# Lagrangian mechanics, infinitesimal movements expressed in coordinate basis

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.
• As a note I would suggest learning Lagrangian mechanics from Goldstein's book if you want rigorous treatments :) Jun 13, 2020 at 11:08

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.

• 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". Jun 10, 2020 at 18:33
• 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. Jun 10, 2020 at 18:37
• @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 $\delta \mathbf r$.
– user258881
Jun 10, 2020 at 18:37
• @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.
– user258881
Jun 10, 2020 at 18:43
• 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 $\delta d\mathbf{r} = d \delta \mathbf{r}$. 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

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