Can you solve this Differential Equation?
Michael Bienstein
reference frame experiences constant acceleration.
I get a(1-x'^2/c^2)^(3/2)=x'' where x(t) is the spatial position at time t
and x' means dx/dt, x'' means d^2x/dt^2 etc.
How do I solve this DE?
Harold Ellis Ensle
1. since you don't have explicit x dependence, you can reduce it to
a non-linear first order equation in v (=x').
2. Use the substitution v/c=sin(a) or cos(a) and this will reduce to a
separable equation.
H.Ellis Ensle
Anthony F. Badalamenti
a negative argument for the the power 3/2 if x' is large enough.
Otherwise you can square both sides and represent x as a Maclauran (or
Taylor) series and hope to get clear recursive relationships for the
coeffieicents by equating like powers on both sides. If that works, then
go backwards and find where the given argument may go negative.
Nathan Urban
> Are you sure this is correctly formulated? The LHS of your equation has
> a negative argument for the the power 3/2 if x' is large enough.
Only if x' exceeds c, which it doesn't. But I'm not sure if it's
the right equation either, I'll have to remember to check when I have
the time.
Tom Roberts
> Hi all. I'm trying to work out the world line of a particle which in it's
> reference frame experiences constant acceleration.
See the section in MTW about an accelerated observer. See the chapter in
Mould about an accelerated observer. See the section in Rindler about
an accelerated observer. See my month-old post to this newsgroup, Subject:
Speed of Light in an Accelerated System.
Misner, Thorne, and Wheeler, _Gravitation_.
Mould, _Basic_Relativity_.
Rindler, _Essential_Relativity_.
> I get a(1-x'^2/c^2)^(3/2)=x'' where x(t) is the spatial position at time t
> and x' means dx/dt, x'' means d^2x/dt^2 etc.
It's easier to use MTW's approach, and solve the 4-vector equations (c=1):
v.v = 1 v = 4-velocity
a.v = 0 a = 4-aceleration
a = dv/d\tau \tau = proper time
a.a = -proper_acceleration^2
v = dx/d\tau x = position
Answer: in inertial coordinates
x = 1/g cosh(g\tau) - 1/g
t = 1/g sinh(g\tau)
Where (x,t) are the inertial coordinates of the particle, \tau is the
particle's proper time, g is the proper acceleration of the particle,
and I applied the initial conditions x(\tau=0)=0, t(\tau=0)=0.
Note that this is a parameterized curve in the (x,t) plane,
with \tau as the Affine parameter of the curve. Your attempt
to directly express x as a function of t is more difficult.
But you ought to be able to eliminate \tau from the two
equations above and verify the result satisfies your equation.
This is called hyperbolic motion, for obvious reasons.
Tom Roberts tjro...@lucent.com
Harold Ellis Ensle
>>Hi all. I'm trying to work out the world line of a particle which in it's
>>reference frame experiences constant acceleration.
>>
>>and x' means dx/dt, x'' means d^2x/dt^2 etc.
>>
>>
>>
>>
>1. since you don't have explicit x dependence, you can reduce it to
> a non-linear first order equation in v (=x').
>2. Use the substitution v/c=sin(a) or cos(a) and this will reduce to a
>separable equation.
>
>H.Ellis Ensle
>
>
v/c=sin(y) or cos(y)
and...@ibm.net
>
> In article <F5v0H...@unx.sas.com>,
> "Michael Bienstein" <zaf...@zaf.sas.com> wrote:
> >Hi all. I'm trying to work out the world line of a particle which in it's
> >reference frame experiences constant acceleration.
> >
> >I get a(1-x'^2/c^2)^(3/2)=x'' where x(t) is the spatial position at time t
> >and x' means dx/dt, x'' means d^2x/dt^2 etc.
> >
> >How do I solve this DE?
> >
> >
> >
> As this might be a homework problem, I won't do it for you, but here's how:
> 1. since you don't have explicit x dependence, you can reduce it to
> a non-linear first order equation in v (=x').
> 2. Use the substitution v/c=sin(a) or cos(a) and this will reduce to a
> separable equation.
>
> H.Ellis Ensle
Not a good recommendation. You want to use hyperbolic functions
sinh and cosh to describe this scenario. I'll do it backwards
going from the solution to the demonstration that it satisfies
the criterion. By the way, this is an ordinary differential equation,
so your use of the term separable is incorrect.
The answer to the problem is
x = sqrt(x0^2 + (c*t)^2) for t > 0
Let x = x0*cosh(alpha)
c*t = x0*sinh(alpha)
where alpha is some parameter. This satisfies
the equations (1) and (2). The proper time,
s, is given by
(ds)^2 = (c*dt)^2 - (dx)^2
c*dt = x0*cosh(alpha)*dalpha
dx = x0*sinh(alpha)*dalpha
So
(ds)^2 = (x0*dalpha)^2
So you can write
x = x0*cosh(s/x0)
c*t = x0*sinh(s/x0)
Calculate the covariant velocity (dt/ds,dx/ds) = (Ut,Ux)
(this is usually called the four velocity for 1 time
and 3 space dimension relativity)
and the covariant acceleration (dUt/ds,dUx/ds) = (At,Ax).
Both of these are Lorentz vectors. They transform
under a Lorentz transformation the same way that
(t,x) does.
Transform (At,Ax) to the local comoving frame of the
accelerated object and you get (0,Acomoving)
where Acomoving is a constant. It's
(c^2)/x0.
I think that the original poster may have been confused
about whether the derivative is with respect to proper time
or coordinate time.
John Anderson
Harold Ellis Ensle
and...@ibm.net wrote:
>Harold Ellis Ensle wrote:
>>
>> In article <F5v0H...@unx.sas.com>,
>> "Michael Bienstein" <zaf...@zaf.sas.com> wrote:
>> >Hi all. I'm trying to work out the world line of a particle which in it's
>> >reference frame experiences constant acceleration.
>> >
>> >I get a(1-x'^2/c^2)^(3/2)=x'' where x(t) is the spatial position at time t
>> >and x' means dx/dt, x'' means d^2x/dt^2 etc.
>> >
>> >How do I solve this DE?
>> >
>> >
>> >
>> As this might be a homework problem, I won't do it for you, but here's how:
>> 1. since you don't have explicit x dependence, you can reduce it to
>> a non-linear first order equation in v (=x').
>> 2. Use the substitution v/c=sin(a) or cos(a) and this will reduce to a
>> separable equation.
>>
>> H.Ellis Ensle
>
>Not a good recommendation. You want to use hyperbolic functions
>sinh and cosh to describe this scenario. I'll do it backwards
>going from the solution to the demonstration that it satisfies
>the criterion. By the way, this is an ordinary differential equation,
>so your use of the term separable is incorrect.
The terminology I use here is appropriate and specific and the fact that
you do not realize it is telling. Let's do it and I will teach you the
proper terminology.
a(1-x'^2/c^2)^(3/2)=x''
This is a non-linear 2nd order ordinary differential equation.
Since there is no explicit x dependence we can reduce this by one
order using v=x' thus
a(1-v^2/c^2)^(3/2)=v'
which is a non-linear 1st order ordinary differential equation.
Let cosy=v/c which yields
(you would not want to use sinh or cosh here as it would not simplify the
left side)
a(1-(cosy)^2)^(3/2)=csiny dy/dt
and thus
a(siny)^2=cdy/dt
In the context of ordinary differential equations "separable" means
that you can separate the dependent and independent variables of the
equation as so:
a/c dt= dy/(siny)^2
You can now simply integrate this:
(a/c)t=-coty+k1
Resubstitute from cosy=v/c
since coty=cosy/sqrt(1-(cosy)^2) we have
(a/c)t=-(v/c)/sqrt(1-v^2/c^2)+k1
solve for v:
v= at/sqrt(1+(at/c-k1)^2)
thus
x'= at/sqrt(1+(at/c-k1)^2)
we need only to do this integral
x=int{at/sqrt(1+(at/c-k1)^2)}dt
which you can look up (or, if you want to do it manually, by
using a tan substitution)
Many have discussed the related physics, but as posed, this is a
purely mathematical problem and the result is obtained by
well known mathematical means.
H.Ellis Ensle