Education

Teaching and learning about math, Maple and MapleSim

  The geometry of the triangle
  Romanova Elena,  8 class,  school 57, Kazan, Russia

       Construction of triangle and calculation its angles

       Construction of  bisectors
      
       Construction of medians
      
       Construction of altitudes


> restart:with(geometry):      

The setting of the height of the triandle and let's call it "Т"
> triangle(T,[point(A,4,6),point(B,-3,-5),point(C,-4,8)]);

                                  T

        Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);

Construction of the triangle АВС

> draw({T(color=gold,thickness=3)},printtext=true,axes=NONE);     
Calculation of the distance between heights А and В - the length of a side АВ

> d1:=distance(A,B);

                           d1 := sqrt(170)

        
        Calculation of the distance between heights В and С - the length of a side ВС
> d2:=distance(B,C);

                           d2 := sqrt(170)

       The setting of line which passes through two points А and В
> line(l1,[A,B]);

                                  l1

       Display the equation of line l1
> Equation(l1);
> x;
> y;

                         -2 + 11 x - 7 y = 0

        The setting of line which passes through two points А and С
> line(l2,[A,C]);

                                  l2

       Display the equation of line l2
> Equation(l2);
> x;
> y;

                          56 - 2 x - 8 y = 0

         The setting of line which passes through two points В and С
> line(l3,[B,C]);

                                  l3

        Display the equation of line l3
> Equation(l3);
> x;
> y;

                          -44 - 13 x - y = 0

        Check the point А lies on line l1
> IsOnLine(A,l1);

                                 true

        Check the point А lies on line l1
> IsOnLine(B,l1);

                                 true

        Calculation of the andle between lines l1 and l2
> FindAngle(l1,l2);

                              arctan(3)

        The conversion of result to degrees
> b1:=convert(arctan(97/14),degrees);

                                      97
                               arctan(--) degrees
                                      14
                     b1 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b2:=evalf(b1);

                      b2 := 81.78721981 degrees

        Calculation of the andle between lines l1 and l3
> FindAngle(l1,l3);

                             arctan(3/4)

       The conversion of result to degrees
> b3:=convert(arctan(97/99),degrees);

                                      97
                               arctan(--) degrees
                                      99
                     b3 := 180 ------------------
                                       Pi

        Calculation of decimal value of this angle
> b4:=evalf(b3);

                      b4 := 44.41536947 degrees

       Calculation of the angle between lines l2 and l3
> FindAngle(l2,l3);

                              arctan(3)

       The conversion of  result to degrees
> b5:=convert(arctan(97/71),degrees);

                                      97
                               arctan(--) degrees
                                      71
                     b5 := 180 ------------------
                                       Pi

        Calculation of decimal value of  this angle
> b6:=evalf(b5);

                      b6 := 53.79741070 degrees

        Check the sum of all the angles of the triangle
> b2+b4+b6;

                         180.0000000 degrees

        Analytical information about the point А
> detail(A);
   name of the object: A
   form of the object: point2d
   coordinates of the point: [4, 6]
          Analytical information about the point В
> detail(B);
   name of the object: B
   form of the object: point2d
   coordinates of the point: [-3, -5]
          Analytical information about the point С
> detail(C);
   name of the object: C
   form of the object: point2d
   coordinates of the point: [-4, 8]

   The setting of heights of the triangle points A,B,C and let's call it "Т"

   with(geometry):
> triangle(ABC, [point(A,7,8), point(B,6,-7), point(C,-6,7)]):
        The setting of the bisector of angle А in triandle АВС
> bisector(bA, A, ABC);

                                  bA

        Analytical information about the bisector of angle А in the triandle
> detail(bA);
   name of the object: bA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (15*170^(1/2)+226^(1/2))*_x+(-13*226^(1/2)-170^(1/2))*_y+97*226^(1/2)-97*170^(1/2) = 0

        Construction of the triangle
> draw(ABC,axes=normal,view=[-8..8,-8..8]);

 Construction of the triangle ABC

> draw({ABC(color=gold,thickness=3)},printtext=true,axes=NONE);     

 Construction of the bisector of angle А

> draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3)},printtext=true,axes=NONE);    

The setting of the bisector of angle В in the triangle АВС

> bisector(bB, B, ABC);

                                  bB

       Analytical information about the bisector of angle B in the triandle
> detail(bB);
   name of the object: bB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (-15*340^(1/2)-14*226^(1/2))*_x+(-12*226^(1/2)+340^(1/2))*_y+97*340^(1/2) = 0

         Construction of the bisector of angle В
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3)},printtext=true,axes=NONE);    



    The setting of the bisector of angle С in the triangle АВС

> bisector(bC, C, ABC);

                                  bC

        Analytical information about the bisector of angle С in the triangle
> detail(bC);
   name of the object: bC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: (14*170^(1/2)-340^(1/2))*_x+(13*340^(1/2)+12*170^(1/2))*_y-97*340^(1/2) = 0

        Construction of the bisector of angle С
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3)},printtext=true,axes=NONE);  

 Calculation of the point of intersection of the bisectors and let's call it "О"

> intersection(O,bA,bB,bC);coordinates(O);

                                  O


     7 sqrt(85) - 3 sqrt(2) sqrt(113) + 3 sqrt(85) sqrt(2)
  [2 -----------------------------------------------------,
       sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

          -16 sqrt(85) - 7 sqrt(2) sqrt(113) + 7 sqrt(85) sqrt(2)
        - -------------------------------------------------------]
             sqrt(85) sqrt(2) + sqrt(2) sqrt(113) + 2 sqrt(85)

       Construction of the bisectors and  marking of the point of intersection  "О" in the triandle
>draw({ABC(color=gold,thickness=3),bA(color=green,thickness=3),bB(color=red,thickness=3),bC(color=blue,thickness=3),O},printtext=true,axes=NONE);
> restart:
> with(geometry):
       The setting of the heights of the triangle points A,B,C and let's call it "Т"
> point(A,7,8),point(B,6,-7),point(C,-6,7);

                               A, B, C

        Let's call "Т1"
> triangle(T1,[A,B,C]);

                                  T1

        Construction of "Т1"
> draw(T1(color=gold,thickness=3),axes=NONE,printtext=true);
  The setting of the median from the point В in the trianglemedian(mB,B,T1,B1);
> median(mb,B,T1);

                                  mB


                                  mb

        Construction of the median from the point В
> draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mb},printtext=true,axes=NONE);

The setting of the median from the point А in the trianglemedian(mA,A,T1,A1);
> median(ma,A,T1);

                                  mA


                                  ma

        Construction of the median from the point А
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),ma},printtext=true,axes=NONE);
The setting of the median from the point С in the trianglemedian(mC,C,T1,C1);
> median(mc,C,T1);

                                  mC


                                  mc

        Costruction of the median from the point С
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=maroon,thickness=3)},printtext=true,axes=NONE);




Calculation of the point of  intersection of the median and let's call it "О"

>intersection(O,ma,mb,mC);coordinates(O);

                                  O


                              [7/3, 8/3]

        Construction of medians and marking of the point of  intersection "О" in the triangle
>draw({T1(color=gold,thickness=3),mB(color=green,thickness=3),mA(color=magenta,thickness=3),mA,mC(color=violet,thickness=3),O},printtext=true,axes=NONE);
> restart:with(geometry):
> _EnvHorizontalName:=x:_EnvVerticalName=y:       The setting of the heights of the triangle points A, B, C  and let's call it "Т"
> triangle(T,[point(A,7,8),point(B,6,-7),point(C,-6,7)]);

                                  T

       Construction of the triangle
> draw(T,axes=normal,view=[-8..8,-8..8]);


The setting of the altitude in the triangle from the point Сaltitude(hC1,C,T,C1);
> altitude(hC,C,T);

                                 hC1


                                  hC

        Analytical information about the altitude hC from the point С in the triangle
> detail(hC);
   name of the object: hC
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -99+_x+15*_y = 0

        Construction of the altitude from the point С
> draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hC},printtext=true,axes=NONE);     

  The setting of the altitude in the triangle from the point Аaltitude(hA1,A,T,A1);
> altitude(hA,A,T);

                                 hA1


                                  hA

        Analytical information about the altitude hA from the point А in the triangle
> detail(hA);
   name of the object: hA
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -28-12*_x+14*_y = 0

        Construction of the altitude from the point А
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hA1},printtext=true,axes=NONE);       The setting of the altitude from the point В

> altitude(hB1,B,T,B1);
> altitude(hB,B,T);

                                 hB1


                                  hB

        Analytical information about the altitude hB from the point В in the triangle
> detail(hB);
   name of the object: hB
   form of the object: line2d
   assume that the name of the horizonal and vertical                    axis are _x and _y
   equation of the line: -71+13*_x+_y = 0

        Consruction of the altitude from the point В
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1},printtext=true,axes=NONE);     
 Calculation of the point of intersection of altitudes and let's call it "О"

>intersection(O,hB,hA,hC);coordinates(O);

                                  O


                               483  608
                              [---, ---]
                               97   97

        Construction of altitudes and marking of the point of intersection "О" in the triangle
>draw({T(color=gold,thickness=3),hC1(color=green,thickness=3),hA1(color=red,thickness=3),hB1(color=blue,thickness=3),hB1,O},printtext=true,axes=NONE);




 

 

 

 

 

 

 

 

 

 

 

 

 

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  Elena, Liya

  "Researching turkish song: the selection of the main element and its graphic transformations",

   Russia, Kazan, school #57

The setting and visualization of the melodic line of the song
> restart:
> with(plots):with(plottools):
> p0:=plot([[0.5,9],[1,7],[2,9],[4,11],[6,9],[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9],[17,7],[18,9]],color=magenta):p1:=plot([[18,9],[20,11],[22,9],[23,11],[24,9],[26,11],[28,11],[29.5,8],[30,11],[32,9],[33.5,8],[34,9],[36,7],[37.5,5],[38,9],[40,7],[42,5],[44,5],[46,4],[47,5],[48,2],[50,4],[51,5],[51.5,4],[52,2],[54,4],[56,4],[56.5,5],[57,4],[58,5],[60,7],[62,5],[64,7],[66,5]],color=cyan):
> p2:=plot([[66,5],[68,5],[69,5],[70,4],[71,5],[71.5,4],[72,2],[73,4],[74,5],[75,7],[76,5],[78,4],[78.5,7],[80,5],[82.5,4],[83.5,4],[84,2],[86,4],[88,4],[90.5,4],[91.5,4]],color=red):
> p3:=plot([[91.5,4],[92,2],[94,4],[96,4],[96.5,9],[97,7],[98,9],[100,11],[100.5,9],[101,11],[102,9],[104,11],[106,9],[108,9],[109,9],[109.5,9],[110,7],[111,9],[112,7],[113,7],[114,9],[116,11],[116.5,9],[117,11],[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9],[126,11],[128,9],[129,7],[130,9],[132,11],[132.5,9],[133,11],[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9],[142,11],[143,7],[143.5,7],[144,9],[144.5,9],[145,7],[146,9],[148,11],[148.5,9],[149,11],[150,9],[151.5,11],[152,9],[154.5,9],[156,9],[156.5,9],[157,11],[157.5,9],[158,11],[160,9],[161,7],[162,9],[164,11],[164.5,9],[165,11],[166,9],[168,11],[168.5,9],[171.5,9],[172,9],[172.5,9],[173.5,11],[174,9],[174.5,11],[175,7],[175.5,7],[176,9],[176.5,9],[177,7],[178,9],[180,11],[180.5,9],[181,11],[182,9],[183.5,11],[184,9],[186.5,9],[188,9],[188.5,9],[189,11],[189.5,9],[190,11],[192,9],[192.5,9],[193,7],[194,9],[196,11],[196.5,9],[197,11],[198,9],[200,11],[201.5,9],[202,11],[203,9],[203.5,8],[204,9],[205,7],[205.5,9],[206,11],[207,9],[208,7],[209,8],[209.5,7],[210,9],[211,7],[212,5],[213,5],[213.5,5],[214,9],[215,7],[216,5],[217,5],[217.5,5],[218,7],[219,5],[220,4],[221,4],[221.5,4],[222,7],[223,5],[224,4],[225,4],[227,4],[227.5,4],[228,2],[230,4]],color=blue):
> p4:=plot([[230,4],[232,4],[232.5,5],[233,4],[234,5],[236,7],[236.5,5],[237,5],[238,9],[240,7],[242.5,5],[244,5],[245,5],[246,4],[246.5,5],[247,4],[248,2],[250,4],[250.5,7],[251,5],[252,4],[254,4],[254.5,7],[255,5],[256,4],[258,4]],color=brown):
> p5:=plot([[258,4],[259,4],[260,2]],color=green):
> plots[display](p0,p1,p2,p3,p4,p5,thickness=2);

 

 

The selection of the main melodic element in graph of whole song. The whole song is divided into separate elements - results of transformationss0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> s1:=plot([[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9]],color=blue):
> s2:=plot([[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9]],color=blue):
> s3:=plot([[150,9],[151.5,11],[152,9],[154.5,9],[156,9],[156.5,9],[157,11],[157.5,9]],color=blue):
> s4:=plot([[166,9],[168,11],[168.5,9],[171.5,9],[172,9],[172.5,9],[173.5,11],[174,9]],color=blue):
> s5:=plot([[182,9],[183.5,11],[184,9],[186.5,9],[188,9],[188.5,9],[189,11],[189.5,9]],color=blue):
> s6:=plot([[250,4],[250.5,7],[251,5],[252,4],[254,4],[254.5,7],[255,5],[256,4]],color=blue):
> plots[display](s0,s1,s2,s3,s4,s5,s6);
> s:=plots[display](s0,s1,s2,s3,s4,s5,s6):

 

Animated display of grafical transformation of the basic element (to click on the picture - on the panel of instruments appears player - to play may step by step).m0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> pm:=plot([[118,9],[119.5,11],[120,9],[122.5,9],[124,9],[124.5,9],[125,11],[125.5,9]],color=red,style=line,thickness=4):
> iop:=plots[display](m0,pm,insequence=true):
> plots[display](iop,s0);

> m0_t:=translate(m0,110,0):
> m0_r:=reflect(m0_t,[[0,9],[24,9]]):
> plots[display](m0,m0_r,insequence=true);
> m0r:=plots[display](m0,m0_r,insequence=true):

> pm0:=plots[display](pm,m0):
> plots[display](pm0,m0r);

> m0:=plot([[7,11],[8,7],[10,9],[12,9],[14,9],[16,7],[16.5,9]],color=blue):
> pn:=plot([[134,9],[136,11],[136.5,9],[138.5,9],[140,9],[140.5,9],[141,11],[141.5,9]],color=blue,thickness=3):
> iop:=plots[display](m0,pn,insequence=true):
> plots[display](iop,s0);

> m0_t1:=translate(m0,126,0):
> m0_r1:=reflect(m0_t1,[[0,9],[24,9]]):
>
> plots[display](m0,m0_r1,insequence=true);
> m0r1:=plots[display](m0,m0_r1,insequence=true):

> pm01:=plots[display](pn,m0):
> plots[display](pm01,m0r1);

 

> pm2:=plots[display](pn,pm,m0):
> plots[display](pm0,m0r,pm01,m0r1);

> pt_i_1:=seq(translate(pm,5*11*i,0),i=0..4):
> plots[display](pt_i_1);

> pm_i:=seq(translate(pm,5*11*i,0),i=0..4):
> plots[display](pm_i);
> iop1:=plots[display](pm_i,insequence=true):
> plots[display](iop1,s0);

 

> pm_i_0:=seq(translate(m0_r,5*11*i,0),i=0..4):
> plots[display](pm_i_0);
> iop2:=plots[display](pm_i_0,insequence=true):
> plots[display](iop2,s0);

 

 

 

 

 

 

Construction of arabesques of melodic line BACH

Elena, Liya "Construction of arabesques of melodic line BACH", Kazan, Russia, school#57
       
> restart:
> with(plots):with(plottools):

      The setting and visualization of line BACH: B - note b-flat, A - note la, C - note do, H - note si.
> p0:=plot([[0,1],[2,0],[4,1.5],[6,1]],thickness=4,color=cyan,scaling=constrained);
>
>   p0 := PLOT(
>
>         CURVES([[0, 1.], [2., 0], [4., 1.500000000000000], [6., 1.]])
>
>         , SCALING(CONSTRAINED), THICKNESS(4), AXESLABELS( ,  ),
>
>         COLOUR(RGB, 0, 1.00000000, 1.00000000),
>
>         VIEW(DEFAULT, DEFAULT))
>
> plots[display](p0);
> r_i:=seq(rotate(p0,i*Pi/4),i=1..8):
> p1:=display(r_i,p0):plots[display](p1,scaling=constrained);

> c1:=circle([0,0],6,color=blue,thickness=2):
> plots[display](c1,p1,scaling=constrained);
> p_c:=plots[display](c1,p1,scaling=constrained):

> pt_i_2:=seq(translate(p1,0,2*6*i),i=0..4):
> plots[display](pt_i_2,scaling=constrained);
> pt_i_22:=seq(translate(p1,0,6*i),i=0..4):
> plots[display](pt_i_22,scaling=constrained);
> pt_i_222:=seq(translate(p1,0,1/2*6*i),i=0..4):
> plots[display](pt_i_222,scaling=constrained);

> pr:=rotate(p1,Pi/8):
> plots[display](pr,scaling=constrained);
> plots[display](p1,pr,scaling=constrained);
> pr_i:=seq(rotate(p1,Pi/16*i),i=0..8):
> plots[display](pr_i,scaling=constrained);


> pt_1:=translate(p1,0,2*6):
> pr_1_i:=seq(rotate(pt_1,Pi/3.5*i),i=0..6):
> plots[display](pr_1_i,scaling=constrained);
> pr_11_i:=seq(rotate(pt_1,Pi/5*i),i=0..10):
> plots[display](pr_11_i,scaling=constrained);
> pr_111_i:=seq(rotate(pt_1,Pi/6.5*i),i=0..12):
> plots[display](pr_111_i,scaling=constrained);


Elena, Liya "Designing of islamic arabesques", Kazan, Russia, school #57


> restart:
      At the theorem of cosines  c^2 = a^2+b^2-2*a*b*cos(phi);
      In our case  c=a0 ,  a=1 ,  a=b , phi; - acute angle of a rhombus (the tip of the kalam).
      s0 calculated at theorem of  Pythagoras.
     (а0 - horizontal diagonal of a  rhombus, s0 - vertical diagonal of a  rhombus)
> a:=1:phi:=Pi/4:
> a0:=sqrt(a^2+a^2-2*a^2*cos(phi));

                       a0 := sqrt(2 - sqrt(2))

> solve((s0^2)/4=a^2-(a0^2)/4,s0);

                sqrt(2 + sqrt(2)), -sqrt(2 + sqrt(2))


      The setting of initial parameters : the size of the tip of the pen-kalam and  depending on its - the main module size - point
       (а0 - horizontal diagonal of a  rhombus, s0 - vertical diagonal of a  rhombus)
> a0:=sqrt(2-sqrt(2)):
> s0:=sqrt(2+sqrt(2)):
      Connection the graphical libraries Maple
> with(plots):with(plottools):
      Construction of unit of measure (point) - rhombus - the tip of the kalam
> p0:=plot([[0,0],[a0/2,s0/2],[0,s0],[-a0/2,s0/2],[0,0]],scaling=constrained,color=gold,thickness=3):
> plots[display](p0);

The setting and construction of altitude of alif - the basis of the rules compilation of the proportions      Example, on style naskh altitude of alif amount five points
> p_i:=seq(plot([[0,0+s0*i],[a0/2,s0/2+s0*i],[0,s0+s0*i],[-a0/2,s0/2+s0*i],[0,0+s0*i]],scaling=constrained,color=black),i=0..4):
> pi:=display(p_i):
> plots[display](p_i);
The setting of appropriate circle of diameter, amount altitude of alifd0:=s0+s0*i:
> i:=4:
> d0:=d0:
> c0:=circle([0,d0/2],d0/2,color=blue):
> plots[display](p_i,c0);


Construction of flower by turning "point"r_i:=seq(rotate(p0,i*Pi/4),i=1..8):
> p1:=display(r_i,p0):plots[display](p1,scaling=constrained);

 The setting of circumscribed circlec1:=circle([0,0],s0,color=blue,thickness=2):
      Construction and the setting of flower inscribed in a circle
> plots[display](c1,p1,scaling=constrained);
> p_c:=plots[display](c1,p1,scaling=constrained):

The setting and construction of arabesque by horizontal parallel transport original flower with different stepspt_i_1:=seq(translate(p1,5*a0*i,0),i=0..4):
> plots[display](pt_i_1);
> pt_i_11:=seq(translate(p1,2*a0*i,0),i=0..4):
> plots[display](pt_i_11);
> pt_i_111:=seq(translate(p1,a0*7*i,0),i=0..4):
> plots[display](pt_i_111);

 The setting and construction of arabesque by vertical parallel transport original flower with different stepspt_i_2:=seq(translate(p1,0,2*s0*i),i=0..4):
> plots[display](pt_i_2);
> pt_i_22:=seq(translate(p1,0,s0*i),i=0..4):
> plots[display](pt_i_22);
> pt_i_222:=seq(translate(p1,0,1/2*s0*i),i=0..4):
> plots[display](pt_i_222);
 Getting arabesques by turning original flower on different anglespr:=rotate(p1,Pi/8):
> plots[display](pr);
> plots[display](p1,pr);

> pr_i:=seq(rotate(p1,Pi/16*i),i=0..8):
> plots[display](pr_i);


> pt_1:=translate(p1,0,2*s0):
> pr_1_i:=seq(rotate(pt_1,Pi/3.5*i),i=0..6):
> plots[display](pr_1_i);
> pr_11_i:=seq(rotate(pt_1,Pi/5*i),i=0..10):
> plots[display](pr_11_i);
> pr_111_i:=seq(rotate(pt_1,Pi/6.5*i),i=0..12):
> plots[display](pr_111_i);


Construction of standard quadrilaterals

      Muchametshina Liya,  8th class,  school № 57, Kazan, Russia


                   Square

                  Rectangle     
                  
                  Rhombus        
 
                  Parallelogram

                   Trapeze

Construction of square

> restart:
> with(plottools):
       Сoordinates (x;y) of the lower left corner of the square and the side "а"
> x:=0;y:=3;a:=6;

                                x := 0


                                y := 3


                                a := 6

      Construction of the square
> P1:=plot([[x,y],[x,y+a],[x+a,y+a],[x+a,y],[x,y]],color=green,thickness=4):
> plots[display](P1,scaling=CONSTRAINED);

The setting of the second square wich moved relative to the first on the vector (2;-3) (vector can be changed) and with side "а-1" (the length of a side can be changed)P2:=plot([[x+2,y-3],[x+2,y-3+a-1],[x+2+a-1,y-3+a-1],[x+2+a-1,y-3],[x+2,y-3]],color=black,thickness=4):
> plots[display](P1,P2,scaling=CONSTRAINED);

Construction of rectangle

> restart:
> with(plottools):
        Сoordinates (x;y) of the lower left corner of the square and the "а" and "b" sides
> x:=0;y:=2;a:=3;b:=9;
>

                                x := 0


                                y := 2


                                a := 3


                                b := 9

       The rectangle is specified by the sequence of vertices with given the lengths "a" and "b"
> l:=plot([[x,y],[x,y+a],[x+b,y+a],[x+b,y],[x,y]]):
> plots[display](l,scaling=CONSTRAINED,thickness=4);
Construction of rhombus

> restart:
> with(plottools):
      The coordinates (x;y) of the initial vertex of the rhombus and the half of the diagonals "a" and "b"
> x:=0;y:=2;a:=3;b:=4;

                                x := 0


                                y := 2


                                a := 3


                                b := 4

       Rhombus is specified by the sequence of vertices with the values "a" and "b"
> ll:=plot([[x,y],[x+a,y+b],[x+a+a,y],[x+a,y-b],[x,y]]):
> plots[display](ll,scaling=CONSTRAINED,thickness=4);

Construction of parallelogram

> restart:
> with(plottools):
      (х;у) - the starting point, (i;j) - the displacement vector of starting point, "а" - the base of the parallelogram
> x:=0;y:=0;i:=4;j:=5;a:=10;

                                x := 0


                                y := 0


                                i := 4


                                j := 5


                               a := 10

     The parallelogram is defined by the sequence of vertices
> P1:=plot([[x,y],[x+i,y+j],[x+i+a,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
 If  i= 0  it turns out the rectangleget.
       If  j= а  it turns out the  square.
       If  a := sqrt(i^2+j^2) it turns out the rhombus. a:=sqrt(i^2+j^2):

Construction of trapeze

Trapeze general form
> restart:
> with(plottools):
>
        (х;у) - the starting point, (i;j) - the displacement vector of starting point, а - the larger base of the trapezoid
> x:=0;y:=2;i:=1;j:=5;a:=11;

                                x := 0


                                y := 2


                                i := 1


                                j := 5


                               a := 11

         The trapez is defined by the sequence of vertices     
> P1:=plot([[x,y],[x+i,y+j],[x+i+j,y+j],[x+i+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Rectangular trapezoid
> restsrt:
> with(plottools):
> x:=0;y:=2;i:=0;j:=6;a:=11;

                                x := 0


                                y := 2


                                i := 0


                                j := 6


                               a := 11

> P1:=plot([[x,y],[x,y+j],[x+j,y+j],[x+a,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);
Isosceles trapezoid
> restart:
> with(plottools):
> x:=0;y:=2;i:=4;j:=6;a:=15;

                                x := 0


                                y := 2


                                i := 4


                                j := 6


                               a := 15

> P1:=plot([[x,y],[x+i,y+j],[x+j+i,y+j],[x+a,y],[x,y]]):
> plots[display](P1,scaling=CONSTRAINED,thickness=4);



 

 

 

> restart;
> a := -10; b := 10; ps := seq(plot([i, t, t = -20 .. 20], x = -10 .. 10, y = -20 .. 20, color = red, style = point), i = a .. b);

plots[display](ps, insequence = true); p := plots[display](ps, insequence = true);

 

restart:
with(plots):
y=sin(x);
p:=implicitplot(y=sin(x),x=-10..10,y=-2..2,thickness=4,color=red,scaling=constrained,numpoints=1000):
plots[display](p);

 

y=sin(3*x);
p0:=implicitplot(y=sin(x),x=-10..10,y=-5..5,thickness=3,color=red,scaling=constrained,numpoints=1000,linestyle=2,style=POINT,symbol=CROSS):
p1:=implicitplot(y=sin(3*x),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p0,p1);
y=sin(1/3*x);
p11:=implicitplot(y=sin(1/3*x),x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p0,p11);

 

 

y=2*sin(x);
p2:=implicitplot(y=2*sin(x),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p0,p2);
y=1/2*sin(x);
p22:=implicitplot(y=1/2*sin(x),x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p0,p22);

 

y=2+sin(x);
p3:=implicitplot(y=2+sin(x),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p0,p3);
y=sin(x)-2;
p33:=implicitplot(y=sin(x)-2,x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p0,p33);

y=sin(x+2);
p4:=implicitplot(y=sin(x+2),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p0,p4);
y=sin(x-2);
p44:=implicitplot(y=sin(x-2),x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p0,p44);

y=-sin(x);
p7:=implicitplot(y=-sin(x),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p0,p7);
y=sin(-x);
p77:=implicitplot(y=sin(-x),x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p0,p77);

 

y=abs(sin(x));
p00:=implicitplot(y=sin(x),x=-10..10,y=-5..5,thickness=3,color=red,scaling=constrained,numpoints=1000,linestyle=2,style=POINT,symbol=BOX):
p5:=implicitplot(y=abs(sin(x)),x=-10..10,y=-5..5,thickness=4,color=blue,numpoints=10000):
plots[display](p00,p5);
plots[display](p5,scaling=constrained);

y=sin(abs(x));
p00:=implicitplot(y=sin(x),x=-10..10,y=-5..5,thickness=3,color=red,scaling=constrained,numpoints=1000,linestyle=2,style=POINT,symbol=BOX):
p6:=implicitplot(y=sin(abs(x)),x=-10..10,y=-5..5,thickness=4,color=navy,numpoints=10000):
plots[display](p00,p6);
plots[display](p6,scaling=constrained);

 

 

Post gialid_GEODROMchik - what is this?

Pilot project of Secondary school # 57 of Kazan, Russia

Use of Maple

in Mathematics Education by mathematics teacher Alsu Gibadullina

and in scientific work of schoolchildren

 

Examples made using the Maple

the 6th class

 

              Arina                         Elza                             David    

       

       Book.mws              Kolobok.mws               sn_angl.mws

 

         Artur    

 

 

I'm an educator (physicist) who has migrated to Maple because of the lower "activation barrier" to get something of interest produced by the student. The students in my courses are exposed to several language (Python, C++, Java) and mathematical systems (Mathematica, Maple, MATLAB.) Many claim that unless forced to used a particular language or system, their first choice is Python and Maple for the reason I cite. 

As a consequence, it is my experience that students truly perfer the math-like appearance of the 2-D Math notation as opposed to the Maple notation. They see it as more natural - again with a lower activation barrier. Hence I see no reason to change. However, I would be interested in reasons why it might be beneficial.

My ultimate question is: do I start them with worksheet mode or documents mode? I'm use to worksheet mode and have found the call and response method easy for them to understand. But document mode has many valuable benefits. Is it worth the increase in learning (and frustration) for the benefits if the students use the software only a few times per semester? Or for some, every week?

I would be interested in hearing about the experiences of other educators.

 

Greetings to all. I am writing today to share a personal story / exploration using Maple of an algorithm from the history of combinatorics. The problem here is to count the number of strings over a certain alphabet which consist of some number of letters and avoid a set of patterns (these patterns are strings as opposed to regular expressions.) This counting operation is carried out using rational generating functions that encode the number of admissible strings of length n in the coefficients of their series expansions. The modern approach to this problem uses the Goulden-Jackson method which is discussed, including a landmark Maple implementation from a paper by D. Zeilberger and J. Noonan, at the following link at math.stackexchange.com (Goulden-Jackson has its own website, all the remaining software described in the following discussion is available at the MSE link.) The motivation for this work was a question at the MSE link about the number of strings over a two-letter alphabet that avoid the pattern ABBA.

As far as I know before Goulden-Jackson was invented there was the DFA-Method (Deterministic Finite Automaton also known as FSM, Finite State Machine.) My goal in this contribution was to study and implement this algorithm in order to gain insight about its features and how it influenced its powerful successor. It goes as follows for the case of a single pattern string: compute a DFA whose states represent the longest prefix of the pattern seen at the current position in the string as it is being scanned by the DFA, with the state for the complete pattern doubling as a final absorbing state, since the pattern has been seen. Translate the transitions of the DFA into a system of equations in the generating functions representing strings ending with a given maximal prefix of the pattern, very much like Markov chains. Finally solve the system of equations for the generating functions and thus obtain the sequence of values of strings of length n over the given alphabet that avoid the given pattern.

I have also implemented the DFA method for sets of patterns as opposed to just one pattern. The algorithm is the same except that the DFA does not consist of a chain with backlinks as in the case of a single pattern but a tree of prefixes with backlinks to nodes higher up in the tree. The nodes in the tree represent all prefixes that need to be tracked where obviously a common prefix between two or more patterns is shared i.e. only represented once. The DFA transitions emanating from nodes that are leaves represent absorbing states indicating that one of the patterns has been seen. We run this algorithm once it has been verified that the set of patterns does not contain pairs of patterns where one pattern is contained in another, which causes the longer pattern to be eliminated at the start. (Obviously if the shorter pattern is forbidden the so is the longer.) The number of states of the DFA here is bounded above by the sum of the lengths of the patterns with subpatterns eliminated. The uniqueness property of shared common prefixes holds for subtrees of the main tree i.e. recursively. (The DFA method also copes easily with patterns that have to occur in a certain order.)

I believe the Maple code that I provide here showcases many useful tricks and techniques and can help the reader advance in their Maple studies, which is why I am alerting you to the web link at MSE. I have deliberately aimed to keep it compatible with older versions of Maple as many of these are still in use in various places. The algorithm really showcases the power of Maple in combinatorics computing and exploits many different aspects of the software from the solution of systems of equations in rational generating functions to the implementation of data structures from computer science like trees. Did you know that Maple permits nested procedures as known to those who have met Lisp and Scheme during their studies? The program also illustrates the use of unit testing to detect newly introduced flaws in the code as it evolves in the software life cycle.

Enjoy and may your Maple skills profit from the experience!

Best regards,

Marko Riedel

The software is also available here: dfam-mult.txt

     Example of the equidistant surface at a distance of 0.25 to the surface
x3
-0.1 * (sin (4 * x1) + sin (3 * x2 + x3) + sin (2 * x2)) = 0
Constructed on the basis of universal parameterization of surfaces.

equidistant_surface.mw 


Hi there, fellow primers, it's good to be back after almost 5 years! I just want to share a worksheet on Numerov's algorithm in Maple using procedures as I've recently found out that google could not find any Maple procedure that implements Numerov's algorithm to solve ODEs.   numerov.mw   Reference.pdf 

Below is the worksheet with the whole material presented yesterday in the webinar, “Applying the power of computer algebra to theoretical physics”, broadcasted by the “Institute of Physics” (IOP, England). The material was very well received, rated 4.5 out of 5 (around 30 voters among the more than 300 attendants), and generated a lot of feedback. The webinar was recorded so that it is possible to watch it (for free, of course, click the link above, it will ask you for registration, though, that’s how IOP works).

Anyway, you can reproduce the presentation with the worksheet below (mw file linked at the end, or the corresponding pdf also linked with all the input lines executed). As usual, to reproduce the input/output you need to have installed the latest version of Physics, available in the Maplesoft R&D Physics webpage.

Why computer algebra?

 

 

 

... and why computer algebra?


We can concentrate more on the ideas instead of on the algebraic manipulations

 

We can extend results with ease

 

We can explore the mathematics surrounding a problem

 

We can share results in a reproducible way

 

Representation issues that were preventing the use of computer algebra in Physics

 

 


Notation and related mathematical methods that were missing:


coordinate free representations for vectors and vectorial differential operators,

covariant tensors distinguished from contravariant tensors,

functional differentiation, relativity differential operators and sum rule for tensor contracted (repeated) indices

Bras, Kets, projectors and all related to Dirac's notation in Quantum Mechanics

 

Inert representations of operations, mathematical functions, and related typesetting were missing:

 

inert versus active representations for mathematical operations

ability to move from inert to active representations of computations and viceversa as necessary

hand-like style for entering computations and textbook-like notation for displaying results

 

Key elements of the computational domain of theoretical physics were missing:

 

ability to handle products and derivatives involving commutative, anticommutative and noncommutative variables and functions

ability to perform computations taking into account custom-defined algebra rules of different kinds

(commutator, anticommutator and bracket rules, etc.)

 

 

Examples

 

The Maple computer algebra environment

   

Classical Mechanics

 

Inertia tensor for a triatomic molecule

   

Classical Field Theory

 

*The field equations for the lambda*Phi^4 model

   

*Maxwell equations departing from the 4-dimensional Action for Electrodynamics

   

*The Gross-Pitaevskii field equations for a quantum system of identical particles

   

Quantum mechanics

 

*The quantum operator components of  `#mover(mi("L",mathcolor = "olive"),mo("→",fontstyle = "italic"))` satisfy "[L[j],L[k]][-]=i `ε`[j,k,m] L[m]"

   

Quantization of the energy of a particle in a magnetic field

   

Unitary Operators in Quantum Mechanics

 

*Eigenvalues of an unitary operator and exponential of Hermitian operators

   

Properties of unitary operators

 

 

Consider two set of kets " | a[n] >" and "| b[n] >", each of them constituting a complete orthonormal basis of the same space.


One can always build an unitary operator U that maps one basis to the other, i.e.: "| b[n] >=U | a[n] >"

*Verify that "U=(&sum;) | b[k] >< a[k] |" implies on  "| b[n] >=U | a[n] >"

   

*Show that "U=(&sum;) | b[k] > < a[k] | "is unitary

   

*Show that the matrix elements of U in the "| a[n] >" and  "| b[n] >" basis are equal

   

Show that A and `&Ascr;` = U*A*`#msup(mi("U"),mo("&dagger;"))`have the same spectrum

   

````

Schrödinger equation and unitary transform

 

 

Consider a ket "| psi[t] > " that solves the time-dependant Schrödinger equation:

 

"i `&hbar;` (&PartialD;)/(&PartialD;t) | psi[t] >=H(t) | psi[t] >"

and consider

"| phi[t] > =U(t) | psi[t] >",

 

where U(t) is a unitary operator.

 

Does "| phi[t] >" evolves according a Schrödinger equation

 "i*`&hbar;` (&PartialD;)/(&PartialD;t) | phi[t] >=`&Hscr;`(t) | phi[t] >"

and if yes, which is the expression of `&Hscr;`(t)?

 

Solution

   

Translation operators using Dirac notation

 

In this section, we focus on the operator T[a] = exp((-I*a*P)*(1/`&hbar;`))

Settings

   

The Action (translation) of the operator T[a]"=(e)^(-i (a P)/(`&hbar;`))" on a ket

   

Action of T[a] on an operatorV(X)

   

General Relativity

 

*Exact Solutions to Einstein's Equations  Lambda*g[mu, nu]+G[mu, nu] = 8*Pi*T[mu, nu]

   

*"Physical Review D" 87, 044053 (2013)

 

Given the spacetime metric,

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -exp(lambda(r)), (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -r^2, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -r^2*sin(theta)^2, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = exp(nu(r))}))

a) Compute the Ricci and Weyl scalars

 

b) Compute the trace of

 

"Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

 

where `&equiv;`(Phi, Phi(r)) is some function of the radial coordinate, R[alpha, `~beta`] is the Ricci tensor, `&Dscr;`[alpha] is the covariant derivative operator and T[alpha, `~beta`] is the stress-energy tensor

 

T[alpha, beta] = (Matrix(4, 4, {(1, 1) = 8*exp(lambda(r))*Pi, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 8*r^2*Pi, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 8*r^2*sin(theta)^2*Pi, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 8*exp(nu(r))*Pi*epsilon}))

c) Compute the components of "W[alpha]^(beta)"" &equiv;"the traceless part of  "Z[alpha]^(beta)" of item b)

 

d) Compute an exact solution to the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)" obtained in c)

 

Background: paper from February/2013, "Withholding Potentials, Absence of Ghosts and Relationship between Minimal Dilatonic Gravity and f(R) Theories", by P. Fiziev.

 

a) The Ricci and Weyl scalars

   

b) The trace of "  Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

   

b) The components of "W[alpha]^(beta)"" &equiv;"the traceless part of " Z[alpha]^(beta)"

   

c) An exact solution for the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)"

   

*The Equivalence problem between two metrics

 

 

From the "What is new in Physics in Maple 2016" page:

  

In the Maple PDEtools package, you have the mathematical tools - including a complete symmetry approach - to work with the underlying [Einstein’s] partial differential equations. [By combining that functionality with the one in the Physics and Physics:-Tetrads package] you can also formulate and, depending on the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation.

Example from: A. Karlhede, "A Review of the Geometrical Equivalence of Metrics in General Relativity", General Relativity and Gravitation, Vol. 12, No. 9, 1980

   

*Equivalence for Schwarzschild metric (spherical and Krustal coordinates)

   

Tetrads and Weyl scalars in canonical form

 

 

Generally speaking a canonical form is obtained using transformations that leave invariant the tetrad metric in a tetrad system of references, so that theWeyl scalars are fixed as much as possible (conventionally, either equal to 0 or to 1).

 

Bringing a tetrad in canonical form is a relevant step in the tackling of the equivalence problem between two spacetime metrics.

The implementation is as in "General Relativity, an Einstein century survey", edited by S.W. Hawking (Cambridge) and W. Israel (U. Alberta, Canada), specifically Chapter 7 written by S. Chandrasekhar, page 388:

 

 

`&Psi;__0`

`&Psi;__1`

`&Psi;__2`

`&Psi;__3`

`&Psi;__4`

Residual invariance

Petrov type I

0

"<>0"

"<>0"

1

0

none

Petrov type II

0

0

"<>0"

1

0

none

Petrov type III

0

0

0

1

0

none

Petrov type D

0

0

"<>0"

0

0

`&Psi;__2`  remains invariant under rotations of Class III

Petrov type N

0

0

0

0

1

`&Psi;__4` remains invariant under rotations of Class II

 

 

The transformations (rotations of the tetrad system of references) used are of Class I, II and III as defined in Chandrasekar's chapter - equations (7.79) in page 384, (7.83) and (7.84) in page 385. Transformations of Class I can be performed with the command Physics:-Tetrads:-TransformTetrad using the optional argument nullrotationwithfixedl_, of Class II using nullrotationwithfixedn_ and of Class III by calling TransformTetrad(spatialrotationsm_mb_plan, boostsn_l_plane), so with the two optional arguments simultaneously.

 

The determination of appropriate transformation parameters to be used in these rotations, as well as the sequence of transformations happens all automatically by using the optional argument, canonicalform of TransformTetrad .

 

restart; with(Physics); with(Tetrads)

`Setting lowercaselatin letters to represent tetrad indices `

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

[IsTetrad, NullTetrad, OrthonormalTetrad, PetrovType, SimplifyTetrad, TransformTetrad, e_, eta_, gamma_, l_, lambda_, m_, mb_, n_]

(7.4.1)

Petrov type I

   

Petrov type II

   

Petrov type III

   

Petrov type N

   

Petrov type D

   

 

 

Physics_2016_IOP_webinar.mw     Physics_2016_IOP_webinar.pdf


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

Rotational motion mechanism with quasi stops      
02rep.pdf
DIMA.mw

Formulating and solving the equivalence problem for Schwarzschild metric in a simple case

 

In connection with the digitizing in Maple 2016 of the database of solutions to Einstein's equations of the book Exact Solutions to Einstein Field Equations. I was recently asked about a statement found in the "What is new in Physics in Maple 2016" page:

  

In the Maple PDEtools package, you have the mathematical tools - including a complete symmetry approach - to work with the underlying [Einstein’s] partial differential equations. [By combining that functionality with the one in the Physics and Physics:-Tetrads package] you can also formulate and, depending on the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation.

This question posed is a reasonable one: "could you please provide one example?" This post provides that example.

 

First of all the existing science behind: in my opinion, the main reference regarding the equivalence problem is at the paper "A Review of the Geometrical Equivalence of Metrics in General Relativity", General Relativity and Gravitation, Vol. 12, No. 9, 1980, by A. Karlhede (University of Stockholm). This approach got refined later by others and, generally speaking, it is currently know as the Cartan-Karlhede method, summarized in chapter 9.2 of the book Exact Solutions to Einstein Field Equations. whose solutions were all digitized within the Physics and DifferentialGeometry packages for Maple 2016. This method of Chapter 9.2 (see also Tetrads and Weyl scalars in canonical form, Mapleprimes post), however, is not the only approach to the problem, and sometimes simpler methods can handle the problem faster, or just in simpler forms.

 

The example worked out below is actually the example from Karlhede's paper just mentioned, on pages 704 - 706: "Show that the Schwarzschild metric and its form written in terms of isotropic spherical coordinates are equivalent, and derive the transformation that relates them". Because this problem happens to be simple for nowadays computer algebra, below I also tackle it modified, slightly more difficult variants of it. The approach shown works for more complicated cases as well.

 

Below we tackle Karlhede's paper-problem using: one PDEtools command, the Physics:-TransformCoordinates, the Physics:-Weyl command to compute the Weyl scalars and the Physics:-Tetrads:-PetrovType to see the Petrov type of the metrics involved. The transformation resolving the equivalence is explicitly derived.

 

Start loading the Physics and Tetrads package. To reproduce the computations below, as usual, update your Physics library with the one available for download at the Maplesoft R&D Physics webpage

with(Physics); with(Tetrads); Setup(auto = true, tetradmetric = null, signature = `+---`)

`Setting lowercaselatin letters to represent tetrad indices `

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

0, "%1 is not a command in the %2 package", Tetrads, Physics

 

`* Partial match of  'auto' against keyword 'automaticsimplification'`

 

[automaticsimplification = true, signature = `+ - - -`, tetradmetric = {(1, 2) = 1, (3, 4) = -1}]

(1)

To formulate the problem, set first some symbols to represent the changed metric, changed mass and changed coordinates - no mathematics at this point

gt, mt, tt, rt, thetat, phit := `&gfr;`, `&mfr;`, `&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`

`&gfr;`, `&mfr;`, `&tfr;`, `&rfr;`, vartheta, varphi

(2)

Set now a new coordinates system, call it Y, involving the new coordinates (in the paper they are represented with a tilde on top of the letters)

Coordinates(Y = [tt, rt, thetat, phit])

`Default differentiation variables for d_, D_ and dAlembertian are: `*{Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)}

 

`Systems of spacetime Coordinates are: `*{Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)}

 

{Y}

(3)

According to eq.(7.6) of the paper, the line element of Schwarzschild solution in isotropic spherical coordinates is given by

`#msup(mi("ds"),mn("2"))` := ((1-mt/(2*rt))/(1+mt/(2*rt)))^2*d_(tt)^2-(1+mt/(2*rt))^4*(d_(rt)^2+rt^2*d_(thetat)^2+rt^2*sin(thetat)^2*d_(phit)^2)

(-2*`&rfr;`+`&mfr;`)^2*Physics:-d_(`&tfr;`)^2/(2*`&rfr;`+`&mfr;`)^2-(1/16)*(2*`&rfr;`+`&mfr;`)^4*(Physics:-d_(`&rfr;`)^2+`&rfr;`^2*Physics:-d_(vartheta)^2+`&rfr;`^2*sin(vartheta)^2*Physics:-d_(varphi)^2)/`&rfr;`^4

(4)

Set this to be the metric

Setup(metric = `#msup(mi("ds"),mn("2"))`)

Check it out

g_[]

Physics:-g_[mu, nu] = Matrix(%id = 18446744078306516254)

(5)

In connection with the transformation used further below, compute now the Petrov type and the Weyl scalars for this metric, just to have an idea of what is behind this metric.

PetrovType()

"D"

(6)

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = -64*`&rfr;`^3*`&mfr;`/(2*`&rfr;`+`&mfr;`)^6, psi__3 = 0, psi__4 = 0

(7)

We see that the Weyl scalars are already in canonical form (see post in Mapleprimes about canonical forms): only `&Psi;__2` <> 0 and the important thing: it depends on only one coordinate, `&rfr;` .

 

Now: we want to see if this metric (5) is equivalent to Schwarzschild metric in standard spherical coordinates

g_[sc]

`Systems of spacetime Coordinates are: `*{X = (t, r, theta, phi), Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)}

 

`Default differentiation variables for d_, D_ and dAlembertian are: `*{X = (t, r, theta, phi)}

 

`The Schwarzschild metric in coordinates `[t, r, theta, phi]

 

`Parameters: `[m]

 

Physics:-g_[mu, nu] = Matrix(%id = 18446744078795590102)

(8)

The equivalence we want to resolve is regarding an arbitrary relationship `&mfr;`(m)between the masses used in (5) and (8) and a generic change of variables from X to Y

TR := {phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(Y)}

{phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(Y)}

(9)

Using a differential equation mindset, the formulation of the equivalence between (8) and (5) under the transformation (9) is actually simple: change variables in (8), using (9) and the Physics:-TransformCoordinates command (this is the command that changes variables in tensorial expressions), then equate the result to (5), then try to solve the problem for the unknowns `&mfr;`(m), Phi(Y), R(Y), Theta(Y) and Tau(Y).

 

We note at this point, however, that the Weyl scalars for Schwarzschild metric in this standard form (8) are also in canonical form of Petrov type D and also depend on only one variable, r 

PetrovType()

"D"

(10)

Weyl[scalars]

psi__0 = 0, psi__1 = 0, psi__2 = -m/r^3, psi__3 = 0, psi__4 = 0

(11)

The fact that the Weyl scalars in both cases ((7) and (11)) are in canonical form (only `&Psi;__2` <> 0 ) and in both cases this scalar depends on only one coordinate is already an indicator that the transformation involved changes only one variable in terms of the other one. So one could just search for a transformation of the form r = R(`&rfr;`) and resolve the problem instantly. Still, to make the problem slightly more general, consider instead a generic transformation for r in terms of all of Y = (`&tfr;`, `&rfr;`, `&vartheta;`, `&varphi;`)

tr := r = R(Y)

r = R(Y)

(12)

PDEtools:-declare(r = R(Y))

R(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*R

(13)

Transform the  coordinates in the metric (because of having used PDEtools:-declare, derivatives of the unknowns R are displayed indexed, for compact notation)

TransformCoordinates(tr, g_[mu, nu])

Matrix(%id = 18446744078873927542)

(14)

Proceed equating (14) to (5) to obtain a set of equations that entirely formulates the problem

"convert(rhs(?)=? ,setofequations)"

{0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+`&mfr;`)^2/(2*`&rfr;`+`&mfr;`)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+`&mfr;`)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}

(15)

This problem, shown in Karlhede's paper as the example of the approach he summarized, is solvable using the differential equation commands of PDEtools (in this case casesplit) in one go and no time, obtaining the same solution shown in the paper with equation number (7.10), the problem actually admits two solutions

PDEtools:-casesplit({0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+`&mfr;`)^2/(2*`&rfr;`+`&mfr;`)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+`&mfr;`)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+`&mfr;`)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}, [R, mt])

`casesplit/ans`([R(Y) = -(1/4)*(m-2*`&rfr;`)^2/`&rfr;`, `&mfr;` = -m], []), `casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;` = m], [])

(16)

By all means this does not mean this differential equation approach is better than the general approach mentioned in the paper (also in section 9.2 of the Exact Solutions book). This presentation above only makes the point of the paragraph mentioned at the beginning of this worksheet "... [in Maple 2016] you can also formulate and, depending on the the metrics also resolve, the equivalence problem; that is: to answer whether or not, given two metrics, they can be obtained from each other by a transformation of coordinates, as well as compute the transformation." 

 

In any case this problem above is rather easy for the computer. Consider a slightly more difficult problem, where `&mfr;` <> m. For example:

"subs(mt = 1/(mt^(2)),?)"

Physics:-g_[mu, nu] = Matrix(%id = 18446744078854733566)

(17)

Tackle now the same problem

"convert(rhs(?)=? ,setofequations)"

{0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}

(18)

The solutions to the equivalence between (17) and (5) are then given by

PDEtools:-casesplit({0 = (diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)/(-R(Y)+2*m), 0 = (diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)/(-R(Y)+2*m), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = ((diff(R(Y), `&tfr;`))^2*R(Y)^2-4*(-(1/2)*R(Y)+m)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (diff(R(Y), `&rfr;`))^2*R(Y)/(-R(Y)+2*m), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = -(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = 2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)/(-R(Y)+2*m)}, [R, mt])

`casesplit/ans`([R(Y) = -(1/4)*(m-2*`&rfr;`)^2/`&rfr;`, `&mfr;`^2 = -1/m], [`&mfr;` <> 0]), `casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m], [`&mfr;` <> 0])

(19)

Moreover, despite that the Weyl scalars suggest that a transformation of only one variable is sufficient to solve the problem, one could also consider a more general transformation, of more variables. Provided we exclude theta (because there is cos(theta) around and that would take us to solve differential equations for Theta(theta), that involve things like cos(Theta(theta))), and also to speed up matters let's remove the change in phi, consider an arbitrary change in r and t

TR := select(has, {phi = Phi(Y), r = R(Y), t = Tau(Y), theta = Theta(Y)}, {r, t})

{r = R(Y), t = Tau(Y)}

(20)

PDEtools:-declare({r = R(Y), t = Tau(Y)})

R(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*R

 

Tau(`&tfr;`, `&rfr;`, vartheta, varphi)*`will now be displayed as`*Tau

(21)

So our transformation now involve two arbitrary variables, each one depending on all the four coordinates, and a more complicated function `&mfr;`(m). Change variables (because of having used PDEtools:-declare, derivatives of the unknowns R and Tau are displayed indexed, for compact notation)

TransformCoordinates(TR, g_[mu, nu])

Matrix(%id = 18446744078309268046)

(22)

Construct the set of Partial Differential Equations to be tackled

"convert(rhs(?)=?,setofequations)"

{0 = (-4*(diff(Tau(Y), `&rfr;`))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), vartheta))+(diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))^2+(diff(R(Y), `&tfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))^2+(diff(R(Y), `&rfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = (diff(Tau(Y), vartheta))^2*(R(Y)-2*m)/R(Y)-(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), varphi))^2+2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)^2)/(R(Y)*(-R(Y)+2*m))}

(23)

Solve the problem running a differential elimination (actually without solving any differential equations): there are more than two solutions

sol := PDEtools:-casesplit({0 = (-4*(diff(Tau(Y), `&rfr;`))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), `&rfr;`))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), varphi))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), varphi))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), vartheta))+(diff(R(Y), varphi))*(diff(R(Y), vartheta))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&rfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), 0 = (-4*(diff(Tau(Y), vartheta))*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))+(diff(R(Y), vartheta))*(diff(R(Y), `&tfr;`))*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), (-2*`&rfr;`+1/`&mfr;`^2)^2/(2*`&rfr;`+1/`&mfr;`^2)^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&tfr;`))^2+(diff(R(Y), `&tfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^4 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), `&rfr;`))^2+(diff(R(Y), `&rfr;`))^2*R(Y)^2)/(R(Y)*(-R(Y)+2*m)), -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4/`&rfr;`^2 = (diff(Tau(Y), vartheta))^2*(R(Y)-2*m)/R(Y)-(diff(R(Y), vartheta))^2*R(Y)/(R(Y)-2*m)-R(Y)^2, -(1/16)*(2*`&rfr;`+1/`&mfr;`^2)^4*sin(vartheta)^2/`&rfr;`^2 = (-4*(-(1/2)*R(Y)+m)^2*(diff(Tau(Y), varphi))^2+2*((1/2)*(diff(R(Y), varphi))^2+(cos(vartheta)-1)*R(Y)*(cos(vartheta)+1)*(-(1/2)*R(Y)+m))*R(Y)^2)/(R(Y)*(-R(Y)+2*m))}, [R, mt])

`casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m, diff(Tau(Y), `&tfr;`) = -1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0]), `casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m, diff(Tau(Y), `&tfr;`) = 1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0]), `casesplit/ans`([R(Y) = -(1/4)*(m-2*`&rfr;`)^2/`&rfr;`, `&mfr;`^2 = -1/m, diff(Tau(Y), `&tfr;`) = -1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0]), `casesplit/ans`([R(Y) = -(1/4)*(m-2*`&rfr;`)^2/`&rfr;`, `&mfr;`^2 = -1/m, diff(Tau(Y), `&tfr;`) = 1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0])

(24)

Consider for instance the first one

sol[1]

`casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m, diff(Tau(Y), `&tfr;`) = -1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0])

(25)

Compute the actual solution behind this case :

pdsolve(`casesplit/ans`([R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, `&mfr;`^2 = 1/m, diff(Tau(Y), `&tfr;`) = -1, diff(Tau(Y), `&rfr;`) = 0, diff(Tau(Y), vartheta) = 0, diff(Tau(Y), varphi) = 0], [`&mfr;` <> 0]), {R, Tau, mt})

{`&mfr;` = -1/m^(1/2), R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, Tau(Y) = -`&tfr;`+_C1}, {`&mfr;` = 1/m^(1/2), R(Y) = (1/4)*(2*`&rfr;`+m)^2/`&rfr;`, Tau(Y) = -`&tfr;`+_C1}

(26)

The fact that the time t appears defined in terms of the transformed time Tau(Y) = -`&tfr;`+_C1 involving an arbitrary constant is expected: the time does not enter the metric, it only enters through derivatives of Tau(Y) entering the Jacobian of the transformation used to change variables in tensorial expressions (the metric) in (22).

 

Summary: the approach shown above, based on formulating the problem for the transformation functions of the equivalence and solving for them the differential equations using the commands in PDEtools, after restricting the generality of the transformation functions by looking at the form of the Weyl scalars, works well for other cases too, specially now that, in Maple 2016, the Weyl scalars can be expressed also in canonical form in one go (see previous Mapleprimes post on "Tetrads and Weyl scalars in canonical form").  Also important: in Maple 2016 it is present the functionality necessary to implement the approach of section 9.2 of the Exact solutions book as well.

  

 

 

Download Equivalence_-_Schwarzschild.mw

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

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