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My goal is to reduce two equations eq1 and eq2. According to the documentation, output of code1 condition for $V_{\text{eff}}(r)$ involves $m$. However, code2 doesn't involve $l$. It looks like substitution is already occurred even though I didn't specify the Backsubstitution->True option. Are there somthing that I have made mistake in my code1 and code2?

Reduce[expr,{x1,x2,…},…] effectively writes expr as a combination of conditions on x1, x2, … where each condition involves only the earlier .

eq1 = Refine[
  Reduce[D[-((G m M)/r) + l^2/(2 r^2 \[Mu]), r] == 0, 
   r], {G m M \[Mu] != 0, l != 0, M != 0}]
eq2 = -((G m M)/r) + l^2/(2 r^2 \[Mu]) == Subscript[V, eff][r]

(*** Code1 ***)
Reduce[{eq1, eq2}, {m, Subscript[V, eff][r]}](* As Doccument, conditon for Subscript[V, eff][r] involves m *)

(*** Code2 ***)
Reduce[{eq1, eq2}, {l, Subscript[V, eff][r]}](* Conditon for Subscript[V, eff][r] not involves l. Already backsubstitued? *)
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Perhaps Solve used with two arguments (not documented as far as I know) is what you are lokking for:

eq1 = Refine[
  Reduce[D[-((G m M)/r) + l^2/(2 r^2 \[Mu]), r] == 0, 
   r], {G m M \[Mu] != 0, l != 0, M != 0}]
eq2 = -((G m M)/r) + l^2/(2 r^2 \[Mu]) == Veff[r]  

(***Code1***)
Solve[{eq1, eq2}, {m}, { 
  Veff[r]}   ](*{{m -> l^2/(G M r \[Mu])}}*)

(***Code2***)
Solve[{eq1, eq2}, { l}, { 
  Veff[r]}   ](*{{l -> -Sqrt[G] Sqrt[m] Sqrt[M] Sqrt[r] Sqrt[\[Mu]]}, {l ->Sqrt[G] Sqrt[m] Sqrt[M] Sqrt[r] Sqrt[\[Mu]]}}*)

This approach works with Reduce instead of Solve too!

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  • $\begingroup$ Thank you for letting me know the hidden usage of Reduce[expr,vars,dom]; Reduce[{eq1, eq2}, { Veff[r]}, { l} ] showed similar outputs as Reduce[{eq1, eq2}, { l, Veff[r]}]. I guess that cancling variable could be located at dom. ※I think this usage suit for me; using Reduce[expr, vars] with only one variable that I want in vars, then observing the output, and then substituting varables;condition=Reduce[{eq1, eq2}, Veff[r] ] Flatten[ToRules/@Reduce[condition[[2]], l][[1]]/. Or->List] condition[[4]]/. % $\endgroup$ Commented Aug 18 at 6:46
  • $\begingroup$ @Soon You're welcome! $\endgroup$ Commented Aug 18 at 9:08

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