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In field theory, I've seen a free theory described as

  • A field with the specific Lagrangian density ${\cal L}=|\partial\phi|^2+m^2\phi^2$
  • A field whose equation of motion yields a linear set of solutions
  • A field with non-interacting i.e. free normal modes

The first seems too specific, the second seems too general, and the third seems ill-defined. I was hoping that these three could be extended to solve any of those problems or if there is some way to unify, say, the first and the second then maybe that final description would strike right.

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    $\begingroup$ The first one makes no sense as a definition because it is for a spin-0 field and there are free field theories with nonzero spin. $\endgroup$
    – G. Smith
    Oct 26 '20 at 0:11
  • $\begingroup$ but I suppose the first one really means a Lagranian that is quadratic in the fields. With that, I think they're all the same $\endgroup$
    – innisfree
    Oct 26 '20 at 0:13
  • $\begingroup$ You can read what you want into it, but that’s not what it says. However, “a Lagrangian density that is quadratic in the fields” is a reasonable definition of a free field theory, because it leads to linear field equations. $\endgroup$
    – G. Smith
    Oct 26 '20 at 0:15
  • $\begingroup$ Ah okay, but why is that the definition? I always felt like the linear definition was the best motivated so is there a way for quadratic Lagrangian terms to imply a linear solution set? $\endgroup$ Oct 26 '20 at 0:19
  • $\begingroup$ They’re equivalent as far as I know. $\endgroup$
    – G. Smith
    Oct 26 '20 at 0:20
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I believe the problem with all the definitions is that they are not well defined. I can always make a highly non linear field redefinition and make a free quadratic Lagrangian appear interacting. That's why the best way to define a free field theory is to say that it's S matrix must be unity.

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I think these might all be equivalent up to change-of-basis. The definition I have heard is "Lagrangian is bilinear in the fields", which I think is also the same.

In the basis in which the bilinear operator is diagonal, the equations of motion are linear, so (2) and (3) are the same up to change-of-basis.

If the field is scalar and all the terms in your bilinear operator involve 0 or 2 derivatives, you can use integration by parts to put the operator in the form of (1) (up to scaling). I think all Lorentz-invariant self-adjoint bilinear operators on scalar fields have to be of this form. However, there might be non-Lorentz-invariant operators which satisfy (2) and (3) but not (1). I'm not sure.

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  • $\begingroup$ What, why bilinear? I thought bilinear was linear in two arguments but what are the two arguments in this Lagrangian or in a general Lagrangian? $\endgroup$ Oct 26 '20 at 0:32
  • $\begingroup$ I mean we can write $\mathcal{L}(x) = A(x,x)$, where $A(x,y)$ is a bilinear operator. This is the same as "quadratic", but calling it a bilinear operator makes the diagonalization perspective more natural to me. $\endgroup$
    – Daniel
    Oct 26 '20 at 0:40
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By defining the action of a free moving object.

By free I mean it is not confined in a potential but I'm sure someone will argue self action. I mean we need a point mass or something which carries charge which implies a field. Moving charge implies work done. Which implies force which implies accelration which implies radiation which implies self action but don't think about it too much.

At least in string theory we start by defining the action of a free string first.

For me the action came before the lagrangian and from it one can create a lagrangian.

Why? One can have an action without defining a lagrangian but the opposite is not true.

Edit: I guess people disagree.

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    $\begingroup$ The problem is not that I specifically think your answer is wrong, but that it seems to be addressing something pretty different from what the question asked. $\endgroup$
    – Daniel
    Oct 26 '20 at 0:42
  • $\begingroup$ >How is a free theory defined? By deriving an action of some object free of potentials. Last I checked gravity is not linear yet it has a free action has it not? A field with non-interacting i.e. free normal modes is the answer. The denisty definition is pure bullshit for most theories. $\endgroup$
    – user220348
    Oct 26 '20 at 0:47
  • $\begingroup$ Flagged as a very low quality answer. $\endgroup$ Oct 26 '20 at 20:38

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