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> This is definitely not an introductory level book I disagree. I used Kleppner and Kolenkow as a freshman, and it has pretty much the same content as this bo

> This is definitely not an introductory level book I disagree. I used Kleppner and Kolenkow as a freshman, and it has pretty much the same content as this book, and I would consider Kleppner and Kolenkow to be an introductory book. Sure, it's more advanced than more common freshmen books like Halliday and Resnick, but it is still very elementary mechanics. Newton's laws is just the tip of the iceberg that is mechanics. Most advanced books, including upper-division undergraduate books, would at least introduce analytical mechanics, whereas this book doesn't. If you want more advanced treatments of mechanics you should look at Landau and Lifshitz, or Arnold, or Marsden, or even Lanczos (the latter three making moderate to extensive use of Riemannian geometry and can be daunting for that reason alone). > but the key to this system is in essence that you leave any kind of coordinate projections till the very end That is not the definition of coordinate invariance, at least as defined in physics. Newton's laws aren't coordinate invariant because the acceleration involves the second derivative of coordinate basis vectors, and that is zero (or constant) only in very few coordinate systems, like Cartesian coordinates for instance. Of course, the statement F = ma still holds true, but becomes very cumbersome to use if, say non-orthogonal coordinate systems are used. As a simple example, try finding the equation of motion of a particle constrained on a smooth surface, say a paraboloid z = x^2 + y^2 in Cartesian coordinates using Newton's laws. Of course, this problem can be solved using Newton's laws by introducing Lagrange multipliers, but that essentially amounts to finding the constraint force. But it would be much better if you could avoid finding the constraint force altogether. > A vector is the direction and magnitude per sé, devoid of the concept of an origin or any particular unit vectors with which you could represent that vector numerically. You're right that it is important to distinguish between the geometrical meaning of polar vectors and their representations in a particular coordinate system. It's even more confusing when axial-/pseudo-vectors, e.g., the cross product of two polar vectors, are introduced since now they do depend on the handedness of the coordinate system used. And why should nature care about handedness? (It actually does, but that's a story for another day.) And this is precisely why analytical mechanics, which is an inherently geometrical subject, was invented in the first place! All equations in analytical mechanics are scalar equations and look the same irrespective of the coordinates you use to describe the system.

ironmagma 7y ago
If you have a bone to pick about the term “coordinate free”, I suggest you take it up with Dr. Rao. I am just forwarding along the message, and making a recommendation. > try finding the equation of motion of a particle constrained on a smooth surface, say a paraboloid z = x^2 + y^2 in Cartesian coordinates using Newton's laws The method described in this book makes this kind of problem very straightforward. > Newton's laws aren't coordinate invariant because the acceleration involves the second derivative of coordinate basis vectors I find this very doubtful. The acceleration of an object in real life does not change depending on how you measure its position. Why would expressing the quantities in math be any different? I think you are making a false assumption that any given vector necessarily has a numerical representation. For instance, if gravity acts down, and there is an object of mass m with no other forces on it, Newton’s law says that F = m * g. Since g is in the down direction, F is also down. Note that both F and g have direction and magnitude in this word problem even though there are no basis vectors to speak of and thus no way we can represent any of this numerically without defining more mathematical objects. What would your coordinates be? We don’t have an origin and we only have one axis, not enough to construct a right handed system or any 3D system. Sure, we could do it one dimensionally but we’re talking 3D Euclidean space for the purposes of this book.
- b215826 7y ago
  > I find this very doubtful. I'm sorry to say this, but I don't think you understand the meaning of coordinate invariance. Coordinate invariance in physics means that if you go from one set of coordinates, say (q1, q2, ..., qn) to (s1, s2, ..., sn), the form of the equations remain invariant (assuming the transformation is nice and smooth). Newton's laws aren't invariant under a general coordinate transformation of that sort, and this is precisely because the acceleration involves the second derivatives of the basis vectors. E.g., Newton's laws when expressed in polar coordinates would involve centrifugal and Coriolis terms, which are absent in Cartesian coordinates. Equations in analytical mechanics (e.g., Lagrange's or Hamilton's equations) are coordinate invariant however [1]. > The method described in this book makes this kind of problem very straightforward. Perhaps we have different definitions of "straightforward", but most physicists I know would not consider using Newton's laws in Cartesian coordinates to find the equations of a particle constrained to move on a smooth surface straightforward. At the very least, one should try to introduce a local coordinate system on the surface. It can be done, no doubt, but it's way more cumbersome than writing Lagrange's equations involving the generalized coordinates on that surface. > Note that both F and g have direction and magnitude in this word problem even though there are no basis vectors to speak of and thus no way we can represent any of this numerically without defining more mathematical objects. No one is claiming that Newton's laws are invalid in different coordinates! Of course they are valid and F = ma still holds true. The invariance you're talking about is the general invariance of equations involving (polar) vectors. That's obvious since vectors are inherently geometrical objects. But that is in no way the same thing as "coordinate invariance". I'm not sure how the book defines "coordinate free", and I certainly don't have a bone to pick with anyone. But perhaps also look at how physicists define coordinate invariance, since after all it's something they've been doing for a while? I still stand by the claim that this book is a pretty elementary one. Sure, you might've learned new things from this book, and I can see how many people would benefit from it. But if you think this is what counts for "advanced mechanics", then you're very very mistaken. [1]: https://www.physicspages.com/pdf/Shankar/Shankar%20Exercises%2002.07.08%20(1-3).pdf https://www.physicspages.com/pdf/Shankar/Shankar%20Exercises...

ironmagma 7y ago

If you have a bone to pick about the term “coordinate free”, I suggest you take it up with Dr. Rao. I am just forwarding along the message, and making a recommendation. > try finding the equation of motion of a particle constrained on a smooth surface, say a paraboloid z = x^2 + y^2 in Cartesian coordinates using Newton's laws The method described in this book makes this kind of problem very straightforward. > Newton's laws aren't coordinate invariant because the acceleration involves the second derivative of coordinate basis vectors I find this very doubtful. The acceleration of an object in real life does not change depending on how you measure its position. Why would expressing the quantities in math be any different? I think you are making a false assumption that any given vector necessarily has a numerical representation. For instance, if gravity acts down, and there is an object of mass m with no other forces on it, Newton’s law says that F = m * g. Since g is in the down direction, F is also down. Note that both F and g have direction and magnitude in this word problem even though there are no basis vectors to speak of and thus no way we can represent any of this numerically without defining more mathematical objects. What would your coordinates be? We don’t have an origin and we only have one axis, not enough to construct a right handed system or any 3D system. Sure, we could do it one dimensionally but we’re talking 3D Euclidean space for the purposes of this book.

b215826 7y ago
> I find this very doubtful. I'm sorry to say this, but I don't think you understand the meaning of coordinate invariance. Coordinate invariance in physics means that if you go from one set of coordinates, say (q1, q2, ..., qn) to (s1, s2, ..., sn), the form of the equations remain invariant (assuming the transformation is nice and smooth). Newton's laws aren't invariant under a general coordinate transformation of that sort, and this is precisely because the acceleration involves the second derivatives of the basis vectors. E.g., Newton's laws when expressed in polar coordinates would involve centrifugal and Coriolis terms, which are absent in Cartesian coordinates. Equations in analytical mechanics (e.g., Lagrange's or Hamilton's equations) are coordinate invariant however [1]. > The method described in this book makes this kind of problem very straightforward. Perhaps we have different definitions of "straightforward", but most physicists I know would not consider using Newton's laws in Cartesian coordinates to find the equations of a particle constrained to move on a smooth surface straightforward. At the very least, one should try to introduce a local coordinate system on the surface. It can be done, no doubt, but it's way more cumbersome than writing Lagrange's equations involving the generalized coordinates on that surface. > Note that both F and g have direction and magnitude in this word problem even though there are no basis vectors to speak of and thus no way we can represent any of this numerically without defining more mathematical objects. No one is claiming that Newton's laws are invalid in different coordinates! Of course they are valid and F = ma still holds true. The invariance you're talking about is the general invariance of equations involving (polar) vectors. That's obvious since vectors are inherently geometrical objects. But that is in no way the same thing as "coordinate invariance". I'm not sure how the book defines "coordinate free", and I certainly don't have a bone to pick with anyone. But perhaps also look at how physicists define coordinate invariance, since after all it's something they've been doing for a while? I still stand by the claim that this book is a pretty elementary one. Sure, you might've learned new things from this book, and I can see how many people would benefit from it. But if you think this is what counts for "advanced mechanics", then you're very very mistaken. [1]: https://www.physicspages.com/pdf/Shankar/Shankar%20Exercises%2002.07.08%20(1-3).pdf https://www.physicspages.com/pdf/Shankar/Shankar%20Exercises...