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explained why initial conditions aren't conserved quantities
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user1379857
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"equations of motion," because they are equations that tell us how the variables of our system (here $q_i$) evolve in time. Indeed, in general, the solution to $n$ second order differential equations involves $2n$ integration constants (or initial conditions) in the solution. However, most people would not call these integration constants "conservation laws." In general usage, a "conserved quantity" $Q$ is a function of the configuration variables (here $q_i$ and $\dot q_i$) that does not change in time when the configuration variables evolve according to the equations of motion:

$$\frac{d}{dt} Q(q_i, \dot q_i) = 0.$$

Note that $Q(q_i, \dot q_i)$ does not depend on $t$ explicitly; it only depends on $t$ insofar as $q_i$ and $\dot q_i$ do. However, an initial condition depends on $q_i$, $\dot q_i$, and $t$. You need to know $t$ in order to know "how far to turn back the clock" to find the initial position and velocity.

"equations of motion," because they are equations that tell us how the variables of our system (here $q_i$) evolve in time. Indeed, in general, the solution to $n$ second order differential equations involves $2n$ integration constants in the solution. However, most people would not call these integration constants "conservation laws." In general usage, a "conserved quantity" $Q$ is a function of the configuration variables (here $q_i$ and $\dot q_i$) that does not change in time when the configuration variables evolve according to the equations of motion:

$$\frac{d}{dt} Q(q_i, \dot q_i) = 0.$$

"equations of motion," because they are equations that tell us how the variables of our system (here $q_i$) evolve in time. Indeed, in general, the solution to $n$ second order differential equations involves $2n$ integration constants (or initial conditions) in the solution. However, most people would not call these integration constants "conservation laws." In general usage, a "conserved quantity" $Q$ is a function of the configuration variables (here $q_i$ and $\dot q_i$) that does not change in time when the configuration variables evolve according to the equations of motion:

$$\frac{d}{dt} Q(q_i, \dot q_i) = 0.$$

Note that $Q(q_i, \dot q_i)$ does not depend on $t$ explicitly; it only depends on $t$ insofar as $q_i$ and $\dot q_i$ do. However, an initial condition depends on $q_i$, $\dot q_i$, and $t$. You need to know $t$ in order to know "how far to turn back the clock" to find the initial position and velocity.

changed fermion to lepton and z to xy
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user1379857
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and an infinitesimal rotation in the $z$$xy$-plane would be given by

In quantum field theory, quantum fields are also governed by Lagrangians. However, it is often difficult to figure out exactly what the Lagrangian of quantum fields should be based off of experimental data. Something that is straightforward to ascertain from experimental data, however, are conserved quantities, like charge, fermionlepton number, baryon number, weak hyper change, and many others. Experimentalists can figure out what these conserved quantities are, and then theorists will cook up Lagrangians with symmetries that have the right conserved quantities. This greatly aids theorists in figuring out the fundamental laws of physics. Considerations of symmetries and conserved quantities historically played a large role in piecing together the standard model, and continue to play a crucial role in theorists trying to figure out what lies beyond it.

and an infinitesimal rotation in the $z$-plane would be given by

In quantum field theory, quantum fields are also governed by Lagrangians. However, it is often difficult to figure out exactly what the Lagrangian of quantum fields should be based off of experimental data. Something that is straightforward to ascertain from experimental data, however, are conserved quantities, like charge, fermion number, baryon number, weak hyper change, and many others. Experimentalists can figure out what these conserved quantities are, and then theorists will cook up Lagrangians with symmetries that have the right conserved quantities. This greatly aids theorists in figuring out the fundamental laws of physics. Considerations of symmetries and conserved quantities historically played a large role in piecing together the standard model, and continue to play a crucial role in theorists trying to figure out what lies beyond it.

and an infinitesimal rotation in the $xy$-plane would be given by

In quantum field theory, quantum fields are also governed by Lagrangians. However, it is often difficult to figure out exactly what the Lagrangian of quantum fields should be based off of experimental data. Something that is straightforward to ascertain from experimental data, however, are conserved quantities, like charge, lepton number, baryon number, weak hyper change, and many others. Experimentalists can figure out what these conserved quantities are, and then theorists will cook up Lagrangians with symmetries that have the right conserved quantities. This greatly aids theorists in figuring out the fundamental laws of physics. Considerations of symmetries and conserved quantities historically played a large role in piecing together the standard model, and continue to play a crucial role in theorists trying to figure out what lies beyond it.

Added a dot to the q_i variables in the leftmost part of first equation
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$$\frac{d}{dt} \frac{\partial L}{\partial q_i} - \frac{\partial L}{\partial q_i} = 0$$$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0$$

$$\frac{d}{dt} \frac{\partial L}{\partial q_i} - \frac{\partial L}{\partial q_i} = 0$$

$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q_i}} - \frac{\partial L}{\partial q_i} = 0$$

added a paragraph answering the question better
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user1379857
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fixed math typos
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