Derivation of Euler Lagrange's equations from D'alemberts principle

  • #1
In the derivation given in Goldstein's book it is given
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I can't understand from where it comes. It's not at all trivial for me but it's presented as if it's trivial.
 

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  • #2
vanhees71
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The point is that you introduce a set of generalized configuration-space coordinates, ##q_j##, describing the position of your particles as
$$\vec{r}_i=\vec{r}_i(q).$$
Then you have
$$\vec{v}_i=\frac{\mathrm{d} \vec{r}_i}{\mathrm{d} t} = \sum_j \frac{\partial \vec{r}}{\partial q_j} \dot{q}_j.$$
Now you have to know that in the Lagrangian version of the Hamilton principle you consider the space ##(q_j,\dot{q}_j)## with the ##q_j## and ##\dot{q}_j## as independent (!) variables, i.e., whenever you write down a partial derivative with respect to ##q_j## or the ##\dot{q}_j## you consider these variables as the independent variables, and the partial derivative means in taking that derivative you consider all variables fixed except the one with respect to which you differentiate. Then from the above formula, it's immediately clear that
$$\frac{\partial \vec{v}_i}{\partial \dot{q}_j}=\frac{\partial \vec{r}_i}{\partial q_j}.$$
 
  • #3
The point is that you introduce a set of generalized configuration-space coordinates, ##q_j##, describing the position of your particles as
$$\vec{r}_i=\vec{r}_i(q).$$
Then you have
$$\vec{v}_i=\frac{\mathrm{d} \vec{r}_i}{\mathrm{d} t} = \sum_j \frac{\partial \vec{r}}{\partial q_j} \dot{q}_j.$$
Now you have to know that in the Lagrangian version of the Hamilton principle you consider the space ##(q_j,\dot{q}_j)## with the ##q_j## and ##\dot{q}_j## as independent (!) variables, i.e., whenever you write down a partial derivative with respect to ##q_j## or the ##\dot{q}_j## you consider these variables as the independent variables, and the partial derivative means in taking that derivative you consider all variables fixed except the one with respect to which you differentiate. Then from the above formula, it's immediately clear that
$$\frac{\partial \vec{v}_i}{\partial \dot{q}_j}=\frac{\partial \vec{r}_i}{\partial q_j}.$$
Sorry but there are a lot of dollar signs and hashtags in your post . I'm not able to read it properly.
 
  • #4
vanhees71
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Then, something's wrong with your browser. You should see formulae instead of the LaTeX source code.
 
  • #5
Then, something's wrong with your browser. You should see formulae instead of the LaTeX source code.
I was using physics forum app on my android phone. When I opened your reply in chrome browser it showed proper notations and not the source code. Thanks for your answer. The independence of q and q dots is the key here (?). It'll take some time to sink in. Thanks again.
 
  • #6
vanhees71
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Yes, the only problem is that I don't know of any textbook that mentions this definition, and when I've first seen the Hamiltonian principle, I stumbled over the very same problem. I don't know, why no textbook writer thinks that you just should tell your readers your definitions properly! :-((. Don't worry, you'll get used to that soon.
 
  • #7
Yes, the only problem is that I don't know of any textbook that mentions this definition, and when I've first seen the Hamiltonian principle, I stumbled over the very same problem. I don't know, why no textbook writer thinks that you just should tell your readers your definitions properly! :-((. Don't worry, you'll get used to that soon.
[emoji4] i hope so
 
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