Lagrangian for a free particle

[RohanJ]

TL;DR

Discussion leading to Lagrangian for a free particle being (mv^2)/2

In Landau mechanics it's been given that multiple Lagrangians can be defined for a system which differ by a total derivative of a function.
This statement is further used for the following discussion.

I understand that the term for L has been expanded as a Taylor series but I can't understand the statement given just after that claims that partial derivative of L w.r.t v^2 is independent of the velocity.Can anyone lead me through it mathematically?

 

[strangerep]

We want $L'$ and $L$ to be physically equivalent, i.e., yielding the same equations of motion (EoM). For this to be true, the 2nd term in the Taylor-expanded Lagrangian must be a total derivative. That's only true if the term is at most linear in the velocity (do you understand why this is so?).

Since there's already a linear factor in that term, i.e., the $v \cdot \epsilon$, the factor $\partial L/\partial v^2$ can't have any $v$ terms.

(I'll stop here for now, since I'm not quite sure which bit you're not understanding.)

 

[RohanJ]

We want $L'$ and $L$ to be physically equivalent, i.e., yielding the same equations of motion (EoM). For this to be true, the 2nd term in the Taylor-expanded Lagrangian must be a total derivative. That's only true if the term is at most linear in the velocity (do you understand why this is so?).

Since there's already a linear factor in that term, i.e., the $v \cdot \epsilon$, the factor $\partial L/\partial v^2$ can't have any $v$ terms.

(I'll stop here for now, since I'm not quite sure which bit you're not understanding.)

I don't understand why the second term must be linear in velocity and the part after that.

 

[vanhees71]

It is easy to see that a Lagrangian
$$\tilde{L}(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,t)$$
gives precisely the same equations of motion as $L$. To see this, note that
$$\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,t) = \dot{q}^k \frac{\partial \Lambda}{\partial q^k} + \partial_t \Lambda.$$
From this you find
$$\frac{\partial}{\partial \dot{q}^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t} = \frac{\partial \Lambda}{\partial q^k}$$
and thus
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial}{\partial \dot{q}^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t} \right) = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \Lambda}{\partial q^k} = \dot{q}^j \frac{\partial^2 \Lambda}{\partial q^k \partial q^j} + \frac{\partial \partial_t \Lambda}{\partial q^k}.$$
On the other hand
$$\frac{\partial}{\partial q^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t}=\dot{q}^j \frac{\partial^2 \Lambda}{\partial q^j \partial q^k} + \frac{\partial \partial_t \Lambda}{\partial q^k}.$$
This implies that the Euler-Lagrange equations for
$$\Omega(q,\dot{q},t)=\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,\dot{q},t),$$
i.e.,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \Omega}{\partial \dot{q}^k}=\frac{\partial \Omega}{\partial q^k},$$
are fulfilled identically, and thus $\tilde{L}$ and $L$ lead to the same equations of motion through Hamilton's least-action principle.

The crucial point is that $\Omega$ is a total time derivative of a function, that only depends on $q$ and $t$ but NOT on $\dot{q}$.

 

[RohanJ]

It is easy to see that a Lagrangian
$$\tilde{L}(q,\dot{q},t)=L(q,\dot{q},t)+\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,t)$$
gives precisely the same equations of motion as $L$. To see this, note that
$$\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,t) = \dot{q}^k \frac{\partial \Lambda}{\partial q^k} + \partial_t \Lambda.$$
From this you find
$$\frac{\partial}{\partial \dot{q}^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t} = \frac{\partial \Lambda}{\partial q^k}$$
and thus
$$\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial}{\partial \dot{q}^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t} \right) = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \Lambda}{\partial q^k} = \dot{q}^j \frac{\partial^2 \Lambda}{\partial q^k \partial q^j} + \frac{\partial \partial_t \Lambda}{\partial q^k}.$$
On the other hand
$$\frac{\partial}{\partial q^k} \frac{\mathrm{d} \Lambda}{\mathrm{d} t}=\dot{q}^j \frac{\partial^2 \Lambda}{\partial q^j \partial q^k} + \frac{\partial \partial_t \Lambda}{\partial q^k}.$$
This implies that the Euler-Lagrange equations for
$$\Omega(q,\dot{q},t)=\frac{\mathrm{d}}{\mathrm{d} t} \Lambda(q,\dot{q},t),$$
i.e.,
$$\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial \Omega}{\partial \dot{q}^k}=\frac{\partial \Omega}{\partial q^k},$$
are fulfilled identically, and thus $\tilde{L}$ and $L$ lead to the same equations of motion through Hamilton's least-action principle.

The crucial point is that $\Omega$ is a total time derivative of a function, that only depends on $q$ and $t$ but NOT on $\dot{q}$.

I understand that addition of a total time derivative does not affect the Euler Lagrange equation.
My question is in the discussion in Landau mechanics ,why does the second term of the taylor expansion being the total time derivative leads to the conclusion that $∂L/∂v^2$ is a linear function of velocity?

 

[Gaussian97]

I understand that addition of a total time derivative does not affect the Euler Lagrange equation.
My question is in the discussion in Landau mechanics ,why does the second term of the taylor expansion being the total time derivative leads to the conclusion that $∂L/∂v^2$ is a linear function of velocity?

$$\frac{\partial L}{\partial v^2}$$ is not linear in velocity it's constant, that's why $L\propto v^2$

 

[RohanJ]

$$\frac{\partial L}{\partial v^2}$$ is not linear in velocity it's constant, that's why $L\propto v^2$

Sorry I made a mistake.I meant why the term containing $∂L/∂v^2$ or the "second term" i.e. $2v.e(∂L/∂v^2)$ must be linear in velocity?

 

[Gaussian97]

Well, as @vanhees71 said you can add a total derivative to your Lagrangian $\frac{d}{dt}\Lambda$, but $\Lambda$ cannot be any function, must be function only of $x$ and $t$. So then, using the chain rule, we have $$\frac{d}{dt}\Lambda=\frac{\partial \Lambda}{\partial x}\frac{d x}{d t}+\frac{\partial \Lambda}{\partial t}=v\frac{\partial \Lambda}{\partial x}+\frac{\partial \Lambda}{\partial t}$$.
So it's a linear term with the velocity.