nu]

(66)

to get equation (11.98) in Jackson's book it suffices to introduce (the customary notation)

1/sqrt(-beta^2+1) = gamma

1/(-beta^2+1)^(1/2) = gamma

(67)

Lambda[`~μ`, nu] = Matrix(%id = 36893488151911556148)

(68)

This is equation (11.98) in Jackson's book.

Finally, to get the form of this general Lorentz transformation excluding 3D rotations, directly expressed in terms of the relative velocity of the two inertial systems of references, introduce

(69)

At this point it suffices to Define (69) as tensors

${Lambda, `&Lopf;`[mu, nu], Physics:-Dgamma[mu], K[i], Physics:-Psigma[mu], S[i], Zeta[i], beta[i], Physics:-d_[mu], Physics:-g_[mu, nu], Physics:-gamma_[a, b], omega[i], v[i], Physics:-LeviCivita[alpha, beta, mu, nu], `#mover(mi("β",fontstyle = "normal"),mo("&circ;"))`[j]}$

(70)

and remove and from the formulation using

gamma = 1/(-beta^2+1)^(1/2), beta = v/c

(71)

Lambda[`~μ`, nu] = Matrix(%id = 36893488153289646316)

(72)

References

[1] J.D. Jackson, "Classical Electrodynamics", third edition, 1999.

[2] L.D. Landau, E.M. Lifshitz, "The Classical Theory of Fields", Course of Theoretical Physics V.2, 4^threvised English edition, 1975.

Download Deriving_the_mathematical_form_of_Lorentz_transformations.mw

Deriving_the_mathematical_form_of_Lorentz_transformations.pdf

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

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