# Least action principle for a free relativistic particle (Landau)

• provolus
In summary, the conversation discusses doubts in understanding the steps for deriving the formula for the variation of the action for a relativistic free particle, using the invariant element of measure and varying the coordinates. The conversation specifically addresses the second step of the formula and clarifies the calculation of the variation of x_i x^i. The final expression from Landau's book is obtained by canceling out the 2 in the numerator and denominator.
provolus
Reading the Landau's "The classical theory of fields" (chapter 2, section 9 ) I have some doubts in explaining the steps in derivig the formula for the variation of the action for the relativistic free particle http://books.google.it/books?id=QIx...age&q="to set up the expression for"&f=false". Given the invariant element of measure:

$$ds=\sqrt{dx_idx^i}$$

where $$x^i$$ ( $$x_i$$ ) are the four contravariant (covariant) coordinates which parametrize the world line of the free particle, I have to vary respect $$x^i$$, that is I make the variation $$\delta x^i$$. So my doubts are about the second step of the formula before the 9.10, that is why:

$$\delta(ds)=\frac{d x_i \delta d x^i}{ds}$$

is obtained, instead of (IMH and erroneous O):

$$\delta(ds)=\frac{d x_i \delta d x^i}{2 \cdot ds}$$

?

Can someone be so kind to show me the steps?

Last edited by a moderator:
What is $\delta \left(x_{i}x^{i}\right) = ?$

It should be:

$$\delta (x_i x^i) = \delta (c^2t^2-r^2) = 2 (c^2 t \delta x^0 - r \delta x^i$$)

but, sorry, I don't get the point... that is... should I calculate

$$\delta (dx_i dx^i)$$

?

The "d" in the brackets is not important. That 2 you have obtained in front cancels the one in the denominator, thus giving you the final expression from Landau's book.

thx for the moment. I hope to need no more help in covariant variation calculus... ;-)

## 1. What is the least action principle for a free relativistic particle?

The least action principle states that a free relativistic particle will follow the path of least action, where action is defined as the integral of the Lagrangian over the particle's trajectory.

## 2. What is the significance of the least action principle?

The least action principle is significant because it provides a fundamental principle for describing the motion of particles in classical mechanics. It also has applications in quantum mechanics, where it is known as the path integral formulation.

## 3. How does the least action principle differ from Newton's laws of motion?

The least action principle is a variational principle, meaning it takes into account all possible paths a particle could take and finds the one that minimizes the action. In contrast, Newton's laws of motion only consider the forces acting on a particle and how they affect its motion.

## 4. Can the least action principle be applied to systems with multiple particles?

Yes, the least action principle can be extended to systems with multiple particles by considering the action as a sum of the individual particle actions. This allows for the determination of the trajectories of all particles in the system.

## 5. What is the role of relativity in the least action principle?

The least action principle is a relativistic principle, meaning it takes into account the effects of special relativity on the motion of a particle. This is important for accurately describing the motion of particles moving at high speeds.

• Classical Physics
Replies
4
Views
772
• Special and General Relativity
Replies
11
Views
3K
• Calculus and Beyond Homework Help
Replies
15
Views
2K
Replies
2
Views
3K
• Special and General Relativity
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Special and General Relativity
Replies
9
Views
2K
• Special and General Relativity
Replies
6
Views
1K
• Special and General Relativity
Replies
2
Views
1K