I can't find my mistake in using the chain rule here

[etotheipi]

Homework Statement

In S and S' the trajectory of a particle is $(x,t)$ and $(x',t')$ respectively. The coordinates are related by $x'=f(x,t)$ and $t'=t$. Give the most general form of $f$ if S and S' are inertial frames.

Relevant Equations

N/A

I literally don't know what's going wrong today, I can't seem to get anything right . The velocity in S' is easy enough $$v' = \frac{dx'}{dt'} = \frac{\partial f}{\partial t} \frac{\partial t}{\partial t'} + \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}\frac{\partial t}{\partial t'} = \frac{\partial f}{\partial t} + v \frac{\partial f}{\partial x}$$ That means that $v' = v'(x,t)$, and the next derivative should also go like $$a' = \frac{dv'}{dt'} = \frac{\partial v'}{\partial t}\frac{\partial t}{\partial t'} + \frac{\partial v'}{\partial x}\frac{\partial x}{\partial t}\frac{\partial t}{\partial t'}$$ Now $$\frac{\partial v'}{\partial t} = \frac{\partial^2 f}{\partial t^2} + a\frac{\partial f}{\partial x}$$ and $$\frac{\partial v'}{\partial x} = v\frac{\partial^2 f}{\partial x^2} + \frac{\partial v}{\partial x}\frac{\partial f}{\partial x}$$ So that leaves me with $$a' = \frac{\partial^2 f}{\partial t^2} + a\frac{\partial f}{\partial x} + v^2 \frac{\partial^2 f}{\partial x^2} + v\frac{\partial f}{\partial x} \frac{\partial v}{\partial x}$$ I was quite happy with that, but apparently it should actually be $$a' = \frac{\partial^2 f}{\partial t^2} + a\frac{\partial f}{\partial x} + v^2 \frac{\partial^2 f}{\partial x^2} + 2v\frac{\partial^2 f}{\partial x \partial t}$$ Maybe I need to go back to sleep and look at this later? Have I done something stupid? Thanks

 

[DrClaude]

$$\frac{\partial v'}{\partial t} = \frac{\partial^2 f}{\partial t^2} + a\frac{\partial f}{\partial x}$$

You are missing a term here. What is
$$
\frac{\partial}{\partial t} \left( v \frac{\partial f}{\partial x} \right)
$$
?

You are missing a similar term when calculating $\partial v' / \partial x$.

 

[etotheipi]

You are missing a term here. What is
$$
\frac{\partial}{\partial t} \left( v \frac{\partial f}{\partial x} \right)
$$
?

You are missing a similar term when calculating $\partial v' / \partial x$.

I would say $$\frac{\partial}{\partial t} \left( v\frac{\partial f}{\partial x} \right) = v\frac{\partial^2 f}{\partial t \partial x} + a\frac{\partial f}{\partial x}$$ And as for $\partial v' / \partial x$, a second attempt might also be $$\frac{\partial}{\partial x} \left (\frac{\partial f}{\partial t} + v\frac{\partial f}{\partial x} \right) = \frac{\partial^2 f}{\partial x \partial t} + \frac{\partial v}{\partial x} \frac{\partial f}{\partial x} + v \frac{\partial^2 f}{\partial x^2}$$ I'm still not so certain about the rules for partial differentiation so there could be some more blunders here...

This seems okay apart from the $\frac{\partial v}{\partial x} \frac{\partial f}{\partial x}$ term...

 

[PeroK]

I would say $$\frac{\partial}{\partial t} \left( v\frac{\partial f}{\partial x} \right) = v\frac{\partial^2 f}{\partial t \partial x} + a\frac{\partial f}{\partial x}$$ And as for $\partial v' / \partial x$, a second attempt might also be $$\frac{\partial}{\partial x} \left (\frac{\partial f}{\partial t} + v\frac{\partial f}{\partial x} \right) = \frac{\partial^2 f}{\partial x \partial t} + \frac{\partial v}{\partial x} \frac{\partial f}{\partial x} + v \frac{\partial^2 f}{\partial x^2}$$ I'm still not so certain about the rules for partial differentiation so there could be some more blunders here...

This seems okay apart from the $\frac{\partial v}{\partial x} \frac{\partial f}{\partial x}$ term...

Are you sure you need all this differentiation?

 

[etotheipi]

Are you sure you need all this differentiation?

I was just going to say that if the two frames are inertial, $a = a'$ and that implies that all of the double derivatives are zero whilst $\frac{\partial f}{\partial x} = 1$, which basically leaves you with the Galilean transformation.

 

[PeroK]

I was just going to say that if the two frames are inertial, $a = a'$ and that implies that all of the double derivatives are zero whilst $\frac{\partial f}{\partial x} = 1$, which basically leaves you with the Galilean transformation.

Even then, you didn't really use $t'=t$. That ought to have simplified things.

 

[etotheipi]

Even then, you didn't really use $t'=t$. That ought to have simplified things.

I guess you're right, it's fairly self evident what the transformation has to be just by looking at it. The context of the question is a partial differentiation exercise, however.

I think I might just pack up shop for today, I'm making mistakes all over the place and hopefully tomorrow I won't be as clumsy...

 

[etotheipi]

I think $\frac{\partial v}{\partial x} \frac{\partial f}{\partial x}$ is zero since $v$ has no dependency on $x$. This gives the required result.

 

[PeroK]

I think $\frac{\partial v}{\partial x} \frac{\partial f}{\partial x}$ is zero since $v$ has no dependency on $x$. This gives the required result.

This didn't quite fit into the tutorial (which is now out for review), as this is really about the total derivative. The key is to recognise that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial t}$ are well-defined functions of two variables. Let's call them $g_1(x, t)$ and $g_2(x, t)$ to emphasise this.

We start with $x'(t) = f(x(t), t)$, which describes the trajectory of the particle.

We have the total derivative of $x'$:
$$v' = \frac{dx'}{dt} = \frac{d}{dt}f(x(t), t) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial t} = g_1(x(t), t)v(t) + g_2(x(t), t)$$
And:
$$a' = \frac{dv'}{dt} = \frac{dg_1}{dt}v(t) + g_1\frac{dv}{dt} + \frac{dg_2}{dt}$$
Now, the key is to apply the chain rule for the total derivative of $g_1, g_2$ exactly as you did for $f$:
$$a' = (\frac{\partial g_1}{\partial x}\frac{dx}{dt} + \frac{\partial g_1}{\partial t})v + g_1 a + (\frac{\partial g_2}{\partial x}\frac{dx}{dt} + \frac{\partial g_2}{\partial t})$$
And, finally, replacing $g_1$ and $g_2$ we have:
$$a' = \frac{\partial^2 f}{\partial x^2}v^2 + 2\frac{\partial^2 f}{\partial x \partial t}v + \frac{\partial f}{\partial x}a + \frac{\partial^2 f}{\partial t^2} $$

 

[etotheipi]

That route is a lot more straightforward. It's useful to note that $f = f(x,y)$ then $f_x = g(x,y)$ and $f_y = h(x,y)$; it seems obvious when stated like that but can be easy to forget the dependencies of the derivatives when focused on the computations.

Also, if $f = f(x,y)$ has explicit dependencies on $x$ and $y$, do expressions like $\frac{\partial y}{\partial x}$ make sense? I believe $\frac{\partial y}{\partial x}$ would be zero, but we don't know the dependencies of $y$ (i.e. it could be $y = y(t,s)$, or alternatively $y = y(x)$ in which case $\frac{\partial y}{\partial x}$ might not be zero!).

And then you also have strange constructions like $\frac{\partial}{\partial x} \frac{dx}{dt}$. In that case, $\frac{dx}{dt} = f(t)$. And this is now similar to the previous example, since $\frac{\partial t}{\partial x}$ in the chain rule expansion is again ambiguous, as we don't know whether $t$ has an explicit dependence on $x$!

Luckily common sense is quite a useful tool, but there are a lot of subtleties to it with I assume can only be fully understood with practice!

 

[PeroK]

One further subtlety is that you need to distinguish between the trajectory of a particle and the underlying coordinates. If we reserve $x, x', t$ for the coordinates themselves, then $x, t$ and $x', t$ are sets of independent variables. The partial derivative $\frac{\partial}{\partial x}$ is defined by holding $t$ constant etc.

When you have a trajectory (or surface, say), then you have a relationship between the coordinate values that define that trajectory. You could have $x = x_0 + vt$, for example. This allows you to do something like:
$$\frac{dx}{dt} = v$$
You have to be careful to distinguish this from something like:
$$\frac{\partial}{\partial x} f(t) = 0$$

In your example, you could use $X(t)$, say, for the trajectory of the particle. Then you have the relationship $X'(t) = f(X(t), t)$. Then:
$$V' = \frac{d X'}{dt} = \frac{d}{dt}f(X(t), t) = \frac{\partial f}{\partial x}\frac{dX}{dt} + \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}V + \frac{\partial f}{\partial t} $$
Where all these functions are evaluated at parameter $t$ and $X(t)$.

 

[haruspex]

Also, if $f = f(x,y)$ has dependencies explicit on $x$ and $y$, do expressions like $\frac{\partial y}{\partial x}$ make sense? I believe $\frac{\partial y}{\partial x}$ would be zero,

Partial derivatives written in the form $\frac{\partial y}{\partial x}$ are a shorthand. In the full form the variable(s) to be held constant would be specified.
If the context is functions of x and y, and nothing else, then clearly it is y that is held constant so $\frac{\partial y}{\partial x}$ is zero.

Wrt $\frac{\partial }{\partial x}\frac{dx}{dt}$, again, what is being held constant? If this is purely in respect of x=x(t) then there is no other variable, so it is equivalent to $\frac d{dx}\frac{dx}{dt}$. If it is in the context of the curve y=y(t), x=x(t) and we take it to mean y is constant then, unless we happen to be at a point where the curve is parallel to the x axis, x must be constant too.

 

[etotheipi]

It's the ones on the same level of the dependency tree that you have to worry about

If we have $z = z(x,y)$, $x=x(s,t)$ and $y=y(s,t)$, then it's fairly obvious that if we partially one variable with respect to one of its dependencies, we will hold the other dependencies of that variable constant. Or, as a rule, if $f = f(x_1, x_2, \dots, x_n)$, then $$\frac{\partial f}{\partial x_i} = \left (\frac{\partial f}{\partial x_i} \right)_{x_1, x_2, \dots, x_{i-1}, x_{i+1}, \dots x_n}$$ And if the variable with respect to which we partially differentiate is not an explicit dependency, we use the chain rule.

But the process is not as clear cut for $\frac{\partial y}{\partial x}$, or $\frac{\partial s}{\partial t}$. Like you say, in this context it's very likely $\left( \frac{\partial y}{\partial x} \right)_{y}$ since what other choice do we have! But no guarantees.

The need to be careful with all these dependencies reminds me of the example of the condition for energy conservation if $L$ is not a function of $t$, i.e. with $$H=\sum_{i=1}^n\dot q_i\frac{\partial L}{\partial \dot q_i}-L$$ If I throw a ball in the air, $L$ will vary with $t$ but that does not mean that it is a function of $t$, i.e. $\frac{\partial L}{\partial t} = 0$, there is no explicit dependence.

This sort of reasoning is distinct to normal derivatives, i.e. $\frac{dv}{dx}$ can be non-zero even if $v$ has no explicit dependence on $x$. I think this is one of the key differences with partial differentiation.

 

[etotheipi]

Actually, I think that last part was exactly what you were referring to here

When you have a trajectory (or surface, say), then you have a relationship between the coordinate values that define that trajectory. You could have $x = x_0 + vt$, for example. This allows you to do something like: $$\frac{dx}{dt} = v$$ You have to be careful to distinguish this from something like: $$\frac{\partial}{\partial x} f(t) = 0$$

I.e. $0 = \frac{\partial v(t)}{\partial x} \neq \frac{dv(t)}{dx}$

Conceptually, I think this is because $t$ is not held constant in the ordinary derivative example, so the $$\frac{dv}{dx} = \lim_{\Delta x \rightarrow 0 }\frac{v(x+ \Delta x, t + \Delta t)-v(x, t)}{\Delta x}$$ also captures the variation in $v$ due to the time elapsed during that interval. Whilst the partial definition is $$\frac{\partial v}{\partial x} = \lim_{\Delta x \rightarrow 0 }\frac{v(x+ \Delta x, t)-v(x, t)}{\Delta x}$$

 

[PeroK]

Actually, I think that last part was exactly what you were referring to here

I.e. $0 = \frac{\partial v(t)}{\partial x} \neq \frac{dv(t)}{dx}$

Conceptually, I think this is because $t$ is not held constant in the ordinary derivative example, so the $$\frac{dv}{dx} = \lim_{\Delta x \rightarrow 0 }\frac{v(x+ \Delta x, t + \Delta t)-v(x, t)}{\Delta x}$$ also captures the variation in $v$ due to the time elapsed during the that interval. Whilst the partial definition is $$\frac{\partial v}{\partial x} = \lim_{\Delta x \rightarrow 0 }\frac{v(x+ \Delta x, t)-v(x, t)}{\Delta x}$$

Velocity is a function of time. It makes no sense to take the partial derivative of velocity wrt the x-coordinate. You have two different types of function here. The multi-variable functions and the single-variable functions. Everything is driven by the total time derivative.

For a single-variable function of time, like $x(t), v(t), a(t), x'(t)$, the total derivative wrt $t$ is just the ordinary derivative.

But, if a function like $f(x(t), t)$ gets into the equations (via some coodinate transformation), then its total derivative wrt time is more complicated and involves its partial derivatives (wrt its first and second arguments). And, in fact, if you read my Insight on the chain rule, the $x$ that appears in the partial derivatives has nothing to do with the $x$ as in $x(t)$.

To see this, you could imagine that $f$ is a fixed function and its partial derivatives are also fixed functions. They are not related in any way to the trajectory of your particle.

But, what you are doing is evaluating its partial derivatives along the trajectory of your particle. This is the critical thing. Plus the notational overload (on the symbol $x$ in this case) that I mention in that Insight.

 

[etotheipi]

Ah okay, understood, it only makes sense to speak of the ordinary derivative $\frac{dv}{dx}$. I understand the part about the notational overload but forget to differentiate between $x$ and $X$ (with the usages you stated) in writing. Thanks for the help!

 

[PeroK]

Ah okay, understood, it only makes sense to speak of the ordinary derivative $\frac{dv}{dx}$.

You can really only write that if you assume that you have expressed $v$ as a function of $x$. In general, that gets problematic.

 

[etotheipi]

You can really only write that if you assume that you have expressed $v$ as a function of $x$. In general, that gets problematic.

Interesting. I'd always assumed the interconversion was just $\frac{dv}{dt} = v\frac{dv}{dx}$, but of course there is actually an implicit change in argument to i.e. $\frac{dv(t)}{dt} = v(x)\frac{dv(x)}{dx}$. It seems obvious now since there would be no way of doing anything with $\frac{dv(t)}{dx}$ (I don't think that even makes sense!). Had never really thought about the implications.

For a specific motion (i.e. mass on a spring, particle decelerating in a straight line) it is I suppose justified because for each $x$ we can associate a unique $v$ and that gives us the mapping $v(x)$, but for general planar motion (e.g. accelerating around in a circle over and over) I can see how it is problematic.