Numbers and Naturalty, part 3

After we leaned a bit about p-adic numbers, it's about time to return to the naturalty. Let me just discuss some hypotheses about the structure of the world we live in

Assume we live in the world that made of p-adic numbers. Yep, instead of $\mathbb{R}^3$, for instance we could live in $\mathbb{Z}_p^3$. Sounds crazy, because we are so accustomed to thinking about lines, axes and so on which in our geometrical imagination are tightly connected with the real numbers.

But can we treat it (our geometrical imagination) as the argument to reject the hypothesis of p-adic space? Well, we can't. There are many examples in which our geometrical intuition fails to discover the real world (non-Euclidean geometry is probably the most vivid one).

But there is another argument.
Let's look at special relativity theory. The key moment is we have spacetime instead of time separately and space separately. The space and the time have similar nature, but the time can't be p-adic variable, because it should be ordered (otherwise we can't talk about causality principle).

To be more precise, the elements of spacetime are partially ordered, but this ordering is based on comparison distance in Minkowski space (which is obtained using arithmetic operations) and zero. Anyway, ordering is needed, and we don't have it in p-adic numbers.

I wrote much about simplicity of real numbers and the simplicity of their definition. It's about time to talk about why they can't substitute real numbers (which doesn't imply they are useless).

1. Real numbers do not depend on the radix you use, and this is essential. If I state that p-adic numbers should be used in some physical theory, I should also name the prime p. 2? 3? 5? I can't see any reason to prefer one number to another. The real numbers are the only.
2. Ordering once again. The first step in theoretical physics is least action principle, which states that some function (action, helps Captain the Obvious) should be (locally) minimal on the solution trajectory. If you don't have ordering, the assertion like this cannot be stated.
3. Every time we talk about optimization, we mean real values. The optimization criteria result should be in ordered field, and, based on the optimization theory, it's much better if the set or results is complete (otherwise there is no guarantee that solution exist).
   How many fields are there that are both ordered and complete?
   
   The answer is known: there is only one. It is $\mathbb{R}$
   (but note: the result doesn't have to belong to some field)

Legendre transformation without Legendre transformation

If you heard something of theoretical mechanics, you definitely know that the most important transition in mechanics is one from Lagrangian Mechanics to Hamiltonian. And its name is Legendre transform.

This transition is some kind of magic that to my mind is considered usually as some secret recipe. The recipe is

• take the momenta: $p_i=\frac{\partial L}{\partial \dot{q_i}}$.
• prove that Hamilton equations hold: $$\frac{\partial H}{\partial q_j}=-\dot{p}_j,\qquad\frac{\partial H}{\partial p_j}=\dot{q}_j$$ where  $$ H\left(q,\;p,\;t\right)=\sum_i\dot{q}_i p_i-L(q,\;\dot{q},\;t) $$

And that is all the idea. But why should I take this Hamiltonian not some other function? Why should I consider Lagrangian derivatives as new variables instead of $q_i$ ? I didn't meet the answers in mechanics courses, though there is one simple intuitive justification. That's Pasha Gavrilenko who told me about it.

What do we have initially? An action on some interval of time, this is the integral 

$$ S = \min_{q(\cdot)} \int_{t_1}^{t_2}  L(q, \dot{q}, t) dt$$ which should be minimized (locally), that what Hamilton's principle states. (There are conditions on the endpoints which I will omit) 

Ok, the only trouble is $q$ and $\dot{q}$ aren't independent, otherwise the Lagrange equations would be much simpler: $ \frac{\partial L}{\partial q} = \frac{\partial L}{\partial \dot{q}} = 0 $

Let us try to replace $\dot(q)$ with $v$, assuming they are equal: $$ S = \min_{q(\cdot)} \int_{t_1}^{t_2}  L(q, v, t)\bigg|_{v=\dot q} dt$$

Hmhmhm. Seems nothing changed. Now the trick. Let's add summand
$$\delta(\dot{q},v) = \begin{cases} 0 & \dot{q} = v \\ +\infty & \text{otherwise} \end{cases}$$ and now we can minimize over all possible trajectories of $q$ and of $v$.  $$ S = \min_{q(\cdot), v(\cdot)} \left[ \int_{t_1}^{t_2}  L(q, v, t) dt + \delta[q,v] \right] $$

See? We have now $q$ and $v$ independent by the cost of additional summand. Now we can write $\delta$ in the following form (make sure you understand it):
$$\delta[q,v] = \max_{p(\cdot)} \int_{t_1}^{t_2}  p(t) (\dot{q}(t) - v(t)) dt $$

After substitution we have the problem on finding saddle point of function:  $$ S = \min_{q(\cdot), v(\cdot)} \max_{p(\cdot)} \int_{t_1}^{t_2}   p(t) (\dot{q}(t) - v(t)) +  L(q, v, t) dt $$ Pay attention that all variables $p,q,v$ are independent now. The solution we need is trajectory $q(\cdot)$, but it has corresponding trajectories $v(\cdot)$ and $p(\cdot)$ which form a saddle point together with $q(\cdot)$.

As we know, at the saddle point all the partial derivatives are zero (assuming the function is differentiable). Calculating variational derivatives with respect to $p,v,q$ gives respectively $$ \dot{q} = v \\ p = \frac{\partial L}{\partial \dot{q}} \\ \dot{p} = \frac{\partial L}{\partial q} $$
Note that energy function $H(q,v,t)$ also appeared in a natural way as well as least action principle in Hamiltonian mechanics $$p(t) (\dot{q}(t) - v(t)) +  L(q, v, t) = p(t) \dot{q}(t) - \left[  p(t)v(t) - L(q,v,t)  \right]  =  \\ = \{\text{changing the variables} \} =  p(t) \dot{q}(t) - H(p,q,t) $$
You may have noticed that the thing I did is just added Lagrange multiplier ho make the condition $\dot{q} = v$ hold.

This way Legendre transformation looks more accessible to my mind. Starting from that moment I understood that Lagrange multipliers is a very powerful tool.