Applications, Examples and Libraries

Autosizing TextArea

by: Mike McDermott 95 Product: Maple

November 11 2020

6 0

I created a little procedure to automatically size text areas based on content. It sizes the text area based on wraparound and tab characters, something that the autosize for the code edit region does not do. (Hint to Maple developers)

Enjoy.

    AutosizeTextArea:=proc(TextAreaName, {intMinRows::nonnegint:=5, intMinChars::nonnegint:=50, intMaxChars::nonnegint:=140})
        description "Autosizes the TextArea based on content",
                  "Parameters",
                  "1) TextAreaName__The name of the textarea",
                  "Optional Parameters",
                  "intMinRows________Minimum number of visible rows",
                  "intMinChars_______Minimum character width",
                  "intMaxChars_______Maximum character width";
        uses DocumentTools, StringTools;          
        local strLines, intLongestLine, nLines;
        strLines := Split(GetProperty(TextAreaName,'value'),"\n");
        intLongestLine := max('numelems'~(strLines));
        # Count the characters in each line (add 7 extra characters for each tab). Determine the number of lines to display each line due to wraparound, then add all these together
        #   to determine the number of rows to display.
        nLines := add(ceil~(('numelems'~(strLines) + StringTools:-CountCharacterOccurrences~(strLines, "\t")*~7)/~intMaxChars));
        SetProperty(TextAreaName, 'visibleRows', max(nLines, intMinRows), 'refresh' = true);
        SetProperty(TextAreaName, 'visibleCharacterWidth', min(max(intLongestLine, intMinChars),intMaxChars), 'refresh' = true);
    
    end proc:

A digression about the US electoral college tie

by: mmcdara 4505 Product: Maple 2015

November 06 2020

3 2

A fascinating race is presently running (even if the latest results seem to have put an end to it).
I'm talking of course about the US presidential elections.

My purpose is not to do politics but to discuss of a point of detail that really left me puzzled: the possibility of an electoral college tie.
I guess that this possibility seems as an aberration for a lot of people living in democratic countries. Just because almost everywhere at World electoral colleges contain an odd number of members to avoid such a situation!

So strange a situation that I did a few things to pass the time (of course with the earphones on the head so I don't miss a thing).
This is done with Maple 2015 and I believe that the amazing Iterator package (that I can't use thanks to the teleworking :-( ) could be used to do much more interesting things.

>

restart:

>	with(Statistics):

>	ElectoralCollege := Matrix(51, 2, [

>	Alabama, 9, Kentucky, 8, North_Dakota, 3,

>	Alaska, 3, Louisiana, 8, Ohio, 18,

>	Arizona, 11, Maine, 4, Oklahoma, 7,

>	Arkansas, 6, Maryland, 10, Oregon, 7,

>	California, 55, Massachusetts, 11, Pennsylvania, 20,

>	Colorado, 9, Michigan, 16, Rhode_Island, 4,

>	Connecticut, 7, Minnesota, 10, South_Carolina, 9,

>	Delaware, 3, Mississippi, 6, South_Dakota, 3,

>	District_of_Columbia, 3, Missouri, 10, Tennessee, 11,

>	Florida, 29, Montana, 3, Texas, 38,

>	Georgia, 16, Nebraska, 5, Utah, 6,

>	Hawaii, 4, Nevada, 6, Vermont, 3,

>	Idaho, 4, New_Hampshire, 4, Virginia, 13,

>	Illinois, 20, New_Jersey, 14, Washington, 12,

>	Indiana, 11, New_Mexico, 5, West_Virginia, 5,

>	Iowa, 6, New_York, 29, Wisconsin, 10,

>	Kansas, 6, North_Carolina, 15, Wyoming, 3 ]):

$ElectoralCollege := Vector(4, {(1) = ` 51 x 2 `*Matrix, (2) = `Data Type: `*anything, (3) = `Storage: `*rectangular, (4) = `Order: `*Fortran_order})$

(1)

>	add(ElectoralCollege[..,2]): tie := %/2;

(2)

>	ec := convert(ElectoralCollege, listlist):

>

# Sets of states that form an electoral college tie

R      := 10^5:
nbties := 0:
states := NULL:
for r from 1 to R do
  poll := combinat:-randperm(ec):
  cpoll := CumulativeSum(op~(2, poll)):
  if tie in cpoll then
    nbties := nbties+1;
    place := ListTools:-Search(tie, cpoll);
    states := states, op~(1, poll)[1..place]:   # see below
  end if:
end do:

>

# electoral college tie is not so rare an event
# (prob of occurrence about 9.4 %).
#
# Why the hell the US constitution did not decide to have an odd
# number or electors to avoid ths kind of situation instead of
# introducing a complex mechanism when tie appears????

nbties;
evalf(nbties/R);

states := [states]:

(3)

>	# What states participate to the tie? names := sort(ElectoralCollege[..,1]): all_states_in_ties := [op(op~(states))]: howoften := Vector( 51, i -> ListTools:-Occurrences(names[i], all_states_in_ties) ): ScatterPlot(Vector(51, i->i), howoften);

>

# All the states seem to appear equally likely in an electoral college tie.
# Why? Does someone have a guess?
#
# The reason is obvious, as each state must appear in the basket of a candidate,
# then in case of a tie each state is either in op~(1, poll)[1..place] (candidate 1)
# or either in op~(1, poll)[place+1..51] (candidate 2);
# So, as we obtained 9397 ties, each states appears exactly 9397 times (with
# different occurences in the baskets of candidate 1 and 2).

>	# Lengths of the configurations that lead to a tie. # # Pleas refer to the answer above to understand why Histogram(lengths) should be # symmetric. lengths := map(i -> numelems(states[i]), [$1..nbties]): sort(Tally(lengths))

(4)

>	Histogram(lengths, range=min(lengths)..max(lengths), discrete=true)

>	ShortestConfigurations := map(i -> if lengths[i]=min(lengths) then states[i] end if, [$1..nbties]): print~(ShortestConfigurations):

(5)

>	LargestConfigurations := map(i -> if lengths[i]=max(lengths) then states[i] end if, [$1..nbties]): print~(LargestConfigurations):

[Alaska, Tennessee, North_Carolina, South_Carolina, District_of_Columbia, Colorado, Minnesota, Georgia, South_Dakota, New_Hampshire, Wyoming, Ohio, Rhode_Island, Arizona, Delaware, Montana, West_Virginia, Vermont, Michigan, Kentucky, Louisiana, Arkansas, Maine, Missouri, New_Mexico, Virginia, Maryland, Oregon, Wisconsin, Iowa, Kansas, Connecticut, North_Dakota, Nevada, Hawaii, Oklahoma]

[West_Virginia, Maryland, Massachusetts, Colorado, South_Dakota, Kentucky, Kansas, Wyoming, North_Dakota, Indiana, Michigan, Utah, Louisiana, Ohio, Alabama, Nebraska, Connecticut, Illinois, Oklahoma, Alaska, New_Jersey, District_of_Columbia, Oregon, Nevada, Missouri, Delaware, Washington, New_Hampshire, Arizona, Maine, South_Carolina, Hawaii, Vermont, Montana, Rhode_Island, Idaho]

(6)

>

# What could be the largest composition of a basket in case of a tie?
# (shortest composition is the complementary of the largest one)

ecs := sort(ec, key=(x-> x[2]));
csecs := CumulativeSum(op~(2, ecs)):

# Where would the break locate?

tieloc := ListTools:-BinaryPlace(csecs, tie);

csecs[tieloc..tieloc+1]

(7)

>	# This 40 states coniguration is not a tie. # # But list all the states in basket of candidate 1 and look to the 41th state (which is # in the basket of candidate 2) ecs[1..tieloc]; print(): ecs[tieloc+1]

(8)

>	# It appears that exchanging Virginia and New_Jersey increases by 1 unit the college of candidate 1 # and produces a tie. LargestBasketEver := [ ecs[1..tieloc-1][], ecs[tieloc+1] ]; add(op~(2, LargestBasketEver))

(9)

>	# The largest electoral college tie contains 40 states (the shortest 11)

Download ElectoralCollegeTie.mw

A little about controlled platforms (parallel...

by: one man 1105 Product: Maple 17

October 28 2020

3 2

Controlled platform with 6 degrees of freedom. It has three rotary-inclined racks of variable length:

and an example of movement parallel to the base:

Perhaps the Stewart platform may not reproduce such trajectories, but that is not the point. There is a way to select a design for those specific functions that our platform will perform. That is, first we consider the required trajectories of the platform movement, and only then we select a driving device that can reproduce them. For example, we can fix the extreme positions of the actuators during the movement of the platform and compare them with the capabilities of existing designs, or simulate your own devices.
In this case, the program consists of three parts. (The text of the program directly for the first figure : PLATFORM_6.mw) In the first part, we select the starting point for the movement of a rigid body with six degrees of freedom. Here three equations f6, f7, f8 are responsible for the six degrees of freedom. The equations f1, f2, f3, f4, f5 define a trajectory of motion of a rigid body. The coordinates of the starting point are transmitted via disk E for the second part of the program. In the second part of the program, the trajectory of a rigid body is calculated using the Draghilev method. Then the trajectory data is transferred via the disk E for the third part of the program.
In the third part of the program, the visualization is executed and the platform motion drive device is modeled.
It is like a sketch of a possible way to create controlled platforms with six degrees of freedom. Any device that can provide the desired trajectory can be inserted into the third part. At the same time, it is obvious that the geometric parameters of the movement of this device with the control of possible emergency positions and the solution of the inverse kinematics problem can be obtained automatically if we add the appropriate code to the program text.
Equations can be of any kind and can be combined with each other, and they must be continuously differentiable. But first, the equations must be reduced to uniform variables in order to apply the Draghilev method.
(These examples use implicit equations for the coordinates of the vertices of the triangle.)

What to take care of when entering a tetrad

by: Freddy Baudine 70 Product: Maple 2020

October 27 2020

6 1

In the study of the Gödel spacetime model, a tetrad was suggested in the literature [1]. Alas, upon entering the tetrad in question, Maple's Tetrad's package complained that that matrix was not a tetrad! What went wrong? After an exchange with Edgardo S. Cheb-Terrab, Edgardo provided us with awfully useful comments regarding the use of the package and suggested that the problem together with its solution be presented in a post, as others may find it of some use for their work as well.

The Gödel spacetime solution to Einsten's equations is as follows.

>

(1)

>

((`Defined as spacetime tensors representing the NP null vectors of the tetrad formalism `*`see <a href='http://www.maplesoft.com/support/help/search.aspx?term=Physics,tetrads`*`,' target='_new'>?Physics,tetrads`*`,</a> `*l[mu]*`, `)*n[mu]*`, `*m[mu]*`, `)*conjugate(m[mu])

(2)

Working with Cartesian coordinates,

>

(3)

the Gödel line element is

>

(4)

Setting the metric

>

Physics:-g_[mu, nu] = Matrix(%id = 18446744078354506566)

$[metric = {(1, 1) = -1, (2, 2) = -1, (3, 3) = (1/2)*exp(2*q*y), (3, 4) = exp(q*y), (4, 4) = 1}, spaceindices = lowercaselatin_is]$

(5)

The problem appeared upon entering the matrix M below supposedly representing the alleged tetrad.

>

Matrix(%id = 18446744078162949534)

(6)

Each of the rows of this matrix is supposed to be one of the null vectors . Before setting this alleged tetrad, Maple was asked to settle the nature of it, and the answer was that M was not a tetrad! With the Physics Updates v.857, a more detailed message was issued:

>

`Warning, the given components form a`*null*`tetrad, `*`with a contravariant spacetime index`*`, only if you change the signature from `*`- - - +`*` to `*`+ - - -`*`.
You can do that by entering (copy and paste): `*Setup(signature = "+ - - -")

(7)

So there were actually three problems:

1.	The entered entity was a null tetrad, while the default of the Physics package is an orthonormal tetrad. This can be seen in the form of the tetrad metric, or using the library commands:

>

Physics:-Tetrads:-eta_[a, b] = Matrix(%id = 18446744078354552462)

(8)

>

(9)

>

(10)

2.

The matrix M would only be a tetrad if the spacetime index is contravariant. On the other hand, the command IsTetrad will return true only when M represents a tetrad with both indices covariant. For instance, if the command IsTetrad is issued about the tetrad automatically computed by Maple, but is passed the matrix corresponding to with the spacetime index contravariant, false is returned:

>

Physics:-Tetrads:-e_[a, `~μ`] = Matrix(%id = 18446744078297840926)

(11)

>

(12)

3.	The matrix M corresponds to a tetrad with different signature, (+---), instead of Maple's default (---+). Although these two signatures represent the same physics, they differ in the ordering of rows and columns: the timelike component is respectively in positions 1 and 4.

The issue, then, became how to correct the matrix M to be a valid tetrad: either change the setup, or change the matrix M. Below the two courses of action are provided.

First the simplest: change the settings. According to the message (7), setting the tetrad to be null, changing the signature to be (+---) and indicating that M represents a tetrad with its spacetime index contravariant would suffice:

>

(13)

The null tetrad metric is now as in the reference used.

>

Physics:-Tetrads:-eta_[a, b] = Matrix(%id = 18446744078298386174)

(14)

Checking now with the spacetime index contravariant

>

Physics:-Tetrads:-e_[a, `~μ`] = Matrix(%id = 18446744078162949534)

(15)

At this point, the command IsTetrad provided with the equation (15), where the left-hand side has the information that the spacetime index is contravariant

>

(16)

Great! one can now set the tetrad M exactly as entered, without changing anything else. In the next line it will only be necessary to indicate that the spacetime index, , is contravariant.

>

$[tetrad = {(1, 1) = -(1/2)*2^(1/2), (1, 3) = (1/2)*2^(1/2)*exp(q*y), (1, 4) = (1/2)*2^(1/2), (2, 1) = (1/2)*2^(1/2), (2, 3) = (1/2)*2^(1/2)*exp(q*y), (2, 4) = (1/2)*2^(1/2), (3, 2) = -(1/2)*2^(1/2), (3, 3) = ((1/2)*I)*exp(q*y), (3, 4) = 0, (4, 2) = -(1/2)*2^(1/2), (4, 3) = -((1/2)*I)*exp(q*y), (4, 4) = 0}]$

(17)

The tetrad is now the matrix M. In addition to checking this tetrad making use of the IsTetrad command, it is also possible to check the definitions of tetrads and null vectors using TensorArray.

>

(18)

>

Matrix(%id = 18446744078353048270)

(19)

For the null vectors:

>

(20)

>

[0 = 0, 1 = 1, 0 = 0, 0 = 0, Matrix(%id = 18446744078414241910)]

(21)

From its Weyl scalars, this tetrad is already in the canonical form for a spacetime of Petrov type "D": only

>

(22)

>

(23)

Attempting to transform it into canonicalform returns the tetrad (17) itself

>

Matrix(%id = 18446744078396685478)

(24)

Let's now obtain the correct tetrad without changing the signature as done in (13).

Start by changing the signature back to

>

(25)

So again, M is not a tetrad, even if the spacetime index is specified as contravariant.

>

`Warning, the given components form a`*null*`tetrad, `*`with a contravariant spacetime index`*`, only if you change the signature from `*`- - - +`*` to `*`+ - - -`*`.
You can do that by entering (copy and paste): `*Setup(signature = "+ - - -")

(26)

By construction, the tetrad M has its rows formed by the null vectors with the ordering . To understand what needs to be changed in M, define those vectors, independent of the null vectors (with underscore) that come with the Tetrads package.

>

and set their components using the matrix M taking into account that its spacetime index is contravariant, and equating the rows of M using the ordering :

>

[l[`~μ`] = Vector[row](%id = 18446744078368885086), n[`~μ`] = Vector[row](%id = 18446744078368885206), m[`~μ`] = Vector[row](%id = 18446744078368885326), mb[`~μ`] = Vector[row](%id = 18446744078368885446)]

(27)

>

${Physics:-D_[mu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-Ricci[mu, nu], Physics:-Riemann[mu, nu, alpha, beta], Physics:-Weyl[mu, nu, alpha, beta], Physics:-d_[mu], Physics:-Tetrads:-e_[a, mu], Physics:-Tetrads:-eta_[a, b], Physics:-g_[mu, nu], Physics:-gamma_[i, j], Physics:-Tetrads:-gamma_[a, b, c], l[mu], Physics:-Tetrads:-l_[mu], Physics:-Tetrads:-lambda_[a, b, c], m[mu], Physics:-Tetrads:-m_[mu], mb[mu], Physics:-Tetrads:-mb_[mu], n[mu], Physics:-Tetrads:-n_[mu], Physics:-Christoffel[mu, nu, alpha], Physics:-Einstein[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}$

(28)

Check the covariant components of these vectors towards comparing them with the lines of the Maple's tetrad

>

l[mu] = Array(%id = 18446744078298368710), n[mu] = Array(%id = 18446744078298365214), m[mu] = Array(%id = 18446744078298359558), mb[mu] = Array(%id = 18446744078298341734)

(29)

This shows the null vectors (with underscore) that come with Tetrads package

>

Physics:-Tetrads:-l_[mu] = Vector[row](%id = 18446744078354520414), Physics:-Tetrads:-n_[mu] = Vector[row](%id = 18446744078354520534), Physics:-Tetrads:-m_[mu] = Vector[row](%id = 18446744078354520654), Physics:-Tetrads:-mb_[mu] = Vector[row](%id = 18446744078354520774)

(30)

So (29) computed from M is the same as (30) computed from Maple's tetrad.

But, from (30) and the form of Maple's tetrad

>

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078297844182)

(31)

for the current signature

>

(32)

we see the ordering of the null vectors is , not used in [1] with the signature (+ - - -). So the adjustment required in M, resulting in , consists of reordering M's rows to be

>

Matrix(%id = 18446744078414243230)

(33)

>

(34)

Comparing with the tetrad computed by Maple ((24) and (31), they are actually the same.

References

[1]. Rainer Burghardt, "Constructing the Godel Universe", the arxiv gr-qc/0106070 2001.

[2]. Frank Grave and Michael Buser, "Visiting the Gödel Universe", IEEE Trans Vis Comput GRAPH, 14(6):1563-70, 2008.

Download Godel_universe_and_Tedrads.mw

Maple Conference 2020 - It's Almost Here!

by: Kat 232 Product: Maple

October 22 2020

7 3

The 2020 Maple Conference is coming up fast! It is running from November 2-6 this year, all remotely, and completely free.

The week will be packed with activities, and we have designed it so that it will be valuable for Maple users of all skill and experience levels. The agenda includes 3 keynote presentations, 2 live panel presentations, 8 Maplesoft recorded presentations, 3 Maple workshops, and 68 contributed recorded presentations.

There will be live Q&A’s for every presentation. Additionally, we are hosting what we’re calling “Virtual Tables” at every breakfast (8-9am EST) and almost every lunch (12-1 EST). These tables offer attendees a chance to discuss topics related to the conference streams of the day, as well as a variety of special topics and social discussions. You can review the schedule for these virtual tables here.

Attendance is completely free, and we’re confident that there will be something there for all Maple users. Whether you attend one session or all of them, we’d love to see you there!

You can register for the Maple Conference here.

Computing a tetrad in canonical form - automatically...

by: ecterrab 12679 Product: Maple

October 20 2020

5 0

In a recent question in Mapleprimes, a spacetime (metric) solution to Einstein's equations, from chapter 27 of the book of Exact Solutions to Einstein's equations [1] was discussed. One of the issues was about computing a tetrad for that solution [27, 37, 1] such that the corresponding Weyl scalars are in canonical form. This post illustrates how to do that, with precisely that spacetime metric solution, in two different ways: 1) automatically, all in one go, and 2) step-by-step. The step-by-step computation is useful to verify results and also to compute different forms of the tetrads or Weyl scalars. The computation below is performed using the latest version of the Maplesoft Physics Updates.

>

(1)

The starting point is this image of page 421 of the book of Exact Solutions to Einstein's equations, formulas (27.37)

Load the solution [27, 37, 1] from Maple's database of solutions to Einstein's equations

>

$`Default differentiation variables for d_, D_ and dAlembertian are:`*{X = (z, zb, r, u)}$

Physics:-g_[mu, nu] = Matrix(%id = 18446744078276690638)

(2)

>

(3)

The assumptions on the metric's parameters are

>

The line element is as shown in the second line of the image above

>

2*r^2*Physics:-d_(z)*Physics:-d_(zb)/P(z, zb, u)^2-2*Physics:-d_(r)*Physics:-d_(u)-2*H(X)*Physics:-d_(u)^2

(4)

Load Tetrads

>

((`Defined as spacetime tensors representing the NP null vectors of the tetrad formalism `*`see <a href='http://www.maplesoft.com/support/help/search.aspx?term=Physics,tetrads`*`,' target='_new'>?Physics,tetrads`*`,</a> `*l[mu]*`, `)*n[mu]*`, `*m[mu]*`, `)*conjugate(m[mu])

(5)

The Petrov type of this spacetime solution is

>

(6)

The null tetrad computed by the Maple system using a general algorithms is

>

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078178770326)

(7)

According to the help page TransformTetrad , the canonical form of the Weyl scalars for each different Petrov type is

So for type II, when the tetrad is in canonical form, we expect only and different from 0. For the tetrad computed automatically, however, the scalars are

>

(8)

The question is, how to bring the tetrad (equation (7)) into canonical form. The plan for that is outlined in Chapter 7, by Chandrasekhar, page 388, of the book "General Relativity, an Einstein centenary survey", edited by S.W. Hawking and W.Israel. In brief, for Petrov type II, use a transformation of to make , then a transformation of making , finally use a transformation of making . For an explanation of these transformations see the help page for TransformTetrad . This plan, however, is applicable if and only if the starting tetrad results in , which we see in (8) it is not the case, so we need, in addition, before applying this plan, to perform a transformation of making

In what follows, the transformations mentioned are first performed automatically, in one go, letting the computer deduce each intermediate transformation, by passing to TransformTetrad the optional argument canonicalform. Then, the same result is obtained by transforming the starting tetrad one step at at time, arriving at the same Weyl scalars. That illustrates well both how to get the result exploiting advanced functionality but also how to verify the result performing each step, and also how to get any desired different form of the Weyl scalars.

Although it is possible to perform both computations, automatically and step-by-step, departing from the tetrad (7), that tetrad and the corresponding Weyl scalars (8) have radicals, making the readability of the formulas at each step less clear. Both computations, can be presented in more readable form without radicals departing from the tetrad shown in the book, that is

>	$e_[a, mu] = (Matrix(4, 4, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = -1, (2, 1) = 0, (2, 2) = r/P(z, zb, u), (2, 3) = 0, (2, 4) = 0, (3, 1) = r/P(z, zb, u), (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = -1, (4, 4) = -H(X)}))$

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078621688766)

(9)

>

(10)

The corresponding Weyl scalars free of radicals are

>

(11)

So set this tetrad as the starting point

>

(12)

All the transformations performed automatically, in one go

To arrive in one go, automatically, to a tetrad whose Weyl scalars are in canonical form as in (31), use the optional argument canonicalform:

>

psi__0 = 0, psi__1 = 0, psi__2 = -(1/6)*(-(diff(diff(H(X), r), r))*r^2+2*(diff(H(X), r))*r-2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))+2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)-2*H(X))/r^2, psi__3 = 1, psi__4 = 0

(13)

Note the length of

>

(14)

That length corresponds to several pages long. That happens frequently, you get Weyl scalars with a minimum of residual invariance, at the cost of a more complicated tetrad.

The transformations step-by-step leading to the same canonical form of the Weyl scalars

Step 0

As mentioned above, to apply the plan outlined by Chandrasekhar, the starting point needs to be a tetrad with , not the case of (9), so in this step 0 we use a transformation of making . This transformation introduces a complex parameter E and to get any value of E suffices. We use :

>

Matrix(%id = 18446744078634914990)

(15)

>

Matrix(%id = 18446744078634940646)

(16)

Indeed, for this tetrad, :

>

(17)

Step 1

Next is a transformation of to make , that in the case of Petrov type II also implies on .According to the the help page TransformTetrad , this transformation introduces a parameter B that, according to the plan outlined by Chandrasekhar in Chapter 7 page 388, is one of the two identical roots (out of the four roots) of the principalpolynomial. To see the principal polynomial, or, directly, its roots you can use the PetrovType command:

>

(18)

The first two are the same and equal to -1

>

(19)

So the transformed tetrad is

>

Matrix(%id = 18446744078641721462)

(20)

Check this result and the corresponding Weyl scalars to verify that we now have and

>

(21)

>

(22)

Step 2

Next is a transformation of that makes . This transformation introduces a parameter E, that according to Chandrasekhar's plan can be taken equal to one of the roots of Weyl scalar that corresponds to the transformed tetrad. So we need to proceed in three steps:

a.	transform the tetrad introducing a parameter E in the tetrad's components

b.	compute the Weyl scalars for that transformed tetrad

c.	take and solve for E

d.	apply the resulting value of E to the transformed tetrad obtained in step a.

a.Transform the tetrad and for simplicity take E real

>

Matrix(%id = 18446744078624751238)

(23)

>

(24)

b. Compute for this tetrad

>

(25)

c. Solve discarding the case which implies on no transformation

>

$simplify(solve({rhs(psi__4 = (r^2*P(z, zb, u)*(E-1)^2*(diff(diff(H(X), r), r))-2*r*P(z, zb, u)^2*(E-1)*(diff(diff(H(X), r), z))+P(z, zb, u)^3*(diff(diff(H(X), z), z))-2*P(z, zb, u)^2*(E-1)^2*(diff(diff(P(z, zb, u), z), zb))+2*r*P(z, zb, u)*(E-1)*(diff(diff(P(z, zb, u), u), z))-2*r*P(z, zb, u)*(E-1)^2*(diff(H(X), r))+4*P(z, zb, u)^2*(E+(1/2)*(diff(P(z, zb, u), z))-1)*(diff(H(X), z))+2*((P(z, zb, u)*(E-1)*(diff(P(z, zb, u), zb))-(diff(P(z, zb, u), u))*r)*(diff(P(z, zb, u), z))+H(X)*P(z, zb, u)*(E-1))*(E-1))/(r^2*P(z, zb, u))) = 0, E <> 0}, {E}, explicit)[1])$

${E = ((diff(diff(H(X), r), r))*r^2*P(z, zb, u)+(diff(diff(H(X), r), z))*P(z, zb, u)^2*r-2*(diff(H(X), r))*P(z, zb, u)*r-(diff(diff(P(z, zb, u), u), z))*r*P(z, zb, u)+(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))*r-2*P(z, zb, u)^2*(diff(H(X), z))-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))*P(z, zb, u)+2*H(X)*P(z, zb, u)+(-P(z, zb, u)^4*((diff(diff(H(X), r), r))*r^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))-2*(diff(H(X), r))*r-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)+2*H(X))*(diff(diff(H(X), z), z))+P(z, zb, u)^4*(diff(diff(H(X), r), z))^2*r^2+(-2*r^2*(diff(diff(P(z, zb, u), u), z))*P(z, zb, u)^3+2*r^2*P(z, zb, u)^2*(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))-4*r*(diff(H(X), z))*P(z, zb, u)^4)*(diff(diff(H(X), r), z))-2*(diff(P(z, zb, u), z))*(diff(H(X), z))*P(z, zb, u)^3*(diff(diff(H(X), r), r))*r^2+P(z, zb, u)^2*(diff(diff(P(z, zb, u), u), z))^2*r^2+(-2*r^2*P(z, zb, u)*(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))+4*r*(diff(H(X), z))*P(z, zb, u)^3)*(diff(diff(P(z, zb, u), u), z))+4*(diff(P(z, zb, u), z))*(diff(H(X), z))*P(z, zb, u)^4*(diff(diff(P(z, zb, u), z), zb))+4*(diff(H(X), z))^2*P(z, zb, u)^4+4*P(z, zb, u)^2*(diff(P(z, zb, u), z))*((diff(H(X), r))*P(z, zb, u)*r-(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))*P(z, zb, u)-(diff(P(z, zb, u), u))*r-H(X)*P(z, zb, u))*(diff(H(X), z))+(diff(P(z, zb, u), u))^2*(diff(P(z, zb, u), z))^2*r^2)^(1/2))/(P(z, zb, u)*((diff(diff(H(X), r), r))*r^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))-2*(diff(H(X), r))*r-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)+2*H(X)))}$

(26)

d. Apply this result to the tetrad (23). In doing so, do not display the result, just measure its length (corresponds to two+ pages)

>

$T__3 := simplify(eval(T__2, {E = ((diff(diff(H(X), r), r))*r^2*P(z, zb, u)+(diff(diff(H(X), r), z))*P(z, zb, u)^2*r-2*(diff(H(X), r))*P(z, zb, u)*r-(diff(diff(P(z, zb, u), u), z))*r*P(z, zb, u)+(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))*r-2*P(z, zb, u)^2*(diff(H(X), z))-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))*P(z, zb, u)+2*H(X)*P(z, zb, u)+(-P(z, zb, u)^4*((diff(diff(H(X), r), r))*r^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))-2*(diff(H(X), r))*r-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)+2*H(X))*(diff(diff(H(X), z), z))+P(z, zb, u)^4*(diff(diff(H(X), r), z))^2*r^2+(-2*r^2*(diff(diff(P(z, zb, u), u), z))*P(z, zb, u)^3+2*r^2*P(z, zb, u)^2*(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))-4*r*(diff(H(X), z))*P(z, zb, u)^4)*(diff(diff(H(X), r), z))-2*(diff(P(z, zb, u), z))*(diff(H(X), z))*P(z, zb, u)^3*(diff(diff(H(X), r), r))*r^2+P(z, zb, u)^2*(diff(diff(P(z, zb, u), u), z))^2*r^2+(-2*r^2*P(z, zb, u)*(diff(P(z, zb, u), u))*(diff(P(z, zb, u), z))+4*r*(diff(H(X), z))*P(z, zb, u)^3)*(diff(diff(P(z, zb, u), u), z))+4*(diff(P(z, zb, u), z))*(diff(H(X), z))*P(z, zb, u)^4*(diff(diff(P(z, zb, u), z), zb))+4*(diff(H(X), z))^2*P(z, zb, u)^4+4*P(z, zb, u)^2*(diff(P(z, zb, u), z))*((diff(H(X), r))*P(z, zb, u)*r-(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))*P(z, zb, u)-(diff(P(z, zb, u), u))*r-H(X)*P(z, zb, u))*(diff(H(X), z))+(diff(P(z, zb, u), u))^2*(diff(P(z, zb, u), z))^2*r^2)^(1/2))/(P(z, zb, u)*((diff(diff(H(X), r), r))*r^2+2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))-2*(diff(H(X), r))*r-2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)+2*H(X)))}[1]))$

>

(27)

Check the scalars, we expect

>

(28)

Step 3

Use a transformation of making . Such a transformation changes , where we need to take A*exp(-I*Omega) = 1/`Ψ__3` , and without loss of generality we can take

Check first the value of in the last tetrad computed

>

(29)

So, the transformed tetrad to which corresponds Weyl scalars in canonical form, with and , is

>

(30)

>

psi__0 = 0, psi__1 = 0, psi__2 = -(1/6)*(-(diff(diff(H(X), r), r))*r^2+2*(diff(H(X), r))*r+2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)-2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))-2*H(X))/r^2, psi__3 = 1, psi__4 = 0

(31)

These are the same scalars computed in one go in (13)

>

psi__0 = 0, psi__1 = 0, psi__2 = -(1/6)*(-(diff(diff(H(X), r), r))*r^2+2*(diff(H(X), r))*r-2*(diff(P(z, zb, u), z))*(diff(P(z, zb, u), zb))+2*(diff(diff(P(z, zb, u), z), zb))*P(z, zb, u)-2*H(X))/r^2, psi__3 = 1, psi__4 = 0

(32)

>

Download The_metric_[27_37_1]_in_canonical_form.mw

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

Uniform point plot

by: Kitonum 19375

October 17 2020

5 1

When we plot a curve with the option style=point , symbols go evenly not along the length of this curve, but along the range of the independent variable. For this reason the plot often looks unattractive. Here are two examples. In the first example, the default option adaptive=true is used, in which Maple adds points in some places.

restart;
plot(surd(x,3), x=-2.5..2.5, style=point, scaling=constrained, symbol=solidcircle, symbolsize=8, numpoints=30, size=[800,300]);
plot(surd(x,3), x=-2.5..2.5, style=point, scaling=constrained, symbol=solidcircle, symbolsize=8, numpoints=30, adaptive=false, size=[800,300]);

The UniformPointPlot procedure allows you to plot curves by symbols (as for style=point), and these symbols go from each other at equal distances, measured along this curve. The procedure uses a well-known formula for the length of a curve in two and three dimensions. The procedure parameters are clear from the three examples below.

UniformPointPlot:=proc(F::{algebraic,list},eq::`=`,n::posint:=40,Opt::list:=[symbol=solidcircle, symbolsize=8, scaling=constrained])
local t, R, P, g, L, step, L1, L2;
uses plots;
Digits:=4:
t:=lhs(eq); R:=rhs(eq);
P:=`if`(type(F,algebraic),[t,F],F); 
g:=x->`if`(F::algebraic or nops(F)=2,evalf(Int(sqrt(diff(P[1],t)^2+diff(P[2],t)^2), t=lhs(R)..x, epsilon=0.001)),evalf(Int(sqrt(diff(P[1],t)^2+diff(P[2],t)^2+diff(P[3],t)^2), t=lhs(R)..x, epsilon=0.001))):
L:=g(rhs(R)); step:=L/(n-1);
L1:=[lhs(R),seq(fsolve(g-k*step, fulldigits),k=1..n-2),rhs(R)];
L2:=map(s->`if`(type(F,algebraic),[s,eval(F,t=s)],eval(F,t=s)), L1):
`if`(F::algebraic or nops(F)=2,plot(L2, style=point, Opt[]),pointplot3d(L2, Opt[]));
end proc:

Examples of use:

UniformPointPlot(surd(x,3), x=-2.5..2.5, 30);

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