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Dimensional analysis of classical action
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Hamilton's action for classical systems has the units Joule seconds ($rm Jcdot s$), which in base units is $rm kg:m^2/s$.
Does the $rm m^2$ have anything to do with area?
I'm having a hard time "reading" the units here.
classical-mechanics action
edited 46 mins ago
Emilio Pisanty
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crastinus
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Hamilton's action for classical systems has the units Joule seconds ($rm Jcdot s$), which in base units is $rm kg:m^2/s$.
Does the $rm m^2$ have anything to do with area?
I'm having a hard time "reading" the units here.
classical-mechanics action
edited 46 mins ago
Emilio Pisanty
75.6k18181372
asked 2 hours ago
crastinus
62
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Hamilton's action for classical systems has the units Joule seconds ($rm Jcdot s$), which in base units is $rm kg:m^2/s$.
Does the $rm m^2$ have anything to do with area?
I'm having a hard time "reading" the units here.
classical-mechanics action
edited 46 mins ago
Emilio Pisanty
75.6k18181372
asked 2 hours ago
crastinus
62
Hamilton's action for classical systems has the units Joule seconds ($rm Jcdot s$), which in base units is $rm kg:m^2/s$.
Does the $rm m^2$ have anything to do with area?
I'm having a hard time "reading" the units here.
classical-mechanics action
classical-mechanics action
edited 46 mins ago
Emilio Pisanty
75.6k18181372
asked 2 hours ago
crastinus
62
edited 46 mins ago
Emilio Pisanty
75.6k18181372
asked 2 hours ago
crastinus
62
edited 46 mins ago
Emilio Pisanty
75.6k18181372
edited 46 mins ago
Emilio Pisanty
75.6k18181372
edited 46 mins ago
Emilio Pisanty
75.6k18181372
75.6k18181372
asked 2 hours ago
crastinus
62
asked 2 hours ago
crastinus
62
asked 2 hours ago
crastinus
62
62
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
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up vote
4
down vote
Not everything has a "nice" explanation in base units. Students often get this idea because a lot of things in introductory mechanics are explained this way ("acceleration has units of $textm/texts^2$, because it's the number of meters per second you speed up per second"). But this works only when you're dealing with very very simple quantities, like a ratio of length to time for speed.
In general you simply can't get an intuition for a general quantity only by looking at the base units. You're not missing anything here!
answered 49 mins ago
knzhou
34k897170
add a comment |Â
up vote
2
down vote
It's not worth trying to interpret the units that way. To give another "what?" example, light intensity as power per area turns out to be kg/s$^3$. What does that mean? Arguably, nothing beyond the more insightful W/m$^2$. You can change the choice of base units in analogy with a change of basis for a vector space. It's probably more natural to use as our base dimensions those of $c, ,h$ etc. rather than the SI units, which are a historical accident. On this view, action has dimension $h^1$.
answered 58 mins ago
J.G.
8,32421124
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up vote
2
down vote
The dimensions of action are generally best understood in the light of a concept known as canonically conjugate pairs of variables. These are important when you formalize newtonian mechanics using the lagrangian and hamiltonian formalisms of analytical mechanics, as well as for quantum mechanics.
Basically, the concept of a conjugate pair of variables is essential to the formulation of the fully-grown equations of motion as Hamilton's equations, as well as to providing a natural geometric structure that underpins classical mechanics, via an operation known as the Poisson bracket. However, knowing what conjugate pairs actually are and do isn't all that important if what you want is a nice look at the dimensionality of action $-$ they're just great at providing nice ways to slice $S$.
So, let me put up some examples. Action has dimensions of $[S] = [M L^2/T]$, and this can be thought of as the products of:
- position $[x]=[L]$ and momentum $[p] = [ML/T]$
- energy $[E] = [ML^2/T^2]$ and time $[t]=[T]$
- angular momentum $[l] = [Ltimes ML/T] = [ML^2/T]$ and angle $[theta]=[1]$
- electric vector potential $[A] = [ML/TQ]$ and electric dipole moment $[d]=[QL]$
The fact that the first two examples crop up elsewhere is, of course, no coincidence.
If you absolutely insisted on interpreting the $L^2$ as a single grouping, then the thing to do would be to find a system that was well characterized by a generalized coordinate with dimensions of area, which would then have a conjugate (vaguely related to surface tension) with dimensions $[M/T]$, though of course that's pretty contrived.
answered 26 mins ago
Emilio Pisanty
75.6k18181372
add a comment |Â
up vote
0
down vote
Actually maybe the simplest thing to do is to think of Joules as energy, i.e. having the same units as $frac12mv^2$. Clearly here the unit of length is not related to the area, but simply to the fact that energy scales likes the displacement squared. You can multiply everything by $s$ without affecting the argument and correctly get kg$cdot$m$^2$s.
answered 48 mins ago
ZeroTheHero
17.2k52556
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Not everything has a "nice" explanation in base units. Students often get this idea because a lot of things in introductory mechanics are explained this way ("acceleration has units of $textm/texts^2$, because it's the number of meters per second you speed up per second"). But this works only when you're dealing with very very simple quantities, like a ratio of length to time for speed.
In general you simply can't get an intuition for a general quantity only by looking at the base units. You're not missing anything here!
answered 49 mins ago
knzhou
34k897170
add a comment |Â
up vote
4
down vote
Not everything has a "nice" explanation in base units. Students often get this idea because a lot of things in introductory mechanics are explained this way ("acceleration has units of $textm/texts^2$, because it's the number of meters per second you speed up per second"). But this works only when you're dealing with very very simple quantities, like a ratio of length to time for speed.
In general you simply can't get an intuition for a general quantity only by looking at the base units. You're not missing anything here!
answered 49 mins ago
knzhou
34k897170
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Not everything has a "nice" explanation in base units. Students often get this idea because a lot of things in introductory mechanics are explained this way ("acceleration has units of $textm/texts^2$, because it's the number of meters per second you speed up per second"). But this works only when you're dealing with very very simple quantities, like a ratio of length to time for speed.
In general you simply can't get an intuition for a general quantity only by looking at the base units. You're not missing anything here!
answered 49 mins ago
knzhou
34k897170
Not everything has a "nice" explanation in base units. Students often get this idea because a lot of things in introductory mechanics are explained this way ("acceleration has units of $textm/texts^2$, because it's the number of meters per second you speed up per second"). But this works only when you're dealing with very very simple quantities, like a ratio of length to time for speed.
In general you simply can't get an intuition for a general quantity only by looking at the base units. You're not missing anything here!
answered 49 mins ago
knzhou
34k897170
answered 49 mins ago
knzhou
34k897170
answered 49 mins ago
knzhou
34k897170
answered 49 mins ago
knzhou
34k897170
34k897170
add a comment |Â
add a comment |Â
up vote
2
down vote
It's not worth trying to interpret the units that way. To give another "what?" example, light intensity as power per area turns out to be kg/s$^3$. What does that mean? Arguably, nothing beyond the more insightful W/m$^2$. You can change the choice of base units in analogy with a change of basis for a vector space. It's probably more natural to use as our base dimensions those of $c, ,h$ etc. rather than the SI units, which are a historical accident. On this view, action has dimension $h^1$.
answered 58 mins ago
J.G.
8,32421124
add a comment |Â
up vote
2
down vote
It's not worth trying to interpret the units that way. To give another "what?" example, light intensity as power per area turns out to be kg/s$^3$. What does that mean? Arguably, nothing beyond the more insightful W/m$^2$. You can change the choice of base units in analogy with a change of basis for a vector space. It's probably more natural to use as our base dimensions those of $c, ,h$ etc. rather than the SI units, which are a historical accident. On this view, action has dimension $h^1$.
answered 58 mins ago
J.G.
8,32421124
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It's not worth trying to interpret the units that way. To give another "what?" example, light intensity as power per area turns out to be kg/s$^3$. What does that mean? Arguably, nothing beyond the more insightful W/m$^2$. You can change the choice of base units in analogy with a change of basis for a vector space. It's probably more natural to use as our base dimensions those of $c, ,h$ etc. rather than the SI units, which are a historical accident. On this view, action has dimension $h^1$.
answered 58 mins ago
J.G.
8,32421124
It's not worth trying to interpret the units that way. To give another "what?" example, light intensity as power per area turns out to be kg/s$^3$. What does that mean? Arguably, nothing beyond the more insightful W/m$^2$. You can change the choice of base units in analogy with a change of basis for a vector space. It's probably more natural to use as our base dimensions those of $c, ,h$ etc. rather than the SI units, which are a historical accident. On this view, action has dimension $h^1$.
answered 58 mins ago
J.G.
8,32421124
answered 58 mins ago
J.G.
8,32421124
answered 58 mins ago
J.G.
8,32421124
answered 58 mins ago
J.G.
8,32421124
8,32421124
add a comment |Â
add a comment |Â
up vote
2
down vote
The dimensions of action are generally best understood in the light of a concept known as canonically conjugate pairs of variables. These are important when you formalize newtonian mechanics using the lagrangian and hamiltonian formalisms of analytical mechanics, as well as for quantum mechanics.
Basically, the concept of a conjugate pair of variables is essential to the formulation of the fully-grown equations of motion as Hamilton's equations, as well as to providing a natural geometric structure that underpins classical mechanics, via an operation known as the Poisson bracket. However, knowing what conjugate pairs actually are and do isn't all that important if what you want is a nice look at the dimensionality of action $-$ they're just great at providing nice ways to slice $S$.
So, let me put up some examples. Action has dimensions of $[S] = [M L^2/T]$, and this can be thought of as the products of:
- position $[x]=[L]$ and momentum $[p] = [ML/T]$
- energy $[E] = [ML^2/T^2]$ and time $[t]=[T]$
- angular momentum $[l] = [Ltimes ML/T] = [ML^2/T]$ and angle $[theta]=[1]$
- electric vector potential $[A] = [ML/TQ]$ and electric dipole moment $[d]=[QL]$
The fact that the first two examples crop up elsewhere is, of course, no coincidence.
If you absolutely insisted on interpreting the $L^2$ as a single grouping, then the thing to do would be to find a system that was well characterized by a generalized coordinate with dimensions of area, which would then have a conjugate (vaguely related to surface tension) with dimensions $[M/T]$, though of course that's pretty contrived.
answered 26 mins ago
Emilio Pisanty
75.6k18181372
add a comment |Â
up vote
2
down vote
The dimensions of action are generally best understood in the light of a concept known as canonically conjugate pairs of variables. These are important when you formalize newtonian mechanics using the lagrangian and hamiltonian formalisms of analytical mechanics, as well as for quantum mechanics.
Basically, the concept of a conjugate pair of variables is essential to the formulation of the fully-grown equations of motion as Hamilton's equations, as well as to providing a natural geometric structure that underpins classical mechanics, via an operation known as the Poisson bracket. However, knowing what conjugate pairs actually are and do isn't all that important if what you want is a nice look at the dimensionality of action $-$ they're just great at providing nice ways to slice $S$.
So, let me put up some examples. Action has dimensions of $[S] = [M L^2/T]$, and this can be thought of as the products of:
- position $[x]=[L]$ and momentum $[p] = [ML/T]$
- energy $[E] = [ML^2/T^2]$ and time $[t]=[T]$
- angular momentum $[l] = [Ltimes ML/T] = [ML^2/T]$ and angle $[theta]=[1]$
- electric vector potential $[A] = [ML/TQ]$ and electric dipole moment $[d]=[QL]$
The fact that the first two examples crop up elsewhere is, of course, no coincidence.
If you absolutely insisted on interpreting the $L^2$ as a single grouping, then the thing to do would be to find a system that was well characterized by a generalized coordinate with dimensions of area, which would then have a conjugate (vaguely related to surface tension) with dimensions $[M/T]$, though of course that's pretty contrived.
answered 26 mins ago
Emilio Pisanty
75.6k18181372
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The dimensions of action are generally best understood in the light of a concept known as canonically conjugate pairs of variables. These are important when you formalize newtonian mechanics using the lagrangian and hamiltonian formalisms of analytical mechanics, as well as for quantum mechanics.
Basically, the concept of a conjugate pair of variables is essential to the formulation of the fully-grown equations of motion as Hamilton's equations, as well as to providing a natural geometric structure that underpins classical mechanics, via an operation known as the Poisson bracket. However, knowing what conjugate pairs actually are and do isn't all that important if what you want is a nice look at the dimensionality of action $-$ they're just great at providing nice ways to slice $S$.
So, let me put up some examples. Action has dimensions of $[S] = [M L^2/T]$, and this can be thought of as the products of:
- position $[x]=[L]$ and momentum $[p] = [ML/T]$
- energy $[E] = [ML^2/T^2]$ and time $[t]=[T]$
- angular momentum $[l] = [Ltimes ML/T] = [ML^2/T]$ and angle $[theta]=[1]$
- electric vector potential $[A] = [ML/TQ]$ and electric dipole moment $[d]=[QL]$
The fact that the first two examples crop up elsewhere is, of course, no coincidence.
If you absolutely insisted on interpreting the $L^2$ as a single grouping, then the thing to do would be to find a system that was well characterized by a generalized coordinate with dimensions of area, which would then have a conjugate (vaguely related to surface tension) with dimensions $[M/T]$, though of course that's pretty contrived.
answered 26 mins ago
Emilio Pisanty
75.6k18181372
The dimensions of action are generally best understood in the light of a concept known as canonically conjugate pairs of variables. These are important when you formalize newtonian mechanics using the lagrangian and hamiltonian formalisms of analytical mechanics, as well as for quantum mechanics.
Basically, the concept of a conjugate pair of variables is essential to the formulation of the fully-grown equations of motion as Hamilton's equations, as well as to providing a natural geometric structure that underpins classical mechanics, via an operation known as the Poisson bracket. However, knowing what conjugate pairs actually are and do isn't all that important if what you want is a nice look at the dimensionality of action $-$ they're just great at providing nice ways to slice $S$.
So, let me put up some examples. Action has dimensions of $[S] = [M L^2/T]$, and this can be thought of as the products of:
- position $[x]=[L]$ and momentum $[p] = [ML/T]$
- energy $[E] = [ML^2/T^2]$ and time $[t]=[T]$
- angular momentum $[l] = [Ltimes ML/T] = [ML^2/T]$ and angle $[theta]=[1]$
- electric vector potential $[A] = [ML/TQ]$ and electric dipole moment $[d]=[QL]$
The fact that the first two examples crop up elsewhere is, of course, no coincidence.
If you absolutely insisted on interpreting the $L^2$ as a single grouping, then the thing to do would be to find a system that was well characterized by a generalized coordinate with dimensions of area, which would then have a conjugate (vaguely related to surface tension) with dimensions $[M/T]$, though of course that's pretty contrived.
answered 26 mins ago
Emilio Pisanty
75.6k18181372
edited 16 mins ago
answered 26 mins ago
Emilio Pisanty
75.6k18181372
answered 26 mins ago
Emilio Pisanty
75.6k18181372
answered 26 mins ago
Emilio Pisanty
75.6k18181372
75.6k18181372
add a comment |Â
add a comment |Â
up vote
0
down vote
Actually maybe the simplest thing to do is to think of Joules as energy, i.e. having the same units as $frac12mv^2$. Clearly here the unit of length is not related to the area, but simply to the fact that energy scales likes the displacement squared. You can multiply everything by $s$ without affecting the argument and correctly get kg$cdot$m$^2$s.
answered 48 mins ago
ZeroTheHero
17.2k52556
add a comment |Â
up vote
0
down vote
Actually maybe the simplest thing to do is to think of Joules as energy, i.e. having the same units as $frac12mv^2$. Clearly here the unit of length is not related to the area, but simply to the fact that energy scales likes the displacement squared. You can multiply everything by $s$ without affecting the argument and correctly get kg$cdot$m$^2$s.
answered 48 mins ago
ZeroTheHero
17.2k52556
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Actually maybe the simplest thing to do is to think of Joules as energy, i.e. having the same units as $frac12mv^2$. Clearly here the unit of length is not related to the area, but simply to the fact that energy scales likes the displacement squared. You can multiply everything by $s$ without affecting the argument and correctly get kg$cdot$m$^2$s.
answered 48 mins ago
ZeroTheHero
17.2k52556
Actually maybe the simplest thing to do is to think of Joules as energy, i.e. having the same units as $frac12mv^2$. Clearly here the unit of length is not related to the area, but simply to the fact that energy scales likes the displacement squared. You can multiply everything by $s$ without affecting the argument and correctly get kg$cdot$m$^2$s.
answered 48 mins ago
ZeroTheHero
17.2k52556
answered 48 mins ago
ZeroTheHero
17.2k52556
answered 48 mins ago
ZeroTheHero
17.2k52556
answered 48 mins ago
ZeroTheHero
17.2k52556
17.2k52556
add a comment |Â
add a comment |Â
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