How to prove Momentum = Gradient of Action?

• Alexandre
In summary: Now, you can calculate its partial derivative with respect to x \frac{\partial S}{\partial x} = m \frac{x - x_{0}}{t - t_{0}} = m v = p. Notice, you also get the Hamiltonian from \frac{\partial S}{\partial t} = - \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} = - H. Now you can back check as follow S = \int d S = \int \left( \frac{\
Alexandre
How can show that momentum is the gradient of the action for the free particle? I tried it like this for one dimensional case:

$$s=\int Ldt$$
$$ds=Ldt$$
$$ds=\frac{mv^2}{2}dt\:$$
Velocity is constant right? So I should be able to to this:
$$\frac{ds}{dx}=\frac{mv^2}{2}\frac{dt}{dx}$$
I'm not sure about the following step where I use: $$\frac{dt}{dx}=\frac{1}{v}$$ is this correct?
$$\frac{ds}{dx}=\frac{mv^2}{2}\frac{1}{v}$$
And finally I get:
$$\frac{ds}{dx}=\frac{mv}{2}$$
It differs from momentum by the factor of 1/2, where did I make the mistake?

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The Lagrangian is to be written as a function of t, q, and qdot (, or t, x, and v in your 1D example). The weird thing about it is that x and v are supposed to be treated as formally distinct independent variables, and v cannot be replaced by dx/dt.

I'm actually not sure what "gradient of the action" means, since action is a functional, depending on the path, rather than depending on the position like a normal function.

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Right. You need to differentiate the potential energy part of the Lagrangian

thierrykauf said:
Right. You need to differentiate the potential energy part of the Lagrangian
Suppose I have no potential, just the kinetic energy of a free particle wandering around (actually moving at a straight line with a constant velocity), the Lagrangian will be equal to kinetic energy only. I've found out a hint why my derivation might be wrong, there's a thing called abbreviated action and it differs from ordinary action which is just an integral of Lagrangian (if it's in units of energy) with time. But I'm clueless what does that mean and how or why it differs from action.

Khashishi said:
I'm actually not sure what "gradient of the action" means, since action is a functional, depending on the path, rather than depending on the position like a normal function.
http://en.wikipedia.org/wiki/Action_(physics)#Abbreviated_action_.28functional.29

$$S_0=\int pdq$$
$$dS_0=pdq$$
$$\frac{dS_0}{dq}=p$$
For more general case I see it's a gradient but what the hell is abbreviated action? I came across this thing in quantum mechanics textbook (in introductory chapter, where classical mechanics was discussed)

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Alexandre said:
How can show that momentum is the gradient of the action for the free particle? I tried it like this for one dimensional case:

$$s=\int Ldt$$
$$ds=Ldt$$
$$ds=\frac{mv^2}{2}dt\:$$
Velocity is constant right? So I should be able to to this:
$$\frac{ds}{dx}=\frac{mv^2}{2}\frac{dt}{dx}$$
I'm not sure about the following step where I use: $$\frac{dt}{dx}=\frac{1}{v}$$ is this correct?
$$\frac{ds}{dx}=\frac{mv^2}{2}\frac{1}{v}$$
And finally I get:
$$\frac{ds}{dx}=\frac{mv}{2}$$
It differs from momentum by the factor of 1/2, where did I make the mistake?
Free particle moves with constant velocity $$v = \frac{x - x_{0}}{t - t_{0}} = \mbox{const.}$$ So the action is $$S = \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} \int^{t}_{t_{0}} d t = \frac{m}{2} \frac{(x - x_{0})^{2}}{t - t_{0}} .$$ Now, you can calculate its partial derivative with respect to $x$ $$\frac{\partial S}{\partial x} = m \frac{x - x_{0}}{t - t_{0}} = m v = p.$$ Notice, you also get the Hamiltonian from $$\frac{\partial S}{\partial t} = - \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} = - H.$$ Now you can back check as follow $$S = \int d S = \int \left( \frac{\partial S}{\partial x} d x + \frac{\partial S}{\partial t} d t \right) = \int \left( p \dot{x} - H \right) \ d t = \int L d t .$$ Of course, to be accurate you must think of the action as function of the end points $S = S ( x , t ; x_{0} , t_{0} )$. For the general case, you can use the variation principle to show that $$\delta S = \int_{0}^{t} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \delta q \right) = p \delta q |_{0}^{t} .$$ Assuming $\delta q (0) = 0$ we find $$\delta S = p(t) \delta q , \ \ \Rightarrow \ p(t) = \frac{\partial S}{\partial q} .$$

Sam

Alexandre and vanhees71
samalkhaiat said:
Sam

Oh, thanks, now I understand. I didn't realize that velocity doesn't depend on time and so Lagrangian goes out of the integral here
$$S = \frac{m}{2} ( \frac{x - x_{0}}{t - t_{0}} )^{2} \int^{t}_{t_{0}} d t$$

1. How is momentum defined in physics?

Momentum is defined as the product of an object's mass and its velocity. It is a measure of an object's motion.

2. What is the gradient of action?

The gradient of action is a mathematical concept in classical mechanics that represents the rate of change of a system's action with respect to its coordinates. It is used to describe the motion of a particle in a potential field.

3. How is momentum related to the gradient of action?

In classical mechanics, it can be shown that the momentum of a particle is equal to the gradient of its action with respect to its coordinates. This relationship is known as the Hamilton-Jacobi equation.

4. What is the significance of proving the momentum = gradient of action?

Proving this relationship is important because it allows us to understand the behavior of particles in potential fields and make predictions about their motion. It also provides a deeper understanding of the fundamental principles of classical mechanics.

5. What are the steps to prove momentum = gradient of action?

To prove this relationship, we first start with the Hamilton-Jacobi equation and use the definition of momentum to substitute for its values. Then, we apply the Euler-Lagrange equations to the action and solve for the momentum. Finally, we compare the resulting equation to the original Hamilton-Jacobi equation to show that they are equivalent.

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