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In a certain textbook a function is given as:


And then this is differentiated w.r.t. $t$ to get:


(Where the notation $f_u=df/du$, $f_{uu}=d^2f/du^2$, etc.)

This is then taken as a functional $A=A(x,\dot{x},t)=\dot{x}f_x$ and differentiated w.r.t. $x$ and $\dot{x}$ and set to zero:

$$A_{\dot{x}}=f_x=0$$ $$A_x=\dot{x}f_{xx}=0$$

My concern is that in doing this the textbook has not completely differentiated $A_{\dot{x}}$ and $A_x$ In particular it has ignored the derivatives $\frac{dx}{d\dot{x}}$ and $\frac{d\dot{x}}{dx}$ If I'm not mistaken the complete differentiation would be:

$$A_{\dot{x}}=f_x+\frac{dx}{d\dot{x}}\dot{x}f_{xx}=0$$ $$A_x=\dot{x}f_{xx}+\frac{d\dot{x}}{dx}f_x=0$$

Multiplying the first of these equations through by $\frac{d\dot{x}}{dx}$ or the second equation by $\frac{dx}{d\dot{x}}$ you get the relation:


and $$\frac{A_x}{A_{\dot{x}}}=\frac{d\ln{\dot{x}}}{dt}$$

Whereas doing it the book's way you get $$\frac{d\ln{f_x}}{dt}=1$$

and $$\frac{A_x}{A_{\dot{x}}}=\frac{d\ln{f_x}}{dt}$$

By ignoring the derivatives of $x$ and $\dot{x}$ w.r.t. each other, then, the book is tacitly assuming that $$-\frac{d\ln{\dot{x}}}{dt}=1$$

EDIT: (and hence that $\frac{A_x}{A_{\dot{x}}}=1$, which introduces a contradiction b/c we also know from the fully executed derivative that $\frac{A_x}{A_{\dot{x}}}=\frac{d\ln{\dot{x}}}{dt}$ which by the said tacit assumption $=-1$)

My question is: is this a safe assumption? What is the physical meaning of this assumption?

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I remember I had this question myself, I couldn't resolve it strictly, but convinced $x$ $\dot x$ are somehow independent. I've tried searching MSE, but couldn't find this question. Maybe you should ask it there, it is really interesting and common question. – Yrogirg Aug 28 '12 at 4:47
@Yrogirg: What is MSE? Please give me the link. – ben Aug 28 '12 at 4:53
Math StackExchange – Yrogirg Aug 28 '12 at 6:15
Comment to the question (v5). Semantically speaking, $A(x,v)=vf^{\prime}(x)$ is a function, not a functional. The main question seems to be if position $x$ and velocity $v$ are independent variables or not, which has also been discussed in this post, and questions linked to it. – Qmechanic Aug 28 '12 at 9:36
@Qmechanic: Thanks for the link to that question. Very helpful discussion there. It does not address one of my main concerns, however, which I mention in my reply to Doug Packard below: the derivative of velocity w.r.t. position is actually logarithmic differentiation of velocity w.r.t. time. – ben Aug 28 '12 at 12:09

One way to see that considering the dependence of $\dot{x}$ on $x$ is problematic is as follows:

$x(t)$ maps a real number $t$ to another real number $x$. So $\dot{x}=dx/dt$ is the derivative of that map, meaning we take $$\lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}$$

So we can see that $dx/dt$ is itself another map from a real number $t$ to a real number $\dot{x}(t)$ ($t$ is the independent variable in the above expression). Now, if $x(t)$ happens to be invertible, we could find a unique map $t(x)$ such that $x(t(x))=x$, and then define $\dot{x}(x) \equiv \dot{x}(t(x))$, and conceive of taking a derivative w.r.t. x, but as we know most functions are not invertible (e.g. $x^2$ does not have an inverse that covers its whole domain, since $x^2$ could get mapped back to either $x$ or $-x$). So things aren't that simple.

The thing is, that when you're dealing with functions of a single variable, you have to be careful about only having one independent variable at a time, even if we're using shorthand for other functions that depend on that independent variable. If we don't do this we will make calculus mistakes. For example, you say $A=A(x,\dot{x},t)=\dot{x}\frac{df}{dx}$, and that is slightly incorrect because you are considering it to be a "functional" instead of the single-variable function $df/dt$, which are actually two different objects. They are different maps (one of two variables, and the other of just one), even though you can write them the same way. In the "functional" case (I'm using scare-quotes because, to me, functional means "map from entire function to a real number"), we simply are declaring $\dot{x}$ and $x$ to be separately independent variables, and asking what real number $A$ they map to (since $df/dx$ is a function of $x$). So it doesn't make sense, within this context, to have $d\dot{x}/dx$ be anything other than zero, unless we have some additional constraint.

Just think of a regular function of two variables $f(x,y)$. Then $dx/dy = 0$, unless we are explicitly making $y$ depend on $x$, such as if we were taking a derivative along some curve in the $x$-$y$ plane, and we can recognize that in this case the chain rule is the same as the formula for a directional derivative (up to normalization).

Hope that helps.

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Ok, but as I allude to in my post, consider that $$ \frac{d \dot{x}}{dx}=\frac{d \dot{x}}{dt} \frac{dt}{dx}=\frac{d\ln{\dot{x}}}{dt} $$ In this sense the derivative of velocity w.r.t. position is in fact a derivative w.r.t. time. Mulitply by $t$ and you get the normalized acceleration. The logarithm restricts $\dot{x}$ to positive values, which I guess implies what you were saying that $t(x)$ is not the inversion of $x(t)$ over the whole domain (only the positive values), but so what? – ben Aug 28 '12 at 12:17
Note also, for that matter, that $$\frac{d\dot{x}}{dx}=\frac{\ddot{x}}{\dot{x}} $$ So if you're going to assume that $ \frac{d\dot{x}}{dx}=0 $ or simply ignore it, then to be consistent you should also ignore $ \frac{\ddot{x}}{\dot{x}} $ wherever it appears. – ben Aug 28 '12 at 13:20

Thinking physically it's not correct to (generically) treat x as a function(al?) of v (velocity). I'm 30 feet above London along the line segment connecting the center of Earth and the Sun; what's my velocity? You might think okay i'll assume he's in free fall and just do some freshman physics to calculate the result. However it's perfectly allowable for me to have a jetpack or I could even be in a building (whether or not a building exists at a specific spatial location appears to be time dependent).

Now from a practical/mathematical standpoint it is sometimes alright to think of things as being functions of other things even though this doesn't universally hold. For instance in the case of free fall under 100% ideal conditions the trajectory is absolutely certain if I tell you how long it's been since I was dropped you could tell me precisely what my velocity is. Or if I told you my position you know how fast i'm falling. Or if I tell you my velocity you know what the radial locations of the hands of my watch are (remember, 100% ideal).

So in this case we can be really sloppy at treat t(x), v(x), x(v), X(t). However this is because we're are dealing with a special case. In essence we have a nice graph which is just a certain very rigidly defined parabola with a definite start a and definite end b. It's a matter of convenience whether or not you choose to think in terms of positions a and b, or times t(a), t(b), or velocities v(a), v(b), etc.

But note that for obvious reasons we don't have an infinite mess of chain rule when we take derivatives. For an example like this velocity, position, and time are in a certain sense different "bases" or different representations of the same curve. I mean by this that each of the graphs of v vs t, x vs t, x vs v, and all inversions thereof (just rotating the graph by 90%) are well defined.

You certainly cannot treat generic positions as well-defined (in the sense of there being a rigid x vs v graph) functions of velocity though. (Remember Galileo!)

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For a particular path $x(t)$, $x$ and $\dot x$ aren't independent of one another because they both depend upon t. On the other hand, you can change the path $x(t)$ to another path $x'(t)$ so that for the same t, the instantaneous velocity and position for one path is different to the other. In this sense for a fixed time t, you can therefore vary the instantaneous position and velocity independently of one another to define a new path at the same t.

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$x'$ is the best notation you could came up with? :) – NikolajK Aug 28 '12 at 21:44
@NickKidman yes :) unless you think it might be confusing for some so I'm open to suggestions. – John McVirgo Aug 28 '12 at 23:51

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