Stability criterion for leapfrog in relativistic physics.
$begingroup$
I am doing a 2D MD simulations of charge carriers in graphene using the Leapfrog algorithm. I am trying to prove that, in some specific cases (when distance between particles is small), the method is unstable.
The Hamiltonian is given as $H = |vec{p_{1}}| + |vec{p_2}| - frac{alpha}{r_{12}}$. With $p$ the momentums, $alpha$ some numerical constant and $r_{12}$ some constants.
The leapfrog iteration would be:
$$
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}\
vec{x_i} to vec{x_i} + Delta t hat{p_i} \
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}
$$
where the vectors with hat's are the direction vectors.
I expect the stability to break down when $r_{ij}$ gets small. Yet, I tried using analysis from this (page 7 onward) source and it didn't really help.
Also, a different way to derive leapfrog is to define coordinate shift operators $L_p$ and $L_r$ that move $p to p +Delta p$ and $x to x + Delta x$. Then the differential equation can be written as:
$$
f(p(t), r(t)) = expleft(t (L_p + L_r) right) f(p(0), r(0)).
$$
Using Trotter's identity we can write:
$$
expleft( Delta t (A + B) right) approx expleft( 0.5Delta t A right) expleft( Delta t B right) expleft(0.5 Delta t A right)
$$
and this is exactly the Leapfrog algorithm. I tried figuring out if things breakdown in 1D case of two particles going straight at each other. To do this I worked out the commutator term $[L_p,L_r]$ that comes up in Zassenhaus formula. This also didn't help.
Anyone has a fresh look or has some derivations that go beyond basic 1D cases where things go nicely? Any help is appreciated.
numerical-methods mathematical-physics
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
$endgroup$
|
show 12 more comments
$begingroup$
I am doing a 2D MD simulations of charge carriers in graphene using the Leapfrog algorithm. I am trying to prove that, in some specific cases (when distance between particles is small), the method is unstable.
The Hamiltonian is given as $H = |vec{p_{1}}| + |vec{p_2}| - frac{alpha}{r_{12}}$. With $p$ the momentums, $alpha$ some numerical constant and $r_{12}$ some constants.
The leapfrog iteration would be:
$$
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}\
vec{x_i} to vec{x_i} + Delta t hat{p_i} \
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}
$$
where the vectors with hat's are the direction vectors.
I expect the stability to break down when $r_{ij}$ gets small. Yet, I tried using analysis from this (page 7 onward) source and it didn't really help.
Also, a different way to derive leapfrog is to define coordinate shift operators $L_p$ and $L_r$ that move $p to p +Delta p$ and $x to x + Delta x$. Then the differential equation can be written as:
$$
f(p(t), r(t)) = expleft(t (L_p + L_r) right) f(p(0), r(0)).
$$
Using Trotter's identity we can write:
$$
expleft( Delta t (A + B) right) approx expleft( 0.5Delta t A right) expleft( Delta t B right) expleft(0.5 Delta t A right)
$$
and this is exactly the Leapfrog algorithm. I tried figuring out if things breakdown in 1D case of two particles going straight at each other. To do this I worked out the commutator term $[L_p,L_r]$ that comes up in Zassenhaus formula. This also didn't help.
Anyone has a fresh look or has some derivations that go beyond basic 1D cases where things go nicely? Any help is appreciated.
numerical-methods mathematical-physics
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
$endgroup$
$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
1
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
1
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
1
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58
|
show 12 more comments
$begingroup$
I am doing a 2D MD simulations of charge carriers in graphene using the Leapfrog algorithm. I am trying to prove that, in some specific cases (when distance between particles is small), the method is unstable.
The Hamiltonian is given as $H = |vec{p_{1}}| + |vec{p_2}| - frac{alpha}{r_{12}}$. With $p$ the momentums, $alpha$ some numerical constant and $r_{12}$ some constants.
The leapfrog iteration would be:
$$
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}\
vec{x_i} to vec{x_i} + Delta t hat{p_i} \
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}
$$
where the vectors with hat's are the direction vectors.
I expect the stability to break down when $r_{ij}$ gets small. Yet, I tried using analysis from this (page 7 onward) source and it didn't really help.
Also, a different way to derive leapfrog is to define coordinate shift operators $L_p$ and $L_r$ that move $p to p +Delta p$ and $x to x + Delta x$. Then the differential equation can be written as:
$$
f(p(t), r(t)) = expleft(t (L_p + L_r) right) f(p(0), r(0)).
$$
Using Trotter's identity we can write:
$$
expleft( Delta t (A + B) right) approx expleft( 0.5Delta t A right) expleft( Delta t B right) expleft(0.5 Delta t A right)
$$
and this is exactly the Leapfrog algorithm. I tried figuring out if things breakdown in 1D case of two particles going straight at each other. To do this I worked out the commutator term $[L_p,L_r]$ that comes up in Zassenhaus formula. This also didn't help.
Anyone has a fresh look or has some derivations that go beyond basic 1D cases where things go nicely? Any help is appreciated.
numerical-methods mathematical-physics
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
$endgroup$
I am doing a 2D MD simulations of charge carriers in graphene using the Leapfrog algorithm. I am trying to prove that, in some specific cases (when distance between particles is small), the method is unstable.
The Hamiltonian is given as $H = |vec{p_{1}}| + |vec{p_2}| - frac{alpha}{r_{12}}$. With $p$ the momentums, $alpha$ some numerical constant and $r_{12}$ some constants.
The leapfrog iteration would be:
$$
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}\
vec{x_i} to vec{x_i} + Delta t hat{p_i} \
vec{p_i} to vec{p_i} - 0.5 Delta t frac{alpha}{r_{ij}^2} hat{r_{ij}}
$$
where the vectors with hat's are the direction vectors.
I expect the stability to break down when $r_{ij}$ gets small. Yet, I tried using analysis from this (page 7 onward) source and it didn't really help.
Also, a different way to derive leapfrog is to define coordinate shift operators $L_p$ and $L_r$ that move $p to p +Delta p$ and $x to x + Delta x$. Then the differential equation can be written as:
$$
f(p(t), r(t)) = expleft(t (L_p + L_r) right) f(p(0), r(0)).
$$
Using Trotter's identity we can write:
$$
expleft( Delta t (A + B) right) approx expleft( 0.5Delta t A right) expleft( Delta t B right) expleft(0.5 Delta t A right)
$$
and this is exactly the Leapfrog algorithm. I tried figuring out if things breakdown in 1D case of two particles going straight at each other. To do this I worked out the commutator term $[L_p,L_r]$ that comes up in Zassenhaus formula. This also didn't help.
Anyone has a fresh look or has some derivations that go beyond basic 1D cases where things go nicely? Any help is appreciated.
numerical-methods mathematical-physics
numerical-methods mathematical-physics
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
edited Jan 14 at 17:27
Piotr Benedysiuk
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
asked Jan 14 at 15:36
Piotr BenedysiukPiotr Benedysiuk
1,269518
1,269518
$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
1
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
1
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
1
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58
|
show 12 more comments
$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
1
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
1
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
1
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58
$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
1
1
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
1
1
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
1
1
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58
|
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$begingroup$
Are you sure about your Hamiltonian? There are squares and a factor 1/2 missing in the kinetic part.
$endgroup$
– LutzL
Jan 14 at 17:52
$begingroup$
Yes. If velocities are relativistic ($v to c$) then kinetic energy no longer equals $frac{p^2}{2m}$. This is the case of $m = 0$.
$endgroup$
– Piotr Benedysiuk
Jan 14 at 18:41
1
$begingroup$
@PiotrBenedysiuk This question should be asked on the physics-site.
$endgroup$
– Peter
Jan 21 at 8:35
1
$begingroup$
Good point. I mirrored it to over there
$endgroup$
– Piotr Benedysiuk
Jan 21 at 10:02
1
$begingroup$
The exact claim, as can be found in the works of Hairer et al., is that the Verlet method in any of its implementation preserves to $O(Δt^4)$ an $O(Δt^2)$ modification of the Hamiltonian. This modification includes derivatives of the Hamiltonian, so that close to singular points of the Hamiltonian the $O(Δt^2)$ modification need not be small at all. // Of course this does not help if the Hamiltonian used is not even a first integral of the system.
$endgroup$
– LutzL
Jan 23 at 13:58