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I have an equation of the type

Exp[a[r]] + Exp[a[r]/2] + Exp[-a[r]]

And I want to rule

Exp[a[r]] -> 1-2*m[r]/r

so that it fits the other terms as well. How can I do this directly?

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  • $\begingroup$ Thanks. I made some changes. a[r] is any function of r $\endgroup$
    – Ramos
    Feb 2, 2023 at 2:17
  • $\begingroup$ You haven't resolved the issue. Exp has a predefined meaning. If it is intended to be a user-defined symbol then use some other symbol. Otherwise, you presumably mean either Exp[a[r]] or E^(a[r]) (they are equivalent). $\endgroup$
    – Bob Hanlon
    Feb 2, 2023 at 2:23
  • $\begingroup$ And on the RHS of the rule, do you mean 1 - 2*m[r]/2 as shown (which is just 1 - m[r]) or did you intend (1 - 2*m[r])/2 $\endgroup$
    – Bob Hanlon
    Feb 2, 2023 at 2:36
  • $\begingroup$ Sorry, I was using the python language. I made the changes you mentioned $\endgroup$
    – Ramos
    Feb 2, 2023 at 2:38

1 Answer 1

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Keep the LHS of rules as simple as possible

expr = Exp[a[r]] + Exp[a[r]/2] + Exp[-a[r]];

expr /. a[r] -> Log[1 - 2*m[r]/r] // Simplify

(* 1 + r/(r - 2 m[r]) - (2 m[r])/r + Sqrt[1 - (2 m[r])/r] *)
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