# 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

(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).

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

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

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

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

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.

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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}##.

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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).

Gold Member
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 ?

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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

(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.

Delta2
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.

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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.

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baw
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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
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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
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?

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

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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.

etotheipi
baw
Ok, thank you

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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".

Delta2
Staff Emeritus
aren't joules and Newton-meters same thing?

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

robphy and vanhees71
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)##.

Staff Emeritus
Yes

No.

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

robphy and vanhees71
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.

Hamiltonian and baw
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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?

PeroK and vanhees71
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

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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!

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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 ;-)).

etotheipi
Staff Emeritus
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.

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.

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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

@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. 😜

weirdoguy
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@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 , 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.

Staff Emeritus
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?

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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