Variation in the action question

[nobraner]

In the Nambu-Goto action, the Lagrange Density is L(dX/dt,dX/dx). When the action is varied, this term becomes a sum of conjugate momentums multiplied by their respective instantaneous velocities, i.e. (Px*dX/dx + Pt*dX/dt). Nowhere, can I find an detailed explanation of how one gets from the first step to the second step. Several references mentioned a Taylor expansion, but otherwise, it seems to be assumed that this step is obvious.

Could anyone direct me to material giving an explicit, step-by-step derivation.

 

[Valerie Park]

The derivation of the equation can be found in many standard texts on classical mechanics, such as Goldstein's Classical Mechanics (2nd ed.). In brief, you start by considering the action S = integral[L(dX/dt,dX/dx)dt], which is a functional depending on the trajectory X(t). You then vary this action with respect to X(t), and obtain a Lagrange equation of the form dP/dt = -∂L/∂X where P = ∂L/∂(dX/dt) is the conjugate momentum to X. This is essentially a Taylor expansion of L around the point X(t). Finally, you substitute the expression for P into the equation and arrive at the desired result.

 

[sir-pinski]

The variation in the action question in the Nambu-Goto action is a common point of confusion for many students. The key to understanding this variation is to first understand the concept of conjugate momentums and how they relate to the Lagrange Density.

In the Nambu-Goto action, the Lagrange Density is given by L(dX/dt,dX/dx). This means that the Lagrange Density is a function of the derivatives of the string coordinates with respect to time and space. In other words, it is a function of the string's velocity and position.

Now, when we vary the action, we are essentially looking for the equations of motion of the string. This means that we are looking for the values of the string's velocity and position that satisfy the Euler-Lagrange equations. In order to do this, we need to express the Lagrange Density in terms of the string's conjugate momentums.

The conjugate momentum of a coordinate is defined as the derivative of the Lagrange Density with respect to the velocity of that coordinate. In the case of the Nambu-Goto action, we have two coordinates: time (t) and space (x). Therefore, we have two conjugate momentums: Pt (conjugate momentum of time) and Px (conjugate momentum of space).

Now, in order to express the Lagrange Density in terms of these conjugate momentums, we use the chain rule. This means that we multiply the derivative of the Lagrange Density with respect to the string's velocity by the string's velocity itself. In other words, we have Pt*dX/dt and Px*dX/dx.

This is where the Taylor expansion comes in. By using the Taylor expansion, we can express the string's position and velocity in terms of its initial position and velocity, and the conjugate momentums. This allows us to write the Lagrange Density as a function of the conjugate momentums.

In summary, the key to understanding the variation in the action question is to understand the concept of conjugate momentums and how they relate to the Lagrange Density. The Taylor expansion is used to express the Lagrange Density in terms of the conjugate momentums, which allows us to find the equations of motion for the string. I hope this explanation helps clarify the steps involved in this variation.