# Energy of a system: multiply by velocity?

1. Jul 18, 2011

### thegreenlaser

I'm reading through some material on vibrating strings. In one book, the author repeatedly uses energy analysis to determine information about a system. What I'm not understanding is how he gets his energy equations. For example, with the equation:

$$\frac{d^2 u}{dt^2}=-\omega_0^2 u$$

He finds the energy by multiplying the whole equation by the velocity,

$$\frac{du}{dt}\frac{d^2 u}{dt^2} + \omega_0^2 \frac{du}{dt}u = 0$$

and then pulling out the derivative, or kind of semi-integrating:

$$\frac{d}{dt}\left( \frac{1}{2} {\left(\frac{du}{dt}\right)}^2 + \frac{\omega_0^2}{2} u^2 \right) = 0$$

So now he says that this represents the derivative of the energy, scaled by a constant with units of mass. i.e. the above equation means that the time derivative of the energy in the system is 0.

I can see the kinetic and potential energy terms in there, but I'm really not understanding this 'multiplying by du/dt' technique. Can anyone explain the justification for this? He uses it constantly throughout the book, and when it gets to partial derivatives he starts using an inner product defined by:

$$\left< f,g \right>_D = \int_D f \cdot g \ dt$$

So to find the derivative of the energy of the system, he takes the inner product of the whole equation with the time derivative of u, and then pulls out the differential operator similar to above to obtain something like (partial / partial t)(expression) = 0.

I haven't done any physics to do with energy since high school, so I might be missing something obvious, but can anyone explain why multiplying by velocity (or more generally taking the inner product with velocity) will yield the derivative of energy scaled by mass?

2. Jul 18, 2011

### matonski

The justification is the result. Multiplying both sides of an equation by the same thing doesn't need any justification since it is a valid algebraic manipulation.

Perhaps you mean more motivation or intuition? Well, Force x Velocity = Power = rate of change of energy, and the first equation you wrote is nothing but F = ma scaled by units of mass.

3. Jul 18, 2011

### thegreenlaser

I know it's a valid algebraic manipulation. I don't understand why the time derivative of energy should be the result. In the example I gave, the terms show up nicely as kinetic energy and potential energy.

Later, he does this: (subscripts denote partial derivatives)

$$u_{tt} + \kappa^2 u_{xxxx}=0 \Rightarrow \left< u_t,u_{tt} \right>_{\mathbb{R}} + \left< u_t,u_{xxxx} \right>_{\mathbb{R}} = 0$$

He then uses some integration by parts trickery to obtain:

$$\left< u_t,u_{tt} \right>_{R} + \left< u_{txx},u_{xx} \right>_{R} = 0 \Rightarrow \frac{d}{dt} \left( \frac{1}{2} || u_t ||^2_{R} + \frac{\kappa^2}{2} || u_{xx} ||^2_{R} \right) = 0$$

(R is the real numbers. Note that the final result is not a partial derivative w.r.t. time, since the inner product cancels out the x component) And then he says that the time derivative of energy equals zero, with $$\frac{1}{2} || u_t ||^2_{R}$$ being the kinetic energy (scaled by mass) and
$$\frac{\kappa^2}{2} || u_{xx} ||^2_{R}$$ being the potential energy (scaled by mass). That's where I'm getting confused. I understand the math for the most part, but I don't understand how to go from the wave equation to the wave equation multiplied by du/dt to the time derivative of energy, which always seems to be nicely split into kinetic and potential. (If there's loss in the system, that always shows up on the right hand side of the equation in place of zero)

Sorry if my confusion still isn't clear. If someone can explain why the inner product technique works, then that will really help, since the ODE case is just a simplification of that.

4. Jul 20, 2011

### thegreenlaser

Bump. I'm still not understanding this, and I still haven't managed to find any info about it...

5. Jul 22, 2011

### Philip Wood

It may help if we home in on one of the energy terms. Let's go for KE.

KE is defined as the amount of work a body can do because of its motion.

Work is defined as $\int$F.ds.

So let's imagine that the body is acted upon by some retarding force, -F, which need not be constant. For a body of constant mass, -F = m dv/dt.

Now by N's third law, the force the body exerts on the thing from which the resistive force comes is F.

So work done by body = $\int$F.ds = -$\int$(m dv/dt).ds = -m$\int$v.dv.

By expanding the dot product in terms of components it's quite easy to show that $\int$v.dv = $\int$v.dv.

The bottom limit is the initial speed of the body, u, and the top limit is 0 (when it's stopped and has no more capacity to do work).

So we get the familiar (1/2) m u2 formula.

Don't know whether this has helped, but I thought that going back to basic definitions of KE and work might be the way forward. Some people prefer to consider the body being brought from rest up to a final speed. This delivers the same result without the fiddly minus sign, but it doesn't go right back to the definition. It assumes that the work done on the body getting it up to speed is the same as the work which the body is then able to do as it comes to rest. This is, of course, true, and you may prefer this method.

Last edited: Jul 22, 2011
6. Jul 24, 2011

### Philip Wood

Have you met grad (V) in which V is a scalar field? I ask only because it would help me write a short sequel to my last post.

7. Jul 31, 2011

### thegreenlaser

Sorry for the late reply. Yes, I have met the gradient operator, but I haven't yet learned what it really means. Thank you for your first post, though, it helped me a fair bit.

8. Aug 1, 2011

### Philip Wood

$\frac{\partial V}{\partial x}$

Last edited: Aug 1, 2011