- #71
homeomorphic
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I just think Minkowski doesn't get enough credit, particularly in the sentence I was referring to. He came up with space-time, completing Einstein's theory.
Since this thread is still open, I'd like to ask sth relevant too!
So let's get back at page 1, shall we? I mean before all the exterior algebra discussion came up.
What is actually the physical meaning of having a field, both solenoidal and conservative? Is there any famous field with that property in classical theory? How does it feel to possesses two potentials (of different nature) after all?
Let us stick to the 3D case...
To my knowledge, the static electric field has never been ascribed a vector potential even in those regions. Neither does a vector potential for the temperature gradient make any sense at all. I am asking about a physically observable vector field with the property of TWO potentials, one scalar and one vectorial.homeomorphic said:Yes, that means the scalar potential is a solution of the Laplace equation. The scalar potential exists because it's conservative and solenoidal then says it's a solution of the Laplace equation. The Laplace equation is pretty interesting, physically. Basically, it comes up when there are no sources or sinks for your vector field. So, for example, the steady state solutions (wait until it approaches an equilibrium) of the heat equation satisfy the Laplace equation, except at places where heat is being pumped in or flowing out of your system. Alternatively, you could think of it as the velocity vector field of an incompressible fluid, where no fluid is being pumped in or out anywhere. Or a static electric field, outside of the places where the charge lies.
To my knowledge, the static electric field has never been ascribed a vector potential even in those regions. Neither does a vector potential for the temperature gradient make any sense at all. I am asking about a physically observable vector field with the property of TWO potentials, one scalar and one vectorial.
Moreover if I unsterstand correctly, your assertion is: a function being harmonic implies that its gradient possesses a vector potential? Can you prove that? I am not sure...
Mea culpa. What I meant to write is: "a function being harmonic implies that its gradient possesses a scalar potential?" And the answer is no, as demonstrated in post #80.homeomorphic said:Yes. If it's harmonic, that just means the gradient is divergence-free. The Laplacian is div grad of the function. In ℝ^3, that means it has a vector potential. However, the mere fact that there is a vector potential doesn't imply that we should care about or use the vector potential, unless there is some reason to care about it, which is the case for the magnetic field.
Post #72 :TrickyDicky said:I think trifis might be confused by the fact that in R^3 there are harmonic functions but no harmonic forms, so if he is thinking in terms of the usual scenario like that of Maxwell equations or classical physics there are no physical examples of laplacian vector fields. That's why this question is well placed in the math subforums.
Trifis said:Once again, not every harmonic field is conservative. Consider another trivial example:
(x+y,y-z,z+x)
This field has obviously a non vanishing curl BUT it is harmonic, since its laplacian is the 0 vector.
Trifis said:@Muphrid hmmm aren't monogenic functions just the generalization of analytic ones in higher dimensions? I might have missed sth, but where exactly do you prove that both of their derivatives must be zero in this decomposition?