# Stationary Solutions

An unbelievably basic question, but it's something I've never been taught. Am I right in thinking that the following defines a stationary solution?

Let $\phi$ be some dynamical variable satisfying a differential equation $D(\phi)=0$ with respect to some parameter $t$. Then $\phi$ is called a stationary solution or a stationary state if all $t$ derivatives of $\phi$ are zero.

Could someone point me to a resource where this is defined? A quick google threw up nothing useful, but perhaps I was looking in the wrong place!

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Here's a quote from section 10.2 of "A first course in general relativity" by Schutz:

We define a static spacetime to be one in which we can find a time coordinate $t$ with two properties: (i) all metric components are independent of $t$, and (ii) the geometry is unchanged by time reversal, $t \rightarrow -t$.

...

(A spacetime with property (i) but not necessarily (ii) is said to be stationary.)

Then ϕ is called a stationary solution or a stationary state if all t derivatives of ϕ are zero.

certainly seems consistent with Schutz's description of stationary.

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Schutz doesn't think of Kerr as stationary, that's all. The two definitions are not compatible because of this--- if you have a dt d\theta metric term, this is non-stationary in Schutz's definition, but this is a quibble over the meaning of words. –  Ron Maimon Sep 6 '12 at 1:47
@RonMaimon, but Shutz does think of Kerr as stationary. I have the book right in front of me. Schutz writes, in the section "Kerr black hole": "The coordinates are called Boyer-Lindquist coordinates: $\phi$ is the angle around the axis of symmetry, $t$ is the time coordinate in which everything is stationary.." –  Alfred Centauri Sep 6 '12 at 1:58
Then he isn't consistent. The definition above requires t->-t symmetry, so no cross terms in the metric, and excludes Kerr by definition. The meaning of stationary can be "time independent metric" or "time independent metric with time reversal symmetry", depending on your whim. –  Ron Maimon Sep 6 '12 at 2:16
@RonMaimon, it's the static solution that requires the t -> -t symmetry, not the stationary solution. Isn't that clear from the quote: "(A spacetime with property (i) but not necessarily (ii) is said to be stationary.)"? –  Alfred Centauri Sep 6 '12 at 2:24
My bad, didn't read it closely enough. Ignore my comments. –  Ron Maimon Sep 6 '12 at 5:36

Independent of the interpretation of a differential equation, we have the following:

A stationary solution of an autonomous differential equation $F(y(t),\dot y(t))=0$ (not depending explicitly on time) is a solution that doesn't depend on time.

Thus the stationary solutions are precisely the solutions of the form $y(t)=y_0$, where $y_0$ solves the nonlinear equation $F(y_0,0)=0$. One doens't need to look at higher derivatives than those in the differential equation, although all derivatives are of course zero for a constant solution).

For a second order differential equation $F(y(t),\dot y(t),\ddot y(t))=0$ the corresponding condition on $y_0$ is $F(y_0,0,0)=0$.

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I think Wikipedia has a fairly good definition for this:

A stationary state is called stationary because a particle remains in the same state as time elapses, in every observable way.

What this means is that every observable quantity which can be computed from the state is constant in time.

In classical mechanics, this is kind of trivial because the state is just given by the positions and velocities (or momenta) of all particles in the system. In other words, the state is itself an observable quantity, and therefore, as you said, a stationary state is necessarily constant. But it's not particularly common to use the term "stationary state" in classical mechanics.

Where it really becomes useful is in quantum mechanics. Here, the state is more than just position and momentum. A quantum state contains some non-observable information, and that information can change over time even if all the observable quantities are constant. So in quantum mechanics, "stationary state" does not mean that all $t$ derivatives of the state are zero. The correct mathematical criterion is that these states are eigenstates of the Hamiltonian, $H\lvert\psi\rangle \propto \lvert\psi\rangle$.

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Thanks David. But is my definition right? I think it is pretty usual to talk about stationary solutions for a generalised coordinate $q$ in classical mechanics. I'm pretty sure I am right, but confirmation would be marvellous! –  Edward Hughes Sep 5 '12 at 21:36
If the state is an observable quantity, then I suppose yes, your definition works. But that's only the case in classical mechanics. And I've practically never heard the term used in a classical context. (I looked in a few classical mechanics books and there was no reference to it.) –  David Z Sep 5 '12 at 21:53
Perhaps it's non-standard terminology that my lecturers have used! Thanks for clarifying that though. –  Edward Hughes Sep 6 '12 at 13:10