Timeline for Why does elastic energy only depend on first derivatives?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2016 at 9:32 | vote | accept | Jules | ||
| Apr 5, 2012 at 13:45 | comment | added | Zo the Relativist | @Jules: Think about it in terms of a discretization approximation--the value of a function (obviously) only depends on the value of the function at a point. The first derivative depends on the value of the function at a point, and the value of the function at it's neighboring point, so that you can calculate $\frac{\Delta f}{\Delta x}$. The second derivative depends on the derivative here, and the derivative at the neighboring point, which needs three points, and so on. Higher order derivatives are necessarily less local, so if your disturbances are small, they are less important. | |
| Apr 5, 2012 at 12:19 | history | edited | leftaroundabout | CC BY-SA 3.0 |
added 1108 characters in body
|
| Apr 5, 2012 at 11:41 | comment | added | Jules | [As an example of reason (1), take $u_{i,j}^4$. This does not appear because we assumed that $u$ is small, so this will be negligible compared to others. As an example of reason (2), take $u_i$. Its coefficient is zero because of the requirement that the energy is translation invariant.] | |
| Apr 5, 2012 at 11:38 | comment | added | Jules | Sure, but you could say exactly the same about comparing $\Delta x^5$ with $\Delta x^4$. The point is that the coefficient in $F$ is just a constant. So asymptotically you can still compare them. As far as I can see there can be two reasons why $u_{i,jk}$ does not appear in $F$: (1) it is asymptotically smaller than the other derivatives because of some assumption or (2) its coefficient in $F$ is zero for some reason. My question is: which is it, and why? | |
| Apr 5, 2012 at 11:23 | comment | added | leftaroundabout | Well, there is no reason for this, because it isn't true! It's actually not even possible to compare those derivatives directly, since they have different physical dimensions. Only together with the coefficients this is meaningful. | |
| Apr 5, 2012 at 10:51 | comment | added | Jules | Right! But the reason that we can ignore the later terms in a Taylor expansion is that each derivative gets multiplied by a power of $\Delta x$, which is small. It's not that the derivatives themselves are small. What is the reason that $u_{i,jk} \ll u_{i,j}$? | |
| Apr 5, 2012 at 10:46 | comment | added | leftaroundabout | It's pretty much equivalent: one way to define the derivatives is to consider the closest-matching polynomial function, and interpret this as a Taylor series. | |
| Apr 5, 2012 at 10:20 | comment | added | Jules | I understand how this implies that we don't see higher powers of terms like $u_{i,j}^5$ (because this is small compared to, say $u_{i,j}^4$), but I don't understand how this implies that we don't see higher derivatives, like $u_{i,jk}$. Can you elaborate on this? | |
| Apr 5, 2012 at 9:58 | history | answered | leftaroundabout | CC BY-SA 3.0 |