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:

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


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

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.
  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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

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

  • 2
    $\begingroup$ +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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