# Lagrangian symmetry for the bullet and gun

• I
• gionole
In summary, you are asking about a situation where there is no interaction, and Lagrangian mechanics does not change.
gionole
TL;DR Summary
Emma noether’s conservation law
I have been learning emma noether’s theorem where every symmetry results in conservstion law of something, sometimes momentum, sometimes angular momentum, sometimes energy or sometimes some kind of quantity.

Every explanation that I have listened to or learned from talks about homogeneity of the space and says that if we have a system of particles and we shift each one of them by some infinetisemal distance(we dont have to shift all of them by same amount, but could be we shift first particle by 2epsilon, where second particle by 3epsilon). This all depends what their lagrangian is and whether this shift gives the possibility that dL=0.

While I understood all that, I have a hard time imagining this in real world usecase. In real life, we dont care about shifting them, they just move without our interaction.

I want to solve the following: imagine a gun which pulls a bullet and bullet hits a box. As we know from school physics, impulse is conserved - total momentum before pulling was 0 and the final momentum of the gun and bullet must be 0 as well. So we write p1+p2=0. I want to see/prove that lagrangian does not change for my example because if we have momentum conservation, there must have been symmetry such as dL=0. I dont even know how to start. Before pulling, L seems 0 as no kinetic and potential energy exists in the system but do I still have to write them in the formula ? Would appreciate a nudge or even better, thorough analysis.

gionole said:
TL;DR Summary: Emma noether’s conservation law

I have been learning emma noether’s theorem where every symmetry results in conservstion law of something, sometimes momentum, sometimes angular momentum, sometimes energy or sometimes some kind of quantity.

Every explanation that I have listened to or learned from talks about homogeneity of the space and says that if we have a system of particles and we shift each one of them by some infinetisemal distance(we dont have to shift all of them by same amount, but could be we shift first particle by 2epsilon, where second particle by 3epsilon). This all depends what their lagrangian is and whether this shift gives the possibility that dL=0.

While I understood all that, I have a hard time imagining this in real world usecase. In real life, we dont care about shifting them, they just move without our interaction.
I suspect you have misunderstood the concept of spatial symmetry. You say you understand all that, but your example shows that you haven't.
gionole said:
I want to solve the following: imagine a gun which pulls a bullet and bullet hits a box. As we know from school physics, impulse is conserved - total momentum before pulling was 0 and the final momentum of the gun and bullet must be 0 as well. So we write p1+p2=0. I want to see/prove that lagrangian does not change for my example because if we have momentum conservation, there must have been symmetry such as dL=0. I dont even know how to start.
If you understand the concept, why do you not even know where to start.
gionole said:
Before pulling, L seems 0 as no kinetic and potential energy exists in the system but do I still have to write them in the formula ? Would appreciate a nudge or even better, thorough analysis.
You should study some more appropriate examples of Lagrangian mechanics. And post your working if you get stuck. There must be some in your textbook.

vanhees71
I suspect you have misunderstood the concept of spatial symmetry. You say you understand all that, but your example shows that you haven't
Can you give me a pointer what I dont understand, so that I can try to work on it first ?

gionole said:
Can you give me a pointer what I dont understand, so that I can try to work on it first ?
You haven't posted anything to show what you do and don't understand. You have simply picked a bizarre example of an explosion fuelled by the release of chemical energy. Is that a typical example of Lagrangian mechanics? Where does the release of chemical energy fit into what you have learned about Lagrangian mechanics?

vanhees71
Well, As I stated, in the symmetry, we say: "would we get dL = 0 if we shift the system by some epsilon. Well, I'm trying to see in real world use case where without our interaction(without our shift), Lagrangian doesn't change. Symmetry is something you can make on the system so that action doesn't change which in turn means L doesn't change. I'm not sure what it is you think that I don't understand till now.

But most of the time, things happen on its own. Car hits another car and ofc, I wasn't sitting in it and calculating Lagrangian when it would give no change in 2 different coordinates of space. It just happened on its own. I'm trying to see in such cases that conservation law of momentum trully comes from the symmetry(dL=0). To me, this seems a legit question and I just picked an example of gun/bullet and want to see why momentum is conserved in terms of Lagrangian. @PeroK

gionole said:
Well, As I stated, in the symmetry, we say: "would we get dL = 0 if we shift the system by some epsilon. Well, I'm trying to see in real world use case where without our interaction(without our shift), Lagrangian doesn't change. Symmetry is something you can make on the system so that action doesn't change which in turn means L doesn't change. I'm not sure what it is you think that I don't understand till now.

But most of the time, things happen on its own. Car hits another car and ofc, I wasn't sitting in it and calculating Lagrangian when it would give no change in 2 different coordinates of space. It just happened on its own. I'm trying to see in such cases that conservation law of momentum trully comes from the symmetry(dL=0). To me, this seems a legit question and I just picked an example of gun/bullet and want to see why momentum is conserved in terms of Lagrangian. @PeroK
I suggest you find an example problem from your textbook and post it in the homework forum. Note that "I don't even know where to start" is not acceptable in terms of the Homework guidelines:

Here are some common mistakes to avoid:
• Don't simply say "I have no clue," "I have no idea where to start," or "I'm completely lost."
These don't qualify as attempts. Instead, it suggests you haven't put much effort into reading and understanding your textbook and lecture notes, going over similar examples, etc.

vanhees71
PeroK said:
I suggest you find an example problem from your textbook and post it in the homework forum.
I am curious about gun/bullet. Are you saying this cant be explained with lagrangian ? If so, why does the momentum get conserved there ? Well, you might explain this in terms of newtonian but trying to see it with Lagrangian.

I will wait maybe someone else can understand this.

The point of Noether's theorem is that momentum conservation must hold for all of physics as long as it's describable by Newtonian (or special relativistic) physics since the space time model of both Newtonian ans special-relativstic physics implies that space (in SR for an inertial observer) is homogeneous, i.e., spatial translations must be a symmetry of all equations describing closed systems. So not only mechanics but also the chemistry behind the explosion of your gun powder (which is an application of non-relativsitic quantum mechanics when viewed from a fundamental point of view) implies that momentum conservation holds true. Of course this cannot be described by Lagrangian mechanics.

The point of Emmy (sic!) Noehter's theorem is that based on the given symmetry group of Newtonian spacetime you can pretty much derive the form of the Lagrangian for a closed system of particles. Assuming in addition that all fundamental forces are pair forces only you get
$$L=\sum_{k=1}^{N} \frac{m_k}{2} \dot{\vec{x}}^2 - \frac{1}{2} \sum_{j=1}^N \sum_{k \neq j} V(|\vec{x}_j-\vec{x}_k|).$$
It's a bit more simple to use this Lagrangian to derive the conservation laws from the space-time symmetry, i.e., the full group of Galilei transformations. The spatial translations are given by
$$\vec{x}_k'=\vec{x}_k+\delta \vec{a}, \quad t'=t, \quad \delta \vec{a}=\text{const}.$$
Since the ##\delta \vec{a}## are arbitrary, and since the Lagrangian is obviously invariant under these transformation, the analysis a la Noether leads to the conservation of the total momentum
$$\vec{P}=\sum_{k=1}^N m_k \dot{\vec{x}}_k.$$

## What is Lagrangian symmetry in the context of a bullet and gun system?

Lagrangian symmetry refers to the invariance of the Lagrangian function, which describes the dynamics of a system, under certain transformations. In the context of a bullet and gun system, this involves analyzing how the system's behavior remains unchanged under specific transformations like translations, rotations, or boosts, which correspond to conservation laws like momentum, angular momentum, and energy.

## How does Lagrangian symmetry relate to conservation laws in a bullet and gun system?

Lagrangian symmetry is directly related to conservation laws through Noether's theorem. For a bullet and gun system, if the Lagrangian is invariant under time translation, it implies conservation of energy. If it is invariant under spatial translations, it implies conservation of linear momentum. Similarly, rotational invariance implies conservation of angular momentum. These symmetries help in understanding the system's behavior during firing and after the bullet is shot.

## What are the typical symmetries considered in a bullet and gun system?

Typical symmetries considered in a bullet and gun system include translational symmetry (both in time and space), rotational symmetry, and sometimes boost symmetry (related to changes in velocity). Translational symmetry in space and time leads to conservation of linear momentum and energy, respectively, while rotational symmetry leads to conservation of angular momentum. These symmetries help simplify the analysis of the system's dynamics.

## How does the recoil of the gun relate to Lagrangian symmetry?

The recoil of the gun is a manifestation of the conservation of momentum, which is a result of translational symmetry of the Lagrangian. When the bullet is fired, the gun experiences an equal and opposite momentum to conserve the total momentum of the system. This recoil is a direct consequence of the symmetry properties of the system's Lagrangian, ensuring that the total momentum before and after firing remains constant.

## Can Lagrangian symmetry be used to predict the behavior of a bullet and gun system under external forces?

Yes, Lagrangian symmetry can be extended to include external forces by modifying the Lagrangian to account for these forces. The modified Lagrangian can then be used to derive equations of motion that predict the system's behavior. However, the presence of external forces may break some of the original symmetries, leading to changes in the corresponding conservation laws. For example, an external force might break translational symmetry and thus affect the conservation of momentum.

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