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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?

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    $\begingroup$ 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. $\endgroup$ – CuriousOne Sep 17 '15 at 13:21
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    $\begingroup$ 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 $\endgroup$ – user83548 Sep 17 '15 at 18:20
  • $\begingroup$ What do you mean that the "Lagrangian [is] less specific?" $\endgroup$ – Kyle Kanos Sep 18 '15 at 11:47
  • $\begingroup$ @kyle: by less specific I mean less input would give the less output compared to Newtonian. Anyways my query has been resolved! $\endgroup$ – Manish Kumar Singh Sep 18 '15 at 17:14
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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.)

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  • $\begingroup$ 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 wheras in lagrangian I need only 2 function. Doesn't that make Lagrangian less specific? $\endgroup$ – Manish Kumar Singh Sep 17 '15 at 19:29
  • $\begingroup$ @ManishKumarSingh No, because you haven't defined what "specific" means, and in fact you can't define it in any way that is meaningful. There exist systems which are much harder to model with Lagrangians, particularly dissipative systems (Lagrangians usually preserve total energy). There exist systems which are much harder to model with Newton's laws, particularly constrained systems (if a constraint force's direction and magnitude are changing with the dynamics of the system, everything becomes nasty). But in the usual case the Euler-Lagrange equations are equivalent to the Newton equations. $\endgroup$ – CR Drost Sep 17 '15 at 20:05
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In order to describe newton's law in a 2d universe, you need a lagrangian with two generalized co-ordinates, x and y, and their repspective velocities.

You'll get two differential equations when you solve it, one for x, and one for y. In your question, you only mentioned the one for x.

So your statement that the two equations are equivalent is incorrect. The vector equation contains information from the equation you wrote, as well as a similar one for the second dimension.

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