# A problem about decomposing a Lagrangian

• I
Haorong Wu
Hi, there. Suppose I write down a Langrangian as ##L(t, x, y, z, f)## where ##t##, ##x##, ##y## and ##z## are identified as variables, and ##f=f(t, x, y, z)##. Now the Euler-Lagrange equations$$\sum_{\mu=0} ^3 \frac d {dx^{\mu}}\left (\frac {\partial L }{\partial (\partial_\mu f)} \right ) -\frac {\partial L}{\partial f} =0$$ give the right equations of motion. The corresponding action is then ##S=\int dx^4 L(t, x, y, z, f)##. Let us suppose that ##t## is coupled to other variables in ##L(t, x, y, z, f)##.

Next, we will utilize the 3+1 decomposition on the action to separate the time variable ##t##, yielding $$S=\int dt \int dx^3 L(t, x, y, z, f).$$

My problem is that could we identify ##\int dx^3 L(t, x, y, z, f)## as a new Lagrangian ##L'(t, x, y, z, f)## such that only ##t## is identified as the variable and ##x##, ##y##, ##z##, ##f## are functions on ##t##. If this is valid, then we can write the canonical momentum conjugate to ##f## as ##\pi_f=\frac {\partial L'}{\partial (\partial_t f)}##.

cf. A master equation for gravitational decoherence: probing the textures of spacetime section 2.1 to 2.2.

Many thanks.