Is the d'Alambert principle universal?

[r4nd0m]

In a lecture on classical mechanics, the professor derived a formula, which is a part of the D'Alambert principle: $$\nabla \Phi_{\alpha} \cdot \delta \vec{r} = 0$$ where $$\Phi_{\alpha}$$ are the restraints. He derived it in a strange way from the Taylor's formula:
$$\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r}) + \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...$$
He said that the element $$\delta \vec{r}$$ is infinitesimal and that $$\vec{r} , \vec{r} + \delta \vec{r}$$ satisfy the restraints.
I understand that the term on the left side of Taylor's formula and the first term on the right side must be therefore equal to zero i.e:
$$0 = \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...$$.
And here comes the critical point. He said that since $$\delta \vec{r}$$ is very small $$\delta \vec{r} \cdot \nabla \Phi_{\alpha} = 0$$
My question is why - there are also the higher terms in the formula. For "very small" elements it might be almost true, but only approximately. So is this principle only approximately true? Can someone derive the last step in a more mathematical way? Thanks a lot.

 

[OlderDan]

In a lecture on classical mechanics, the professor derived a formula, which is a part of the D'Alambert principle: $$\nabla \Phi_{\alpha} \cdot \delta \vec{r} = 0$$ where $$\Phi_{\alpha}$$ are the restraints. He derived it in a strange way from the Taylor's formula:
$$\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r}) + \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...$$
He said that the element $$\delta \vec{r}$$ is infinitesimal and that $$\vec{r} , \vec{r} + \delta \vec{r}$$ satisfy the restraints.
I understand that the term on the left side of Taylor's formula and the first term on the right side must be therefore equal to zero i.e:
$$0 = \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...$$.
And here comes the critical point. He said that since $$\delta \vec{r}$$ is very small $$\delta \vec{r} \cdot \nabla \Phi_{\alpha} = 0$$
My question is why - there are also the higher terms in the formula. For "very small" elements it might be almost true, but only approximately. So is this principle only approximately true? Can someone derive the last step in a more mathematical way? Thanks a lot.

I'm not sure, but here is what comes to mind.

It seems to me you are saying you have no problem believing that
$$\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r})$$
which justifies the first cancellation of terms. If that is true, then the gradient of $$\Phi_{\alpha}$$ must be perpendicular to $$\delta \vec{r}$$ which makes the dot product vanish because of orthogonality. The higher order terms are all that is left.

 

[r4nd0m]

If that is true, then the gradient of $$\Phi_{\alpha}$$must be perpendicular to $$\delta \vec{r}$$

Why? Can you prove it?

 

[OlderDan]

Why? Can you prove it?

Yes. It is proved by $$\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r})$$
The gradient is always perpendicular to a hypersurface of constant value of a scalar. If the allowed displacement does not change the value of $$\Phi_{\alpha}$$, then the gradient must be perpendicular to the displacement. The same argument that leads to the first cancellation of terms, necessarily implies that the first remaining term in the expansion is zero.