# Commutativity of partial and total derivative

• RohanJ
In summary: 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.
RohanJ
Homework 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?

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}##

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

haruspex said:
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}##?

RohanJ said:
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?

PeroK said:
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:

RohanJ said:
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?

PeroK said:
If you don't know what ##\frac{d}{dt}## means, how can you prove anything about it?
PeroK said:
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.

RohanJ said:
##\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.

PeroK said:
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.

Delta2 and PeroK
RohanJ said:
##\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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PeroK said:
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?

RohanJ said:
##\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?

PeroK said:
This is not right.
Why not right that ##\frac{\partial }{\partial q}\dot q=0##?
PeroK said:
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.
PeroK said:
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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haruspex said:
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.

PeroK
strangerep said:
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.

haruspex said:
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 modeled 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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haruspex said:
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.

PeroK said:
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?

strangerep said:
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?

RohanJ said:
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##.

PeroK said:
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.

PeroK said:
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.

haruspex said:
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..

RohanJ said:
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.

## 1. What is the commutativity property of partial and total derivatives?

The commutativity property of partial and total derivatives states that the order in which the derivatives are taken does not affect the final result. In other words, the derivative of a function with respect to one variable followed by the derivative of the resulting function with respect to another variable is equal to the derivative of the original function with respect to the second variable followed by the derivative of the resulting function with respect to the first variable.

## 2. Why is the commutativity property important in calculus?

The commutativity property is important in calculus because it allows us to simplify the process of taking derivatives. Instead of having to consider the order in which the derivatives are taken, we can simply take the derivatives in any order and still arrive at the same result. This makes solving complex problems involving multiple variables much easier.

## 3. Can the commutativity property be applied to all types of functions?

Yes, the commutativity property can be applied to all types of functions. It is a fundamental property of derivatives and holds true for all differentiable functions.

## 4. How is the commutativity property related to the chain rule?

The commutativity property is closely related to the chain rule in calculus. The chain rule states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This can be seen as a specific application of the commutativity property, where the inner and outer functions are treated as separate variables.

## 5. Are there any exceptions to the commutativity property?

Yes, there are some exceptions to the commutativity property. For example, if a function is not differentiable at a certain point, then the commutativity property may not hold at that point. Additionally, the commutativity property may not hold for functions with discontinuities or other irregularities.

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