Understanding Partial Differentiation in Classical Mechanics

[nonequilibrium]

Hello,

I got confused in my Classical Mechanics class (on a mathematical issue). So let $$L$$ denote a function dependent on x and its derivative explicitly, such that its image is $$L(x,x*)$$ (NOTE: I'm using * as the overdot-Leibniz notation for the derivative) and x is a function of t.

To make it easy, I'll give an explicit form $$L(x,x*) = Ax + Bx*$$. Now I found it odd that $$\frac{\partial L}{\partial x} = A$$ instead of $$\frac{\partial L}{\partial x} = A + \frac{\partial x*}{\partial x}$$... What is the reasoning behind this? Is it because L is an explicit function of x* too? Or is it because x* is not explicitly dependent of x? Well it's quite probable that x* is physically dependent on x (only not the case when x is a linear function of degree 1 or 0), but for some reason mathematically it isn't(?)

 

[tiny-tim]

Hello mr. vodka!

Well it's quite probable that x* is physically dependent on x (only not the case when x is a linear function of degree 1 or 0), but for some reason mathematically it isn't(?)

Yup, that's it …

L is specially defined mathematically …

when you differentiate, x and x* are treated as independent variables.

 

[nonequilibrium]

So, for example, in general it's true that

$$\frac{\partial f(x,y,z)}{\partial \frac{dx}{dt}} = 0$$

?

EDIT: Hm, re-reading my post, my question doesn't seem to make a lot of sense: you can't take a partial derivative wrt something that isn't explicitly a variable. By definition of partial derivative: I'm holding x,y,z constant and letting dx/dt variate... I don't know if that is even consistent, but the first part of my sentence would imply that ... = 0. On the other hand, I could also write down $$\frac{\mathrm{d} f(x,y,z)}{\mathrm{d} \frac{dx}{dt}}$$; is this also zero?

EDIT2: In my classical mechanics course, v_i is a function of q_j's. Then it makes a statement about $$\frac{\partial v_i}{\partial \frac{\mathrm d q_j}{\mathrm d t}}$$ Is this even well-defined? A partial derivative is only defined wrt explicit variables...

 

[tiny-tim]

If f isn't a function of ∂x/∂t (as well as of x y and z), that doesn't make much sense.

oh … just seen your edit … you beat me to it!

(and messing about with d instead of ∂ is never gong to happen in practice. )

If v is expressed as a function of qs only, that really means it's a function of qs and dq/dts, but it's constant in the dq/dts.

(i'll post quickly before he does another edit! )

 

[nonequilibrium]

Thanks for the diligent help!

I'm not sure what you mean with "but it's constant in the dq/dts".
Wouldn't that imply that $$\frac{\partial v_i}{\partial \frac{\mathrm d q_j}{\mathrm d t}} = 0$$? However, in my course, there is stated that it's equal to $$\frac{\partial r_i}{\partial q_j}$$ (called dot cancellation, apparently -- I'll assume that by now you've figured out r is the position vector and v the velocity)

 

[tiny-tim]

Now I'm totally confused as to what v is.

 

[nonequilibrium]

My apologies: v is the velocity vector, i.e. $$\vec v_i = \frac{\mathrm d \vec r_i}{\mathrm d t}$$. The q_j's are the (independent) variables on which r_i (and thus v_i) are dependent (along with explicitly/implicitly time t). Note: I left out the vector notation for v and r in previous posts due to laziness.

EDIT: perhaps this is a question best asked in a physics forum?

 

[tiny-tim]

I'm used to seeing ps and qs as the variables …

which are the ps in your notation?

(sorry for the delay … i thought i'd pressed "Submit Reply", but i hadn't )

 

[nonequilibrium]

Hm, did I use p's somewhere? There are no p's, only the q's are the variables.

i.e.:

$$\vec r_i : R^n \to R^3: (q_1,q_2, ...,q_n) \to \vec r_i(q_1,q_2, ...,q_n)$$

And by definition of derivative:

$$\vec v_i : R^n \to R^3: (q_1,q_2, ...,q_n) \to \frac{\mathrm d \vec r_i}{\mathrm d t}(q_1,q_2, ...,q_n)$$