# Please Prove that a Force that is Proportional to Velocity Must be Nonconservative

1. Apr 16, 2005

I tried to do this myself, but I was unsuccessful.

Thank you.

2. Apr 16, 2005

Ok, I was able to do this after all.

My first method was to substitute in -bv in for F and try to show that the closed line integral of F with dr is not equal to 0. This did not get me anywhere, although I imagine that some masters of vector calculus could get the result this way.

What was successful for me was exploiting Stokes' Theorem, and also solving the differential equation

m dv/dt = -bv.

I got v from this, plugged it back into my equation for force, and then computed the curl, which is not equal to zero so long as the velocity is not equal to zero.

Could someone verify that this is correct?

Thank you.

Last edited: Apr 16, 2005
3. Apr 16, 2005

### jdavel

No fair! You didn't really say that the force had to be in the same direction as the velocity! In that case it's obvious; the velocity of the mass is always increasing. So if it travels ina closed loop, it's going faster at the end than it was at the beginning. KE isn't conserved.

4. Apr 17, 2005

### Galileo

Consider the work done on a particle in going from a to b.
Now consider going along the same path, but with twice the velocity.

5. Apr 17, 2005

### Andrew Mason

For the differential equation of motion:

$$\frac{dv}{dt} + \frac{b}{m}v = 0$$

the integrating factor is $$e^{\frac{b}{m}t$$

So:
$$e^{\frac{b}{m}t}\frac{dv}{dt} + e^{\frac{b}{m}t}\frac{b}{m}v = \frac{d}{dt}(ve^{\frac{b}{m}t}) = 0$$

This means that:

$$v = v_0e^{{\frac{-b}{m}t}$$

If the force is conservative, the work done is a function of position only, so the line integral of Fds along a closed path must always be 0:

$$\oint m\frac{dv}{dt}ds = - \oint bvds = W$$

Edit: oops - integration corrected as per subsequent posts:

$$W = -b \oint vds = -b\oint v^2dt = -b\int_{t_0}^{t_1} v_0^2e^{\frac{-2b}{m}t}dt + -b\int_{t_1}^{t_2} v_0^2e^{\frac{-2b}{m}t}dt$$

$$W = -b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_1}-e^{\frac{-2b}{m}t_0}) + (-b(\frac{-m}{2b}v_0^2(e^{\frac{-2b}{m}t_2}-e^{\frac{-2b}{m}t_1}))$$

$$W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_1} - e^{\frac{-2b}{m}t_0}) + \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_1})$$

$$W = \frac{1}{2}mv_0^2(e^{\frac{-2b}{m}t_2} - e^{\frac{-2b}{m}t_0}) = 0$$ only if $t_2 = t_0$, which is impossible or if v_0 = 0.

In other words, the line integral along any path is time dependent, not position dependent and cannot be 0, so it is not a conservative force,

AM

Last edited: Apr 17, 2005
6. Apr 17, 2005

Thanks, Andrew.

It looks like you integrated with respect to time, although the line above, you have a "ds." I think this might change the answer, although the salient feature is the same.

7. Apr 17, 2005

### dextercioby

Andrew,your analysis and computations are incorrect.

$$W_{1\rightarrow 2}=:\int_{1}^{2} \vec{F}\cdot d\vec{s}=-b\int_{1}^{2}\vec{v}\cdot d\vec{s}$$

Since $\vec{v}\uparrow\uparrow \vec{r}$,then

$$W_{1\rightarrow 2}=-b\int_{1}^{2} v \ ds=-b\int_{t_{1}}^{t_{2}} \left(v\frac{ds}{dt}\right)dt=-b\int_{t_{1}}^{t_{2}} v^{2} dt$$

Do everything again.

Daniel.

8. Apr 17, 2005

### Andrew Mason

Thanks for pointing that out - of course E_0 = mv^2/2 and not mv. See above correction. The bottom line is that W is not position dependent so it is not conservative.

AM

9. Apr 17, 2005

### PBRMEASAP

Another way of saying it, entirely equivalent to AM's analysis, is that $$\vec{F} \cdot \vec{ds}$$ has the same sign along the entire path of the particle. Further, this dot product is only zero where V is zero, the velocity has the same direction as $$\vec{ds}$$. Thus the only way a closed loop integral could be zero is if it has zero length--the particle doesn't move.

10. Apr 19, 2005

### vinter

Why isn't anyone considering this elegant solution?

11. Apr 19, 2005

### Galileo

Thank you vinter.