A problem about decomposing a Lagrangian

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In summary, the conversation discusses the use of the Euler-Lagrange equations and the 3+1 decomposition to describe the dynamics of a system. It is possible to identify the Lagrangian density as a new Lagrangian, with only the time variable considered as a variable. The canonical momentum conjugate to a field can also be defined as the canonical momentum density at a given point in space and time.
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Haorong Wu
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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.
 
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Hello,

Thank you for your interesting question. It seems like you have a good grasp on the Euler-Lagrange equations and the 3+1 decomposition of the action. To answer your question, yes, it is possible to identify ##\int dx^3 L(t, x, y, z, f)## as a new Lagrangian ##L'(t, x, y, z, f)##. This is known as the Lagrangian density, and it is often used in field theories to describe the dynamics of a system.

In this case, the Lagrangian density is a function of all the variables ##t##, ##x##, ##y##, ##z##, and ##f##, but only the time variable ##t## is considered as a variable in the Lagrangian density. The other variables are considered as functions of ##t##. So, in a sense, the Lagrangian density encapsulates all the information about the system's dynamics at a given time ##t##.

As for your question about the canonical momentum conjugate to ##f##, yes, it can be defined as ##\pi_f = \frac{\partial L'}{\partial (\partial_t f)}##. This is known as the canonical momentum density, and it is the momentum conjugate to the field ##f## at a given point in space and time.

I hope this clarifies your doubts. Feel free to ask any further questions if needed. Good luck with your research!
 

Related to A problem about decomposing a Lagrangian

1. What is a Lagrangian?

A Lagrangian is a mathematical function that describes the dynamics of a physical system in terms of its position and velocity.

2. Why is decomposing a Lagrangian important?

Decomposing a Lagrangian allows us to break down a complex system into smaller, more manageable parts, making it easier to analyze and understand.

3. What are the steps involved in decomposing a Lagrangian?

The first step is to identify the degrees of freedom in the system. Then, we use the Euler-Lagrange equations to derive the equations of motion for each degree of freedom. Finally, we combine these equations to get the overall dynamics of the system.

4. Can a Lagrangian be decomposed for any physical system?

Yes, a Lagrangian can be decomposed for any physical system, regardless of its complexity. However, the process may become more challenging for highly complex systems.

5. How is decomposing a Lagrangian relevant in scientific research?

Decomposing a Lagrangian is relevant in many fields of science, including physics, engineering, and mathematics. It allows us to understand the behavior of complex systems and make predictions about their future behavior.

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