Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm solving an exercise about Hamiltonian equations. I have followed the proceeding below. The results given by the book are different to mine because its first result is the half of mine (and the second one linked to the first one is different to mine). I think that my proceeding is correct and so I can't understand...

Given these two Hamiltonian equations:

$$\tag{1} \dot p ~=~ - \alpha pq,$$ $$\tag{2} \dot q ~=~\frac{1}{2} \alpha q^2.$$

Find $q(t)$ and p$(t)$, considering initial conditions $p_0$ and $q_0$.

I have integrated the second equation and obtained:

$$\tag{3} q(t)~=~\frac{2q_0}{2-q_0 \alpha (t-T_0)}$$

Then I have pugged this, in the second canonical eq, and I have obtained:

$$\tag{4} p(t)~=~p_0(2-q_0 \alpha (t-t_0))^2.$$

The solutions given by the book are:

$$\tag{5} q(t)~=~\frac{q_0}{1- \frac{1}{2} \alpha q_0 (t-t_0)},$$ $$\tag{6} p(t)~=~p_0[1-\frac{1}{2} \alpha q_0 (t-t_0)]^2.$$

I can obtain the solutions of the book if I divide numerator and denominator of $q$ for 2.. but.. can I do it?

Is my proceeding correct?

share|cite|improve this question
Why do you feel it's wrong?. try taking 2 common from both n/d and then cancel it out, if you like it that way. – ABC May 31 '13 at 17:56
@userØØ7 but the solutions of motion (p(t) and q(t)) are different! which pair is correct? – sunrise May 31 '13 at 17:59
I'm a Newtonian Mechanics person. was just helping with math. I don't understand all this :P – ABC May 31 '13 at 18:02
@userØØ7 ;) I have understood the mathematical trick, but I haven't understood if it is "legal" ;) – sunrise May 31 '13 at 18:04
It's completely legal. – user1504 Jun 7 '13 at 20:19

Yes, all equations (1)-(6) are correct.

The only problem is that the symbol $p_0$ usually is (implicitly) assumed to denote the momentum

$$\tag{7} p_0~:=~p(t=t_0)$$

at $t=t_0$. Equations (4) and (7) would then lead to the contradiction $p_0=4p_0$. For this reason it is better if you call your integration constant from eq. (1) something differently, say $p_1$. Then your fourth equation becomes

$$\tag{4'} p(t)~=~p_1(2-q_0 \alpha (t-t_0))^2.$$

Now deduce that $p_0:=p(t=t_0)=4p_1$, and proceed to derive eq. (6).

share|cite|improve this answer
thanks for your answer! I'm sorry but I haven't understood why in eq. (4) I have to call $p_1$ instead of $p_0$.. when I integrate eq. (1), I integrate between $p_0$ and $p$.. or am I wrong? – sunrise May 31 '13 at 20:19
I updated the answer. – Qmechanic May 31 '13 at 20:24
I got it! :) But if I say that eq (3) and (4) are solutions of motion, I'm not wrong. Am I? – sunrise May 31 '13 at 20:27
No, you are right as I wrote in my first sentence. You are only wrong if you simultaneously (implicitly) assume eq. (7). – Qmechanic May 31 '13 at 20:31
Thanks a lot!!! – sunrise May 31 '13 at 20:33

Yes, you can divide q's numerator and denominator by 2

share|cite|improve this answer
I can't understand which pair of solutions (p(t) and q(t)) is correct... they give different results! – sunrise May 31 '13 at 18:00
seems to me that your solutions are the same – Jim May 31 '13 at 18:03
the same? how is it possible? can you tell me more clearly why they are the same? thanks for your help! – sunrise May 31 '13 at 18:05
your q(t) equations are the same; identical. Plug in any number for t and I guarantee they will give the same result. Your p(t) is found using q(t). It looks like you've done the procedure correctly, thus it stands to reason that (once you get rid of that factor of 2 in your p(t) that shouldn't be there) the p(t)'s will also be the same; give the exact same result for any t – Jim May 31 '13 at 18:10
sure! but if I want the same result for the p(t)'s, I have to multiply my p(t) for 2.. if I don't moltiply, I haven't the same result. And so, how can I say that the solutions are the same? – sunrise May 31 '13 at 18:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.