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I started a lecture on differential equation with following example. If a body is moving in a straight line in plane with constant speed, how can we describe this motion mathematically?

To answer this, put a coordinate system in plane. Then the path the body is following is given by equation of straight line $ax+by+c=0$. If the coordinate system is changed then equation of line also gets changed; it will be as $a'x+b'y+c'=0$. One can observe that for both the coordinate systems, the path of motion is solution of the differential equation $\frac{d^2y}{dx^2}=0$. Then I said

The motion of body is expressed by the differential equation $\frac{d^2y}{dx^2}=0$, rather than linear equation, since it is coordinate-free.

The following question raised then in the class:

What $y$ and $x$ represents in the differential equation? Are they coordinates?

I couldn't answer this. Can one help me? What is the correct way to say the equation of motion of body (in straight line, with constant speed), which is coordinate-free?

p Groups
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  • Are differential equations of motion actually coordinate free? If I choose to transform my space from $x, y$ to some $u, v$ where, say, $u = ln x$ and $v = y$, the differential equation will change, correct? – Siddharth Bhat Apr 27 '16 at 09:39
  • Oh, does it mean that it is independent of "points" in the same coordinate system? as in, the differential equation does use the usual euclidian geometry, but does not explicitly refer to points $(x_0, y_0)$ in this geometry? – Siddharth Bhat Apr 27 '16 at 09:40
  • Coordinate-free, or component-free, means "without reference to any particular coordinate system" – alexjo Apr 27 '16 at 10:18

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Given that the $Oxy$ coordinate system is an inertial reference frame, the equation of motion for the body (whose mass is $m$) is given by Newton's $2$nd law of motion, i.e. $$\sum F=m\frac{d^2y}{dx^2}=0$$ and its solution reads: $\frac{d^2y}{dx^2}=0\Leftrightarrow \frac{dy}{dx}=a\Leftrightarrow y=ax+b$ which represents a straight line in $Oxy$ (that is the trajectory of the free particle in an inertial reference system), with $a$, $b$ being the integration constants to be determined from the two initial conditions: the position $x(t_0)$ at $t=t_0$ and the velocity $u(t_0)=\frac{dx}{dt}\Big|_{t_0}$ at $t=t_0$.

On the other hand, if the coordinate system is changed from $Oxy$ to $O'x'y'$, and the new coordinate system $O'x'y'$ keeps a constant velocity with respect to $Oxy$, i.e. if the new reference frame is again an inertial reference frame, then in the $O'x'y'$ coordinate system, the equation of motion is given again by Newton's $2$nd law of motion, but now with respect to the new coordinates $x',y'$, i.e.: $$\sum F'=m\frac{d^2y'}{dx'^2}=0$$ and its solution reads: $\frac{d^2y'}{dx'^2}=0\Leftrightarrow \frac{dy'}{dx'}=a\Leftrightarrow y'=a'x'+b'$ which represents a straight line in $O'x'y'$ (i.e. the trajectory of the free particle in an inertial reference system), with $a'$, $b'$ being the integration constants (now, with respect to the coordinates $x'$, $y'$).

P.S.: Newton's $2$nd law of motion actually provides a coordinate free description of the motion of a body, as long as we keep ourselves confined to inertial observers: $$ \sum\vec{F}=m\vec{a} $$ where, $\vec{F}$ is the resultant force exerted on the body and $\vec{a}$ is the acceleration of the body, both measured with respect to an inertial reference frame.

KonKan
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