The definition of generalised momentum

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In Lagrangian mechanics, the change in generalized momentum over time is calculated using the expression d/dt(∂T/∂ṡq) rather than dT/dq because the latter does not yield a meaningful result in multi-variable contexts. The kinetic energy T is defined as the work required to accelerate a body, and its partial derivative with respect to velocity gives the generalized momentum. The use of partial derivatives is crucial as it ensures the equations remain valid across different coordinate systems, unlike total derivatives which can lead to inconsistencies. The Lagrangian formulation aligns with Newtonian mechanics, providing a consistent framework for deriving equations of motion. This approach emphasizes the importance of correctly identifying the relationship between kinetic energy and generalized forces in the context of Lagrangian dynamics.
  • #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.
 
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  • #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
 
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  • #35
Valuable guide,
Physics is not Algebra
Goes mostly ignored
 
  • #36
Roses are red, 🌹
violets are blue, 💙
I think 50V is being a little condescending,
how about you?
 
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  • #37
PeroK said:
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.

BMS said:
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.
 
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  • #38
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.
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.
 
  • #39
robphy said:
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.
 
  • #40
vanhees71 said:
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?
 
  • #41
robphy said:
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.
 
  • #42
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.
 
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  • #43
##\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.
 
  • #44
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)
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.
 

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