Is the d'Alambert principle universal?

In summary, the professor derived a formula for the D'Alambert principle in a lecture on classical mechanics, using the Taylor's formula. He explained that the element \delta \vec{r} is infinitesimal and that \vec{r} , \vec{r} + \delta \vec{r} satisfy the restraints. The first term on the right side of the formula is cancelled out, leaving only the higher order terms. The professor stated that this is because the gradient of \Phi_{\alpha} must be perpendicular to \delta \vec{r}. This is proved by the fact that \Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\
  • #1
r4nd0m
96
1
In a lecture on classical mechanics, the professor derived a formula, which is a part of the D'Alambert principle: [tex]\nabla \Phi_{\alpha} \cdot \delta \vec{r} = 0[/tex] where [tex]\Phi_{\alpha}[/tex] are the restraints. He derived it in a strange way from the Taylor's formula:
[tex]\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r}) + \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...[/tex]
He said that the element [tex]\delta \vec{r}[/tex] is infinitesimal and that [tex]\vec{r} , \vec{r} + \delta \vec{r}[/tex] 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:
[tex] 0 = \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...[/tex].
And here comes the critical point. He said that since [tex] \delta \vec{r}[/tex] is very small [tex] \delta \vec{r} \cdot \nabla \Phi_{\alpha} = 0[/tex]
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.
 
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  • #2
r4nd0m said:
In a lecture on classical mechanics, the professor derived a formula, which is a part of the D'Alambert principle: [tex]\nabla \Phi_{\alpha} \cdot \delta \vec{r} = 0[/tex] where [tex]\Phi_{\alpha}[/tex] are the restraints. He derived it in a strange way from the Taylor's formula:
[tex]\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r}) + \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...[/tex]
He said that the element [tex]\delta \vec{r}[/tex] is infinitesimal and that [tex]\vec{r} , \vec{r} + \delta \vec{r}[/tex] 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:
[tex] 0 = \delta \vec{r} \cdot \nabla \Phi_{\alpha} + ...[/tex].
And here comes the critical point. He said that since [tex] \delta \vec{r}[/tex] is very small [tex] \delta \vec{r} \cdot \nabla \Phi_{\alpha} = 0[/tex]
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
[tex]\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r}) [/tex]
which justifies the first cancellation of terms. If that is true, then the gradient of [tex]\Phi_{\alpha}[/tex] must be perpendicular to [tex]\delta \vec{r} [/tex] which makes the dot product vanish because of orthogonality. The higher order terms are all that is left.
 
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  • #3
If that is true, then the gradient of [tex]\Phi_{\alpha}[/tex]must be perpendicular to [tex]\delta \vec{r} [/tex]

Why? Can you prove it?
 
  • #4
r4nd0m said:
Why? Can you prove it?
Yes. It is proved by [tex]\Phi_{\alpha} (\vec{r} + \delta \vec{r}) = \Phi_{\alpha} (\vec{r})[/tex]
The gradient is always perpendicular to a hypersurface of constant value of a scalar. If the allowed displacement does not change the value of [tex]\Phi_{\alpha}[/tex], 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.
 

1. What is the d'Alambert principle?

The d'Alambert principle, also known as the principle of virtual work, is a fundamental principle in classical mechanics that states that the sum of the external forces acting on a system is equal to the product of the mass of the system and its acceleration.

2. Is the d'Alambert principle universal?

No, the d'Alambert principle is not universal. It is limited to systems that have a finite number of degrees of freedom and are in a state of mechanical equilibrium.

3. How is the d'Alambert principle applied in science?

The d'Alambert principle is commonly used in the field of classical mechanics to solve problems related to the motion of particles and rigid bodies. It is also used in other branches of science, such as structural engineering and fluid mechanics, to analyze the behavior of systems under external forces.

4. Are there any limitations to the d'Alambert principle?

Yes, the d'Alambert principle has some limitations. It cannot be applied to systems that are undergoing non-conservative forces, such as friction. It also does not take into account the effects of rotational motion.

5. Can the d'Alambert principle be derived from other principles?

Yes, the d'Alambert principle can be derived from the principle of least action, which states that the path taken by a system between two points is the one that minimizes the action integral. It can also be derived from Newton's second law of motion, F=ma, under certain conditions.

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