How Is the Equation of Motion Derived from the Classical String Model?

[Phymath]

Homework Statement

I'm starting QFT and many books I've started to read start with the introduction of a field in a classical string model

with a Lagrange equation

$$L(q,\dot{q}) = \sum[\frac{m}{2}\dot{q}_{j}^{2}-\frac{k}{2}(q_{j}-q_{j+1})^{2}]$$

the equation of motion becomes

$$m \ddot{q_j}-k(q_{j+1}-2q_j + q_{j-1}) = 0$$

my question is how do we get this equation of motion its a discrete derivative but I recognize this as the 3 point second derivative formula, but Lagrange formulation is only first order for the potential part

$$f'' = \frac{f(x+h)-2f(x)+f(x-h)}{h^2}$$

any help?

 

[Kurdt]

If you remember from classical mechanics, the equations of motion formed from the Lagrangian will always give you $n$ second order differential equations for a system with $n$ degrees of freedom.

$$\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right) = \frac{\partial L}{\partial q_i}$$

 

[Phymath]

I agree with you on that however the second order only appears on the

$$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q_i}})$$

the potential term is only first order so how do I evaluate

$$\frac{\partial L}{\partial q_i}$$ to give

$$\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{j}-q_{j+1})^{2}) = k(q_{j+1}-2q_j + q_{j-1})$$

 

[George Jones]

how do I evaluate

$$\frac{\partial L}{\partial q_i}$$ to give

$$\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{j}-q_{j+1})^{2}) = k(q_{j+1}-2q_j + q_{j-1})$$

What do you get for

$$\frac{\partial L}{\partial q_i}$$

when

$$L = \sum_j \left[\frac{m}{2}\dot{q}_{j}^{2}-\frac{k}{2} \left(q_{j}-q_{j+1} \right)^{2} \right] ?$$

 

[Phymath]

i don't know that's exactly what I'm asking...

$$<br /> \frac{\partial L}{\partial q_j} = -k(q_j-q_{j+1}) (1-\frac{\partial}{\partial q_j}q_{j+1})<br />$$

 

[badphysicist]

There are two terms that have $$q_j$$ in them because of the sum. You are taking the derivative of:
$$\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{j}-q_{j+1})^{2}-\frac{k}{2}(q_{j-1}-q_{j})^{2})\equiv -k(q_{j}-q_{j+1})+k(q_{j-1}-q_{j}) \equiv k(q_{j-1}-2q_{j}+q_{j+1})$$.
I hope that helps

 

[George Jones]

i don't know that's exactly what I'm asking...

$$<br /> \frac{\partial L}{\partial q_j} = -k(q_j-q_{j+1}) (1-\frac{\partial}{\partial q_j}q_{j+1})<br />$$

Careful; in my post the derivative was with respect to $i$ and the sum is over $j$. Let me expand on what badphysicist wrote. Use

$$\frac{\partial q_j}{\partial q_i} = \delta_{ij}.$$

For example, $\partial x / \partial x = 1$ and $\partial y / \partial x = 0.$

After differentiating, there is still a sum over $j$, i.e.,

$$\frac{\partial}{\partial q_i} \sum_j = \sum_j \frac{\partial}{\partial q_i},$$

but, because of the $\delta$'s, all but a few terms will be zero.

 

[badphysicist]

Oh yeah, I see the typo I made. Here's the corrected version..
$$\frac{\partial}{\partial q_i}(-\frac{k}{2}(q_{i}-q_{i+1})^{2}-\frac{k}{2}(q_{i-1}-q_{i})^{2})\equiv -k(q_{i}-q_{i+1})+k(q_{i-1}-q_{j}) \equiv k(q_{i-1}-2q_{i}+q_{i+1})$$

 

[Phymath]

thanks that definitely helps I didn't think of the sum thanks again