# How to add physics to the transformation of generalised coordinates

I've been introduced to the conversion between generalized coordinates, where $x_j = f(q_1,q_2,q_3,t_q)$ a scalar function where we are converting from $(q_1,q_2,q_3,t_q)$ to $(x_1,x_2,x_3,t_x)$.

Because $q_1,q_2,q_3$ can be written as functions of $t_q$ we have the derivative

$$\dot{\mathbf{x}}_{j}=\sum_{k}\frac{\partial\mathbf{x}_{j}}{\partial q_{k}}\dot{q}_k+\frac{\partial\mathbf{x}_{j}}{\partial t_q}$$

for $k,j$ ranging from $1$ to $3$. There is a similar equation for $t_x$.

There are a couple of things which confuse me. First of all, so far, other than the limits on the number of coordinates there is nothing physical about these equations. The coordinates are essentially supposed to describe position and time but at the moment they describe arbitrary functions of four variables. For example, at this point it seems that $\frac{\partial\mathbf{x}_{j}}{\partial t_q}$ could well be a non zero term. In real life a change in position in one coordinate system must follow a change in position in another, but this final term implies that $x$ can change independent of the position in the other coordinate system. The only counterexample I can find is the galilean transformation.

Another example of physics is that time between coordinate systems can only be shifted by a constant. This will have its own implications.

So I'm trying to justify all these intuitions by introducing laws of physics into this equation. what is the physical significance of the final term $\frac{\partial\mathbf{x}_{j}}{\partial t_q}$? If it is usually zero, then what is the physical reason for this, and if not, where does it come from?

In real life a change in position in one coordinate system must follow a change in position in another [...]

Your view of what constitutes a change in coordinates is too limited, I think. Consider the case of a bead which slides along a frictionless circular hoop of radius $R$, which is being rotated at a constant angular velocity $\omega$. The position of the bead can be described by specifying the coordinates $x(t),y(t)$, and $z(t)$, but such coordinates would be subject to a very irritating constraint.

Instead, the position could be specified by the angle the bead makes from the vertical ($\theta=0$ denoting the bottom of the hoop, $\theta=\pi$ denoting the top) and the azimuthal angle $\phi$ which is changing with time ($\dot \phi = \omega$). Furthermore, because $\omega$ is constant, we are perfectly entitled to switch to the variable $\psi = \phi - \omega t$ which is constant in time ($\dot \psi = 0$).

It should be clear that a bead with a constant $\theta$ and $\psi$-coordinates will generically correspond to non-constant $x$ and $y$-coordinates. In some sense, this is the entire point of changing to a generalized coordinate system.

To be explicit, given $x,y,$ and $z$ (Cartesian coordinates), we can switch to our $\theta,\psi$ coordinates as follows:

$$x = R\sin(\theta)\cos(\psi+\omega t)$$ $$y = R\sin(\theta)\sin(\psi+\omega t)$$ $$z = R(1-\cos(\theta))$$ and $$\theta = \cos^{-1} \left(1-\frac{z}{R}\right)$$ $$\psi = \tan^{-1}\left(\frac{y}{x}\right)-\omega t$$

As you can see, we can let $\theta$ and $\psi$ be constants (where $\theta \neq 0$) and still have $x(t)$ and $y(t)$ which change with time. In general, this situation arises whenever we introduce generalized coordinates with explicit time dependence.

To address the overall theme of the question, the physics is not present in the form of the coordinates themselves, but rather in the equations which govern their evolution in time. Coordinate changes may or may not introduce explicit time dependence to the Lagrangian, but the underlying time evolution of the system will be governed by the Lagrangian equations of motion, which is where the physics actually resides.

• how can the variables be constant when they depend on time? – user86425 Aug 8 '17 at 14:41
• @lucky-guess Consider the simple case of a particle moving in one direction with constant velocity $v$. Define $f(x,t)=x-vt$. The total derivative of $f$ with respect to time is $\frac{df}{dt} = \frac{\partial f}{\partial x} \dot x + \frac{\partial f}{\partial t} = \dot x - v = v - v = 0$, so the value of $f$ is constant despite possessing explicit time dependence. – J. Murray Aug 8 '17 at 14:47
• right but is there a specific example where dq/dt is 0 but the partial of x with respect to time is non zero? – user86425 Aug 8 '17 at 19:50
• @lucky-guess That's exactly what I just wrote. $\frac{df}{dt}=0$ but $\frac{\partial f}{\partial t} = -v$. – J. Murray Aug 8 '17 at 19:52
• Yes, that's also arbitrary. You could scale the time by a constant factor, add a constant to it, raise it to the third power, or anything else you want that would result in valid, invertible change of coordinates. Such changes are seldom useful in my experience, but you're perfectly entitled to make them. That's why they're called generalized coordinates. – J. Murray Aug 9 '17 at 12:09