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In this page explaining the foundation of the Lagrangian mechanics, we can see that starting at:

$$ m \left[ \frac {d} {dt} \left( \frac {d \mathbf{r}} {dt} \delta \mathbf{r} \right) - \frac {d \mathbf{r}} {dt} \frac {d \delta \mathbf{r}} {dt} \right] $$

it is possible to obtain:

$$ m \sum_{k} \left[ \frac {d} {dt} \left( \frac {d \mathbf{r}} {dt} \frac{\partial \mathbf{r}}{\partial r_k }\delta r_k\right) - \frac {d \mathbf{r}} {dt} \frac{\partial \dot{\mathbf{r}}}{\partial r_k } \delta r_k \right] $$

where I suppose the following equalities have been applied:

$$ \delta \mathbf{r} = \sum_{k} \frac{\partial \mathbf{r}}{\partial r_k }\delta r_k $$

and

$$ \frac {d \delta \mathbf{r}} {dt} = \sum_{k} \frac{\partial \dot{\mathbf{r}}}{\partial r_k } \delta r_k. $$

could some one help me to clarify these replacements?

Clarification of the doubts, after the useful answer given by @FakeMod:

First, what are the $r_k$ that appears in these expression ? How they relate to $\mathbf{r}$ and/or $\delta \mathbf{r}$ ?

About $\delta \mathbf{r}$, the text says:

an arbitrary, infinitesimal displacement of the position of our particle under the net force 
[...]
This statement is known as d'Alembert's principle and is the jumpoff point for our efforts. 

When talking about d'Alembert's principle, wikipedia defines $\delta \mathbf{r}$ it as:

is the virtual displacement of the particle, consistent with the constraints.  

note nowhere it is said that $\delta \mathbf{r}$ was the real displacement done by the particle.

  • we can say that $d \delta \mathbf{r} = \delta d \mathbf{r} $ ?
  • meaning of $d \, ( \delta \mathbf{r} )$ or meaning of $\frac{ d }{dt}\delta \mathbf{r}$, if necessary in these expressions.
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    $\begingroup$ As a note I would suggest learning Lagrangian mechanics from Goldstein's book if you want rigorous treatments :) $\endgroup$
    – TaeNyFan
    Jun 13, 2020 at 11:08

2 Answers 2

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The first equation

$$\delta \mathbf{r} = \sum_{k} \frac{\partial \mathbf{r}}{\partial r_k }\delta r_k\tag{1}$$

is just a consequence of chain rule for multivariable functions. If you check the linked Wikipedia page, you will get to know that we can write the following equation for a function $f$ which depends on the variables $a_1,a_2,a_3,\dots,a_n$

$$\mathrm df(a_1,a_2,a_3,\dots,a-n)=\sum_i^n \left(\frac{\partial f}{\partial a_i}\right)\mathrm d a_i\tag{2}$$

This can be generalized to vectors as well, and thus used to derive equation $(1)$.

The derivation of second equation in your question

$$\frac {\mathrm d (\delta \mathbf{r})} {\mathrm d t} = \sum_{k} \frac{\partial \dot{\mathbf{r}}}{\partial r_k } \delta r_k.\tag{3}$$

is a bit more subtle. Here, you can rewrite $\displaystyle \frac{\mathrm d (\delta \mathbf{r})}{\mathrm dt}$ as $\displaystyle \delta \left(\frac{\mathrm d \mathbf r}{\mathrm d t}\right)$ which is equivalent to $\delta \dot{\mathbf r}$. Now again applying the chain rule (see equation $(1)$ and $(2)$), we can prove equation $(3)$ as well.

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  • $\begingroup$ Are you so kind of clarify the phrase "This can be generalized to vectors as well" ?. Taken into account what is in this text $\delta \mathbf{r}$: "an arbitrary, infinitesimal displacement of the position of our particle". $\endgroup$ Jun 10, 2020 at 18:33
  • $\begingroup$ About second equation, if we pass from $d (\delta \mathbf{r} )$ to $\delta (d \mathbf{r} )$, we make a relation between the infinitesimal displacement with the object position, two concepts that I do not see directly linked (in fact, I wondering why the author writes $\delta \mathbf{r}$ and not $\delta \mathbf{u}$ to make a clear difference between the displacement and the position. $\endgroup$ Jun 10, 2020 at 18:37
  • $\begingroup$ @pasabaporaqui Yes, sure. I meant that in the equation $(2)$, you could also use a vector function $f$ (which spits out vectors) and still the equation will hold true. In your case, the $\mathrm df$ is analogous to $\delta \mathbf r$. $\endgroup$
    – user258881
    Jun 10, 2020 at 18:37
  • $\begingroup$ @pasabaporaqui $$\mathrm d (\delta \mathbf r)=\delta(\mathbf r+\mathrm d \mathbf r)-\delta(\mathbf r)=\delta(\mathbf r)+\delta(\mathrm d \mathbf r)-\delta(\mathbf r)=\delta(\mathrm d \mathbf r)$$ To be honest, I don't know a more rigorous proof of the above equation. However asking on Mathematics SE might help. $\endgroup$
    – user258881
    Jun 10, 2020 at 18:43
  • $\begingroup$ My main confusion is the meaning of $\mathbf{r}$ in these equations. By example, looking at the first equation of my question, in $d\mathbf{r}$ it is the (differential) of the mass position, while in $\delta \mathbf{r}$ it is a (infinetissimal) displacement in any direction coherent with the constraints. Don't see they are the same ( position and work displacement ), thus, I do not see why we can say $\delta d\mathbf{r} = d \delta \mathbf{r}$. $\endgroup$ Jun 10, 2020 at 18:59
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First, what are the $r_k$ that appears in these expression ? How they relate to $\mathbf{r}$ and/or $\delta \mathbf{r}$ ?

The $r_k$ that appear in the given equations are generalized coordinates. It is related to $\mathbf{r}$ by $$\mathbf{r}=\mathbf{r}(r_1,r_2,...r_k,t).$$ For a system with $k$ degrees of freedom, there will be $k$ generalized coordinates. This can be contrasted with $\mathbf{r}$ expressed in terms of Cartesian coordinates, which is usually given by $$\mathbf{r}=\mathbf{r}(x,y,z,t).$$

$\delta\mathbf{r}$ is the virtual displacement. It is defined as any arbitrary infinitesimal change of the position of the particle, consistent with forces and constraints imposed on the system at the given instant $t$. Compare this with $d\mathbf{r}$, which is as an infinitesimal displacement of the particle in time $dt$, without any regard for the constraint conditions.

The virtual displcement $\delta \mathbf{r}$ is hence given by $$\delta \mathbf{r} = \sum_{k} \frac{\partial \mathbf{r}}{\partial r_k }\delta r_k,$$ where $\delta r_k$ are the virtual displacements of the generalized coordinates. Note that $\delta\mathbf{r}$ is to be evaluated at the given instant t, i.e. $\delta t = 0$, hence giving $$\frac{\partial \mathbf{r}}{\partial t }\delta t=0.$$ This can be contrasted with the infinitesimal displacement $d\mathbf{r}$ which is given by $$d \mathbf{r} = \sum_{k} \frac{\partial \mathbf{r}}{\partial r_k }d r_k + \frac{\partial \mathbf{r}}{\partial t }dt.$$

  • we can say that $d \delta \mathbf{r} = \delta d \mathbf{r} $ ?
  • meaning of $d \, ( \delta \mathbf{r} )$ or meaning of $\frac{ d }{dt}\delta \mathbf{r}$, if necessary in these expressions.

To show that $$ \frac {d \delta \mathbf{r}} {dt} = \sum_{k} \frac{\partial \dot{\mathbf{r}}}{\partial r_k } \delta r_k, $$ start with the definition for $\delta\mathbf{r}$ given by$$\delta \mathbf{r} = \sum_{k} \frac{\partial \mathbf{r}}{\partial r_k }\delta r_k.$$ Differentiating both sides with respect to $t$, we get $${d\delta \mathbf{r} \over dt} = \sum_{k} (\frac{d}{dt}\frac{\partial \mathbf{r}}{\partial r_k } )\delta r_k +\frac{\partial \mathbf{r}}{\partial r_k } \frac{d\delta r_k}{dt}. $$ Since the vitual displacement $\delta r_k$ does not depend on time $t$, we have $\frac{d\delta r_k}{dt}=0$. This simplifies the RHS to$${d\delta \mathbf{r} \over dt} = \sum_{k} (\frac{d}{dt}\frac{\partial \mathbf{r}}{\partial r_k } )\delta r_k.$$ Next to evaluate$\frac{d}{dt}\frac{\partial \mathbf{r}}{\partial r_k } $, we note that $\frac{\partial \mathbf{r}}{\partial r_k }$ is a function given by $$\frac{\partial \mathbf{r}}{\partial r_k } = \frac{\partial \mathbf{r}}{\partial r_k }(r_1,r_2,...,r_{k'},t).$$ The corresponding differential equation is hence given by $$d(\frac{\partial \mathbf{r}}{\partial r_k })=\sum_{k'} \frac{\partial(\frac{\partial \mathbf{r}}{\partial r_k })}{\partial r_{k'}}dr_{k'} + \frac{\partial(\frac{\partial \mathbf{r}}{\partial r_k })}{\partial t}dt.$$ Dividing throughout by $dt$ gives $${d(\frac{\partial \mathbf{r}}{\partial r_k })\over dt}=\sum_{k'} \frac{\partial(\frac{\partial \mathbf{r}}{\partial r_{k'} })}{\partial k'}{dr_{k'}\over dt} + \frac{\partial(\frac{\partial \mathbf{r}}{\partial r_k })}{\partial t}\frac{dt}{dt}=\frac{\partial(\frac{\partial \mathbf{r}}{\partial r_k })}{\partial t} =\frac{\partial \dot{\mathbf{r}}}{\partial r_k }, $$ making use of ${dr_{k'}\over dt} =0$ because the generalized coordinates $r_k'$ are independent of time $t$.

We hence obtain $${d\delta \mathbf{r} \over dt} = \sum_{k} (\frac{d}{dt}\frac{\partial \mathbf{r}}{\partial r_k } )\delta r_k = \sum_{k} \frac{\partial \dot{\mathbf{r}}}{\partial r_k } \delta r_k$$ as required.

References: Goldstein, Classical Mechanics, Chapter 1

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    $\begingroup$ +1. Great answer. Thanks for the solely needed proof of the required identity. $\endgroup$
    – user258881
    Jun 13, 2020 at 10:50
  • $\begingroup$ In generalized coordinates, we must say $\mathbf{r}=r(q_1,...,q_k,t)$ or $\mathbf{r}=r(q1(t),...,qk(t))$ or $\mathbf{r}=( x(q1,...,qk,t)), y(q1,...,qk,t)), z(q1,...,qk,t)) )$ ? $\endgroup$ Jun 30, 2020 at 19:22
  • $\begingroup$ @pasaba The first one. $\endgroup$
    – TaeNyFan
    Jul 1, 2020 at 3:23

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