# Commutativity of partial and total derivative

#### RohanJ

Problem Statement
Use the definition of the total time derivative to
a) show that $(∂ /∂q)(d/dt)f(q,q˙,t) = (d /dt)(∂/∂q)f(q,q˙,t)$ i.e. these derivatives commute for any function $f = f(q, q˙,t)$.
Relevant Equations
My approach is given below. Please tell if it is correct and if not , then what is the correct approach?
Problem Statement: Use the definition of the total time derivative to
a) show that $(∂ /∂q)(d/dt)f(q,q˙,t) = (d /dt)(∂/∂q)f(q,q˙,t)$ i.e. these derivatives commute for any function $f = f(q, q˙,t)$.
Relevant Equations: My approach is given below. Please tell if it is correct and if not , then what is the correct approach?

I have got the terms for the left hand side but am stuck on the right hand side.

How to proceed after this if my approach is correct?

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#### PeroK

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It's difficult to see what you've done. It looks like you have too many terms. It should be quite simple. Can you put anything in Latex? For example, what is the defintion of the total time derivative?

Hint. To help to see how this works out you could let $g(q, \dot q, t) = \frac{\partial f}{\partial q}$

#### haruspex

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Think what is meant by $\frac{\partial }{\partial q}$ and the implication of that for $\frac{\partial }{\partial q}\frac {dq}{dt}$.

#### RohanJ

Think what is meant by $\frac{\partial }{\partial q}$ and the implication of that for $\frac{\partial }{\partial q}\frac {dq}{dt}$.
Since $\dot q , q$ and $t$ are independent variables $\frac{\partial }{\partial q}\frac {dq}{dt}$ is zero but what about the term $\frac{\partial}{\partial q}\frac{d \dot q}{dt}$?

#### PeroK

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Since $\dot q , q$ and $t$ are independent variables $\frac{\partial }{\partial q}\frac {dq}{dt}$ is zero but what about the term $\frac{\partial}{\partial q}\frac{d \dot q}{dt}$?
This is not right. You should answer my question in post #2. What is the definition of $\frac{d}{dt}$ in this context?

#### RohanJ

This is not right. You should answer my question in post #2. What is the definition of $\frac{d}{dt}$ in this context?
I don't understand what you asked about the definition of $\frac{d}{dt}$ but here is the question:

#### PeroK

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I don't understand what you asked about the definition of $\frac{d}{dt}$ but here is the question:
View attachment 244872
If you don't know what $\frac{d}{dt}$ means, how can you prove anything about it?

#### RohanJ

If you don't know what $\frac{d}{dt}$ means, how can you prove anything about it?
If you don't know what $\frac{d}{dt}$ means, how can you prove anything about it?
$\frac{d}{dt}$ means differentiating with respect to time. I have used this definition to get the results I got.

#### PeroK

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$\frac{d}{dt}$ means differentiating with respect to time. I have used this definition to get the results I got.
But, $f$ is a function of three variables. You can take the partial derivative of $f$ with respect to $t$, but there is no way to take any other derivative with respect to $t$.

In this case $\frac{d}{dt}$ looks like the regular single-variable derivative, but it can't be. It's called the "total" derivative in this context. It uses the same notation as the single-variable derivative, but it isn't a single-variable derivative in this context. So, you need to tell me how it's defined.

#### RohanJ

But, $f$ is a function of three variables. You can take the partial derivative of $f$ with respect to $t$, but there is no way to take any other derivative with respect to $t$.

In this case $\frac{d}{dt}$ looks like the regular single-variable derivative, but it can't be. It's called the "total" derivative in this context. It uses the same notation as the single-variable derivative, but it isn't a single-variable derivative in this context. So, you need to tell me how it's defined.
$\frac{df}{dt}$ will be equal to $\frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\frac{dq}{dt} + \frac{\partial f}{\partial \dot q}\frac{d\dot q}{dt}$ ,I think.

#### PeroK

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$\frac{df}{dt}$ will be equal to $\frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\frac{dq}{dt} + \frac{\partial f}{\partial \dot q}\frac{d\dot q}{dt}$ ,I think.
In fact, it is defined to be that. But that's a subtlety that is not critical. Now, using that it should be straightforward to show it commutes with the partial derivative with respect to $q$...

PS for the right-hand-side, you might use my hint in post #2 to introduce a new function $g$ to help make the derivatives clear.

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#### RohanJ

In fact, it is defined to be that. But that's a subtlety that is not critical. Now, using that it should be straightforward to show it commutes with the partial derivative with respect to $q$...

PS for the right-hand-side, you might use my hint in post #2 to introduce a new function $g$ to help make the derivatives clear.
I don't see how it is straightforward. Can you elaborate?

#### PeroK

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$\frac{df}{dt}$ will be equal to $\frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\frac{dq}{dt} + \frac{\partial f}{\partial \dot q}\frac{d\dot q}{dt}$ ,I think.
Hang on. I was a bit quick to agree to that! Note that $q$ is not a function of $t$ in this context, so you must write instead:

$\frac{df}{dt} \equiv \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial \dot q} \ddot q$

And if you ask what $\ddot q$ is, let's just say it's another independent variable that we've introduced. Another subtlety that you needn't worry about for now.

Now, does that make things more straightfoward?

#### haruspex

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This is not right.
Why not right that $\frac{\partial }{\partial q}\dot q=0$?
Hang on. I was a bit quick to agree to that! Note that $q$ is not a function of $t$ in this context, so you must write instead:

$\frac{df}{dt} \equiv \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial \dot q} \ddot q$.
The distinction is too subtle for me.
And if you ask what $\ddot q$ is, let's just say it's another independent variable that we've introduced.
If that is valid then $\frac{\partial }{\partial q}\ddot q=0$ and @RohanJ has the solution, but I don't buy it.
It helps me to think of this geometrically. f is a function of position in a 3D space with coordinates $q, \dot q, t$. So it is meaningful to consider the partial derivatives of f wrt these coordinates.
The total derivative, d/dt, is only meaningful for some chosen q=q(t), i.e. a trajectory through this space. What does $\frac{\partial g}{\partial q}$ mean here? It means we are stepping off that trajectory by a small distance in the q direction and asking how g changes. But if g is df/dt then the answer depends on the trajectory at the new point, and there is nothing that determines that. We can choose any trajectory we like through that point.
The appearance of the $\frac{\partial \ddot q}{\partial q}$ term illustrates this. It says that to find $\frac{\partial }{\partial q}\frac{df}{dt}$ we need more information about the difference in the trajectories.

You could elect to restrict to a family of trajectories that only differ by a shift in the q direction. In that case $\frac{\partial \ddot q}{\partial q}=0$ and the difficulty goes away, but this strikes me as artificial and not implied by the question.

The right hand side of the equivalence does not have this problem. It represents how the gradient of f in the q direction varies with time along a chosen trajectory.

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#### strangerep

The total derivative, d/dt, is only meaningful for some chosen q=q(t), i.e. a trajectory through this space.
No -- the total derivative is meaningful anywhere in the (extended) (velocity) phase space. A priori, the variables $t, q, \dot q, \ddot q, \dots$ are independent variables. (It's also known as a "jet space" in other contexts.)

Only after you impose the EoM (Euler-Lagrange equations) is there an inter-dependence among those variables.

As a simpler analogy, consider the 2D Cartesian x-y plane. A priori, x and y are independent. (This is called "off-shell" in the phase space context). But if we impose an equation, e.g., $y = x^2$, only then is there a dependence between x and y. That's called "on-shell".

In a sense, the original problem statement is a bit silly, imho. The commutation holds trivially because of the definition of phase space and the total derivative thereon.

#### haruspex

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No -- the total derivative is meaningful anywhere in the (extended) (velocity) phase space. A priori, the variables $t, q, \dot q, \ddot q, \dots$ are independent variables. (It's also known as a "jet space" in other contexts.)

Only after you impose the EoM (Euler-Lagrange equations) is there an inter-dependence among those variables.

As a simpler analogy, consider the 2D Cartesian x-y plane. A priori, x and y are independent. (This is called "off-shell" in the phase space context). But if we impose an equation, e.g., $y = x^2$, only then is there a dependence between x and y. That's called "on-shell".

In a sense, the original problem statement is a bit silly, imho. The commutation holds trivially because of the definition of phase space and the total derivative thereon.
I'm still confused.
I see there is something called a "total derivative", but this seems to be a vector; essentially the grad, $(\frac{\partial f}{\partial q}, \frac{\partial f}{\partial \dot q}, \frac{\partial f}{\partial t})$. In this, t plays no special role.
The question refers to a "total time derivative" (so this is the terminology I should have used in post #14) and writes it as df/dt. I cannot see how these can be the same thing.
I can get the second from the vector by dotting it with $(\dot q, \ddot q, 1)$, but as soon as I do that I have restricted to a specific trajectory.

#### strangerep

The question refers to a "total time derivative" (so this is the terminology I should have used in post #14) and writes it as df/dt. I cannot see how these can be the same thing.
The notion and use of such total derivative wrt $t$ is perhaps better understood in the context of (say) a 2nd-order ODE like
$$\ddot q ~=~ \omega(t,q,\dot q) ~,~~~~~~ (1)$$ where $q=q(t)$ and $\dot q := dq/dt$ . The time $t$ is the independent variable in this ODE, and $q$ (and hence $\dot q$) are dependent variables.

A particular function $q(t)$ which satisfies (1) is said to be "on-shell", or "on the solution variety".

The entire problem can be modelled in a phase space framework, consisting of independent variables $t$, $q$, $v \equiv \dot q$, and $a \equiv \ddot q$ . Consider an arbitrary function on this phase space, denoted $f=f(t,q,\dot q)$. The total derivative of such an $f$ is defined via the chain rule:
$$D_t f ~:=~ \frac{df}{dt} ~\equiv~ \frac{\partial f}{\partial t} ~+~ \dot q \, \frac{\partial f}{\partial q} ~+~ \ddot q \, \frac{\partial f}{\partial \dot q} ~.~~~~~~ (2)$$
If we wish to find constants of the motion (a.k.a. "first integrals"), we solve
$$0 ~=~ D_t f \Big|_{\ddot q ~=~ \omega} =~ \frac{\partial f}{\partial t} ~+~ \dot q \, \frac{\partial f}{\partial q} ~+~ \omega \, \frac{\partial f}{\partial \dot q} ~.~~~~~~ (3)$$
We then usually want to analyze symmetries and conserved quantities applicable to the system characterized by $\omega(t,q,\dot q)$. By finding all solutions $f$ of eq(3) we find the conserved quantities for the system.

For deeper analysis, it is useful to define the operator $$A ~:=~ \partial_t + \dot q \partial_q + \omega \partial_{\dot q} ~\equiv~ D_t f \Big|_{\ddot q ~=~ \omega} ~.~~~~~~ (4)$$ Indeed, we can find all the dynamical symmetries of the system (i.e., all the differential operators which map solutions among themselves), by finding all differential operators $$X ~:=~ \xi\partial_t + \eta \partial_q + \lambda \partial_{\dot q} ~,~~~~~~ (5)$$ (where $\xi,\eta,\lambda$ are functions of $t,q,\dot q$), such that $X$ satisfies $$[ X , A] ~=~ \Lambda(t,q,\dot q) \, A ~,~~~~~~ (6)$$ where the $\Lambda$ function is arbitrary.

HTH.

["Someone" should really write an Insight about all this.]

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#### PeroK

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I'm still confused.
I see there is something called a "total derivative", but this seems to be a vector; essentially the grad, $(\frac{\partial f}{\partial q}, \frac{\partial f}{\partial \dot q}, \frac{\partial f}{\partial t})$. In this, t plays no special role.
The question refers to a "total time derivative" (so this is the terminology I should have used in post #14) and writes it as df/dt. I cannot see how these can be the same thing.
I can get the second from the vector by dotting it with $(\dot q, \ddot q, 1)$, but as soon as I do that I have restricted to a specific trajectory.
There are actually two different uses of the notation $\frac{d}{dt}$ (in addition to its normal use as a single-variable derivative).

The "total derivative" of a specific particle trajectory. Here we must take the step of assuming that each of the other variables is a specific function of $t$. In this case we could define:

$F(t) = f(q(t), \dot{q}(t), t)$

And $\frac{df}{dt} \equiv \frac{dF}{dt}$ is simply a single variable derivative.

Note that in this case you can actually define $\frac{df}{dt}$ as a limit in the usual way.

As mentioned above, in the theory of Lagrangian mechanics the total derivative is also often used while we are still working in "configuration" space and we have no specific particle trajectory.

In this case, there is no single variable function of $t$ to invoke and, in fact, the total derivative in this case cannot be expressed in the form of a limit. In this sense it's not really a"derivative" at all. Instead, it is a symbolic shorthand:

$\frac{df}{dt} \equiv \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial \dot q} \ddot q$

Most physics texts move from this "symbolic" derivative to the particle-trajectory "total" derivative without mentioning any of this.

#### RohanJ

Hang on. I was a bit quick to agree to that! Note that $q$ is not a function of $t$ in this context, so you must write instead:

$\frac{df}{dt} \equiv \frac{\partial f}{\partial t} + \frac{\partial f}{\partial q}\dot q + \frac{\partial f}{\partial \dot q} \ddot q$

And if you ask what $\ddot q$ is, let's just say it's another independent variable that we've introduced. Another subtlety that you needn't worry about for now.

Now, does that make things more straightfoward?
We don't know anything about $\ddot q$. How do we know it is an independent variable?

No -- the total derivative is meaningful anywhere in the (extended) (velocity) phase space. A priori, the variables $t, q, \dot q, \ddot q, \dots$ are independent variables. (It's also known as a "jet space" in other contexts.)

Only after you impose the EoM (Euler-Lagrange equations) is there an inter-dependence among those variables.

As a simpler analogy, consider the 2D Cartesian x-y plane. A priori, x and y are independent. (This is called "off-shell" in the phase space context). But if we impose an equation, e.g., $y = x^2$, only then is there a dependence between x and y. That's called "on-shell".

In a sense, the original problem statement is a bit silly, imho. The commutation holds trivially because of the definition of phase space and the total derivative thereon.
Can you provide any text references for further reading into this topic?

#### PeroK

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We don't know anything about $\ddot q$. How do we know it is an independent variable?

Can you provide any text references for further reading into this topic?
In general all the derivatives of $q$ exist as independent variables in configuration space. The assumption however is that the Lagrangian depends on $q$ and $\dot q$ only.

The total time derivative of the Lagrangian, however, depends also on $\ddot q$.

#### RohanJ

In general all the derivatives of $q$ exist as independent variables in configuration space. The assumption however is that the Lagrangian depends on $q$ and $\dot q$ only.

The total time derivative of the Lagrangian, however, depends also on $\ddot q$.
Okay,got it now. Now I will do the second part to confirm the concept.
Thank you very much.

#### haruspex

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In general all the derivatives of q exist as independent variables in configuration space.
So post #1 basically got there. It was only necessary to plug in that the partial derivatives of $\dot q$ and $\ddot q$ wrt $q$ are zero.

#### RohanJ

So post #1 basically got there. It was only necessary to plug in that the partial derivatives of $\dot q$ and $\ddot q$ wrt q are zero.
Yes and that was my only concern that why should $\ddot q$ be independent..

#### strangerep

Can you provide any text references for further reading into this topic?
H. Stephani, "Differential Equations -- Their solution using symmetries",
Cambridge University Press, 1989C, ISBN 0-521-36689-5.

"Commutativity of partial and total derivative"

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