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Make the x axis circular?

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

Attachment

4 Replies

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.

POSTED BY: Michael Rogers

If by choosing the coordinates on contourplot3d to be 2Pi i get the picture on the attachment.

Attachment

Attachments:
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}]
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