Linearity of a lot of mathematical physics ecuations

>>8322515
Because physicists are sexist and only study straight pole linear equations where feminine curved equations are ignored.

>>8322515
because non-linear equations are hard.
Also cause variational principles

Is it just because they are constructed as simplifications or there is a deeper reason?

>>8322523
care to elaborate a little about variational principles and linearity in Physics? I know the basics of calculus of variations

>>8322515
B/c for the more fundamental equations, physicists tend to assume things are just moving in vacuum with no external things effecting the system.

So the base theories tend to look very nice.

If you start trying to model a more realistic, more complicated, physical system. Then you will probably get a few extra terms making the equation nonlinear.

>>8322546
But that's wrong you fucking faggot. Complex systems don't change the fundamental equations.
[math]\nabla E[/math] is REALLY equal to [math]\frac{\rho}{\epsilon}[/math] no matter how "real life" your system is.

>>8322515
The physical reason is something is always conserved. If that's the case then you're gonna have a linear relation somewhere.

>>8322566
>Complex systems don't change the fundamental equations.
>∇E∇E is REALLY equal to ρϵρϵ no matter how "real life" your system is.


Yeah you get maxwells equations from looking at just the usual (-1/2)Tr(F^2) lagrangian. Now add in terms involving interactions with other fields along with terms for interactions of those fields with themselves too. Are you equations of motion still linear?

>>8322606
Linear in terms of what? Yeah no matter how many terms you add energy conservation isn't going away.

>>8322613
Energy conservation does not mean the equations of motion are linear.

>>8322515

What the fuck are you doing op, you're going to ruin this racket we have going for everyone. Do you have any idea how much more obnoxious physics will become if we have to start dealing with three or more dimensions? Delete this thread.

>>8322515
Because they're easy to solve? Lmao non-linear equations require theory of distributions to handle and they often don't even have existence and uniqueness of solutions.
>>8322530
Dirichlet's principle states that minima of the energy functional [math]E[u] = \int_\Omega \left|\nabla u\right|^2 d^nx[/math] satisfies the Laplace equation [math]\nabla^2 u = 0[/math] on [math]\Omega[/math] with [math]u\mid_{\partial \Omega} = 0[/math]. One of Hilbert's 24 problems asks whether there is a one to one correspondence between linear PDEs and variational problems.

>>8322614
As long as you have conservation of energy (or other quantities), you're gonna be able to write linear equations.
Don't know what your deal is with adding other fields, yeah the standard model is linear too. Adding a bunch of linear terms can't make a system nonlinear.

>>8322624
No. Chern-Simons terms give you non-linear equations of motion for the gauge fields, and they show up in condensed matter systems all the time.
Also energy conservation is related to time-translation symmetry. It has no relation to whether the equations of motion are linear or not.
>undergrads talking out of his ass
Typical.

>>8322624
>As long as you have conservation of energy (or other quantities), you're gonna be able to write linear equations.

The obvious counter example here is gravity.

>>8322633
How is this a counter example?

>>8322623
interesting

>>8322636
Einstein's equations are ridiculously nonlinear.

>>8322623
Wait a sec, doesn't the nonlinear Poisson equation [math]\Delta u= F(u)[/math] admit a variational formulation? The Lagrangian density is [math]\mathcal{L}=\int_\Omega\left [\frac{(\nabla u)^2}{2}-\int_{u_0}^u F(v)dv\right ] d\tau[/math]

File: 1472517041924s.jpg (2KB, 125x78px) Image search: [Google]

2KB, 125x78px

>>8322606
equal to Pepe??? like the frog>????? haha !!111!!! xDDDDDDD lol :0

>>8323423
not density, full fledged Lagrangian, pardon

>>8322613
Linearity of equations has nothing to do with energy conservation. You can have nonlinear equations in which energy is conserved; see Noether's theorem. All you need is time-translation-invariance in its Lagrangian (density)

>>8323423
No. If the integral over the functional [math]F[/math] is a functional integral, then [math]F[/math] has to be well-behaved enough so that its integral converges (at least in the mean). If not (as in if the integral is a line integral), then the limits on your integral [math]u_0[/math] and [math]u[/math] are not defined prior to solving the variational problem, i.e. the EOM determins a curve [math]\gamma: [0,1] \rightarrow \Omega[/math] by [math]t \rightarrow u(t)[/math] such that your functional integral can be written as
[eqn]
\int_{t_0}^{t}F(u(t'))u'(t') dt'.
[/eqn]
This means that you saying that the Lagrangian gives the Poisson equation is circular, since you need the well-posedness of the Poisson problem to define the limits of your Lagrangian to begin with.

In addition, if you want your Lagrangian to have a solution curve defined on all of [math]\Omega[/math], you need the condition that the Lagrangian be regular [math]\operatorname{det}\left(\frac{\partial \mathcal{L}}{\partial x^a \partial x^b}\right) = 0[/math], and this usually restricts [math]F[/math] to be a quadratic form.

>>8322515
Enjoy your Navier-Stokes. Fluid mechanics is a bitch.

>>8322515

One answer is that you can approximate nonlinear operators in some local neighborhood of solutions by a linear operator.

>>8322521
kek, what's your problem

>>8323426
please fuck off

>>8323496
I'm really sorry, I messed up a load of things (it was morning for me).
To be absolutely precise, Poisson's non-linear problem
[math]
\left \{ \begin{split}
&\Delta u = F(u),\; x\in\Omega\\
&u|_{\partial \Omega}=0
\end{split}
\right .
[/math]
is equivalent to finding the (Gateaux) stationary points of the functional
[math]
J:H^1(\Omega)\to\mathbb{R}
[/math]
defined by
[math]
J[u]=\int_\Omega \left [ \frac{|\nabla u|^2}{2}+\int_\bar{u}^u F(\lambda)\mathrm{d}\lambda\right ] \mathrm{d}\tau
[/math]
It is very easy to see that
[math]
\frac{\mathrm{d}}{\mathrm{d}\mu}J[u(\cdot)+\mu h(\cdot)]=0\; \forall h\in H^1_0
[/math]
where [math]H^1_0[/math] is the loop-space of [math]H^1[/math], reproduces Poisson's equation.
There is no problem of circularity since you don't have to know [math]u[/math] beforehands to stationarize [math]J[/math]! All you do is use the fact that [math]h|_{\partial\Omega}=0[/math], divergence theorem and the fundamental theorem of calculus (for the [math]u(x)+\mu h(x)[/math] in the upper limit of the integral).
>and this usually restricts F to be a quadratic form.
A quadratic form is still nonlinear in [math]u[/math].
I'm not sure about the requirements on the functional in this variational formulation. Probably it's (more than) enough for it to be Fréchet-differentiable.

>>8322521
Luce Irigaray please