My text shows the following operation. Where did this come from? where did the q[sub]k[/sub] come from? What kind of math course covers this stuff, so far I’ve been able to follow this kind of stuff but this one tottaly stumps me.
S = sigma
d = partial symbol except for the first d/dt
qdot = the time rate of change of q
Oops that should be
d/dt(dx[sub]I[/sub]/dq[sub]j[/sub]) = S[sub]k[/sub] (d[sup]2[/sup]x[sub]i[/sub]/dq[sub]k[/sub]dq[sub]j[/sub])qdot[sub]k[/sub] + d[sup]2[/sup]x[sub]I[/sub]/(dq[sub]j[/sub]dt)
The memory of the physics degree that I took is a little hazy. However, I seem to remember that:
d/dt(dxI/dqI) means the second derivative of dxI/dqI. dxI/dqI means the differential of x and q whilst holding I constant. It is just notation and is used primarily to describe a 3D surface (whereas ‘normal’ differentiation is usually just a line). Also, I seem to remember that a partial differential is written with a squiggly d (can’t write it here) to differentiate it from ‘normal’ differentiation.
The formula above seems to resemble something called the ‘chain rule’ which is used when working out partial differentials but, I believe, is also used in regular differentials.
I did A-Level maths in the UK and did not come across partial differentiation until i did my physics degree. A-Level maths is probably the most advanced course in maths available in the world (because the Brits only specialise in three/four subjects from the ages of 16 to 18). I imagine that you would need to be on a fairly advanced maths course to be taught about this (at least degree level). What stage are you at?
As I said, I remember very little about this so I am expecting somebody to come along and tell me that I am completely wrong.
Let y=dxi/dqj, and ’ denote derivative wrt t and dt denote partial derivative.
Then we are saying y’=(dy/dq1)q1+…+dy/dt.
Hopefully someone who’s done this more recently’ll come along, but it looks to me like y is a function of q1 etc and t, so the derivative wrt t is the standard partial derivative of a function with serveral parameters, ie the sum of the partial derivative wrt each parameter (q1 and t) times the derivative of that paramter wrt t.
Does that make sense? You should have covered this somewhere before coming across something like that.
You can use the symbol tag to make things like [symbol]S, ¶[/symbol]. Just bracket the symbols with [**symbol] and [/**symbol] and use the appropriate codes between the tags. The partial differentiation symbol, for example, is alt+0182 (That is, press and hold the alt key while pressing 0,1,8, and 4 in succession on the keypad. See this thread for other codes.
This is the Chain Rule. The form being used here is ( differentiability conditions not explicitly stated to avoid muddying the issue):
Suppose f is a function of the n variables x[sub]1[/sub], …, x[sub]n[/sub] and that each x is a function of t. Then
df/dt = [symbol]S¶[/symbol]f/[symbol]¶[/symbol]x[sub]i[/sub][symbol]´[/symbol]dx[sub]i[/sub]/dt
To take an example, suppose
f(x,y,z) = x + 2y + 3z and that
x = t, y = t[sup]2[/sup], z = t[sup]3[/sup].
Then the Chin Rule says that
df/dt = 1.1 + 2.2t + 3.3t[sup]2[/sup] = 1 + 4t + 9t[sup]2[/sup]
and this is easily checked by substituting explicitly for x, y and z and then differentiating.
The case you have presented is where f = [symbol]¶[/symbol]x[sub]I[/sub]/[symbol]¶[/symbol]q[sub]I[/sub].
Thanks Jabba but I guess my question is more in line with interpreting the notation than it is the actual math. (I’m pretty familiar with the chain rule)
Why is it that when you differentiate the expression wrt to t you wind up with a term that has both the original q[sub]j[/sub] and an added q[sub]k[/sub] in the denominator of the differential? Why isn’t it just [symbol]¶[/symbol]q[sub]j[/sub][sup]2[/sup]?
I know that the second partial derivative (f[sub]xy[/sub]) of a function of two variables has dxdy in the denominator of the differential but in the this case j is just an indices. Why is an extra k indices needed? Also why is the partial wrt t added to equation.
OK, d/dt tells you how the system actually changes with time. Part of that time dependence will be direct, or explicit: That’s the part covered by the [symbol]¶[/symol]/[symbol]¶[/symbol]t . But the thing you’re differentiating might also depend on, say, q[sub]j[/sub], and maybe that depends on time. So you need a term for that, too. Likewise, a term for every variable that your differand depends on.
Okay, I was slightly sloppy above and did not notice the final term (!) but this is actually included in what I posted above ( I should just have stated it more explicitly). To simplify matters I shall work with 3 spatial and 1 time dimension. If you grasp this the extension to the general case is trivial ( I bet you hear that a lot in lectures ).
Our system is defined in terms of 3 spatial coordinates x,y,z and time t. We might be interested in one of the derivatives, say [symbol]¶[/symbol]x/[symbol]¶[/symbol]y. In particular, we may be interested in how this function varies with time t. Since x depends on the 4 available variables, so does [symbol]¶[/symbol]x/[symbol]¶[/symbol]y, so we write
[symbol]¶[/symbol]x/[symbol]¶[/symbol]y = [symbol]¶[/symbol]x/[symbol]¶[/symbol]y (x,y,z,t).
We want to differentiate this with respect to t ( and we are taking the full derivative, not the partial one). The chain rule says: differentiate [symbol]¶[/symbol]x/[symbol]¶[/symbol]y with respect to each of the 4 independent variables, multiply each of these partial derivatives by the corresponding full derivative and sum ( you said you understand this bit).
The ( partial) derivative of [symbol]¶[/symbol]x/[symbol]¶[/symbol]y with respect to x is [symbol]¶[/symbol][sup]2[/sup]x/[symbol]¶[/symbol]x[symbol]¶[/symbol]y. We multiply this by the full derivative of x with respect to t i.e. x-dot.
We do the same for the other 2 spatial dimensions: this gives us the term
[symbol]S[/symbol][symbol]¶[/symbol][sup]2[/sup]x/[symbol]¶[/symbol]x[sub]i[/sub][symbol]´[/symbol]x[sub]j[/sub]-dot.
For the final term, note thet dt/dt = 1, so applying the above technique to the variable t gives [symbol]¶[/symbol][sup]2[/sup]x/[symbol]¶[/symbol]y[symbol]¶[/symbol]t.
As for the interpretation: ask the philosophers, I’m just a humble (?) mathematician. My take is: to see how much a variable depends on t, see how much the separate variables depend on t ( the full derivatives) and weight these according to how much the original function depends on each variable ( the partial derivatives).
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