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## Main Question or Discussion Point

I'm trying to derive the equations of motion for a particle falling in a uniform gravitational fill with air drag proportional to the square of velocity. However I'm getting the velocity as a complex number. Here is what I've done

The force of friction is [tex]F=-k\left(\dfrac{dx}{dt}\right)^2\dfrac{\textbf{v}}{||\textbf{v}||}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

We put the particle stationary at [tex]x=0[/tex]

The Lagrangian is [tex]\mathcal{L}=\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^2-mgx[/tex]

[tex]\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

[tex]m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

If we put [tex]c=\dfrac{k}{m}[/tex]

[tex]\dfrac{d^2x}{dt^2}+c\left(\dfrac{dx}{dt}\right)^2+g=0[/tex]

We put [tex]p=\dfrac{dx}{dt}[/tex]

We have [tex]\dfrac{d^2x}{dt^2}=\dfrac{dp}{dt}=\dfrac{dp}{dx}\dfrac{dx}{dt}=\dfrac{dp}{dx}p[/tex]

The differential equation becomes

[tex]\dfrac{dp}{dx}p+cp^2+g=0[/tex]

[tex]\dfrac{p}{cp^2+g}\dfrac{dp}{dx}=-1[/tex]

[tex]\int\dfrac{p}{cp^2+g}\,dp=-x[/tex]

To solve the integral we put [tex]u=cp^2+g[/tex]

[tex]\dfrac{du}{dp}=2cp[/tex]

[tex]p=\dfrac{1}{2c}\dfrac{du}{dp}[/tex]

[tex]\dfrac{1}{2c}\int \dfrac{1}{u}\,du=-x[/tex]

[tex]\int \dfrac{1}{u}\,du=\ln |u|=\ln u[/tex] because [tex]u>0[/tex]

[tex]\dfrac{1}{2c}\ln (cp^2+g)+A=-x[/tex]

[tex]A(cp^2+g)=e^{-2cx}[/tex]

At [tex]t=0,x=0,p=0[/tex]

[tex]Ag=1[/tex]

[tex]A=\dfrac{1}{g}[/tex]

[tex]\dfrac{c}{g}p^2+1=e^{-2cx}[/tex]

[tex]p^2=\dfrac{g(e^{-2cx}-1)}{c}[/tex]

Now the sign of [tex]\dfrac{g(e^{-2cx}-1)}{c}[/tex] is determined by [tex]e^{-2cx}-1[/tex] which is not necessary positive definite.

In fact if we put [tex]c=1,x=1[/tex] we get [tex]e^{-2}-1<0[/tex] which means that [tex]p\in \mathbb{C}[/tex]

But [tex]p=\dfrac{dx}{dt}[/tex] which makes no sense at all. Did I do something wrong?

The force of friction is [tex]F=-k\left(\dfrac{dx}{dt}\right)^2\dfrac{\textbf{v}}{||\textbf{v}||}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

We put the particle stationary at [tex]x=0[/tex]

The Lagrangian is [tex]\mathcal{L}=\dfrac{1}{2}m\left(\dfrac{dx}{dt}\right)^2-mgx[/tex]

[tex]\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

[tex]m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2[/tex]

If we put [tex]c=\dfrac{k}{m}[/tex]

[tex]\dfrac{d^2x}{dt^2}+c\left(\dfrac{dx}{dt}\right)^2+g=0[/tex]

We put [tex]p=\dfrac{dx}{dt}[/tex]

We have [tex]\dfrac{d^2x}{dt^2}=\dfrac{dp}{dt}=\dfrac{dp}{dx}\dfrac{dx}{dt}=\dfrac{dp}{dx}p[/tex]

The differential equation becomes

[tex]\dfrac{dp}{dx}p+cp^2+g=0[/tex]

[tex]\dfrac{p}{cp^2+g}\dfrac{dp}{dx}=-1[/tex]

[tex]\int\dfrac{p}{cp^2+g}\,dp=-x[/tex]

To solve the integral we put [tex]u=cp^2+g[/tex]

[tex]\dfrac{du}{dp}=2cp[/tex]

[tex]p=\dfrac{1}{2c}\dfrac{du}{dp}[/tex]

[tex]\dfrac{1}{2c}\int \dfrac{1}{u}\,du=-x[/tex]

[tex]\int \dfrac{1}{u}\,du=\ln |u|=\ln u[/tex] because [tex]u>0[/tex]

[tex]\dfrac{1}{2c}\ln (cp^2+g)+A=-x[/tex]

[tex]A(cp^2+g)=e^{-2cx}[/tex]

At [tex]t=0,x=0,p=0[/tex]

[tex]Ag=1[/tex]

[tex]A=\dfrac{1}{g}[/tex]

[tex]\dfrac{c}{g}p^2+1=e^{-2cx}[/tex]

[tex]p^2=\dfrac{g(e^{-2cx}-1)}{c}[/tex]

Now the sign of [tex]\dfrac{g(e^{-2cx}-1)}{c}[/tex] is determined by [tex]e^{-2cx}-1[/tex] which is not necessary positive definite.

In fact if we put [tex]c=1,x=1[/tex] we get [tex]e^{-2}-1<0[/tex] which means that [tex]p\in \mathbb{C}[/tex]

But [tex]p=\dfrac{dx}{dt}[/tex] which makes no sense at all. Did I do something wrong?