Studying Is Richard Feynman's Teaching Style Hard to Follow?

[Chenkel]

I was wondering if anyone else had trouble with reading Richard Feynman's lectures on physics. I think he's a good man and had fundamental contributions to science, but has anyone noticed that it is sometimes hard to follow what he is saying? I was reading his chapter about psuedo forces and came away wondering what he was getting at. I find I can generally come away with something useful from watching a Walter Lewin lecture, but when Mr Feynman is talking about an advanced concept I'm lucky if I can grasp anything; his book seems to have such good reviews, what am I missing?

 

[berkeman]

Can you post a link to such a lecture? I wonder if I would have the same problem/issue...

 

[Chenkel]

Can you post a link to such a lecture? I wonder if I would have the same problem/issue...

I haven't read the whole chapter, just the the Section on psuedo forces. I feel I understand some of it, but as I read the book, the number of questions I have quickly out number the number of answers. I always feel I'm missing some key information that's not provided and the points that need emphasis are implicitly talked about.

 

[dlgoff]

I made some DVDs years ago. I couldn't get a very good picture, but here's one:

Part 1=Photons Corpuscles of Light
Part 2=Reflection
Part 3=Electron Interactions
Part 4=Problems in QED and The Standard Model of Particle Physics

Edit: I also have 7 DVDs of The Character of Physical Laws

 

[Borek]

There is no such thing as "a lecture for everyone". Just because many of us like Feynman's lectures and find them clear, doesn't mean they have to fit your learning style. These are very individual things.

 

[Chenkel]

From the book:

$$ \frac {dx} {dt} = \frac {dx'} {dt} + \frac {ds} {dt} $$

"Previously, we considered the case where s was constant, and we found that s made no difference in the laws of motion, since ds/dt=0; ultimately, therefore, the laws of physics were the same in both systems. But another case we can take is that s=ut, where u is a uniform velocity in a straight line. Then s is not constant, and ds/dt is not zero, but is u, a constant. However, the acceleration d2x/dt2 is still the same as d2x′/dt2, because du/dt=0. This proves the law that we used in Chapter 10, namely, that if we move in a straight line with uniform velocity the laws of physics will look the same to us as when we are standing still."

After reading this I am still perplexed, how do the laws of physics ever change? I don't see a proof for the laws of physics looking the same if we move in uniform velocity vs when we do not, do the laws of physics ever not look the same? What does it mean for them to look the same? What if they do? What are the implications? It's hard for me to know what Mr Feynman is getting at. This is a small example of what I mean, it's a very difficult read, seeing a text as popular as this one, I wish I could read it, but each paragraph raises countless questions.

 

[pbuk]

I don't see a proof for the laws of physics looking the same if we move in uniform velocity vs when we do not, do the laws of physics ever not look the same?

You need to read on to the text starting "But we wish to discuss the interesting case where s is still more complicated, say $s=\frac{at^2}{2}$".

each paragraph raises countless questions.

... which are usually answered in the next paragraph. This is admittedly part of Feynman's style and it may not suit you.

 

[fresh_42]

To all who are interested in a printed version:
https://www.amazon.com/dp/0465023827/

 

[codelieb]

You can read The Feynman Lectures on Physics, as well as hear (the tapes) and see (photos and videos) of Feynman lecturing at https://www.feynmanlectures.caltech.edu .

 

[Chenkel]

You need to read on to the text starting "But we wish to discuss the interesting case where s is still more complicated, say $s=\frac{at^2}{2}$".

... which are usually answered in the next paragraph. This is admittedly part of Feynman's style and it may not suit you.

From Joe's perspective (centered at x') the force on the object is $m\frac {d^2x} {dt^2} = F_x$ and from Moe's perspective Mr Feynman writes

$$ m\frac {d^2x'} {dt^2} = F_x - ma $$

How is this from Moe's perspective? The object does not move relative to Moe, so what does it mean to say "from Moe's perspective."

 

[pbuk]

How is this from Moe's perspective? The object does not move relative to Moe, so what does it mean to say "from Moe's perspective."

The copy I am reading doesn't use the word perspective, it says:

This means that although the laws of motion from the point of view of Joe would look like $$ m\,\frac{d^2x}{dt^2}=F_x, $$ the laws of motion as looked upon by Moe would appear as $$ m\,\frac{d^2x'}{dt^2}=F_{x'}=F_x-ma. $$

Note also that Joe is not "centered at $x'$". Neither $x$ nor $x'$ is a place, it is a coordinate system. It doesn't matter where Joe is, what matters is that Joe is not accelerating. That means that Joe's view of the world has coordinates like $x$. Because Moe is accelerating, his view of the world has coordinates like $x'$. Because for Moe $F_{x'}=F_x-ma = 0$ the object appears stationery.

Note that the phrase we actually use instead of 'perspective', 'view of the world', 'point of view' or 'as looked upon by' is frame of reference or reference frame; we say that Joe measures things in an inertial frame of reference (because he is not accelerating) but Moe measures things in a non-inertial frame of reference.

What level of study are you at? IMHO one problem with Feynman is that his mind worked in vector (actually tensor) calculus, and for the rest of us that means his explanations are sometimes not the best introduction to a subject. If you haven't already studied the laws of motion with some introduction to fictitious forces I'm not sure Feynman is the best place to start.

 

[Chenkel]

The copy I am reading doesn't use the word perspective, it says:

This means that although the laws of motion from the point of view of Joe would look like $$ m\,\frac{d^2x}{dt^2}=F_x, $$ the laws of motion as looked upon by Moe would appear as $$ m\,\frac{d^2x'}{dt^2}=F_{x'}=F_x-ma. $$

Note also that Joe is not "centered at $x'$". Neither $x$ nor $x'$ is a place, it is a coordinate system. It doesn't matter where Joe is, what matters is that Joe is not accelerating. That means that Joe's view of the world has coordinates like $x$. Because Moe is accelerating, his view of the world has coordinates like $x'$. Because for Moe $F_{x'}=F_x-ma = 0$ the object appears stationery.

Note that the phrase we actually use instead of 'perspective', 'view of the world', 'point of view' or 'as looked upon by' is frame of reference or reference frame; we say that Joe measures things in an inertial frame of reference (because he is not accelerating) but Moe measures things in a non-inertial frame of reference.

What level of study are you at? IMHO one problem with Feynman is that his mind worked in vector (actually tensor) calculus, and for the rest of us that means his explanations are sometimes not the best introduction to a subject. If you haven't already studied the laws of motion with some introduction to fictitious forces I'm not sure Feynman is the best place to start.

I have studied basic laws of motion. I still am a little confused, is the distance between the coordinate systems (s) the only thing that's changing, because x = x' + s?

I have studied vector calc (is that the same as tensor calc?) although it's been sometime since I studied it seriously.

Perhaps if I don't understand his book, my first assumption should be that his teaching style is not conducive to my learning style, that being said I'm still willing to try to understand what he's getting at!

 

[pbuk]

I have studied basic laws of motion. I still am a little confused, is the distance between the coordinate systems (s) the only thing that's changing, because x = x' + s?

No, if the distance was changing at a constant velocity it wouldn't matter (all inertial frames are equivalent, that is what the first part was about). It doesn't matter whether you are 'stationary' (whatever that is), or moving with constant velocity.

The reason the $x'$ frame has fictitious forces is because it is accelerating.

 

[Chenkel]

No, if the distance was changing at a constant velocity it wouldn't matter (all inertial frames are equivalent, that is what the first part was about). It doesn't matter whether you are 'stationary' (whatever that is), or moving with constant velocity.

The reason the $x'$ frame has fictitious forces is because it is accelerating.

What is s if it's not the distance between the coordinate systems?

 

[hutchphd]

You can read The Feynman Lectures on Physics, as well as hear (the tapes) and see (photos and videos) of Feynman lecturing at https://www.feynmanlectures.caltech.edu .

Thanks to @codelieb for that website. I missed seeing the Messenger lectures at Cornell by six years...one of the few times I now wish I were older.

May I suggest the Messenger video lectures to the OP. If he still dislikes RP Feynman then chacun a sa gout

 

[DaveE]

I think the reason the Feynman Lectures (the books) are so popular is that they are interesting and different from "normal" textbooks. If you did a poll here I'd bet almost everyone learned physics from some other text first and then read Feynman (or maybe in parallel).

When I was an undergrad at Caltech in the late 1970's, everyone had a copy. But, none of the physics classes used it as their main text. I think that pretty much tells the story. They did use Leighton's associated book of problems.

 

[vanhees71]

The Feynman Lectures are indeed special, because Feynman took the effort to really write a new, original textbook (series) and did not copy just all the textbooks around for centuries ;-)). The only problem is that it's definitely not an introductory university-physics textbook, and it's rather a theoretical-physics than an experimental-physics textbook too. That's why I think, one should first study physics at the university level from a somewhat more elementary perspective. Then you'll appreciate the additional insight you can gain from reading about any subject in the Feynman lectures, and there's always some additional insight to almost any apparently simple thing.

I don't know, why the treatment of non-inertial reference frames is so confusing. At least I was very confused when first getting at it in the experimental-physics lecture. My guess is that physicists sometimes are a bit lazy in formulating the mathematics in a crisp way. In the context of (non-inertial) reference frames in Newtonian mechanics it's the lack of distinguishing between invariant vectors (the true physical quantities) and the components used to describe them with respect to a (Cartesian) basis.

In the here considered case of non-rotating non-inertial frames (non-rotating means of course non-rotating with respect to any inertial frame of reference), that's not even the point, because here you can use the same Cartesian basis in both the inertial reference frame, which mathematically is defined by a point $O$ ("the origin of the inertial frame") and a Cartesian (right-handed) basis $\boldsymbol{e}_j$. In the following bold-faced symbols are invariant vectors, and $\boldsymbol{r}$ is the position vector pointing from $O$ to the position $P$ of the particle.

If the non-inertial frame to be considered is not rotating wrt. the inertial frame, it simply means you have a point $O'$, which is used as the origin of the non-inertial frame that may move on any trajectory you like, and you can use the same time-independent Cartesian basis $\boldsymbol{e}_j$ as in the non-inertial frame.

Now let $\overrightarrow{OO'}=\boldsymbol{R}(t)$ the given trajectory of $O'$. and $\boldsymbol{r}'=\overrightarrow{O'P}$ the position vector of you particle under consideration wrt. the non-inertial frame. Then you obviously have
$$\boldsymbol{r}=\overrightarrow{OP}=\overrightarrow{OO'}+\overrightarrow{O'P}=\boldsymbol{R} + \boldsymbol{r}'.$$
You know the equation of motion in the non-inertial frame. According to Newton's 2nd Law it's
$$m \ddot{\boldsymbol{r}}=\boldsymbol{F}(\boldsymbol{r}).$$
From this for the position vector in the non-inertial frame you get
$$m \ddot{\boldsymbol{r}}=m (\ddot{\boldsymbol{R}}+\ddot{\boldsymbol{r}}')=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}'),$$
or to bring the EoM. in the non-inertial frame to the same form as in the inertial one:
$$m \ddot{\boldsymbol{r}}'=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}')-m \ddot{\boldsymbol{R}},$$
i.e., the fact that you describe the motion of the particle within a non-inertial (but non-rotating) reference frame looks like the Newtonian EoM. for a particle under the influence of the force and the additional "inertial force"
$$\boldsymbol{F}_{\text{inertial}}=-m \ddot{\boldsymbol{R}}.$$
As you see, in this case you can easily even avoid to argue with the components of the vectors in the different reference frames.

 

[Chenkel]

From this for the position vector in the non-inertial frame you get
$$m \ddot{\boldsymbol{r}}=m (\ddot{\boldsymbol{R}}+\ddot{\boldsymbol{r}}')=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}'),$$
or to bring the EoM. in the non-inertial frame to the same form as in the inertial one:
$$m \ddot{\boldsymbol{r}}'=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}')-m \ddot{\boldsymbol{R}},$$

I believe understand most of what you wrote up to this point, but now I have a couple basic questions, is F a vector function of position? Also what does EoM mean? Also why is the mass times the second derivative of r' wrt time the same as the force at the position of the particle relative to O minus the mass of the particle times the acceleration of the O' coordinate system.

 

[hutchphd]

Oh I see the notation is a little non-specific.

The mass is multiplied by the position vector but F is a vector valued function of the position vector. Good?

 

[vanhees71]

I believe understand most of what you wrote up to this point, but now I have a couple basic questions, is F a vector function of position? Also what does EoM mean? Also why is the mass times the second derivative of r' wrt time the same as the force at the position of the particle relative to O minus the mass of the particle times the acceleration of the O' coordinate system.

In non-relativistic mechanics usually forces are functions of the positions of the involved particles only. Here I wrote down the most simple case, where the motion of one particle in some external force field is a sufficient approximation. EoM=Equation of Motion. Since $$\boldsymbol{r}=\boldsymbol{R}+\boldsymbol{r}'$$ you have $\boldsymbol{F}(\boldsymbol{r})=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}').$

 

[Chenkel]

In non-relativistic mechanics usually forces are functions of the positions of the involved particles only. Here I wrote down the most simple case, where the motion of one particle in some external force field is a sufficient approximation. EoM=Equation of Motion. Since $$\boldsymbol{r}=\boldsymbol{R}+\boldsymbol{r}'$$ you have $\boldsymbol{F}(\boldsymbol{r})=\boldsymbol{F}(\boldsymbol{R}+\boldsymbol{r}').$

Makes sense! So when O' is accelerating the EoM change so that we have to subtract the mass of the object times the acceleration of the reference frame O' from the normal calculation of force. Also the if O' is moving with constant velocity, the force calculation is not different than when O' is not moving at all. Does all of this make the mass of the object times the acceleration of O' a psuedo force?

Thank you, your explanation makes a lot of sense.

 

[vanhees71]

One calls $-m \ddot{\boldsymbol{R}}$ often a "fictititious force", but I don't like this, because it's not fictitious but just a term in the standard Newtonian equation of motion, expressed in terms of vectors in a non-inertial reference frame, which you bring from the left-hand side "$m a$" to the right-hand side "$F$". They are thus not fictitious but necessary to express the same equation of motion in the non-inertial frame as in the inertial frame. I prefer to call these terms "inertial forces", because it describes better how they come about when translating the equation of motion in terms of coordinates referring to an inertial frame to that in terms of coordinates referring to a non-inertial frame.

 

[haushofer]

One calls $-m \ddot{\boldsymbol{R}}$ often a "fictititious force", but I don't like this, because it's not fictitious but just a term in the standard Newtonian equation of motion, expressed in terms of vectors in a non-inertial reference frame, which you bring from the left-hand side "$m a$" to the right-hand side "$F$". They are thus not fictitious but necessary to express the same equation of motion in the non-inertial frame as in the inertial frame. I prefer to call these terms "inertial forces", because it describes better how they come about when translating the equation of motion in terms of coordinates referring to an inertial frame to that in terms of coordinates referring to a non-inertial frame.

From a purely mathematical point of view, fictitious forces are just terms which smell like normal forces and pop up if you transform a differential equation. From a causal point of view however, they're really fictitious, for they're not caused by an interaction with the object, but by an interaction with the observer. And in the end, a force is by its definition caused by a physical interaction.

That's how I see it.

 

[vanhees71]

Indeed, this makes them in a sense "fictitious", but the derivation shows, they simply belong to the left-hand side of Newton's equation of motion and are not "forces" to begin with. Unfortunately we have to live with the unfortunate established naming. On the other hand these terms are simply there when describing the motion within a non-inertial reference frame and thus they are not fictitious. That's why I find the naming "inertial forces" a bit better than calling them "fictitious", but that's of course just my personal opinion.

 

[haushofer]

I completely agree. The accelerating observer describes the notion of inertia as a force, hence inertial force.

Tell a circling astronaut loosing his brain through his nose and ears in a centrifuge that his passing out is merely "fictitious", and one should see the point :P

An interesting point is that according to Mach, inertia IS the result of some (vaguely described) interaction.

 

[vanhees71]

You can have a lot of discussion about whether GR is "Machian" or not, again thanks to the vagueness of Mach's idea. For physics itself it's pretty irrelevant though.

 

[robphy]

I like Ivey & Hume's view on "ficticious forces" ( &t=14m40s ; punchline: &t=16m10s )


details on the film at: archive.org/details/frames_of_reference

 

[haushofer]

You can have a lot of discussion about whether GR is "Machian" or not, again thanks to the vagueness of Mach's idea.

Why? A single particle in an otherwise empty universe has, according to GR, inertial properties. Hence GR is not Machian, whatever the precise interaction which causes this inertia may be.

 

[vanhees71]

Exactly! Case closed!