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I'm reading Feynman's Thesis, and have some background in math and physics, but I'm not sure where Feynman gets his solution to his harmonic oscillator equation. He gives three different formulations. The first includes an integral. Is this standard Fourier Analysis stuff? Can you give me some kind of link?

One of them is

$$ x(t) = x(0) \cos \omega t + \dot{x}(0) \frac {\sin\omega t}{\omega} + \frac {1}{m \omega}\int_o^t \gamma (s) \sin \omega (t-s) ds $$

$$\gamma = I_y + I_z $$. The "I" is a function involving the coordinates of positions y and z.

$$ m\ddot{x} + m\omega^2 x = [ I_y + I_z ] $$

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Welcome to Physics.SE! This question would be very much improved with both a link to the copy of the document and transcriptions of the equations you're interested in. We have MathJax on the site so you can write LaTeX as in $\frac{p^2}{2m} + kx^2$. –  dmckee Jan 17 '12 at 23:44
    
@dmckee: I don't know LaTex (yet) and the equations, from what I can see, only appear in his 'thesis', and, hence, in the book. The eq'ns are on pp. 18-19. –  Lino Jan 18 '12 at 0:02
    
The above is obtained by putting \frac{p}{2m} + kx^2 inside a pari of single dollar signs $. Searching on LaTeX integral sign should give easy result for things like \int_{x_0}^{x_1} \mathrm{d}x ($\int_{x_0}^{x^1} \mathrm{d}x$). –  dmckee Jan 18 '12 at 0:05
    
You should edit updates into the question as I have done with your comment. To get block formatted equations put them between pairs of double-dollar signs ($$). –  dmckee Jan 18 '12 at 0:38

1 Answer 1

This comes from linearity, the solution of the homogenous equation, plus the response to a delta-function kick.

The homogenous equation gives the first two terms. The equation

$$ \ddot{x} + \omega^2 x = 0 $$

is solved by a combination of cos and sin which reflect the initial position and velocity

$$ x(t) = x_0 cos(\omega t) + {v_0\over \omega} sin(\omega t) $$

You can verify that this is correct by looking near t=0 to see that it starts at $x_0$ with velocity $v_0$.

Now consider adding a delta-function kick at some time $t_0$

$$ \ddot{x} + \omega^2 x = \delta (t-t_0)$$

You want the solution to this equation which only has influence into the future, meaning that for $t<t_0$, $x(t)=0$. The delta function is an impulsive kick which makes the particle move with a velocity 1 at time $t_0+\epsilon$, so that the solution from this point onward is the solution to the equation with initial velocity 1 starting at $t=t_0$, or

$$ x(t) = {1\over \omega} sin(\omega (t-t_0))$$

Now you consider the source term $\gamma(t)$ to be a sum of delta functions at each time, each one producing this response. This leads to a total response (by linearity) of

$$ x(t) = \int_0^t{\gamma(t')\over \omega} sin(\omega(t-t')) dt' $$

And this integral is the last term in Feynman's solution. The forward effects of the sin kernel means that this solution doesn't affect the initial conditions, so that you just add this to the solution with the given initial conditions to find the general solution.

The only things he uses here are linearity of the equation, plus the solution of the homogenous equation for a given initial condition.

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Thanks so much. Physics is not pure math, is it! –  Lino Jan 18 '12 at 15:54
    
@Lino: It isn't pure math, but in this case, both physicists and mathematicians would do the same thing. –  Ron Maimon Jan 18 '12 at 22:32

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