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Im currently during my 2nd year of my Physics degree and I've learned about the Lagrangian and Noether's Theorem. In my HW, I've encountered the following question:

"We will look at a ball with mass $m$, free-falling with constant gravitational acceleration $g$. Write the Lagrangian of the ball in Cartesian coordinates. Is there spatial translation symmetry along the axis parallel to the gravitational force? If not, explain. If yes, write the conserved quantity" (I hope my translation is ok lol).

In my attempt to solve it, I've written the following lagrangian:

$$L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2+\dot{z}^2)-mgz$$ (assume the gravitational force is along the $z$-axis)

Let's apply the following transformation: $$z \to z' = z+ε\quad(ε\ll 1)$$ so that $ż'=ż$. $ε$ is a parameter so we will treat him like a constant.

Substitute in $L$: $$L'=\frac{1}{2}m(ẋ^2+ẏ^2+ż^2)-mg(z+ε)=L-mgε$$

Define $A=-mgεt$ so that $$L'=L+\frac{dA}{dt}$$ And by omitting complete derivatives of time we get: $$L'=L$$

So there is symmetry and by Noether's therom the canonical momentum along the $z$ axis, which in this case is equal to the newtonian momentum $p=m\dot{z}$, is conserved, which completely contradicts Newtonian Mechanics.
What have I done wrong? Please help

Thomas Fritsch
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Ilay
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1 Answers1

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You can not get rid of mg\epsilon

Noether's Theorm says if L is invariant under a continuous transformation, then there exists a corresponding conserved quantity. "Invariant" means the number of L is same after transformation, so you can not get rid of any part of it. L'=L-mg\epsilon is not equal to L, because L here is just number and it actually changes under translation operation in the z-direction.

hhh
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