Why in field theory Lagrangian is an integral of space-time

AI Thread Summary
In field theory, the Lagrangian is an integral over space-time to ensure Lorentz invariance, reflecting the unified nature of space and time in relativity. This integration is necessary for deriving correct equations of motion, as the dynamical variables in field theory depend on four coordinates rather than just time. The distinction between particle Lagrangian theory and field theory lies in the number of parameters they depend on, with field theory involving four parameters. The confusion between the action and the Lagrangian is clarified, as the action is the integral of the Lagrangian over time, while the Lagrangian in field theory is derived from integrating the Lagrangian density over space. Understanding this framework is essential for grasping the principles of quantum field theory.
kof9595995
Messages
676
Reaction score
2
I remember when I learned some basic continuum mechanics, Lagrangian is just a integral of lagrangian density over space, which is quite easy to accept because it's just a continuous version of L=T-U. Now I'm trying to start a bit QFT and notice that Lagrangian is an integral over space-time, and I just can't quite accept the integral over time part.
I suppose in classical relativistic field theory one must include the integral over time to get the correct equation of motion, but I can't find any introductory article about this on the net, so could you guys provide some guidance? I just want to convince myself that we need the integral over time.
 
Physics news on Phys.org
Here is simple intuitive response. The whole point of relativity theory is that we deal with space and time as one entity i.e. space-time. The transformations act on these variables together. The operator doesn't know about "time" or "space" it only knows about "space-time".
We want the Lagrangian to transform as a scalar under Lorentz transformations (otherwise there would be no point), hence we need a Lorentz invariant measure which is we integrate over space-time. (In simple minkowski spacetime the invariant measure is just the integration over spacetime, but when the metric is more complicated then we need some extra stuff to get an invariant measure.)
 
It's a matter of what are the dynamical variables of the system, and on what parameters they depend. In particle theory, the dynamical variables are the positions and velocities of the particles, that depend on just one parameter, time. In field theory, the dynamical variables are...the fields, and they depend on four coordinates. Space is no longer a dynamical variable, but a parameter.

In mathematical terms, particle lagrangian theory is just a special case of field theory. In general, you have a lagrangian that depends on a certain number of fields (and their first partial derivatives), which in turn depend on a certain number (n) of parameters. Relativistic field theory has n=4, lagrangian particle theory (relativistic or not) has n=1. Another example is a lagrangian formulation of a drum (n=2).
 
Thank you both for the reply, I think I simply confused the action S and Lagrangian. S=int L dt
and in field theory L=int (lagrangian density) d^3x, so S is an integral of space time, not L.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top