Space to discuss classical mechanics specific questions/topics (for example, are all orbits in a central potential stable to perturbations?)

Hi everyone,

So I was working on showing that <\phi_i|\phi_j> = 0, and I can see how Iām supposed to set it up (integrate \phi_i \cdot \phi_j over -infinity to infinity), but I just canāt seem to see the exact step I need to do to prove that itās always true. Graphing it makes it visually clear, in the sense that it oscillates between negative and positive in equal measures when i != j, but is always positive in the case when i=j (given that itās squared with no imaginary parts explicitly there). Iāve tried different trig identities for the expression, as well as trying to integrate over just the period itself. I may be making a silly mistake, but is there anything glaring that Iām missing and should try?

Hey Gabe,

Just started looking at this too, seems like your comments about the graphs are exactly right! Just two things I noticed that I think you might be missing:

- the inner product is integrated over the range of the functions themselves, so the limits of the period should be good in your integral.
- the prompt for this is somewhat misleading,
`< phi_i | phi_j > = 0`

**if**i != j, its some non-zero value for i = j as you said.

The reason I think the the second point must be true is because these phi s are from the wave equation and we want to show they are orthogonal. And for orthogonal functions `< f_i | f_j > = 0 if i != j`

.

Take the orthogonal basis of usual x-y coordinate system for example:

`v_1 = (0,1)`

and `v_2 = (1,0)`

.

```
<v_1 | v_2> = 0*1 + 1*0 = 0
<v_1 | v_1> = 0*0 + 1*1 = 1
```

here, `<v_i | v_j> = 0`

and `<v_i | v_i> = 1`

.

Does that make sense? Thatās the angle Iām approaching it for now at least, but I still have to work through the actual integrals

-Holly

I think Iāve figured out a simple trick that may work pretty well with little to no actual integration (remembered an approximation from undergrad E&M).

you can write out the expression as a sum of two different sin terms with equal amplitude using the identity:

\sin(x)sin(y)=\frac{1}{2}[\cos(x-y)-\cos(x+y)]

integrating from -infinity to infinity over a trig function is just going to be 1/2 the amplitude. Therefore in the case where i \neq j they cancel out (after plugging in our own stuff for x and y of course).

Hey @GabeGrauvogel and @hmatt !

Thank you both for having this discussion here, it is so nice to see the forum alive with questions and back and forths .

Super important thing, the inner product that we are defining is only over the length of the rope (from 0 to L). Otherwise things start getting weird because the functions do not have compact support (meaning they are not squared integrable, they donāt go to zero fast enough far away). So, in that sense the types of integrals you have to evaluate are:

<\phi_i(x)|\phi_j(x)>=\int_0^L \text{sin}(\frac{i\pi x }{L})\text{sin}(\frac{j\pi x }{L}) dx

You will find that if i=j the integrals gives you a finite value, while if i\neq j it gives you zero. You can use Wolfram alpha or mathematica to evaluate the integral if you are unsure on how to do so by pen and paper. Once you find the value of the integral for i=j that will give you the ānormā of your basis functions \phi_i(x) and you should divide by it your basis so they are orthonormal with respect of the inner product we defined.

Once that is done, you can get the coefficients for your expansion of your initial condition f(x) by projecting on this set of orthonormal folks .

Hope this helps and if anything is still confusing please let me know and we can chat about it!

Pablo

Hello everyone! Here is the homework for classical mechanics:

I suggest you consider the problem with the āstraight-lineā geometry as we discussed in class first and relate it to the problem shown here. It should be straightforward to do this using the Newtonian-matrix method presented in class but it is worth your time to consider the Lagrangian method as well. We will present a discussion of this next time, but for now I point you to Section 5.5 of Lemosās Analytical Mechanics.

The matrix equation provided there is (**V** - \omega^2 **T**)**q** = 0

Jeremiah

So Iāve been trying to work through a few of the Lagrangian problems on previous subject exams, and on the first on I seem to have gotten to an answer, although it doesnāt feel right to me that the Hamiltonian would somehow have just gotten rid of the potential term from before and remained as just the kinetic energy term. I would have expected the terms with K in the Lagrangian to just flip sign (given the expression H=T+V), but my solution killed them. Itās probably a dumb mistake in there somewhere, but has anyone else taken a stab at it?

I donāt see anything wrong with your math, so that looks good to me! I am also curious why the potential term seems to disappear.

Maybe there is some restriction we are not considering? Are there any forms of the Lagrangian where you canāt use H = sum(p_i * q_i) - L? Or systems where H = T + V isnāt true?

For problem #2 in the 2016 Summer Subject exam, when we get to part 3 and it gives us constraints in x and y for the problem, what are these x and y coordinates with respect to, or where is the center of our system? I would assume that the x and y they are talking about is the center of mass of the rod, but Iām not entirely sure.

Iām also not entirely sure how Iām supposed to use constraints in x and y for a lagrangian they have had us write in terms of r and theta. I know that I could rewrite the lagrangian in terms of x and y, but is there a more elegant way to do this with what they have already given/lead us to?

Hi Gabe, I donāt think #3 is the same problem as the pendulum one. It looks like theyāre different problems

Iāve also been trying to do some of the past exams problems, and Iām not sure if I got the potential energy right. Itās problem 2 from the August 2016 exam.

I did problem #3 from the August 2016 exam. I was a bit confused on how to construct the matrix, so Iām not sure if itās right or completely wrong

Here is the exercise I recommend for next time, modified from Lemosās Analytical Mechanics Example 4.2: Consider an isosceles right triangle in the XY plane with mass M.

- Write an appropriate mass density function such that the total mass is M.
- Using the coordinate system above:

-Guess the principle axes, justify your reasoning

-By symmetry arguments and the perpendicular axis theorem, determine which 2 integrals are necessary to fully determine the moment of inertia tensor (not unique!) Relate your chosen integrals to all the entries of the tensor.

-Evaluate the integrals to determine the moment of inertia tensor

-Use this tensor to verify your guess of principle axes, as well as determining the principle moments. If your guess of principle axes was incorrect, update your reasoning.

Hi Gabe, I think your math is correct to this point, but you still need to express the Hamiltonian in terms of the canonical momenta. This will make K reappear.

Hi Alegranados,

I would revisit this one! I would write everything considering the center mass of the rod. Your potential energy should include the spring energy (should be proportional to (r-r_0)^2 ) and mgh for the center mass. You are not that far off.

As for the kinetic energy I would write the coordinate of the center mass of the rod (hint: you should have the y part in your calculation of h for the potential energy) and take a time derivative. Then you can apply the identity that T_tot = T_cm + T_rot . It will be a pretty ugly expression!

(Following is my response to your normal modes question as it will not let me reply 3 times in a row.):

I get the same result!

Nice work!

I did problem #4 from the August 2016 exam. This is what I got.

Also, I was wondering if you do #5 using eulerian angles?

I tried to do it but Iām still confused on how to find the principle axes so I kinda skipped that part

when a question asks us to use the properties of the hamiltonian and poisson brackets to show that total energy is conserved, isnāt the total energy just the Hamiltonian in cases where H=T+V? In that case isnāt it sufficient to say that H commutes with itself therefore total energy is conserved?