Why I get the error below with this code:


ρ = 1000;
g = 9.80665;
μ = 0.001;
p[x_, y_, z_, t_] = {x^2, Sqrt[y], -z, t^3};
f[x_, y_, z_, t_] = {-x^2, Sqrt[y]/2, z, t^2};

ρ Dt[v, t] = -Gradient[p] + μ Laplacian[v] + f


Set::write: "Tag Times in 1000\ Dt[v,t] is Protected."
  • 2
    $\begingroup$ Are you trying to solve the Navier Stokes's equation? IF so, you should look at NDSolve... If you are just trying to get rid of the error, switching = with == in the final line helps... check ref/character/Equal in Mathematica Help. $\endgroup$
    – dearN
    Commented Oct 10, 2012 at 19:36
  • 1
    $\begingroup$ See also: How do I identify the source of a “Tag times protected” error? $\endgroup$
    – Mr.Wizard
    Commented Feb 3, 2016 at 6:56

1 Answer 1


Any time you try to execute a command of the form a*b=c, you'll generate this error:

enter image description here

Of course, that's exactly what you've done in your last line:

ρ Dt[v, t] = -Gradient[p] + μ Laplacian[v] + f

We can see the issue more clearly if we examine the left side of the equation in FullForm

(* Out: Times[a,b] *)

Furthermore, Times is protected.

(* Out: {Flat, Listable, NumericFunction, OneIdentity, Orderless, Protected} *)

Thus, you're trying to set the value of a protected symbol. As rm-rf points out, it's not just about removing the ρ, as otherwise you're trying to set the value of Dt.

Perhaps more to the point - if you want to represent an equation that you want to solve, you should type:

ρ Dt[v, t] == -Gradient[p] + μ Laplacian[v] + f

Note the double equals sign. This tests for equality, rather than setting a value - which is what you generally want when solving an equation.

  • $\begingroup$ Could you explain a bit more about the why? OP clearly has no clue why it happened, and if he merely takes this suggestion and removes the ρ to not make it of the form a*b=c, then he'll get a Tag Dt is Protected error... back to square 1 :) $\endgroup$
    – rm -rf
    Commented Oct 10, 2012 at 19:49

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