Euler-Lagrange formalism with SymPy
I am trying to use SymPy to apply the Euler-Lagrange formalism to the following Lagrangian function:
$$mathcal{L}(x, dot{x}) = mathcal{K}(x, dot{x}) - mathcal{P}(x) = dfrac{1}{2} mathfrak{m} dot{x} - mathfrak{m} g x.$$
The result should be
$$
f = frac{d}{dt} frac{partialmathcal{L}}{partialdot{x}}
- frac{partialmathcal{L}}{partial x}
= mathfrak{m} ddot{x} + mathfrak{m} g.$$
However, I get $f = mathfrak{m} g$ with SymPy...
Below is the source of my jupyter notebook.
In [1]: from sympy import *
from sympy.physics.mechanics import *
In [2]: m,g = symbols('m g', real=True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
xdd = dynamicsymbols('x', 2)
In [3]: kin_energy = 1/2 * m * xd ** 2
pot_energy = m * g * x
In [4]: L = kin_energy - pot_energy
f = diff(L, xdd) - diff(L, x)
In [5]: pprint(f)
Out[5]: g⋅m
euler-lagrange-equation python
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
add a comment |
I am trying to use SymPy to apply the Euler-Lagrange formalism to the following Lagrangian function:
$$mathcal{L}(x, dot{x}) = mathcal{K}(x, dot{x}) - mathcal{P}(x) = dfrac{1}{2} mathfrak{m} dot{x} - mathfrak{m} g x.$$
The result should be
$$
f = frac{d}{dt} frac{partialmathcal{L}}{partialdot{x}}
- frac{partialmathcal{L}}{partial x}
= mathfrak{m} ddot{x} + mathfrak{m} g.$$
However, I get $f = mathfrak{m} g$ with SymPy...
Below is the source of my jupyter notebook.
In [1]: from sympy import *
from sympy.physics.mechanics import *
In [2]: m,g = symbols('m g', real=True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
xdd = dynamicsymbols('x', 2)
In [3]: kin_energy = 1/2 * m * xd ** 2
pot_energy = m * g * x
In [4]: L = kin_energy - pot_energy
f = diff(L, xdd) - diff(L, x)
In [5]: pprint(f)
Out[5]: g⋅m
euler-lagrange-equation python
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
add a comment |
I am trying to use SymPy to apply the Euler-Lagrange formalism to the following Lagrangian function:
$$mathcal{L}(x, dot{x}) = mathcal{K}(x, dot{x}) - mathcal{P}(x) = dfrac{1}{2} mathfrak{m} dot{x} - mathfrak{m} g x.$$
The result should be
$$
f = frac{d}{dt} frac{partialmathcal{L}}{partialdot{x}}
- frac{partialmathcal{L}}{partial x}
= mathfrak{m} ddot{x} + mathfrak{m} g.$$
However, I get $f = mathfrak{m} g$ with SymPy...
Below is the source of my jupyter notebook.
In [1]: from sympy import *
from sympy.physics.mechanics import *
In [2]: m,g = symbols('m g', real=True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
xdd = dynamicsymbols('x', 2)
In [3]: kin_energy = 1/2 * m * xd ** 2
pot_energy = m * g * x
In [4]: L = kin_energy - pot_energy
f = diff(L, xdd) - diff(L, x)
In [5]: pprint(f)
Out[5]: g⋅m
euler-lagrange-equation python
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
I am trying to use SymPy to apply the Euler-Lagrange formalism to the following Lagrangian function:
$$mathcal{L}(x, dot{x}) = mathcal{K}(x, dot{x}) - mathcal{P}(x) = dfrac{1}{2} mathfrak{m} dot{x} - mathfrak{m} g x.$$
The result should be
$$
f = frac{d}{dt} frac{partialmathcal{L}}{partialdot{x}}
- frac{partialmathcal{L}}{partial x}
= mathfrak{m} ddot{x} + mathfrak{m} g.$$
However, I get $f = mathfrak{m} g$ with SymPy...
Below is the source of my jupyter notebook.
In [1]: from sympy import *
from sympy.physics.mechanics import *
In [2]: m,g = symbols('m g', real=True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
xdd = dynamicsymbols('x', 2)
In [3]: kin_energy = 1/2 * m * xd ** 2
pot_energy = m * g * x
In [4]: L = kin_energy - pot_energy
f = diff(L, xdd) - diff(L, x)
In [5]: pprint(f)
Out[5]: g⋅m
euler-lagrange-equation python
euler-lagrange-equation python
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
edited Nov 26 '18 at 20:23
gt6989b
33.1k22452
33.1k22452
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
asked Nov 26 '18 at 20:21
Henrique Ferrolho
1233
1233
add a comment |
add a comment |
1 Answer
1
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oldest
votes
There are a couple of errors in your code
1/2
Depending on the version of python you're using, this could be interpreted as either '0' or '1/2'. So To avoid problems just use
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
${rm d}/{rm d}t(partial L/ partial dot{x})$
Note that
$$
frac{{rm d}}{{rm d}t}frac{partial mathcal{L}}{partial dot{x}} not= frac{partial mathcal{L}}{partial ddot{x}}
$$
If you implement these two things, you should get
m = symbols('m', real = True)
g = symbols('g', real = True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
L = kin_energy - pot_energy
f = diff(diff(L, xd), 't') - diff(L, x)
pprint(f)
>>> 2
d
g⋅m + 1.0⋅m⋅───(x(t))
2
dt
answered Nov 26 '18 at 21:02
caverac
13.8k21130
I am using Python 3, so the1 / 2should not pose a problem. I guess what I did wrong was really thediff(diff(L, xd), 't'). Thank you for your answer!
– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are a couple of errors in your code
1/2
Depending on the version of python you're using, this could be interpreted as either '0' or '1/2'. So To avoid problems just use
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
${rm d}/{rm d}t(partial L/ partial dot{x})$
Note that
$$
frac{{rm d}}{{rm d}t}frac{partial mathcal{L}}{partial dot{x}} not= frac{partial mathcal{L}}{partial ddot{x}}
$$
If you implement these two things, you should get
m = symbols('m', real = True)
g = symbols('g', real = True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
L = kin_energy - pot_energy
f = diff(diff(L, xd), 't') - diff(L, x)
pprint(f)
>>> 2
d
g⋅m + 1.0⋅m⋅───(x(t))
2
dt
answered Nov 26 '18 at 21:02
caverac
13.8k21130
I am using Python 3, so the1 / 2should not pose a problem. I guess what I did wrong was really thediff(diff(L, xd), 't'). Thank you for your answer!
– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
add a comment |
There are a couple of errors in your code
1/2
Depending on the version of python you're using, this could be interpreted as either '0' or '1/2'. So To avoid problems just use
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
${rm d}/{rm d}t(partial L/ partial dot{x})$
Note that
$$
frac{{rm d}}{{rm d}t}frac{partial mathcal{L}}{partial dot{x}} not= frac{partial mathcal{L}}{partial ddot{x}}
$$
If you implement these two things, you should get
m = symbols('m', real = True)
g = symbols('g', real = True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
L = kin_energy - pot_energy
f = diff(diff(L, xd), 't') - diff(L, x)
pprint(f)
>>> 2
d
g⋅m + 1.0⋅m⋅───(x(t))
2
dt
answered Nov 26 '18 at 21:02
caverac
13.8k21130
I am using Python 3, so the1 / 2should not pose a problem. I guess what I did wrong was really thediff(diff(L, xd), 't'). Thank you for your answer!
– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
add a comment |
There are a couple of errors in your code
1/2
Depending on the version of python you're using, this could be interpreted as either '0' or '1/2'. So To avoid problems just use
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
${rm d}/{rm d}t(partial L/ partial dot{x})$
Note that
$$
frac{{rm d}}{{rm d}t}frac{partial mathcal{L}}{partial dot{x}} not= frac{partial mathcal{L}}{partial ddot{x}}
$$
If you implement these two things, you should get
m = symbols('m', real = True)
g = symbols('g', real = True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
L = kin_energy - pot_energy
f = diff(diff(L, xd), 't') - diff(L, x)
pprint(f)
>>> 2
d
g⋅m + 1.0⋅m⋅───(x(t))
2
dt
answered Nov 26 '18 at 21:02
caverac
13.8k21130
There are a couple of errors in your code
1/2
Depending on the version of python you're using, this could be interpreted as either '0' or '1/2'. So To avoid problems just use
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
${rm d}/{rm d}t(partial L/ partial dot{x})$
Note that
$$
frac{{rm d}}{{rm d}t}frac{partial mathcal{L}}{partial dot{x}} not= frac{partial mathcal{L}}{partial ddot{x}}
$$
If you implement these two things, you should get
m = symbols('m', real = True)
g = symbols('g', real = True)
x = dynamicsymbols('x')
xd = dynamicsymbols('x', 1)
kin_energy = 0.5 * m * xd ** 2
pot_energy = m * g * x
L = kin_energy - pot_energy
f = diff(diff(L, xd), 't') - diff(L, x)
pprint(f)
>>> 2
d
g⋅m + 1.0⋅m⋅───(x(t))
2
dt
answered Nov 26 '18 at 21:02
caverac
13.8k21130
answered Nov 26 '18 at 21:02
caverac
13.8k21130
answered Nov 26 '18 at 21:02
caverac
13.8k21130
answered Nov 26 '18 at 21:02
caverac
13.8k21130
13.8k21130
I am using Python 3, so the1 / 2should not pose a problem. I guess what I did wrong was really thediff(diff(L, xd), 't'). Thank you for your answer!
– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
add a comment |
I am using Python 3, so the1 / 2should not pose a problem. I guess what I did wrong was really thediff(diff(L, xd), 't'). Thank you for your answer!
– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
I am using Python 3, so the
1 / 2 should not pose a problem. I guess what I did wrong was really the diff(diff(L, xd), 't'). Thank you for your answer!– Henrique Ferrolho
Nov 27 '18 at 0:15
I am using Python 3, so the
1 / 2 should not pose a problem. I guess what I did wrong was really the diff(diff(L, xd), 't'). Thank you for your answer!– Henrique Ferrolho
Nov 27 '18 at 0:15
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
@HenriqueFerrolho happy to help
– caverac
Nov 27 '18 at 0:43
add a comment |
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