Hi Jakob, Thanks for writing these great books. They really help with getting the big picture of these subjects. I found a typo in the CM book, on page 157 in the derivation, in the step labelled ‘expanding the products,’ the term $ \frac{\partial \bar F}{\partial Q}\frac{\partial Q}{\partial p}\frac{\partial \bar H}{\partial Q}\frac{\partial Q}{\partial q}$ appears twice, but one of them should be $ \frac{\partial \bar F}{\partial Q}\frac{\partial Q}{\partial p}\frac{\partial \bar H}{\partial P}\frac{\partial P}{\partial q}$. The incorrect one is then retained in the cancellation in the next step. Do you have somewhere online where you post errata? That could help so… Read more »

Guest

Griff

Oh, I thought the LaTeX would compile automagically :/

Guest

Griff

On page 161, equation 7.88, you also have a $\frac{\partial}{\dot q}$ where it should be $\frac{\partial}{\partial \dot q}$.

thanks a lot for reporting these typos. I will correct them as soon as possible. Currently, I have no errata list online but it’s a great idea and will try to put one online as soon as I find the time.

Best,
Jakob

Guest

Phil

Hello there. On p 84 the boxed comment states “All terms first order in epsilon must vanish.” I’m not really sure what the author means by this. I can see that if x = a is the lowest point, then for any epsilon (positive or negative), f(a) < f(a + epsilon). However, I don't really understand what the author means by "vanish". I suspect it's something quite straightforward that I'm not seeing, but if someone could flesh this out for me a bit, I would appreciate it.

Hi Phil, the idea is that if we consider a minimum $x=a$, it necessarily goes up if we move away from there. Therefore, if we inspect the neighborhood of the minimum, $x= a + \epsilon$, we know that all terms proportional to $\epsilon$ are necessarily zero if $a$ is indeed a minimum. (But all terms containing $\epsilon^2$ are not problematic and can be neglected if $\epsilon$ is infinitesimal.) If this wouldn’t be true, we could chose $\epsilon$ negative and find a location this way that corresponds to an ever lower value of the function. Here’s how Richard Feynman explains this… Read more »

Guest

Phil

Many thanks, Jakob. That helps a lot.

Guest

Lucas

Hi Jakob! First of all, thank you VERY MUCH for creating these wonderful books. Reading them is a joy. I just wanted to mention that I, too, have found this section to be the first part of the book I had trouble understanding and had to look for an explanation elsewhere. Thank you again for these great books.

The concept is straightforward, but possibly the notation isn’t. If epsilon (e) vanishes then e=0, but 3x2ae =0. That is 0=0 Rather factor e out, e(3x2a +1)=0 ,etc then a=-1/6. This makes more sense to me.

About Classical Mechanics
Not a very good book, an exceptional book, really!
It goes quite deep but you are helped all along with the margin notes: looks like the author reads our mind and he answers in these notes.
Also repeating an equation from previous pages is really convenient.
I am ordering right away the Quantum Mechanics book from the same author

Compared to Susskind Theoretical Minimum, i think this books goes into more details and is easier to follow.

Guest

Tamir

P.77 – “the ball cannot stay at the top indefinitely since we consider paths between a fixed initial position at fixed initial time and a fixed final position at a fixed final time.”
I dont get it. don’t thake into account the posibility of a constant function q(t)? Is there no stable equilibrium in the world?
sure I can see why it’s not even an equilibrium from newton perspective, but as for the lagrangian formulation it’s not clear…

Hi Tamir, thanks for your question! In the situation describes on page 77, we throw the ball into the air from fixed position and want to find its trajectory if it returns after a given number of seconds to its starting position. In other words, we are dealing with a concrete situation which is specified by appropriate initial conditions. In this case, the initial conditions are given by an initial location and initial velocity. As a result of the nonzero initial velocity (since we throw the ball into the air), a constant q(t) is not a valid possibility. I hope… Read more »

Guest

Robert Herman

Great book for me! I am trying to relearn CM after decades, and the pace and clarity fit me well.

I’m pretty sure I know, but on page 34, (2.12) has force = to the first derivative of momentum, p,
or F= m*a.

Am I correct in assuming that when we take the first derivative of p, m just carries over, since the exponent of v time m = m?

I find this better than the Theoretical Minimum too, and plan on buying Mr. Schwichtenberg’s books!

Yes, in many situations the mass $m$ is constant and therefore $\dot p = \frac{dp}{dt} = \frac{d (mv)}{dt}= m\frac{d v}{dt} = m a = m \dot v$. I hope this answers your question and please let me know if you have any further questions!

Guest

Roy

Hello, I will be using John Taylor’s CM next semester in a class. I am hoping this will allow me to get ahead instead of being behind.
Feel free to provide any advice.

I hope so too! Let me know if you have any questions or find something confusing in the book.

Guest

Roy

I purchased all three of your books.

Guest

Ronnie Webb

Hello and thank you for writing an amazing book on Classical Mechanics that even a self learner like myself can finally learn from! I have a question about the derivations of the q and p transformations via the Poisson Bracket (page 171). In the q derivation the bracket is {q, G}, but in the p derivation the bracket is {G, p}. The Poisson Bracket calculation on the next line (for the p transformation) under this portrays a different calculation i.e. it looks like it should be {p, G}. But, above all of this the equations say q+eG o q and… Read more »

yes you’re right that’s a typo. It should read ${p,G}$ in the third line. I will fix it as as soon as possible.

Best,
Jakob

Guest

Lucas

Jakob, on page 370 it seems that in order to obtain equation D.16 you had to assume that the partial derivative of (dx/dt)2 with respect to x and the partial derivative of x2 with respect to dx/dt are both zero. This makes sense if we assume that x and dx/dt are independent variables, but in page 95, when deriving Hamilton’s equations, you state clearly that it is “not possible” to treat x and dx/dt as “completely independent variables” “because the velocity is always simply the rate of change of the location”. If this is so, why can we assume that… Read more »

Our goal in the Lagrangian formalism is to figure out the correct path in configuration space between two fixed locations. A path is characterized by a location and velocity at each point in time. We want to stay as general as possible and consider really all possible paths. In particular, this means that we consider all possible pairings of location and velocity functions. Now the physical path that we are eventually interested in is special for two reasons: it’s a solution of the Euler-Lagrange equation (= extremum of the action) the locations and velocities at each moment in time are… Read more »

Guest

Lucas

Beautiful! Thank you so much.

Guest

Lucas

Jakob, what is the order you recommend when reading your books? I intend to do Classical Mechanics, Electrodynamics, Quantum Mechanics and then Physics from Symmetry? Would you agree?

Hi Jakob, At the bottom of page 164 to arrive at (7.98) you cancel a partial derivative of Q with a total derivative of Q. Is that allowed? Although final answer is correct it seems that is because the partial (p) of F/pQ x Qdot sums to zero with its equivalent in the parentheses. Your thoughts? Thank you. BTW loving the book.

Yes, $\frac{\partial F }{\partial Q} \dot Q$ cancels with $\frac{\partial F }{\partial Q} \frac{d Q }{d t}$ in the parentheses. Hope this helps! (And if not, just ask again since I’m not entirely sure I understood your question correctly.)

Guest

Edi

Congratulations to an excellent book! If you don’t mind, I have a question about the classical Lagrangian T-V. In Section 12.3.6 you show how the gravitational potential affects the proper time and thus gives rise to the –V term. However, my understanding is that the classical Lagrangian T-V hold for any type of potential, gravitational or non-gravitational. So, the following question arises: Do all kinds of potential, e.g., the electrostatic potential, affect the proper time? I know from your other book (Physics from Symmetry) that the Lagrangian of electromagnetic interactions can be obtained by making the U(1) symmetry of the… Read more »

That’s an excellent question and something I haven’t thought about so far. But I think that the electrostatic potential affects the proper time in a similar way as any other form of energy.

While the electromagnetic Lagrangian can be understood using the U(1) gauge symmetry, it also holds in the classical case. However, you’re probably right that there is some deeper insight hidden in the connection between the quantum mechanical derivation and the classical derivation. But I’m not sure if anyone has understood it yet.

Hi Jakob, Thanks for writing these great books. They really help with getting the big picture of these subjects. I found a typo in the CM book, on page 157 in the derivation, in the step labelled ‘expanding the products,’ the term $ \frac{\partial \bar F}{\partial Q}\frac{\partial Q}{\partial p}\frac{\partial \bar H}{\partial Q}\frac{\partial Q}{\partial q}$ appears twice, but one of them should be $ \frac{\partial \bar F}{\partial Q}\frac{\partial Q}{\partial p}\frac{\partial \bar H}{\partial P}\frac{\partial P}{\partial q}$. The incorrect one is then retained in the cancellation in the next step. Do you have somewhere online where you post errata? That could help so… Read more »

Oh, I thought the LaTeX would compile automagically :/

On page 161, equation 7.88, you also have a $\frac{\partial}{\dot q}$ where it should be $\frac{\partial}{\partial \dot q}$.

Hi Griff,

thanks a lot for reporting these typos. I will correct them as soon as possible. Currently, I have no errata list online but it’s a great idea and will try to put one online as soon as I find the time.

Best,

Jakob

Hello there. On p 84 the boxed comment states “All terms first order in epsilon must vanish.” I’m not really sure what the author means by this. I can see that if x = a is the lowest point, then for any epsilon (positive or negative), f(a) < f(a + epsilon). However, I don't really understand what the author means by "vanish". I suspect it's something quite straightforward that I'm not seeing, but if someone could flesh this out for me a bit, I would appreciate it.

Hi Phil, the idea is that if we consider a minimum $x=a$, it necessarily goes up if we move away from there. Therefore, if we inspect the neighborhood of the minimum, $x= a + \epsilon$, we know that all terms proportional to $\epsilon$ are necessarily zero if $a$ is indeed a minimum. (But all terms containing $\epsilon^2$ are not problematic and can be neglected if $\epsilon$ is infinitesimal.) If this wouldn’t be true, we could chose $\epsilon$ negative and find a location this way that corresponds to an ever lower value of the function. Here’s how Richard Feynman explains this… Read more »

Many thanks, Jakob. That helps a lot.

Hi Jakob! First of all, thank you VERY MUCH for creating these wonderful books. Reading them is a joy. I just wanted to mention that I, too, have found this section to be the first part of the book I had trouble understanding and had to look for an explanation elsewhere. Thank you again for these great books.

Thanks for your feedback! I will try to improve the explanation in future editions of the book.

The concept is straightforward, but possibly the notation isn’t. If epsilon (e) vanishes then e=0, but 3x2ae =0. That is 0=0 Rather factor e out, e(3x2a +1)=0 ,etc then a=-1/6. This makes more sense to me.

About Classical Mechanics

Not a very good book, an exceptional book, really!

It goes quite deep but you are helped all along with the margin notes: looks like the author reads our mind and he answers in these notes.

Also repeating an equation from previous pages is really convenient.

I am ordering right away the Quantum Mechanics book from the same author

Compared to Susskind Theoretical Minimum, i think this books goes into more details and is easier to follow.

P.77 – “the ball cannot stay at the top indefinitely since we consider paths between a fixed initial position at fixed initial time and a fixed final position at a fixed final time.”

I dont get it. don’t thake into account the posibility of a constant function q(t)? Is there no stable equilibrium in the world?

sure I can see why it’s not even an equilibrium from newton perspective, but as for the lagrangian formulation it’s not clear…

Hi Tamir, thanks for your question! In the situation describes on page 77, we throw the ball into the air from fixed position and want to find its trajectory if it returns after a given number of seconds to its starting position. In other words, we are dealing with a concrete situation which is specified by appropriate initial conditions. In this case, the initial conditions are given by an initial location and initial velocity. As a result of the nonzero initial velocity (since we throw the ball into the air), a constant q(t) is not a valid possibility. I hope… Read more »

Great book for me! I am trying to relearn CM after decades, and the pace and clarity fit me well.

I’m pretty sure I know, but on page 34, (2.12) has force = to the first derivative of momentum, p,

or F= m*a.

Am I correct in assuming that when we take the first derivative of p, m just carries over, since the exponent of v time m = m?

I find this better than the Theoretical Minimum too, and plan on buying Mr. Schwichtenberg’s books!

Yes, in many situations the mass $m$ is constant and therefore $\dot p = \frac{dp}{dt} = \frac{d (mv)}{dt}= m\frac{d v}{dt} = m a = m \dot v$. I hope this answers your question and please let me know if you have any further questions!

Hello, I will be using John Taylor’s CM next semester in a class. I am hoping this will allow me to get ahead instead of being behind.

Feel free to provide any advice.

I hope so too! Let me know if you have any questions or find something confusing in the book.

I purchased all three of your books.

Hello and thank you for writing an amazing book on Classical Mechanics that even a self learner like myself can finally learn from! I have a question about the derivations of the q and p transformations via the Poisson Bracket (page 171). In the q derivation the bracket is {q, G}, but in the p derivation the bracket is {G, p}. The Poisson Bracket calculation on the next line (for the p transformation) under this portrays a different calculation i.e. it looks like it should be {p, G}. But, above all of this the equations say q+eG o q and… Read more »

Hi Ronnie,

yes you’re right that’s a typo. It should read ${p,G}$ in the third line. I will fix it as as soon as possible.

Best,

Jakob

Jakob, on page 370 it seems that in order to obtain equation D.16 you had to assume that the partial derivative of (dx/dt)2 with respect to x and the partial derivative of x2 with respect to dx/dt are both zero. This makes sense if we assume that x and dx/dt are independent variables, but in page 95, when deriving Hamilton’s equations, you state clearly that it is “not possible” to treat x and dx/dt as “completely independent variables” “because the velocity is always simply the rate of change of the location”. If this is so, why can we assume that… Read more »

Our goal in the Lagrangian formalism is to figure out the correct path in configuration space between two fixed locations. A path is characterized by a location and velocity at each point in time. We want to stay as general as possible and consider really all possible paths. In particular, this means that we consider all possible pairings of location and velocity functions. Now the physical path that we are eventually interested in is special for two reasons: it’s a solution of the Euler-Lagrange equation (= extremum of the action) the locations and velocities at each moment in time are… Read more »

Beautiful! Thank you so much.

Jakob, what is the order you recommend when reading your books? I intend to do Classical Mechanics, Electrodynamics, Quantum Mechanics and then Physics from Symmetry? Would you agree?

Yes this order would be perfect.

Hi Jakob, At the bottom of page 164 to arrive at (7.98) you cancel a partial derivative of Q with a total derivative of Q. Is that allowed? Although final answer is correct it seems that is because the partial (p) of F/pQ x Qdot sums to zero with its equivalent in the parentheses. Your thoughts? Thank you. BTW loving the book.

Yes, $\frac{\partial F }{\partial Q} \dot Q$ cancels with $\frac{\partial F }{\partial Q} \frac{d Q }{d t}$ in the parentheses. Hope this helps! (And if not, just ask again since I’m not entirely sure I understood your question correctly.)

Congratulations to an excellent book! If you don’t mind, I have a question about the classical Lagrangian T-V. In Section 12.3.6 you show how the gravitational potential affects the proper time and thus gives rise to the –V term. However, my understanding is that the classical Lagrangian T-V hold for any type of potential, gravitational or non-gravitational. So, the following question arises: Do all kinds of potential, e.g., the electrostatic potential, affect the proper time? I know from your other book (Physics from Symmetry) that the Lagrangian of electromagnetic interactions can be obtained by making the U(1) symmetry of the… Read more »

Thanks!

That’s an excellent question and something I haven’t thought about so far. But I think that the electrostatic potential affects the proper time in a similar way as any other form of energy.

While the electromagnetic Lagrangian can be understood using the U(1) gauge symmetry, it also holds in the classical case. However, you’re probably right that there is some deeper insight hidden in the connection between the quantum mechanical derivation and the classical derivation. But I’m not sure if anyone has understood it yet.