# Air Drag in 1D

## 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 $$F=-k\left(\dfrac{dx}{dt}\right)^2\dfrac{\textbf{v}}{||\textbf{v}||}=-k\left(\dfrac{dx}{dt}\right)^2$$

We put the particle stationary at $$x=0$$

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

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

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

If we put $$c=\dfrac{k}{m}$$

$$\dfrac{d^2x}{dt^2}+c\left(\dfrac{dx}{dt}\right)^2+g=0$$

We put $$p=\dfrac{dx}{dt}$$

We have $$\dfrac{d^2x}{dt^2}=\dfrac{dp}{dt}=\dfrac{dp}{dx}\dfrac{dx}{dt}=\dfrac{dp}{dx}p$$

The differential equation becomes

$$\dfrac{dp}{dx}p+cp^2+g=0$$

$$\dfrac{p}{cp^2+g}\dfrac{dp}{dx}=-1$$

$$\int\dfrac{p}{cp^2+g}\,dp=-x$$

To solve the integral we put $$u=cp^2+g$$

$$\dfrac{du}{dp}=2cp$$

$$p=\dfrac{1}{2c}\dfrac{du}{dp}$$

$$\dfrac{1}{2c}\int \dfrac{1}{u}\,du=-x$$

$$\int \dfrac{1}{u}\,du=\ln |u|=\ln u$$ because $$u>0$$

$$\dfrac{1}{2c}\ln (cp^2+g)+A=-x$$

$$A(cp^2+g)=e^{-2cx}$$

At $$t=0,x=0,p=0$$

$$Ag=1$$

$$A=\dfrac{1}{g}$$

$$\dfrac{c}{g}p^2+1=e^{-2cx}$$

$$p^2=\dfrac{g(e^{-2cx}-1)}{c}$$

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

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

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

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$$\dfrac{d}{dt}\left(\dfrac{\partial \mathcal{L}}{\partial \dot{x}}\right)-\dfrac{\partial \mathcal{L}}{\partial x}=-k\left(\dfrac{dx}{dt}\right)^2$$
The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.

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$$m\dfrac{d^2 x}{dt^2}+mg=-k\left(\dfrac{dx}{dt}\right)^2$$
I think you have a sign issue here. Shouldn't it be:

$$m\dfrac{d^2 x}{dt^2}=mg-k\left(\dfrac{dx}{dt}\right)^2$$

Mass*acceleration equals the sum of the forces. The net force is gravity minus drag.

The sign of the drag force in the above equation is wrong.
By the way, I don't see why we have to use Lagrangian here, since Newtonian method is much more simple.

EDIT: Actually you may change either the sign of the drag force or the sign of the potential energy, but only one of them. Both yield the same correct equation.
Yes I see that now. Can you explain where I went wrong in my derivation of the friction.This is what I thought

$$F=-k\dot{x}^2 \hat{\textbf{v}}$$

$$\hat{\textbf{v}}=\dfrac{\textbf{v}}{||\textbf{v}||}=\dfrac{(\dot{x})}{\dot{x}}=(1)$$

Where the brackets denote a vector in the x direction.

Now obviously $$\hat{\textbf{v}}=(1)$$ is wrong, because the particle accelerates in the negative direction, but why?

Edit
Never mind. I find out on my own.

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There is nothing wrong with $$\vec{F}=-kv^2\hat{v}$$. But you must be careful when jotting down $$F = -kv^2$$ (*) (that means F<0). Since the ball is falling, the force must act upwards. So when you write (*), that means the positive direction of x axis is downward.
Now in order that potential energy V = mgx, the x axis must point upwards, which is inconsistent with (*) as explained above.