Feynman Lectures on Physics Vol 2: Understanding "The Next Approximation

  • Thread starter Thread starter jackiefrost
  • Start date Start date
  • Tags Tags
    Curl
AI Thread Summary
The discussion centers on understanding Feynman's reference to "the next approximation" in the context of Stokes' Theorem in the Feynman Lectures on Physics Vol 2. The question arises from a specific equation (3.33) that appears to be a truncated Taylor series, which typically includes higher-order terms like (delta y)^2. Participants clarify that the next approximation involves considering these higher-order terms in a power series expansion for a vector function. The conversation highlights the complexity of multi-variable functions and the necessity of power series approximations in this analysis. Overall, the discussion emphasizes the importance of recognizing higher-order terms in mathematical physics.
jackiefrost
Messages
140
Reaction score
1
I didn't want to overload the last topic, "The Meaning of Curl in Electrodynamics", but I have a question so I'll do it as a new thread.

I'm studying The Feynman Lectures on Physics - Vol 2, Sections 3-5 and 3-6: "The Circulation of a vector field" and "The circulation around a square:Stokes' Theorem". I've include a scan of these two sections at http://home.comcast.net/~perion_666/stuff/feynman1.jpg and http://home.comcast.net/~perion_666/stuff/feynman2.jpg .

In the second section, after eq. 3.33, Feynman says:
If we included the next approximation, it would involve terms in (delta y)^2 ...

My question is - what is "the next approximation" he's referring to. I don't see where any higher order terms like (delta y)^2 terms would come from in this analysis.

jf
 
Last edited by a moderator:
Physics news on Phys.org
Equation 3.33 is basically a Taylor series, truncated at the linear term. For a function of one variable,

f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{1}{2} f''(x_0) (x - x_0)^2 + ...

For more than one variable, things get messier, but it's the same general idea.
 
Ok, yes - I see it. I wasn't thinking about power series approximation. I guess he had to use that since he used a generalized form for the vector function C(x,y) , so C could be about anything. Thanks.

jf
 
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top