Adding a variable to a Lagrangian

[MyoPhilosopher]

Homework Statement

$$ L = m\vec{\dot r}$$

Relevant Equations

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Would the following: $$ L = m(\vec{\dot r + \vec v})$$ (constant velocity added to above eq.)
lead to equivalent euler-lagrange equations due to the fact that the ratio of T and V is unaffected by an increase in constant velocity?
And would this be an example of energy conservation?

 

[TSny]

$$ L = m\vec{\dot r}$$

$L$ should have dimensions of energy, not momentum.

$L$ can be expressed in terms of $T$ and $V$.

 

[Orodruin]

$L$ should have dimensions of energy, not momentum.

Even more importantly, it should be a scalar and not a vector.

 

[wrobel]

actually the Lagrange equations do not care even if you add
$$\frac{d}{dt}f(t,\boldsymbol r)=\frac{\partial f}{\partial t }+\Big(\frac{\partial f}{\partial \boldsymbol r},\boldsymbol{\dot r}\Big)$$

 

[MyoPhilosopher]

$L$ should have dimensions of energy, not momentum.

$L$ can be expressed in terms of $T$ and $V$.

actually the Lagrange equations do not care even if you add
$$\frac{d}{dt}f(t,\boldsymbol r)=\frac{\partial f}{\partial t }+\Big(\frac{\partial f}{\partial \boldsymbol r},\boldsymbol{\dot r}\Big)$$

What exactly does adding constant velocity do? Would is simply increase the r_dot vectors equally and therefore not affect the equations of motion?

 

[MyoPhilosopher]

$L$ should have dimensions of energy, not momentum.

$L$ can be expressed in terms of $T$ and $V$.

I should have squared my second term, would this be appropriate $$ L = {m(\vec{\dot r + \vec v})}^2$$

 

[Orodruin]

What exactly does adding constant velocity do? Would is simply increase the r_dot vectors equally and therefore not affect the equations of motion?

Adding a constant velocity is effectively doing a Galilean boost, i.e., transforming to a coordinate system that moves at constant velocity relative to your original coordinate system.

The difference between the Lagrangians is $m (\vec v \cdot \dot{\vec r} + v^2/2) = \frac d{dt}(\vec v \cdot \vec r + v^2 t/2)$. Thus, as per #4, the equations of motion are invariant under this transformation (which actually is the transformation $\vec r \to \vec r + \vec v t$).

 

[MyoPhilosopher]

Adding a constant velocity is effectively doing a Galilean boost, i.e., transforming to a coordinate system that moves at constant velocity relative to your original coordinate system.

The difference between the Lagrangians is $m (\vec v \cdot \dot{\vec r} + v^2/2) = \frac d{dt}(\vec v \cdot \vec r + v^2 t/2)$. Thus, as per #4, the equations of motion are invariant under this transformation (which actually is the transformation $\vec r \to \vec r + \vec v t$).

Ahhh thank you that clarified it elegantly!