# I Wanted to see if anyone knew if this was trivial

1. Feb 15, 2016

### alex2066

Wanted to see if anyone knew if this was trivial

Particles exert forces (and therefor the acceleration) partials exert on each other very by the inverse square of the distance

Modeled by this system of seconded order nonlinear differential equations

a(t)’’=(Q/|Q|)(a(t)-b(t))^-2

b(t)’’=(Q/|Q|)(a(t)-b(t))^-2

solved hear

a(t)=-K ln(T)+ AT+C

b(t)=-K ln(T)+ BT+C

a’’=-K/t^2

b'’=-K/t^2
with these variable constricts

C=AR#

-K=(Q/|Q|)(A-B)^-2

1/√|K|=A-B

demonstration

-K*-1/t^2=( (-Kln(T)+AT+C) -(-Kln(T)+BT+C))^-2

-K*-1/t^2=( (-Kln(T)+AT+C) -(-Kln(T)+BT+C))^-2

K/T^2=( ( (A -B)T)^-2

K/T^2=( ( (A -B)T)^-2

K/T^2=(1/√|K|*T)^-2

K/T^2=(1/√|K|*T)^-2

K/T^2= K/T^2

K/T^2= K/T^2

so it solves

=)

I also have it for more 3 and 4 particles

The whole thing is in a XL doc at this link

if your interested I can make a doc on how I solved it
or if i messed up somewhere please tell me

Last edited: Feb 15, 2016
2. Feb 15, 2016

### Staff: Mentor

One of your acceleration terms should get a negative sign, otherwise they violate conservation of momentum.
That is clearly not the general solution. What happens at t=0, for example?

And you cannot generalize it to more than two particles.

3. Feb 16, 2016

### alex2066

thank you
i don't know how i did not see that
=P

still a solution to that differential equation even though it violates conservation of momentum.
i wonder if there is way to write it so that it dose not.

if by generalize you mean solve i degree with you hear.

for 3

a(t)''=X(a(t)-b(t))^-2+Z(a(t)-c(t))^-2
b(t)''=X(a(t)-b(t))^-2+Y(b(t)-c(t))^-2
c(t)''=Z(a(t)-c(t))^-2+Y(b(t)-c(t))^-2

a(t) = -K ln(T) + A T+P
b(t) = -K ln(T) + B T+P
c(t) = -K ln(T) + C T+P

once replaced
K = X (A-B)^-2 + Z (A-C)^-2
K = X (A-B)^-2 + Y (B-C)^-2
K = Z (A-C)^-2 + Y (B-C)^-2

one numerical solution
K = -2
A = 0
B = 1
C = 2
Z = 4
X = 1
Y = 1
P = any real number

and for 4
a(t)'' = X (a(t)-b(t))^-2 + Z (a(t)-c(t))^-2 + W (a(t)-d(t))^-2
b(t)'' = X (a(t)-b(t))^-2 + Y (b(t)-c(t))^-2 + T (d(t)-c(t))^-2
c(t)'' = Z (a(t)-c(t))^-2 + Y (b(t)-c(t))^-2 + U (c(t)-d(t))^-2
d(t)'' = W (a(t)-d(t))^-2 + T (d(t)-c(t))^-2 + U (c(t)-d(t))^-2

a(t) = -K ln(T) + A T+P
b(t) = -K ln(T) + B T+P
c(t) = -K ln(T) + C T+P
d(t) = -K ln(T) + D T+P

once replaced
K = X (A-B)^-2 +Z (A-C)^-2 +W (A-D))^-2
K = X (A-B)^-2 +Y (B-C)^-2 +Q(D-B)^-2
K = Z (A-C)^-2 +Y (B-C)^-2 +U (C-D)^-2
K = W(A-D))^-2 +Q(D-B)^-2 +U (C-D)^-2

any algebraic solution will work
for example

Z (A-C)^-2 = K
U (C-D)^-2 = K
Q (D-B)^-2 = K
Y (B-C)^-2 = - K
X (A-B)^-2 = K
W (A-D))^-2 = - K

or

K = 1
A = 0
B = 2
C = 1
D = -1
Y = -4
Z = 1
Q = 9
X = 1
U = 4

im sure you can see the pattern

4. Feb 16, 2016

### Staff: Mentor

Those are not solutions to the equations of motions.

Let a(0)=0, b(0)=1, c(0)=100. Now find parameters that fit.
Your logarithm doesn't allow a time of zero at all. But even if you restrict it to positive times:
Let a(1)=0, b(1)=1, c(1)=100. Now find parameters where the particles satisfy the equations of motion. You cannot.

5. Feb 16, 2016

### alex2066

i cant do it where momentum is conserved but i think i can do it if that is overlooked

dose it have to be 100 1 and 0 or can it be any 3 values.

hear it is for 2 i think i can do it for 3

a(t)'' = - (a(t)-b(t))^-2
b(t)'' = - (a(t)-b(t))^-2

a(t) = - ln(t+c) + A T + M
b(t) = - ln(t+c) + B T + W
W - M = C
1 = A-B
1 = B-A

- 1/T^2 = ( (-ln(T+C)+AT+M) -(-ln(T+C)+BT+W)^-2
- 1/T^2 = ( (-ln(T+C)+AT+M) -(-ln(T+C)+BT+W)^-2

1/(T+C)^2 = ( AT -BT+C)^-2
1/(T+C)^2 = ( AT -BT+C)^-2
1/(T+C)^2 = ( (A -B)T+C)^-2
1/(T+C)^2 = ( (A -B)T+C)^-2
1/(T+C)^2 = ((T+C)^-2
1/(T+C)^2 = (T+C)^-2
1/(T+C)^2 = 1/(T+C)^2
1/(T+C)^2 = 1/(T+C)^2

i guess that might have been what you meant by 2 is the only one with a general solution. sorry i did not understand what you meant by general i thought you just meant solved. still i think i can do it with 3

im going 2 spend some time thinking about it. your probably right I misunderstood what you meant by general.
i just thought that this was interesting.
i think i can solve it for
a(t)'' = (N(a(t)-b(t))^-2 + (M(c(t)-a(t)))^-2
b(t)'' = (N(b(t)-a(t)))^2 + (Z(b(t)-c(t)))^-2
c(t)'' = (M(c(t)-a(t)))^-2 + (Z(c(t)-b(t)))^-2
M,N,Z are all either 1 or -1(of coerce there is no such thing as negative distance.)

with
a(t) = ln(KT+P) + ln(-QT+R) + A T + X
b(t) = ln(KT+P) + ln(-QT+R) + B T + Y
c(t) = ln(KT+P) + ln(-QT+R) + C T + Z

i still cant even solve or find on the internet solutions to

a(t)'' = (a(t)-b(t))^-2
b(t)'' = (a(t)-b(t))^-2
a''+b''=0

i still think
still at least it solves
a(t)'' = (a(t)-b(t))^-2
b(t)'' = (a(t)-b(t))^-2
a''=b''

I still cant solve it where momentum is conserved but i will keep trying. Im a bit busy right now geting ready for my first your of collage next year.

reagards alex

Last edited by a moderator: Feb 16, 2016
6. Feb 16, 2016

### Staff: Mentor

You can edit your posts if you want to add something.
Not if you want to find solutions to the equations of motion.
The 3 values I set. A general solution has to work for all initial conditions, including those.
You cannot. And you should check your solutions before you make claims like that. Every such check, properly done, will show you that the approach does not work. So I can save both of us some time and close the thread.