The definition of generalised momentum

In summary, the Lagrangian approach in Lagrangian mechanics uses partial derivatives rather than total derivatives, which allows for a more general property that works in various coordinate systems. This approach is based on the definition of kinetic energy and is consistent with Newtonian mechanics. The alternative approach of taking the reverse of the kinetic energy derivative does not work and would not be consistent with other coordinate systems. This is how Lagrange knew that the Lagrangian approach was correct.
  • #1
baw
9
0
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).
 
Physics news on Phys.org
  • #2
[tex]L=L(q,\dot{q},t)[/tex]
[tex]\frac{d}{dt}=\dot{q}\frac{\partial}{\partial q}+\ddot{q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial t}[/tex]

[tex]\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)[/tex]

[tex]\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}[/tex]

[tex]\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)[/tex]

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.
 
Last edited:
  • #3
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}##.
 
Last edited:
  • #4
anuttarasammyak said:
[tex]L=L(q,\dot{q},t)[/tex]
[tex]\frac{d}{dt}=\dot{q}\frac{\partial}{\partial q}+\ddot{q}\frac{\partial}{\partial \dot{q}}+\frac{\partial}{\partial t}[/tex]

[tex]\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)[/tex]

[tex]\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}[/tex]

[tex]\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)[/tex]

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?
 
  • #5
andrewkirk said:
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).
 
  • #6
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 ?
 
  • #7
baw said:
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.
 
  • Like
Likes Delta2
  • #8
anuttarasammyak said:
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.
 
Last edited:
  • #9
If you put ##\frac{dL}{dq}## into LHS of Lagrange equation, the equation would become
[tex]\frac{dL}{dq}=\frac{\partial L}{\partial q}[/tex]
Obviously wrong.

Newton states formula as physics law
[tex]F=\frac{dp}{dt}[/tex]
to SOLVE it. Not solving but applying this law again you get identity
[tex]F=F[/tex]
and
[tex]\frac{dp}{dt}=\frac{dp}{dt}[/tex]
They are useless.
 
Last edited:
  • Like
Likes baw
  • #10
anuttarasammyak said:
If you put ##\frac{dL}{dq}## into LHS of Lagrange equation, the equation would become
[tex]\frac{dL}{dq}=\frac{\partial L}{\partial q}[/tex]
Obviously wrong.
Moreover, the LHS is not well-defined.
 
  • #11
anuttarasammyak said:
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.
 
  • Skeptical
Likes PeroK
  • #12
baw said:
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.
 
  • Like
Likes baw
  • #13
anuttarasammyak said:
If you put ##\frac{dL}{dq}## into LHS of Lagrange equation, the equation would become
[tex]\frac{dL}{dq}=\frac{\partial L}{\partial q}[/tex]
Obviously wrong.

Newton states formula as physics law
[tex]F=\frac{dp}{dt}[/tex]
to SOLVE it. Not solving but applying this law again you get identity
[tex]F=F[/tex]
and
[tex]\frac{dp}{dt}=\frac{dp}{dt}[/tex]
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?
 
  • #14
You will find in formula of (2) of comment #2 they differ by the first and the second RHS terms.
 
  • #15
baw said:
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.
 
  • Like
Likes etotheipi
  • #16
Ok, thank you
 
  • #17
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".
 
  • Like
Likes Delta2
  • #18
baw said:
aren't joules and Newton-meters same thing?

No. Energy is not torque, even if they have the same dimensions.
 
  • Like
Likes robphy and vanhees71
  • #19
baw said:
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)##.
 
  • #20
etotheipi said:
Yes

No.

Energy is not torque, even if they have the same dimensions.
 
  • Like
Likes robphy and vanhees71
  • #21
Vanadium 50 said:
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.
 
  • Like
Likes Hamiltonian and baw
  • #22
etotheipi said:
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?
 
  • Like
Likes PeroK and vanhees71
  • #23
Vanadium 50 said:
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:
andrewkirk said:
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 :-p
 
  • #24
baw said:
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:

PeroK said:
##\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!
 
Last edited:
  • #25
etotheipi said:
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 :-p
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 ;-)).
 
  • Like
Likes etotheipi
  • #26
vanhees71 said:
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.
 
  • #27
Vanadium 50 said:
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.

Vanadium 50 said:
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.
 
  • #29
Vanadium 50 said:
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
 
  • Like
Likes DrStupid, robphy and Vanadium 50
  • #30
@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. 😜
 
  • Like
Likes weirdoguy
  • #31
@etotheipi , that sounds a lot like "Now that I have the last word, I invite you to shut up" An awful lot like that.

PeroK said:
Wikipedia agrees with you

Thanks.
 
  • #32
Vanadium 50 said:
@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. :wink:
 
  • #33
Again, "Now that I have the last word, I invite you to stop with your childish bickering" :confused:

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?
 
  • #34
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
 
  • Love
Likes Vanadium 50
  • #35
Valuable guide,
Physics is not Algebra
Goes mostly ignored
 

Similar threads

Replies
3
Views
996
Replies
4
Views
897
  • Mechanics
Replies
5
Views
1K
Replies
11
Views
1K
Replies
2
Views
232
Replies
5
Views
908
  • Mechanics
Replies
3
Views
1K
Replies
14
Views
2K
Replies
6
Views
2K
Replies
4
Views
1K
Back
Top