I can also do this in Mathematica the following way,
$Assumptions = True; T1 = (t1 - ((u x1)/c^2))/Sqrt[1 - (u^2/c^2)]; T2 = (t2 - ((u x2)/c^2))/Sqrt[1 - (u^2/c^2)]; dT = T2 - T1; FullSimplify[dT /. t2 -> dt + t1] (c^2*dt + u*(x1 - x2))/(c^2*Sqrt[1 - u^2/c^2]) FullSimplify[dT /. t2 -> dt + t1, Element[c, Reals]] (c^2*dt + u*(x1 - x2))/(Sqrt[(c - u)*(c + u)]*Abs[c]) $Assumptions = Element[c, Reals]; FullSimplify[dT /. t2 -> dt + t1] (c^2*dt + u*(x1 - x2))/(Sqrt[(c - u)*(c + u)]*Abs[c]) Still don't know how to do this in Sage :( On Jun 19, 1:32 pm, Jacare Omoplata <[email protected]> wrote: > I found out that in Mathematica this can be done by > PolynomialReduce[dT, dt, {t1, t2}]. Output given below. > > In[26]:= FullSimplify[PolynomialReduce[dT, dt, {t1, t2}]] > > Out[26]= {{1/Sqrt[1 - u^2/c^2]}, (u (x1 - x2))/(c^2 Sqrt[1 - u^2/ > c^2])}, > > But I'd rather use Sage. Does Sage have a counterpart to this > Mathematica function? If not how do get the same result? > > On Jun 19, 11:19 am, Jacare Omoplata <[email protected]> wrote: > > > > > > > > > The following are the expressions, > > > sage: var('x1,t1,x2,t2,u,c',domain=RR);assume(u>0);assume(c>u); > > (x1, t1, x2, t2, u, c) > > sage: T1 = (t1-((u*x1)/(c^2)))/sqrt(1-((u^2)/(c^2))) > > sage: T2 = (t2-((u*x2)/(c^2)))/sqrt(1-((u^2)/(c^2))) > > sage: dT = T2-T1 > > sage: dt = t2-t1 > > > Suppose I know that dT is in this form, > > > dT = a*dt + b, > > > Assuming I DID NOT know that, > > a = 1/sqrt(1-((u^2)/(c^2))) , > > b = (u*(x2 - x1))/((c^2)*sqrt(1-((u^2)/(c^2)))) , > > > is there any way I can find 'a' and 'b' using Sage? > > > What if I didn't know that dT is in the form of a*dt + b, but just > > knew dT in terms of x1,t1,x2,t2,u and c ? > > > Can I still express dT in terms of dt using sage? -- To post to this group, send email to [email protected] To unsubscribe from this group, send email to [email protected] For more options, visit this group at http://groups.google.com/group/sage-support URL: http://www.sagemath.org
