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Einstein's principle of relativity...

Posted: ecterrab 14540 Product: Maple

September 05 2022

9 1

Einstein's principle of relativity

The main difference between Newtonian mechanics and the mechanics based on Einstein's principle of relativity is that in the latter the velocity of light, , is the same in all inertial reference systems. Therefore, when comparing the velocity of an object measured in two reference systems 1 and 2 that are moving relative to each other, the Newtonian rule of addition of velocities, , where is the velocity of one system with respect to the other one, is not valid; if it were, the speed and of light in the systems 1 and 2 would not be the same. This introduces surprising conceptual consequences, and algebraic complications in the formulas relating the values of measurements, in the systems 1 and 2, of time, space and everything else that is related to that.

This post is thus about Einstein's principle of relativity and the consequences of the velocity of light being the same in all inertial reference systems. Although the topic is often considered advanced, the concepts, as shocking as they are, are easy to understand, and the algebra is still tractable in simple terms. The presentation, following Landau & Lifshitz [1], Chapter 1, is at a basic level, with no prerequisite expertise required, and illustrates well how to handle the basic algebraic aspects of special relativity using computer algebra.

Finally, it seems to me not useful to just present the algebra when the concepts behind Einstein's theory are straightforward and surprising. For that reason, the short sections 1 and 2 are all about these concepts, and the algebra only starts in section 3, with the Lorentz transformations (which was recently the topic of a Mapleprimes post at a more advanced level ). To reproduce the computations shown in this worksheet, please install the Maplesoft Physics Updates v.1314 or any subsequent version.

Download Einstens_principle_of_relativity.mw

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

New coordinate-system labels in Physics...

Posted: ecterrab 14540 Product: Maple

August 25 2022

9 0

Notation is one of the most important things to communicate with others in science. It is remarkable how many people use or do not use a computer algebra package just because of its notation. For those reasons, in the context of the Physics package, strong emphasis is put on using textbook notation as much as possible regarding input and output, including, for that purpose, as people here know, significant developments in Maple typesetting.

Still, for historical reasons, when using the Physics package, the labels used to refer to a coordinate system had been a single Capital Letter, as in X, Y, ...It was not possible to use, e.g. X', or x.

That has changed. Starting with the Maplesoft Physics Updates v.1308, any symbol can be used as a coordinate system label. The lines below demo this change.

Download new_coordinates_labels.mw

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

Deriving 4D relativistic Lorentz transfo...

Posted: ecterrab 14540 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

The Electromagnetic Field of Moving Char...

Posted: ecterrab 14540 Product: Maple

June 18 2022

11 1

This is an interesting exercise, the computation of the Liénard–Wiechert potentials describing the classical electromagnetic field of a moving electric point charge, a problem of a 3^rd year undergrad course in Electrodynamics. The calculation is nontrivial and is performed below using the Physics package, following the presentation in [1] (Landau & Lifshitz "The classical theory of fields"). I have not seen this calculation performed on a computer algebra worksheet before. Thus, this also showcases the more advanced level of symbolic problems that can currently be tackled on a Maple worksheet. At the end, the corresponding document is linked and with it the computation below can be reproduced. There is also a link to a corresponding PDF file with all the sections open.

Moving charges:
The retarded and Liénard-Wiechert potentials, and the fields and

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

(1) Retired, passionate about Mathematics and Physics

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

Generally speaking, determining the electric and magnetic fields of a distribution of charges involves determining the potentials and , followed by determining the fields and from

,

In turn, the formulation of the equations for and is simple: they follow from the 4D second pair of Maxwell equations, in tensor notation

where is the electromagnetic field tensor and is the 4D current. After imposing the Lorentz condition

, i.e.

we get

which in 3D form results in

"(∇)^2A-1/(c^2) (((&PartialD;)^2)/(&PartialD;t^2)( A))=-(4 Pi)/c j"

Laplacian(`ϕ`)-(diff(`ϕ`, t, t))/c^2 = -4*Pi*rho/c

where is the current and is the charge density.

Following the presentation shown in [1] (Landau and Lifshitz, "The classical theory of fields", sec. 62 and 63), below we solve these equations for and resulting in the so-called retarded potentials, then recompute these fields as produced by a charge moving along a given trajectory - the so-called Liénard-Wiechert potentials - finally computing an explicit form for the corresponding and .

While the computation of the generic retarded potentials is, in principle, simple, obtaining their form for a charge moving along a given trajectory , and from there the form of the fields and shown in Landau's book, involves nontrivial algebraic manipulations. The presentation below thus also shows a technique to map onto the computer the manipulations typically done with paper and pencil for these problems. To reproduce the contents below, the Maplesoft Physics Updates v.1252 or newer is required.

with(Physics); Setup(coordinates = Cartesian); with(Vectors)

(1)

The retarded potentials and

The equations which determine the scalar and vector potentials of an arbitrary electromagnetic field are input as

(2)

%Laplacian(varphi(X))-(diff(diff(varphi(X), t), t))/c^2 = -4*Pi*rho(X)

(3)

%Laplacian(A_(X))-(diff(diff(A_(X), t), t))/c^2 = -4*Pi*j_(X)

(4)

The solutions to these inhomogeneous equations are computed as the sum of the solutions for the equations without right-hand side plus a particular solution to the equation with right-hand side.

	Computing the solution to the equations for and

The Liénard-Wiechert potentials of a charge moving along

From (13), the potential at the point is determined by the charge , i.e. by the position of the charge e at the earlier time

The quantityis the 3D distance from the position of the charge at the time to the 3D point of observation. In the previous section, the charge was located at the origin and at rest, so , the radial coordinate. If the charge is moving, say on a path , we have

From (13)≡ `ϕ`(r, t) = de(t-r/c)/r and the definition of above, the potential of a moving charge can be written as

When the charge is at rest, in the Lorentz gauge we are working, the vector potential is . When the charge is moving, the form of can be found searching for a solution to "(∇)^2A-1/(c^2) (((&PartialD;)^2)/(&PartialD;t^2)( A))=-(4 Pi)/c j" that gives when . Following [1], this solution can be written as

"A( )^(alpha)=(e u( )^(alpha))/(`R__beta` u^(beta))"

where is the four velocity of the charge, .

Without showing the intermediate steps, [1] presents the three dimensional vectorial form of these potentials and as

`ϕ` = e/(R-`#mover(mi("v"),mo("→"))`/c.`#mover(mi("R"),mo("→"))`) , `#mover(mi("A"),mo("→"))` = e*`#mover(mi("v"),mo("→"))`/(c*(R-`#mover(mi("v"),mo("→"))`/c.`#mover(mi("R"),mo("→"))`))

	Computing the vectorial form of the Liénard-Wiechert potentials

The electric and magnetic fields and of a charge moving along

The electric and magnetic fields at a point are calculated from the potentials and through the formulas

,

where, for the case of a charge moving on a path , these potentials were calculated in the previous section as (24) and (18)

`ϕ`(x, y, z, t) = e/(LinearAlgebra[Norm](`#mover(mi("R"),mo("→"))`)-`#mover(mi("R"),mo("→"))`.(`#mover(mi("v"),mo("→"))`/c))

`#mover(mi("A"),mo("→"))`(x, y, z, t) = e*`#mover(mi("v"),mo("→"))`/(c*(LinearAlgebra[Norm](`#mover(mi("R"),mo("→"))`)-`#mover(mi("R"),mo("→"))`.(`#mover(mi("v"),mo("→"))`/c)))

These two expressions, however, depend on the time only through the retarded time . This dependence is within and through the velocity of the charge . So, before performing the differentiations, this dependence on must be taken into account.

(29)

(30)

The Electric field `#mover(mi("E"),mo("→"))` = -(diff(`#mover(mi("A"),mo("→"))`, t))/c-%Gradient(`ϕ`)

Computation of

	Computation of

Collecting the results of the two previous subsections, we have for the electric field

E_(X) = -(diff(A_(X), t))/c-%Gradient(varphi(X))

(60)

(61)

The book, presents this result as equation (63.8):

`#mover(mi("E"),mo("→"))` = e*(1-v^2/c^2)*(`#mover(mi("R"),mo("→"))`-`#mover(mi("v"),mo("→"))`*R/c)/(R-(`#mover(mi("v"),mo("→"))`.`#mover(mi("R"),mo("→"))`)/c)^3+`&x`(e*`#mover(mi("R"),mo("→"))`/c(R-(`#mover(mi("v"),mo("→"))`.`#mover(mi("R"),mo("→"))`)/c)^6, `&x`(`#mover(mi("R"),mo("→"))`-`#mover(mi("v"),mo("→"))`*R/c, `#mover(mi("a"),mo("→"))`))

where and . To rewrite (61) as in the above, introduce the two triple vector products

(62)

$simplify(E_(X) = -e*(Physics[Vectors][Norm](R_)^2*a_*c-v_*Physics[Vectors][Norm](v_)^2*Physics[Vectors][Norm](R_)-Physics[Vectors][Norm](R_)*Physics[Vectors][`.`](R_, v_)*a_+v_*Physics[Vectors][`.`](R_, a_)*Physics[Vectors][Norm](R_)+c*v_*Physics[Vectors][`.`](R_, v_))/(c*(1-Physics[Vectors][`.`](R_, v_)/(Physics[Vectors][Norm](R_)*c))*(c*Physics[Vectors][Norm](R_)-Physics[Vectors][`.`](R_, v_))^2*Physics[Vectors][Norm](R_))+c*e*(-Physics[Vectors][Norm](v_)^2*R_-Physics[Vectors][Norm](R_)*c*v_+R_*c^2+Physics[Vectors][`.`](R_, a_)*R_+Physics[Vectors][`.`](R_, v_)*v_)/(c*Physics[Vectors][Norm](R_)-Physics[Vectors][`.`](R_, v_))^3, {v_*Physics[Vectors][`.`](R_, a_)-Physics[Vectors][`.`](R_, v_)*a_ = Physics[Vectors][`&x`](R_, Physics[Vectors][`&x`](v_, a_))})$

E_(X) = e*(-Physics:-Vectors:-Norm(R_)*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(v_, a_))+R_*c*Physics:-Vectors:-`.`(R_, a_)-Physics:-Vectors:-Norm(R_)^2*a_*c+(-c^2*v_+v_*Physics:-Vectors:-Norm(v_)^2)*Physics:-Vectors:-Norm(R_)+R_*c^3-R_*c*Physics:-Vectors:-Norm(v_)^2)/(c*Physics:-Vectors:-Norm(R_)-Physics:-Vectors:-`.`(R_, v_))^3

(63)

(64)

$simplify(E_(X) = e*(-Physics[Vectors][Norm](R_)*Physics[Vectors][`&x`](R_, Physics[Vectors][`&x`](v_, a_))+R_*c*Physics[Vectors][`.`](R_, a_)-Physics[Vectors][Norm](R_)^2*a_*c+(-c^2*v_+v_*Physics[Vectors][Norm](v_)^2)*Physics[Vectors][Norm](R_)+R_*c^3-R_*c*Physics[Vectors][Norm](v_)^2)/(c*Physics[Vectors][Norm](R_)-Physics[Vectors][`.`](R_, v_))^3, {Physics[Vectors][`.`](R_, a_)*R_-Physics[Vectors][Norm](R_)^2*a_ = Physics[Vectors][`&x`](R_, Physics[Vectors][`&x`](R_, a_))})$

E_(X) = (c*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(R_, a_))-Physics:-Vectors:-Norm(R_)*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(v_, a_))+(c-Physics:-Vectors:-Norm(v_))*(c+Physics:-Vectors:-Norm(v_))*(R_*c-Physics:-Vectors:-Norm(R_)*v_))*e/(c*Physics:-Vectors:-Norm(R_)-Physics:-Vectors:-`.`(R_, v_))^3

(65)

Split now this result into two terms, one of them involving the acceleration . For that purpose first expand the expression without expanding the cross products

(66)

Introduce the notation used in the textbook, and and proceed with the splitting

E_(X) = e*(-R*c^2*v_+R*v^2*v_+R_*c^3-R_*c*v^2)/(c*R-Physics:-Vectors:-`.`(R_, v_))^3-e*(R*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(v_, a_))-c*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(R_, a_)))/(c*R-Physics:-Vectors:-`.`(R_, v_))^3

(67)

Rearrange only the first term using simplify; that can be done in different ways, perhaps the simplest is using subsop

E_(X) = e*(c-v)*(c+v)*(-R*v_+R_*c)/(c*R-Physics:-Vectors:-`.`(R_, v_))^3-e*(R*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(v_, a_))-c*Physics:-Vectors:-`&x`(R_, Physics:-Vectors:-`&x`(R_, a_)))/(c*R-Physics:-Vectors:-`.`(R_, v_))^3

(68)

By eye this result is mathematically equal to equation (63.8) of the textbook, shown here above before (62) .

	Algebraic manipulation rewriting (68) as the textbook equation (63.8)

The magnetic field

The book does not show an explicit form for , it only indicates that it is related to the electric field by the formula

Thus in this section we compute the explicit form of and show that this relationship mentioned in the book holds. To compute we proceed as done in the previous sections, the right-hand side should be taken at the previous (retarded) time . For clarity, turn OFF the compact display of functions.

We need to calculate

(75)

	Deriving the chain rule

So applying to (75) the chain rule derived in the previous subsection we have

(87)

where is taken as a function of only in . Now that the functionality is understood, turning ON the compact display of functions and displaying the fields by their names,

(88)

The value of is computed lines above as (48)

(89)

The expression for with no dependency is computed lines above, as (28),

A_(x, y, z, t__0) = e*v_/((Physics:-Vectors:-Norm(R_)-Physics:-Vectors:-`.`(R_, v_)/c)*c)

(90)

The expressions for and the velocity in terms of with no dependency are

(91)

(92)

[%Gradient(t__0(X)) = (r_(x, y, z)-r__0_(t__0))/(-c*Physics:-Vectors:-Norm(r_(x, y, z)-r__0_(t__0))+Physics:-Vectors:-`.`(r_(x, y, z)-r__0_(t__0), v_(t__0))), A_(x, y, z, t__0) = e*v_(t__0)/((Physics:-Vectors:-Norm(r_(x, y, z)-r__0_(t__0))-Physics:-Vectors:-`.`(r_(x, y, z)-r__0_(t__0), v_(t__0))/c)*c)]

(93)

Introducing this into "H(X)=`%Curl`(A_(x,y,z,t[`0`]))+(`%Gradient`(t[`0`](X)))*((&PartialD;A)/(&PartialD;`t__0`))" ,

(94)

Before computing the first term , for readability, re-introduce the velocity , the acceleration , then remove the dependency of these functions on , not relevant anymore since there are no more derivatives with respect to . Performing these substitutions in sequence,

(95)

(96)

Activate now the inert curl

(97)

From (34)≡, and reintroducing

(98)

(99)

To conclude, rearrange this expression as done with the one for the electric field at (65), so first expand (99) without expanding the cross products

(100)

Then introduce the notation used in the textbook, and and split into two terms, one of which contains the acceleration

H_(X) = Physics:-Vectors:-`&x`(R_, v_)*e*(-c^2+v^2)/(c*R-Physics:-Vectors:-`.`(R_, v_))^3-e*(Physics:-Vectors:-`&x`(R_, a_)*R*c-Physics:-Vectors:-`&x`(R_, a_)*Physics:-Vectors:-`.`(R_, v_)+Physics:-Vectors:-`.`(R_, a_)*Physics:-Vectors:-`&x`(R_, v_))/(c*R-Physics:-Vectors:-`.`(R_, v_))^3

(101)

	Verifying

References

[1] Landau, L.D., and Lifshitz, E.M. Course of Theoretical Physics Vol 2, The Classical Theory of Fields. Elsevier, 1975.

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

The Standard Model of Particle Physics i...

Posted: ecterrab 14540

March 18 2022

8 9

The Standard Model of Particle Physics in Maple 2022

One of the most important mathematical formulations in human history is that of the Standard Model in particle physics. It describes all the elementary particles (leptons like the electron, quarks, bosons as the Higgs or the photon), which in different arrangements, form all the observable particles in nature. The formulation is not just a tremendous theoretical achievement that rendered Nobel prizes but also a practical one. Basically, all the measurements performed in the particle accelerators at CERN and the Fermilab take this mathematical, abstract formulation as the starting point. However, for computer algebra systems, the complexity of the model is somewhat extreme: is not only the number of terms in the corresponding Lagrangian impressively large but also the mathematical properties of each of these objects that represented a challenge for a long time. With hacks of different kinds, the computer algebra representation of only some aspects of the Standard Model was possible, with restricted computational capabilities.

Hidden among the novelties of Maple 2022, a breakthrough in computer algebra is the introduction of a new, fully computable representation of the Standard Model. This representation includes the accessory commands to calculate related scattering amplitudes (the essence of the computations behind particle collision experiments) and related Feynman integrals . This is a remarkable achievement in computational physics. And from the educational point of view, it brings one more brick of knowledge from "the dark side" of the moon into "the bright side." Making the Standard Model computations be at the tip of one's fingers completely transforms the possible experience we can have with the underlying knowledge.

The illustration below of this new Maple 2022 StandardModel package is advanced in time with regards to the release of Maple 2022 days ago, and introduces a new command, Lagrangian, that increases one level the usability of the package. The so updated StandardModel is distributed as usual, within the Maplesoft Physics Updates for Maple 2022.

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