# Thread: linear algebra conservative system

1. ## linear algebra conservative system

It can be shown that for n-many masses the system of masses and springs can be written in time form M x + K x = 0 ( equation 1)
where x's are vectors and the first x has two dots over it.
[K]_ij = [K]_ji => K^T = K
where K is a matrix of spring constants and M is a matrix of masses

show that equation 1 is a conservative system.
Hint: d/dt(x^T * Mx) = 2x^TMx
where the x^T 's have a dot above them and the lone x has two dots
x with a dot above it = dx/dt where x is a vector

can someone please go through this step by step? I missed this lecture in class and the teacher said he went over this and I can't figure out how to do it.

2. Ok. You have the system

$\displaystyle M\ddot{\vec{x}}+K\vec{x}=0,$

where $\displaystyle M$ and $\displaystyle K$ are both symmetric. For a system to be conservative, what must happen?

I'm going to attempt to write up your hint, here. You're saying that

$\displaystyle \displaystyle{\frac{d}{dt}\left(\dot{\vec{x}}\cdot M\vec{x}\right)=2\dot{\vec{x}}\cdot M\ddot{\vec{x}}}$.

Is that correct?

3. It makes more sense that it should be:

$\displaystyle \displaystyle{\frac{d}{dt}\left(\dot{\vec{x}}^TM\d ot{\vec{x}}\right) = 2\dot{\vec{x}}^T M\ddot{\vec{x}}}$

If so, we have

$\displaystyle$\displaystyle \frac{d}{dt}\left(\dot{\vec{x}}^TM\dot{\vec{x}}\ri ght) = \left(\frac{d}{dt}\dot{\vec{x}}^T\right)M\dot{\vec {x}} + \dot{\vec{x}}^TM\left(\frac{d}{dt}\dot{\vec{x}}\ri ght) = \ddot{\vec{x}}^T M\dot{\vec{x}}+ \dot{\vec{x}}^T M\ddot{\vec{x}}

The multiplication Mx is: an n by n matrix by a vector, say n by 1. This gives an n by 1 vector. Then x^T multiplied by Mx will give a number, since x^T is 1 by n. So, by the same reason $\displaystyle \ddot{\vec{x}} M\dot{\vec{x}}$ $must be just a number. Hence it is equal to its own transpose.$\displaystyle $\displaystyle \ddot{\vec{x}}^T M\dot{\vec{x}} = (\ddot{\vec{x}}^T M\dot{\vec{x}})^T = \dot{\vec{x}}^TM^T\ddot{\vec{x}} = \dot{\vec{x}}^TM\ddot{\vec{x}}$$Combining this with what we had above, we get the result$\displaystyle $\displaystyle \ddot{\vec{x}}^T M\dot{\vec{x}}+ \dot{\vec{x}}^T M\ddot{\vec{x}} = \dot{\vec{x}}^TM\ddot{\vec{x}}+\dot{\vec{x}}^TM\dd ot{\vec{x}} = 2\dot{\vec{x}}^TM\ddot{\vec{x}}$

4. My dot product notation and your transpose notation are entirely equivalent.

5. Thank you for all the responses, sorry I couldn't figure out how to write it out right on my own. So, do the x's with dots above them just mean the derivative? I've never seen that notation before, and also the ones with 2 dots a second derivative?

6. Correct. It's a relatively standard notation in physics: dots mean time derivatives, and primes mean spatial derivatives. The more primes or dots, the more derivatives.

So the question right now is this: what do you suppose has to happen in order for the system to be conservative?

7. Does conservative just mean that no energy is lost? So theres no dampening on the system? Or am I thinking of this incorrectly?

8. Correct: no energy is lost. Can you translate the sentence "no energy is lost" into a mathematical equation? Suppose, say, that we write energy as $\displaystyle E$.

9. umm the energy + energy lost = energy?

10. The problem with that expression is that you would then need to provide a formula for the energy lost. Can you come up with a formula that only uses energy and time?

11. E(t) = E

12. Right. Or you can do $\displaystyle E'(t)=0,$ which is precisely the same thing. So, can you show a formula for $\displaystyle E(t)$, using symbols that show up in the OP?

13. E(dot) = 0, Is that what you meant?

14. Hehe. Good catch. I wasn't following my own notation! It should be $\displaystyle \dot{E}(t)=0.$

So can you write out an expression for $\displaystyle E(t)$ using symbols in the OP?

15. I'm not sure what to do

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