The definition of generalised momentum

[baw]

Why, in lagrangian mechanics, do we calculate: $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}$ to get the (generalised) momentum change in time
instead of $\frac{d T}{dq}$?

(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external forces are applied)

The kinetic energy is defined as the amount of work needed to accelerate some body to its velocity, so it is derived by $\int \frac{dp}{dt} dx$.
Quite intuitive way of obtaining the time rate of change of the generalised momentum would be to take "the reverse": $\frac{d T}{dq}$, but it doesn't seem to work. Why?
Such procedure would be based directly on the definition of T, while the actual, lagrangian approach looks (at least for me) like based on some property in cartesian coordinates. It's not obvious for me, that it would work in other coordinate systems etc. Could someone explain me why this property is so general? (I know it is just a definition but it's not completely arbitraty because it has to comply with Newtonian mechanics).

 

[anuttarasammyak]

L=L(q,\dot{q},t)
\frac{d}{dt}=\dot{q}\frac{\partial}{\partial q}+\ddot{q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial t}

\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\dot{q}\frac{\partial}{\partial q}\frac{\partial L}{\partial \dot{q}}+\ddot{q}\frac{\partial}{\partial \dot{q}}\frac{\partial L}{\partial \dot{q}}+\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{q}}...(1)

\frac{d}{dq}=\frac{dt}{dq}\frac{\partial}{\partial t}+\frac{dt}{d \dot q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial q}=\frac{1}{\dot{q}}\frac{\partial}{\partial t}+\frac{1}{\ddot {q}}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial q}

\frac{dL}{dq}=\frac{dt}{dq}\frac{\partial L}{\partial t}+\frac{dt}{d \dot q}\frac{\partial L}{\partial \dot{q}}+\frac{\partial L}{\partial q}=\frac{1}{\dot{q}}\frac{\partial L}{\partial t}+\frac{1}{\ddot{q}}\frac{\partial L}{\partial \dot{q}}+\frac{\partial L}{\partial q}...(2)

We observe in RHS $(1) \neq (2)$ in mathematics. As physics Lagrange equation states (1) = the third term of (2),i.e. time change or derivative of generalized momentum = generalized force.

 

[andrewkirk]

We can observe straight away that $\frac{dT}{dq_i}$ would not work for polar coordinates because with $q_i$ being the angle $\theta$, its units will be the same as those of KE, ie Joules or $kg\,m^2s^{-2}$, rather than $kg\,ms^{-2}$ as required. The Lagrangian formula uses partial rather than total derivatives, which removes this problem, when we look at the formula for KE expressed in polar coordinates, see here.

Edited 18/1/21 to correct Newton-metres to $kg\,ms^{-2}$.

 

[baw]

L=L(q,\dot{q},t)
\frac{d}{dt}=\dot{q}\frac{\partial}{\partial q}+\ddot{q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial t}

\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\dot{q}\frac{\partial}{\partial q}\frac{\partial L}{\partial \dot{q}}+\ddot{q}\frac{\partial}{\partial \dot{q}}\frac{\partial L}{\partial \dot{q}}+\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot{q}}...(1)

\frac{d}{dq}=\frac{dt}{dq}\frac{\partial}{\partial t}+\frac{dt}{d \dot q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial q}=\frac{1}{\dot{q}}\frac{\partial}{\partial t}+\frac{1}{\ddot {q}}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial q}

\frac{dL}{dq}=\frac{dt}{dq}\frac{\partial L}{\partial t}+\frac{dt}{d \dot q}\frac{\partial L}{\partial \dot{q}}+\frac{\partial L}{\partial q}=\frac{1}{\dot{q}}\frac{\partial L}{\partial t}+\frac{1}{\ddot{q}}\frac{\partial L}{\partial \dot{q}}+\frac{\partial L}{\partial q}...(2)

We observe in RHS $(1) \neq (2)$ in mathematics. As physics Lagrange equation states (1) = the third term of (2),i.e. time change of generalized momentum = generalized force.

I noticed that approaches I mentioned give different results, but my point is: how do we know which of them is correct? or maby... how did Lagrange know that?

 

[baw]

We can observe straight away that $\frac{dT}{dq_i}$ would not work for polar coordinates because with $q_i$ being the angle $\theta$, its units will be the same as those of KE, ie Joules, rather than Newton-metres as required. The Lagrangian formula uses partial rather than total derivatives, which removes this problem, when we look at the formula for KE expressed in polar coordinates, see here.

aren't joules and Newton-meters same thing? When we integrate torque (Nm) with respect to the angle (dimensionless) we get work (J).

 

[anuttarasammyak]

Let me know the point. In "which of them is correct? " you say, are you referring
$\frac{dL}{dq}$ or $\frac{\partial L}{\partial q}$ ?
or
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ or $\frac{\partial L}{\partial q}$ ?
or
all these three ?
;and may I take "correct" to be correct to appear in Lagrange equation ?

 

[PeroK]

Summary:: Why $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}$ instead of $\frac{dT}{dq}$?

Why, in lagrangian mechanics, do we calculate: $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}$ to get the (generalised) momentum change in time
instead of $\frac{d T}{dq}$?

(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external forces are applied)

Example: $$T = \frac 1 2 m \dot x^2, \ \ \frac{\partial T}{\partial \dot{x}} = m \dot x$$ gives the momentum.

Moreover, $\frac{d T}{dq}$ makes no mathematical sense. It must be a partial derivative.

 

[baw]

Let me know the point. In "which of them is correct? " you say, are you referring
$\frac{dL}{dq}$ or $\frac{\partial L}{\partial q}$ ?
or
$\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ or $\frac{\partial L}{\partial q}$ ?
or
all these three ?
;and may I take "correct" to be correct to appear in Lagrange equation ?

I mean the second: $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ or $\frac{dL}{dq}$.

There must be some physical reason that makes the idea of $\frac{dL}{dq}$ incorrect and $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ correct, that could be found befeore we e.g. solve the same problem with Newtonian approach and compare the results.

Yes, I'm wondering which of these (and why) should appear in Lagrange equations to provide a solution which agrees with Newtonian solution of the same case.

 

[anuttarasammyak]

If you put $\frac{dL}{dq}$ into LHS of Lagrange equation, the equation would become
\frac{dL}{dq}=\frac{\partial L}{\partial q}
Obviously wrong.

Newton states formula as physics law
F=\frac{dp}{dt}
to SOLVE it. Not solving but applying this law again you get identity
F=F
and
\frac{dp}{dt}=\frac{dp}{dt}
They are useless.

 

[PeroK]

If you put $\frac{dL}{dq}$ into LHS of Lagrange equation, the equation would become
\frac{dL}{dq}=\frac{\partial L}{\partial q}
Obviously wrong.

Moreover, the LHS is not well-defined.

 

[baw]

If you take dLdq as generalized momentum in Lagrange equation, the equation would become
dLdq=∂L∂q
Obviously wrong.

That's not exactly what I meant. I'm wondering how could Lagrange, deriving his equation, decide which of $\frac{dL}{dq}$ or $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ to choose. I guess the only thing he could compare it to ( in terms of consistency ) was Newton's or d'Alebert's equation.

 

[PeroK]

That's not exactly what I meant. I'm wondering how could Lagrange, deriving his equation, decide which of $\frac{dL}{dq}$ or $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}$ to choose. I guess the only thing he could compare it to ( in terms of consistency ) was Newton's or d'Alebert's equation.

How Lagrange developed the theory in the first place is a historical question. The main motivation in a modern treatment is to look at the simple case in Cartesian and Polar coordinates. And then to prove that, in general, the Euler-Lagrange equations are equivalent to Newton's laws.

 

[baw]

If you put $\frac{dL}{dq}$ into LHS of Lagrange equation, the equation would become
\frac{dL}{dq}=\frac{\partial L}{\partial q}
Obviously wrong.

Newton states formula as physics law
F=\frac{dp}{dt}
to SOLVE it. Not solving but applying this law again you get identity
F=F
and
\frac{dp}{dt}=\frac{dp}{dt}
They are useless.

Ok, that's right, but are $\frac{dL}{dq}$ and $\frac{\partial L}{\partial q}$ exactly same things like F and F?

 

[anuttarasammyak]

You will find in formula of (2) of comment #2 they differ by the first and the second RHS terms.

 

[PeroK]

Ok, that's right, but are $\frac{dL}{dq}$ and $\frac{\partial L}{\partial q}$ exactly same things like F and F?

$\frac{dL}{dq}$ is meaningless. You can only have the total derivative of a multi-variable function with respect to some parameter (e.g. $t$) that all your variables depend on.

 

[baw]

Ok, thank you

 

[vanhees71]

It's simply a definition to call
$$p_k=\frac{\partial L}{\partial \dot{q}^k}$$
the canonically conjugated momentum of the configuration coordinate $q^k$. It's a useful definition, because then the equation of motion reads
$$\dot{p}_k=\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k} = \frac{\partial L}{\partial q^k}.$$
The right-hand side is sometimes called "generalized force component".

 

[Vanadium 50]

aren't joules and Newton-meters same thing?

No. Energy is not torque, even if they have the same dimensions.

 

[etotheipi]

aren't joules and Newton-meters same thing? When we integrate torque (Nm) with respect to the angle (dimensionless) we get work (J).

Yes, $\mathrm{J}$ is identical to $\mathrm{Nm}$, by definition, $\mathrm{J} = \mathrm{Nm} = \mathrm{kg m^2 s^{-2}}$.

IMO this question has nothing to do with units, and I personally don't understand @andrewkirk's argument in post #3 (unless there was a typo). It's simply by definition that $p_i = \partial \mathcal{L} / \partial \dot{q}^i$, and hence also $\dot{p}_i = \partial \mathcal{L} / \partial{q}^i$ by EL. Sometimes, but not always, $p_i$ will coincide with our old/usual definition of momentum.

And, as mentioned above by @PeroK, it only makes sense for it to be a partial derivative, since $\mathcal{L} = \mathcal{L}(q, \dot{q}, t)$.

 

[Vanadium 50]

Yes

No.

Energy is not torque, even if they have the same dimensions.

 

[etotheipi]

No.

Energy is not torque, even if they have the same dimensions.

Yes, but we are not talking about energy and torque. We are talking about Joules and Newton-metres! Different quantities can have the same physical dimensions, energy and torque can be expressed using the same units. 1 Joule is exactly equivalent to 1 Newton-metre.

 

[Vanadium 50]

1 Joule is exactly equivalent to 1 Newton-metre

You keep saying that. It is not true. Energy is not torque, even if they have the same dimensions.

How much water can I boil with 100 pound-feet of torque?

 

[etotheipi]

Energy is not torque, even if they have the same dimensions.

OK well you can keep banging that drum if you like but I never said that energy is torque! You put those words in my mouth. Of course those things are not the same, one is a bivector whilst the other is a scalar. I said that, in the SI, $1\mathrm{J} = 1\mathrm{Nm}$. I was responding to:

its units will be the same as those of KE, ie Joules, rather than Newton-metres as required

which I suspected, might be a simple typo. I'm just talking about the units.

Anyway I did hear some time ago that various users had been calling me 'stubborn' in the advisor forums, and for fear of proving their point I think it's wise that I shut up now

 

[andrewkirk]

aren't joules and Newton-meters same thing? When we integrate torque (Nm) with respect to the angle (dimensionless) we get work (J).

I certainly agree with the second statement. But unfortunately that doesn't help. I admit I stuffed up my attempt to explain that in my post above, by wrongly saying that the time rate of change of momentum had units of Newton-metres, which are $kg\,m^2s^{-2}$, which is the same as for Joules. In fact the time rate of change of momentum has units $kg\,ms^{-2}$ which, unlike Newton-metres, does have a different representation in terms of the fundamental units kg, m, s, from the representation of KE.

I also agree with @PeroK that:

$\frac{d T}{dq}$ makes no mathematical sense

I think the expression can sometimes make physical sense, in the context of a single object that follows a known path, in which case the expression relates to the rate of change of T as the object moves along the path. But even then, if the path intersects itself (eg an orbit), the expression may not be well-defined.

But my point is that, even if we could make sense of $\frac{dT}{dq}$, it couldn't give us generalised momentum because it has the wrong units. Having the right units is a necessary but not sufficient condition for a formula to represent a quantity. Since we can often check units more easily than we can do algebraic manipulations, that makes dimensional analysis (checking units) a very useful test of a claim!

 

[vanhees71]

OK well you can keep banging that drum if you like but I never said that energy is torque! You put those words in my mouth. Of course those things are not the same, one is a bivector whilst the other is a scalar. I said that, in the SI, $1\mathrm{J} = 1\mathrm{Nm}$. I was responding to:which I suspected, might be a simple typo. I'm just talking about the units.

Anyway I did hear some time ago that various users had been calling me 'stubborn' in the advisor forums, and for fear of proving their point I think it's wise that I shut up now

Why stubborn? It's of course correct that by definition $1 \; \text{J}=1 \; \text{N} \text{m}$.

Of course it would be very misleading to give a troque in terms of Joules though it's formally not wrong ;-)).

 

[Vanadium 50]

Why stubborn? It's of course correct that by definition 1 J = 1 Nm

Torqua is not energy and torque and energy do not have the same units. They do have the same dimensions, md²t^-2, but that is not the same thing. Physics is not algebra. One joule (of energy) does not equal one Nm (of torque).

If you want another example, strain (Δx/x) and angle (arc length/radius) are both dimensionless. But you can't equate them, even if 1 = 1.

 

[etotheipi]

torque and energy do not have the same units

This is patently incorrect! In the SI, the units of torque are $\mathrm{kg \, m^2 \, s^{-2}}$, and the units of energy are $\mathrm{kg \, m^2 \, s^{-2}}$. The Newton is defined by $\mathrm{N} = \mathrm{kg \, m \, s^{-2}}$, and the Joule by $\mathrm{J =kg \, m^2 \, s^{-2}}$.

That the different physical quantities torque and energy have the same units in the SI is no more than coincidence.

Physics is not algebra.

We most certainly can treat units algebraically! See, for instance, section 3.12.1 'Dimensional Analysis/Units' of Mathematical Methods for Physics and Engineering, Blennow.

 

[PeroK]

There's a similar discussion here:

https://physics.stackexchange.com/questions/37881/why-is-torque-not-measured-in-joules

 

[PeroK]

Torqua is not energy and torque and energy do not have the same units. They do have the same dimensions, md²t^-2, but that is not the same thing. Physics is not algebra. One joule (of energy) does not equal one Nm (of torque).

If you want another example, strain (Δx/x) and angle (arc length/radius) are both dimensionless. But you can't equate them, even if 1 = 1.

For what it's worth, Wikipedia agrees with you:

"A result of this similarity is that the SI unit for torque is the Newton-metre, which works out algebraically to have the same dimensions as the joule, but are not interchangeable. The General Conference on Weights and Measures has given the unit of energy the name joule, but has not given the unit of torque any special name, hence it is simply the Newton-metre (N⋅m) – a compound name derived from its constituent parts.[24] The use of Newton metres for torque and joules for energy is helpful to avoid misunderstandings and miscommunications.[24] "

And, if you follow the link to the International Standard on Weights and Measures, the joule is exclusively reserved as the unit of energy:

"As an example of a special name, the particular combination of base units m2 kg s−2 for energy is given the special name joule, symbol J, where by definition J = m2 kg s−2."

https://www.bipm.org/utils/common/pdf/si_brochure_8_en.pdf

 

[etotheipi]

@PeroK I don't interpret that to say that energy and torque cannot be expressed using the same units (because they can!), only that the term 'Joule/$\mathrm{J}$' is exclusively used when referring to energy (for obvious reasons, since it would be pathological to do otherwise).

In any case, @Vanadium 50, I grow a little tired of this petty squabble, so let's agree to disagree and go spend the rest of the day doing something actually useful.

 

[Vanadium 50]

@etotheipi , that sounds a lot like "Now that I have the last word, I invite you to shut up" An awful lot like that.

Wikipedia agrees with you

Thanks.

 

[etotheipi]

@etotheipi , that sounds a lot like "Now that I have the last word, I invite you to shut up" An awful lot like that.

Right, well I apologise if you took offence to what I thought was a pretty innocuous statement of intention - that was not intended. You are quite welcome, of course, to continue this discussion without me.

Though let's act like adults, and not degenerate into childish bickering, please.

 

[Vanadium 50]

Again, "Now that I have the last word, I invite you to stop with your childish bickering"

This is important. Lots of students get hung up on this and end up working problems where they equate torque to energy. Angles just disappear, or worse, disappear from the right place and appear in the wrong place. It is, as the International Standard on Weights and Measures makes clear, best to keep energy and torque separate, with different units, even if those units have the same dimensions.

If you need another example, radians and steradians have the same dimensions (dimensionless). If we can say 1 J = 1 Nm, shouldn't we also be able to say 1 rad = 1 sr?

 

[PeroK]

Time for a Haiku or two:

The units are equal?
Energy may be twisted,
By angular force?

We can but discern,
By empirical measure,
The joule from the torque

 

[Vanadium 50]

Valuable guide,
Physics is not Algebra
Goes mostly ignored

 

[etotheipi]

Roses are red,
violets are blue,
I think ⁵⁰V is being a little condescending,
how about you?

 

[robphy]

There's a similar discussion here:

https://physics.stackexchange.com/questions/37881/why-is-torque-not-measured-in-joules

One of the answers given there suggests an important distinction
between the units for energy and the units for torque,
as well as making contact with the OP's question.

https://physics.stackexchange.com/a/93735/148184
Yes, torque has units of joules in SI. But it's more accurate and less misleading to call it joules per radian.
...
\displaystyle \tau=\frac{dE}{d\theta}.

That is, the torque--regarded as a generalized force--has units of Joules/radian.
This complements the distinction to energy being a scalar and torque being thought of as a bivector.

It appears that these above (arguably more fundamental) constructions
are (knowingly or unknowingly) reduced to the simplified forms we see in introductory physics.
(Sometimes, it seems that "taking advantage of symmetries" blurs the real distinctions between these constructions.)

The OP might appreciate the work of Enzo Tonti
https://en.wikipedia.org/wiki/Tonti_diagram
in helping to clarify the structure of physics theories.

 

[vanhees71]

Torqua is not energy and torque and energy do not have the same units. They do have the same dimensions, md²t^-2, but that is not the same thing. Physics is not algebra. One joule (of energy) does not equal one Nm (of torque).

If you want another example, strain (Δx/x) and angle (arc length/radius) are both dimensionless. But you can't equate them, even if 1 = 1.

I didn't claim that torque and energy are the same. In the SI they have the same units, which is $\text{N} \text{m}$, and of course quantities that are measured in the same units need not be the same, as your examples show.

Why should they? It depends on the system of units you use, in which units a quantity is given. In the fully natural system of units (Planck units) all quantities are "dimensionless". That of course doesn't imply that all quantities are the same and can be added.

 

[vanhees71]

One of the answers given there suggests an important distinction
between the units for energy and the units for torque,
as well as making contact with the OP's question.That is, the torque--regarded as a generalized force--has units of Joules/radian.
This complements the distinction to energy being a scalar and torque being thought of as a bivector.

It appears that these above (arguably more fundamental) constructions
are (knowingly or unknowingly) reduced to the simplified forms we see in introductory physics.
(Sometimes, it seems that "taking advantage of symmetries" blurs the real distinctions between these constructions.)

The OP might appreciate the work of Enzo Tonti
https://en.wikipedia.org/wiki/Tonti_diagram
in helping to clarify the structure of physics theories.

Hm, but rad is just 1. Angles, measured in radians, are dimensionless. You can of course also measure them in degrees or gons.

 

[robphy]

Hm, but rad is just 1. Angles, measured in radians, are dimensionless. You can of course also measure them in degrees or gons.

From a practical point of view, I would agree...
However, from a structural point of view,
I would agree that "rad is equal to 1" within some system of units [akin to a choice of coordinates?]
or within some context [where it is declared to be so by introducing additional structure]
... but, at another more fundamental level, they are distinct... which may not be adequately captured by a system of units where "rad is equal to 1".

I think there are seeds of this "structure" (which I can't put my finger on) in this discussion
https://golem.ph.utexas.edu/category/2006/09/dimensional_analysis_and_coord.html
but I haven't had the time to go through it.

We find it convenient to use "rad" in a calculation, then treat it as "1".
But we wouldn't use "1" in another calculation, then treat it as "rad".
So, obviously, there's something else going on.
How can this be formalized?
For example, how would these "rules" be encoded in (say) a computer program?

 

[vanhees71]

From a practical point of view, I would agree...
However, from a structural point of view,
I would agree that "rad is equal to 1" within some system of units [akin to a choice of coordinates?]
or within some context [where it is declared to be so by introducing additional structure]
... but, at another more fundamental level, they are distinct... which may not be adequately captured by a system of units where "rad is equal to 1".

I think there are seeds of this "structure" (which I can't put my finger on) in this discussion
https://golem.ph.utexas.edu/category/2006/09/dimensional_analysis_and_coord.html
but I haven't had the time to go through it.

We find it convenient to use "rad" in a calculation, then treat it as "1".
But we wouldn't use "1" in another calculation, then treat it as "rad".
So, obviously, there's something else going on.
How can this be formalized?
For example, how would these "rules" be encoded in (say) a computer program?

I think there's an endless discussion of this, i.e., whether writing rad in the sense of a unit for angles is consistent with the rules of the SI (or any other system of units of course) or not.

I'd say, you are right in saying that it can be advantageous dependent on the context. E.g., it may be more clear for a beginning student if you write "the right angle is $\pi/2 \text{rad}$", because then it's emphasized that you give the angle in terms of radians. On the other hand it must be clear that rad=1, because otherwise you couldn't calculate the trigonometric functions with their power series, because if rad weren't a dimensionless numbers you couldn't add $\pi/2 \; \text{rad}$ and $-(\pi/2)^3 \text{rad}^3/3!$ to evaluate the power series of the sinus.

On the other hand, if you want to stay strictly formal, you are not allowed to use "rad" as a unit within the SI, because an angle is defined as the ratio of two lengths and are thus just a dimensionless number.

 

[anuttarasammyak]

Multiplying second action and angular momentum have same dimension. $\hbar$ is action in commutation relation or path integral. (a half of ) It is angular momentum for electron spin. We are familiar with this duality. As far as I know unit Joule second is shared for the both. So I am not keen on the distinction.

In the commutation relation
[L_y,L_z]=i\hbar L_x
, is $\hbar$ action ? angular momentum ?

In relativity antisymmetric angular momentum tensor of particles
M^{\mu\nu}:=\frac{1}{2}\sum_{particles}(x^\mu p^\nu-p^\mu x^\nu)
(i,j) component vector shows angular momentum vector and (0,j) component vector shows motion of CoM. They have different meaning but have same dimension and belong to the same tensor.

PS
I have just been reminded that the Planck constant is defined to have the exact value h= 6.62607015×10−34 J⋅s in SI units. It means $\hbar$ is an irrational number.

 

[vanhees71]

$\hbar$ is a unit-defining conversion factor. The new SI is pretty "natural" now! $h$ is defined exactly, and thus is a rational number times the unit Js, which implies that $\hbar=h/(2 \pi)$ is an irrational number times Js.

The most natural place where $h$ or $\hbar$ play a role is that it is the natural measure of phase-space volumes, which was utterly missing in the early days of classical statistical/kinetic gas theory. In this sense it's of course closely related to the action and the quantization condition in old quantum theory. That's where the name "Planck's action quantum" originates from.

 

[vin300]

Summary:: Why $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}$ instead of $\frac{dT}{dq}$?

Why, in lagrangian mechanics, do we calculate: $\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}$ to get the (generalised) momentum change in time
instead of $\frac{d T}{dq}$?

(T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external forces are applied)

Your equation would hold in at least one particular case: When the momentum change in time is zero and there is actually no force field acting on the body, so the potential energy is also zero. In all other cases, it would just be a blunder, I guess.