16
$\begingroup$

In my life I hear/read this statement a lot:

A non-linear equation or theory leads to self-interactions.

For example in GR, we say that gravity can interact with itself because it is non-linear. For some reason I always assumed it was correct. However now I think about it, I can not see a clear reason in the maths for why this statement holds. Can someone help me out? :D

Edit$_1$: As Vadim pointed out. The statement should be the other way around.

A self interacting physical system leads to non-linear equations.

Edit$_2$: The questions is beautifully answered by @gandalf61 for 2 variable system. However, still do not really understand what is going on for 1 variable system, e.g. in general relativity. Could someone maybe also give an example there? Thank you in advance. :D

In the comments on the answer of @gandalf61, you will also find the answer of edit$_2$.

$\endgroup$
5
  • 1
    $\begingroup$ It means that two solutions of a linear equation without sources "do not know" about each other: their sum is also a solution. While for non-linear equations the sum of two solutions is not a solution (generically). $\endgroup$
    – nwolijin
    Dec 15 '20 at 10:48
  • 14
    $\begingroup$ Strictly speaking, it is the other way around: self-interactions lead to non-linear equations. (Self-)interactions are objective reality, which exists regardless of whether the humans wrote an equation to describe it or whether humans exist at all. $\endgroup$ Dec 15 '20 at 10:52
  • 2
    $\begingroup$ @drandran12 If the solution for one thing and another thing together is different from the solution for one thing combined with the solution for another thing, clearly something special happens when they're both there at the same time! $\endgroup$
    – user253751
    Dec 15 '20 at 18:49
  • $\begingroup$ You may enjoy this question about the linearity of the Lorentz transformation in SR: physics.stackexchange.com/q/562263/123208 $\endgroup$
    – PM 2Ring
    Dec 16 '20 at 14:56
  • 1
    $\begingroup$ General tip: Let's not have posts look like revision histories $\endgroup$
    – Qmechanic
    Feb 16 at 11:07
39
$\begingroup$

If I go to a shop and buy $5$ apples and $10$ bananas then I can usually take the price of one apple $a$ and the price of one banana $b$ and add these together to get a total cost of $5a+10b$. And I pay the same total amount if I buy apples and bananas at the same time or I buy apples, then go back to the shop later and buy bananas - my purchases do not interact with one another. This is a linear system.

But if there is an offer of "$5$ apples for the price of $3$" or "one free banana with every $5$ apples" or "$10\%$ off if you spend more than $\$5$" then the cost of $5$ apples and $10$ bananas will no longer be $5a+10b$. This is a non-linear system, and there is an interaction between my different purchases.

$\endgroup$
3
  • 4
    $\begingroup$ In other words, the principle of superposition (which is a fancy way of saying "linearity") means you can decompose the "result" of the "combination" of two scenarios into their respective, noninteracting results: $f(a+b)=f(a)+f(b)$. Any self-interaction manifests as a nonlinearity, which would prevent such decomposition. $\endgroup$
    – HTNW
    Dec 16 '20 at 8:13
  • $\begingroup$ Later when thinking more about this, I was still a little confused. What if we have a non-linear system that has the same variable. Then the analogy is nonsensical. Could you maybe elaborate more on that? $\endgroup$
    – drandran12
    Dec 16 '20 at 10:19
  • 5
    $\begingroup$ @drandran12 An example of a non-linear system with a single variable would be “$10\%$ off if you buy ten or more bananas”. You pay less if you buy ten bananas all at once than if you buy five bananas today and five bananas tomorrow. $\endgroup$
    – gandalf61
    Dec 16 '20 at 10:51
7
$\begingroup$

In the context of a conventional Lagrangian formulation, the main points go as follows:

  • Linear EOMs have a quadratic Lagrangian.

  • $n$ coupled linear EOMs can be diagonalized into $n$ uncoupled linear EOMs in 1 variable each.

  • Non-linear EOMs have a Lagrangian, which has cubic or higher-order terms, aka. interaction terms, which correspond to vertices in Feynman diagrams.

  • If an interaction term only depends on 1 variable, it is called a self-interaction.

  • The perturbative solution to the EOMs can be represented as a directed rooted tree, where branch points/vertices are interactions, cf. e.g. eq. (6) in my Phys.SE answer here.

  • For linear EOMs, the tree has no branch points/vertices/interactions.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.