Return to Answer

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked. Also if $F$ were a function of other variables we could not have dismissed the rate of change so easily.

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked. So, adding to the Lagrangian a total time derivative of a function of $q$ and $t$ does not change the equations of motion

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked.

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked. Also if $F$ were a function of other variables we could not have dismissed the rate of change so easily.

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$.

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked.

If you follow some of the steps in the derivations, you might wonder where the importance of the time derivative of $F$ matters. One of the equations presented in the question, the one under where it says "it is shown to be true because" is the key. This equation says:

$\frac{\partial \dot{F}}{\partial \dot{q}}=\frac{\partial F}{\partial q}$.

This equation says, although not obviously, that the last two terms in the fourth equation in the question, notably this equation:

$ \frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}-\frac{\partial}{\partial q} \frac{d F}{d t}=0 $.

are in fact equal and thus cancel. So you are left with

$\frac{d}{d t} \frac{\partial L}{\partial \dot{q}}-\frac{\partial L}{\partial q} = 0$

From that you can get the equations of motion, just as you would with $L'$. So $L'$ and $L$ give the same equations of motion.

But to get why the time derivative of $F$ is important, and not just $F$, lets start with the third term which is $\frac{d}{d t} \frac{\partial}{\partial \dot{q}} \frac{d F}{d t}$ and write it as $\frac{d}{d t} \frac{\partial}{\partial \frac{d}{dt}{q}} \frac{\frac{d}{dt} F}{1}$. Now you can see that we are taking the partial of the rate of change of $F$ with respect to the rate of change of $q$. Also not stated in the question, it is required that $F$ is a function of $t$ and $q$. That is $F=F(q,t)$

So we can dump the rate of change part and just keep the derivative of $F$ w.r.t. $q$ and get $\frac{d}{d t} \frac{\partial}{\partial {q}} F$, which is $ \frac{\partial}{\partial {q}} \frac{d F}{d t}$ which is the same as the fourth term and therefore they cancel one another.

Without the $d/dt$ in front of $F$ this would not have worked.