# Is there any loss or gain of information if a physics law is changed from one form to another? [closed]

Is there any loss or gain of information if a physics law is changed from one form to another such that the parameter appearing in them is changed from vector to a scalar? For example, consider the Newtons Second Law i.e.

$$\vec{F}=m\vec{a}$$

where $F$ and $a$ are vectors. and similarly consider the lagrange equation

$$\frac{d}{dt}\frac{\partial{L}}{\partial{v}} - \frac{\partial{L}}{\partial{x}} = 0$$

where $x$ and $v$ are generalized coordinates and velocities.

In the first equation it involves a function and a vector(ie $F$) whereas in second it only involves function (ie $L$). My Intuition tells me a vector has more information than a scalar, but these two laws are equivalent, so can someone tell me the loss-gain analysis of information?

P.S For more information, consider the analysis done by me: Suppose my universe is 2-D. From Newtonian viewpoint any force F can be factored into 2 parts fx and fy. These are two simple functions which can be evaluated independently and then added as per vector rule (need to specify coordinate system and basis). Now if I follow Lagrangian approach, i have L = m/2(vx^2 + vy^2), again two function but since the addition is scalar, I can be okay with this. So here is what I conclude, in Newtonian, I need 2 function and one vector addition rule whereas in Lagrangian I need only 2 function. Doesn't that make Lagrangian less specific?

• If you change the representation of quantities in a particular application of a physical law you are going from one system to another. That's comparing apples and oranges. Newtonian physics can be applied to a one-dimensional single particle system or to the 100 billion stars in a galaxy, it still stays Newtonian physics. – CuriousOne Sep 17 '15 at 13:21
• you can map a vector of any finite dimension into a scalar and vice versa without any loss of information. The two formalisms are equivalent – user83548 Sep 17 '15 at 18:20
• What do you mean that the "Lagrangian [is] less specific?" – Kyle Kanos Sep 18 '15 at 11:47
• @kyle: by less specific I mean less input would give the less output compared to Newtonian. Anyways my query has been resolved! – Manish Kumar Singh Sep 18 '15 at 17:14

The Lagrangian is defined over a bunch of different position variables and their derivatives, which is basically why you don't lose any information. That is, for a single particle in 3D, you have $\mathcal L = \frac 12 m \dot x^2 + \frac 12 m \dot y^2 + \frac 12 m \dot z^2 - U(x, y, z)$, which contains all of the information you need to calculate $p_x = \frac{\partial \mathcal L}{\partial \dot x}$ and $F_x = \frac{\partial \mathcal L}{\partial x}$.
There is no reason to believe that the space of smooth functions $\mathbb R^{6N+1} \to\mathbb R$ from our $6N$ position coordinates and $1$ time coordinate is smaller than the space of continuous functions from $1$ time coordinate to the positions and momenta of $N$ particles, $\mathbb R \to \mathbb R^{6N}$. Certainly if we replace $\mathbb R$ with a discrete set like $\{1, 2, 3\}$ we get something very different: the number of functions from $A \to B$ is $|B|^{|A|}$, so even for $N=1$ the "Lagrangian" case contains $3^{3^7}$ different functions $(= 3^{2187} \approx 3\cdot 10^{1043})$ while the "Newtonian" case contains only $(3^6)^3 = 3^{18} \approx 4\cdot 10^{8}$ functions. So in some sense there is a lot more "space" in the latter than the former. (There are technical reasons why this is probably not asymptotically true; there are probably not more pairs of real numbers than there are real numbers, so there is probably some bijection $\mathbb R^{6N} \leftrightarrow \mathbb R,$ and both of these probably have the cardinality $|\mathbb R|^{|\mathbb R|}$, but the point is that the Lagrangian set is no smaller than the Newtonian set.)