Sorry to repost again , i really tried to answer the notification but it had been removed before I could answer, it is really important for me to modify the x and y axis of the usual graphic coordinates so that i can apply to a problem of physcis ( the lorentz gamma factor ) so i will post it again hoping that someone can answer or help me, as it is shown in the drawing I want to prove that by making the x and y coordinates circular a computational value for the gamma factor can be shown graphically.
m1=1.6726*10^-27
m2=9.109*10^-31
s=RandomReal[5.29*10^-11,{200}]
c=RandomReal[300000000,{200}]
n=RandomInteger[100000000000,{200}]
n1=RandomInteger[100000000000,{200}]
n2=RandomInteger[100000000000,{200}]
t=RandomReal[1000000000000,{200}]
G=6.67*10^-11
e=m1*c^2
e1=m2*c^2
s1=s^-1
v2=s1/s
v1=(e-e1)*v2/(m2-m1)*c^2
f=(e1/e)/Sqrt[(e1/e)-((v2/c)*e1/e)^2]*m1-m2
g=(v2*(n-1))/(c*Sqrt[1-v1^2/c^2])
t1=t*c*v2*(n-1)/(1-v1^2)
gama=1/Sqrt[n1^2/Pi^2+(n1^2*v1^2)/(n2^2*c^2)]
gama1=1/Sqrt[n1^2/Pi^2-(n1^2*v1^2)/(n2^2*c^2)]
gama2=gama^-1
gama3= gama+gama1
t2=f/t1
t3=t2*g
ListPlot[v1]
ListPlot[v2]
ListPlot[t1]
ListPolarPlot[f]
ListPolarPlot[g]
ListPolarPlot[t3]
ListLinePlot[gama3]
data=Table[SinIntegral[f*g],{g,0,-Pi,Pi},{f,0,-Pi,Pi}]
ListPlot[data]
ContourPlot3D[gama*gama1==Cos [f] ,{gama,-Pi,Pi},{f,-Pi,Pi},{gama1,-Pi,Pi}]
It follows the sketch of the proposal so that i can get some help in computing
m1=1.672610^-27 10^-3110^-11,{200}] 10^-11c^2 c^2v2/(m2-m1)c^2e1/e)^2]m1-m2 (n-1))/(c Sqrt[1-v1^2/c^2]) c v2*(n-1)/(1-v1^2)v1^2)/(n2^2 c^2)]v1^2)/(n2^2 c^2)]g g],{g,0,-Pi,Pi},{f,0,-Pi,Pi}]
mroge02
July 30, 2019, 4:58pm
3
This is the code I copied, properly formatted:
m1 = 1.6726*10^-27
m2 = 9.109*10^-31
s = RandomReal[5.29*10^-11, {200}]
c = RandomReal[300000000, {200}]
n = RandomInteger[100000000000, {200}]
n1 = RandomInteger[100000000000, {200}]
n2 = RandomInteger[100000000000, {200}]
t = RandomReal[1000000000000, {200}]
G = 6.67*10^-11
e = m1*c^2
e1 = m2*c^2
s1 = s^-1
v2 = s1/s
v1 = (e - e1) v2/(m2 - m1) c^2 f = (e1/e)/
Sqrt[(e1/e) - ((v2/c) e1/e)^2] m1 - m2
g = (v2 (n - 1))/(cSqrt[1 - v1^2/c^2])
t1 = tcv2*(n - 1)/(1 - v1^2)
gama = 1/Sqrt[n1^2/Pi^2 + (n1^2 v1^2)/(n2^2 c^2)]
gama1 = 1/Sqrt[n1^2/Pi^2 - (n1^2 v1^2)/(n2^2 c^2)]
gama2 = gama^-1
gama3 = gama + gama1 t2 = f/t1 t3 = t2*g
ListPlot[v1]
ListPlot[v2]
ListPlot[t1]
ListPolarPlot[f]
ListPolarPlot[g]
ListPolarPlot[t3]
ListLinePlot[gama3]
data = Table[SinIntegral[f*g], {g, 0, -Pi, Pi}, {f, 0, -Pi, Pi}]
ListPlot[data]
ContourPlot3D[
gama*gama1 == Cos[f], {gama, -Pi, Pi}, {f, -Pi, Pi}, {gama1, -Pi, Pi}]
Please check. Also learn how to format code: You can select code blocks and click the code button, which is the leftmost one above the edit window.
I’m sorry, I don’t have an answer to your question.
Thank you very much…i really apreciate maybe someone knows how to change the graph…i posted the code accordingly but it went wrong it is really important to be able to express the x y coordinates in a circular manner as it unimaginates the imaginary value of the Lorentz Gamma factor…proving that the ratio of unimaginating equals the value of the hydrogen Gravity.
If by choosing the coordinates on contourplot3d to be 2Pi i get the picture on the attachment.
