# Why is lagrangian density correct?

The textbooks I have available explain that due to the infinite degrees of freedom of a field, the relevant object in QFT is the lagrangian density. A lagrangian is then obtained for the field by integrating over space.

I find the justification for this procedure unclear. In classical mechanics, the lagrangians of two particles may be added only if the particules do not interract. Does it mean that the lagrangian density concept is only valid for a free field? What happens in the case of interracting particles?

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Even classically, you can add interaction terms. For instance, if you're doing two coupled simple harmonic oscillators (with one fixed to a wall), you can have the Lagrangian $L = \frac{1}{2}m_{1}\dot x_{1}^{2} + \frac{1}{2}m_{2}\dot x_{2}^{2} - \frac{1}{2}k_{0} \left(x_{1}-x_{2}\right)^{2}+\frac{1}{2}k_{1}x_{1}^{2}$, which has a clear interaction term. –  Jerry Schirmer Jan 26 '11 at 4:09
I understand that it is possible to add interaction terms a posteriori, but can it be done a priori? To take an example, suppose the free lagrangian of particle A is La and that of particle B is Lb. The lagrangian of the system will be La+Lb if the particles do not interact. But what if they do? How is this reflected in the initial 'lagrangian density' ? –  Whelp Jan 26 '11 at 4:34
It is reflected in exactly the way that Jerry gave you an example of. –  David Z Jan 26 '11 at 4:45
If they do, then you have interaction terms in the Lagrangian. The a posterori, flimsy concept is not the interaction, it's the notion of a 'free particle'. Literally, you just lop off the part of the Lagrangian that is hard to solve classically, solve the rest, promote the solutions to operators, and call that a 'free particle'. And then you treat the rest as 'small' corrections. But the real theory was always defined by the full Lagrangian. Defining a 'free particle' is just an approximation technique. –  Jerry Schirmer Jan 26 '11 at 5:03

You are confusing the concepts of "interactions" and "nonlocality". In realistic field theories, including all theories we ever used to study phenomena in the world around us, the interactions exist but they keep the physics local.

As David mentioned, the Lagrangian density takes the form $${\mathcal L} = \sum_i \left[ (\partial_\mu \phi_i)^2 + m^2 \phi_i^2 \right] + O(\phi^{3+n})$$ The sum over $i$ of the terms bilinear in $\phi$ or its first derivatives produces the free particles. But the higher-order terms - that I only wrote under the $O$ symbol - which are cubic, quartic, or of even higher orders - are responsible for all the interactions.

In particular, the electromagnetic interaction between two charged objects boils down to their common interaction with the electromagnetic field via the interaction $${\mathcal L_{em}} = j_\mu A^\mu$$ where $j_\mu$ is the 4-vector including the charge density and the flux. Because of this local term at one point, the electromagnetic field is perturbed by the first charge. The electromagnetic field $A^\mu$ continues to propagate to another charge particle, much like if it were a free field, and then the local interaction term of the type above "clicks" again and makes the second particle accelerate according to the first particle's position and charge.

This description is particularly optimized for quantum field theory where the photon is called the virtual particle - or a messenger of the interaction. However, even in classical field theory, one may use a similar language. Even the Lagrangian of a classical field theory that describes the electromagnetic interactions involving charge matter has a local form.

As far as I understand, what you were proposing was that there would be bilocal or otherwise non-local terms in the Lagrangian $$L = \int d^{d-1} x\,{\mathcal L}_{local} + \int d^{d-1} x\, d^{d-1} y\, F(x) G(y)$$ which would directly attract or repel some densities $F,G$ that exist at any pair of points $x,y$, right? This never works in field theory at the fundamental level. The Lagrangian above is not local - it is not an integral of a density - so fields at point $x$ would be immediately influenced by fields at any other point $y$. This would violate locality - an action at a distance - and it would contradict relativity because when combined with the principle of relativity, a violation of locality also means a violation of causality (the rule that causes must precede their effects).

However, the bilocal Lagrangian above may be "approximately derived" by "integrating out the electromagnetic field". I can't explain exactly what it means, especially because this term is only standard in quantum mechanics (in classical physics, it corresponds to "solving $A_\mu$ away from the equations"). However, let me just say the conclusion. The bilocal interaction terms between two charges you are thinking about may be "approximately" derived from the fully local Lagrangian I began with.

Because special relativity is a well-established fact about the reality and we have never observed any "action at a distance" - or interactions between two separated bodies occur because of a "messenger" that has to move along a path connecting the two objects - all Lagrangians of field theories we ever study are written as integrals of a Lagrangian density. This feature is called "locality".

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In classical mechanics, the lagrangians of two particles may be added only if the particules do not interract.

I wouldn't say that. You can always write a Lagrangian $L$ for a system of two particles. In general, it takes the form

$$L = L_1 + L_2 + L_i$$

where $L_i$ is an interaction term that depends on the coordinates and/or velocities of both particles. If and only if the particles don't interact, $L_i = 0$, and only in that case can you write the Lagrangian as the sum of individual particle Lagrangians $L_1$ and $L_2$.

A similar idea applies in quantum field theory. Remember that QFT Lagrangian densities take forms like

$$\mathcal{L}(\phi, \partial\phi) \sim (\partial\phi)^2 - m^2\phi^2 - \sum_n g_n\phi^n$$

Of course there are many different kinds, but in general there is always a kinetic term which involves the derivatives of the fields, and other terms which represent either the mass of the field or interactions between the field and itself or other fields.

Now, in a sense, a derivative is a way of coupling the values of some object at different spacetime points. So it should make sense that the kinetic term of the actual Lagrangian

$$L_\text{kin} \sim \int\mathrm{d}^3\mathbf{x}\ (\partial\phi)^2$$

couples the values of the field $\phi$ at different points in spacetime. This is analogous to the term $L_i$ in the classical Lagrangian which involves the coordinates of multiple particles, except here, coordinates are replaced by fields and particles are replaced by locations. So you have a term that couples the fields at different spacetime points.

Notice, though, that in the rest of the Lagrangian, there are no derivatives. This means that outside of the kinetic term, there is no connection between what happens at different points in spacetime. Specifically, the interaction terms

$$\int\mathrm{d}^3\mathbf{x}\ \sum_n g_n\phi^n$$

are local, which means that all field interactions occur at a single spacetime point. This is a simple way to ensure that interactions don't proceed differently when viewed from different reference frames. So there's no problem with integrating the interaction terms over all of space.

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