Let's consider the simplest case: a **free particle** of mass m moving in one dimension (let's say along the x-axis) with no external forces acting on it.
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## 1. The Lagrangian
The Lagrangian L is defined as the kinetic energy T minus the potential energy V.
- **Kinetic Energy (T)**: For a particle of mass m moving with velocity x˙ (where x˙=dtdx), the kinetic energy is T=21mx˙2.
- **Potential Energy (V)**: Since the particle is free and no forces are acting on it, the potential energy is constant. We can choose this constant to be zero for simplicity, so V=0.
Therefore, the Lagrangian for a free particle is:
L=T−V=21mx˙2−0=21mx˙2
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## 2. Equations of Motion (Euler-Lagrange Equation)
The equation of motion is derived from the Euler-Lagrange equation:
dtd(∂x˙∂L)−∂x∂L=0
Let's calculate the partial derivatives:
- ∂x˙∂L: Differentiating L=21mx˙2 with respect to x˙ (treating x as constant): ∂x˙∂L=mx˙
- dtd(∂x˙∂L): Differentiating mx˙ with respect to time t: dtd(mx˙)=mx¨ (where x¨=dt2d2x is the acceleration).
- ∂x∂L: Differentiating L=21mx˙2 with respect to x (treating x˙ as constant): ∂x∂L=0 (since L does not explicitly depend on x).
Substituting these into the Euler-Lagrange equation:
mx¨−0=0
mx¨=0
Since m=0, we have:
x¨=0
This is the equation of motion. It states that the acceleration of the free particle is zero, which makes intuitive sense (Newton's first law).
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## 3. Solving the Equation of Motion
To find the position x(t) as a function of time, we integrate x¨=0 twice with respect to time:
- First integration:
∫x¨dt=∫0dt
x˙(t)=C1
where C1 is the first constant of integration. This means the velocity is constant.
- Second integration:
∫x˙(t)dt=∫C1dt
x(t)=C1t+C2
where C2 is the second constant of integration. This is the general solution for the position of the particle.
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## 4. Applying Initial Conditions 🚀
To find the specific solution for a particular scenario, we need to apply initial conditions. Let's assume:
- At time t=0, the initial position of the particle is x(0)=x0.
- At time t=0, the initial velocity of the particle is x˙(0)=v0.
Now, let's use these conditions to find C1 and C2:
1. Using x˙(0)=v0:
From x˙(t)=C1, we have:
x˙(0)=C1
So, C1=v0.
2. Using x(0)=x0:
From x(t)=C1t+C2, and knowing C1=v0:
x(t)=v0t+C2
Now substitute t=0 and x(0)=x0:
x(0)=v0(0)+C2
x0=0+C2
So, C2=x0.
Substituting the values of C1 and C2 back into the general solution x(t)=C1t+C2, we get the specific solution:
x(t)=v0t+x0
This equation tells us that for a free particle, its position at any time t is its initial position x0 plus its initial velocity v0 multiplied by time t. This is the familiar equation for motion with constant velocity.
Example Values:
If a particle starts at x0=5 meters with an initial velocity v0=2 m/s, its position at any time t is given by:
x(t)=2t+5 meters.
For instance, at t=3 seconds, its position would be x(3)=2(3)+5=6+5=11 meters.