The definition of generalised momentum

[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.