Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

How do we truncate a polynomial in Maple, that is, obtain only the terms of degree <= n?

For example, if 

P(x)=x^2+2*x^3+x^6

then the truncation [P(x)]4 is

P(x)=x^2+2*x^3

and the truncation [P(x)]is

P(x)=x^2

dAlmbert ode has the form

 

Also from Maple own help page, it agrees with Wikipedia and says:

 

Now, given this ode

ode:=y(x)=ln(cos(diff(y(x),x)))+diff(y(x),x)*tan(diff(y(x),x));
eval(ode,diff(y(x),x)=p)

Then clearly the above is not dAlmbert. Right? it is missing the x. But odeadvisor says it is:

restart;
ode:=y(x)=ln(cos(diff(y(x),x)))+diff(y(x),x)*tan(diff(y(x),x));
DEtools:-odeadvisor(ode)

What Am I missing here?

Update

These are the rules I know about this ode. For y=x f(p)+ g(p). 

g(p) can be zero, yes, but in this case, f(p) has to be nonlinear in p for it to be dAlembert (else it will be either separable or linear.

I did not think f(p) can be zero and it remains dAlermber, even if g(p) remains nonlinear in p. So y=g(p) can not be dAlembert, even if g(p) is nonlinear.

May be Maple uses its own definition of dAlembert?. I do not know. This will be new definition to me. Is there a reference that mentions this case of y=g(p) classified as dAlembert for nonlinear g?

Maple 2022.1 on windows 10


Using plot3d(..., style=surfacecontour, ...) or contourplot3d(...) displays wrong level curves when some axis are switched to a log mode.

Example:

restart:

interface(version)

`Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895`

(1)

X := (0.4000000000e-4*(-R+80.00))/(R*(0.4e-1+M__a)):

plot3d(X, R=0..10, M__a=10^0..10^4, style=surfacecontour, color=gold)

 

plot3d(X, R=0..10, M__a=10^0..10^4, axis[2]=[mode=log], axis[3]=[mode=log], style=surfacecontour, color=gold)

 

plots:-display(
  plots:-contourplot3d(X, R=0..10, M__a=10^0..10^4, axis[2]=[mode=log], axis[3]=[mode=log], color=red),
  plot3d(X, R=0..10, M__a=10^0..10^4, style=surface, color=gold)
)

 

 

Download WrongLevelCurves.mw

 

The problem is not dramatic because there is a workaround.
 

restart:

interface(version)

`Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895`

(1)

X := (0.4000000000e-4*(-R+80.00))/(R*(0.4e-1+M__a)):

fig := plot3d(X, R=0..10, M__a=10^0..10^4, style=surfacecontour, color=gold):
Tr  := plottools:-transform((x, y, z) -> [x, log[10](y), log[10](z)]):
plots:-display(Tr(fig), axis[2]=[tickmarks=[seq(i=10^i, i=0..4)]], axis[3]=[tickmarks=[seq(i=nprintf("%1.0e", 10.^i), i=-7..-1)]])

 

 

Download WrongLevelCurves_Workaround.mw

 

I've attached a maple file (.mw). After solving the first order condition of SW expression, I tried to obtain an optimal point for en. To do that I used solve API, and obtained a solution in rootof. I want to comment on the relationship of en with other parameters like A, theta, zeta, etc. Therefore, I want to understand the interpretation of rootof and how analytically I can simplify it further if possible.

restart

SW := (1/6)*a*(8*sqrt(epsilon[n]*theta[n]*a)*Zeta[n]+3*a*tau-3*a)+(1/3)*e[n]*(-beta[n]*e[n]^2+3*delta[n]*A+4*sqrt(e[n]*theta[n]*A))

(1/6)*a*(8*(varepsilon[n]*theta[n]*a)^(1/2)*Zeta[n]+3*a*tau-3*a)+(1/3)*e[n]*(-beta[n]*e[n]^2+3*delta[n]*A+4*(e[n]*theta[n]*A)^(1/2))

(1)

``

NULL

Opt_effort_FOC := diff(SW, e[n])

-(1/3)*beta[n]*e[n]^2+delta[n]*A+(4/3)*(e[n]*theta[n]*A)^(1/2)+(1/3)*e[n]*(-2*beta[n]*e[n]+2*theta[n]*A/(e[n]*theta[n]*A)^(1/2))

(2)

``

NULL

solve(Opt_effort_FOC = 0, e[n])

RootOf(-A^3*delta[n]*theta[n]^2-2*A^2*_Z*theta[n]^2+_Z^4*beta[n])^2/(theta[n]*A)

(3)

NULL

``

Download RootOf_maple.mw

I know that this is a stupid question but I cannot find how to do the dyadic product of two matrix. Those matrix could be two tensor matrix and I need to do this multiplication:

Thank you in advance for your help.

Mario

Hello,

    I am regularly using tabled data for things like material properties, or other tabled look-up data from induustry or custom datasets.

    I have not used workbooks within Maple and yet see that exel files or other data sets can be "connected(?)" in the workbook tree.  Is this helpful?

    What is the best practice view of connecting to DB data and looking up variables that repeat project to project or worksheet to worksheet - without pulling into memory the full datasets for each ref call?

All thoughts appreciated.   

P.S., Tables vs. dataframes are a slight confusing matter also.  The access to the cell info seems different based on dataframe/table/matrix type.  I have no problems with matrix element access.

Thanks,
Bill

"object at address is binary"

                                                          

The following code is attempting to pass 'debug' through 2 procedures

p1 := proc(a, b, {debug::truefalse := false}) print("p1", debug); a + b; end proc;
p2 := proc(a, b, {debug::truefalse := false}) print("p2", debug); p1(a, b, 'debug' = true); end proc;
p2(1, 2, 'debug' = true);

The result I get is 
 

                           "p2", true

                          "p1", false

                               3

How can I get the value of debug in my call statement to p2 to be passed to p1?

Thanks

Hi 

I wanna export data from maple to excel and I tried this:

ExcelTools:-Export(op([1, 1], plots:-display( convert( ans1[12..14,3], list))));

I was expecting 4 columns including z vector and 3 columns from ans1[12..14,3] but got only 2 columns.Case1_Revised_080922.mw

Please any help?

when i try to save my program maple is close, crashing. how can i fix this?

Hello dear

please guide me how to solve linearly system in explicit form.

In this purpose of this problem is to find  w1, w2 w3  as symblic solution in Matrix form. and I hope that in the futher the vairiables consist of w1 ... wn and will still be applied in those simplied form or explicit form.

Thank you

_____________code________________________________________________________________

Vector[column](3, [0.85*((phi__cs*beta__s*f__con) . (D__pile*w__1)) + Vector[column](1, [-5/6*H - 5/8*P]), 0.85*((phi__cs*beta__s*f__con) . (D__pile*w__2)) + Vector[column](1, [5/6*H - 5/8*P]), 0.85*((phi__cs*beta__s*f__con) . (D__pile*w__3)) + Vector[column](1, [1/2*H - 3/8*P])])


 

New display of arbitrary constants and functions

 

When using computer algebra, first we want results. Right. And textbook-like typesetting was not fully developed 20+ years ago. So, in the name of getting those results, people somehow got used to the idea of "give up textbook-quality computer algebra display". But computers keep evolving, and nowadays textbook typesetting is fully developed, so we have better typesetting in place. For example, consider this differential equation:

 

Download New_arbitrary_constants_and_functions.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

Problem statement:
Determine the relativistic uniformly accelerated motion, i.e. the rectilinear motion for which the acceleration w in the proper reference frame (at each instant of time) remains constant.

As an application of the post presented by Dr Cheb Terrab in MaplePrimes on the principle of relativity ( found here ), we solve the problem stated on page 24 of Landau & Lifshitz book [1], which makes use of the relativistic invariant condition of the constancy of a four-scalar, viz., `w__&mu;`*w^mu where w^mu is the four-acceleration. This little problem exemplify beautifully how to use invariance in relativity. This is the so-called hyperbolic motion and we explain why at the end of this worksheet.

NULL

let's introduce the coordinate system, X = (x, y, z, tau)with tau = c*t 

with(Physics)

Setup(coordinates = [X = (x, y, z, tau)])

[coordinatesystems = {X}]

(1)

%d_(s)^2 = g_[lineelement]

%d_(s)^2 = -Physics:-d_(x)^2-Physics:-d_(y)^2-Physics:-d_(z)^2+Physics:-d_(tau)^2

(2)

NULL

Four-velocity

 

The four-velocity is defined by  u^mu = dx^mu/ds and dx^mu/ds = dx^mu/(c*sqrt(1-v^2/c^2)*dt) 

Define this quantity as a tensor.

Define(u[mu], quiet)

The four velocity can therefore be computing using

u[`~mu`] = d_(X[`~mu`])/%d_(s(tau))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/%d_(s(tau))

(1.1)

NULL

As to the interval d(s(tau)), it is easily obtained from (2) . See Equation (4.1.5)  here with d(diff(tau(x), x)) = d(s(tau)) for in the moving reference frame we have that d(diff(x, x)) = d(diff(y(x), x)) and d(diff(y(x), x)) = d(diff(z(x), x)) and d(diff(z(x), x)) = 0.

 Thus, remembering that the velocity is a function of the time and hence of tau, set

%d_(s(tau)) = d(tau)*sqrt(1-v(tau)^2/c^2)

%d_(s(tau)) = Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2)

(1.2)

subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/%d_(s(tau)))

u[`~mu`] = Physics:-d_(Physics:-SpaceTimeVector[`~mu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))

(1.3)

Rewriting the right-hand side in components,

lhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = Library:-TensorComponents(rhs(u[`~mu`] = Physics[d_](Physics[SpaceTimeVector][`~mu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

u[`~mu`] = [Physics:-d_(x)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(y)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics:-d_(z)/(Physics:-d_(tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]

(1.4)

Next we introduce explicitly the 3D velocity components while remembering that the moving reference frame travels along the positive x-axis

NULL

simplify(u[`~mu`] = [Physics[d_](x)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](y)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), Physics[d_](z)/(Physics[d_](tau)*(-(v(tau)^2-c^2)/c^2)^(1/2)), 1/(-(v(tau)^2-c^2)/c^2)^(1/2)], {d_(x)/d_(tau) = v(tau)/c, d_(y)/d_(tau) = 0, d_(z)/d_(tau) = 0}, {d_(x), d_(y), d_(z)})

u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)]

(1.5)

Introduce now this explicit definition into the system

Define(u[`~mu`] = [v(tau)/(c*((c^2-v(tau)^2)/c^2)^(1/2)), 0, 0, 1/(-(v(tau)^2-c^2)/c^2)^(1/2)])

{Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-d_[mu], Physics:-g_[mu, nu], u[mu], w[`~mu`], w__o[`~mu`], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}

(1.6)

NULL

Computing the four-acceleration

 

This quantity is defined by the second derivative w^mu = d^2*x^mu/ds^2 and d^2*x^mu/ds^2 = du^mu/ds and du^mu/ds = du^mu/(c*sqrt(1-v^2/c^2)*dt)

Define this quantity as a tensor.

Define(w[mu], quiet)

Applying the definition just given,

w[`~mu`] = d_(u[`~mu`])/%d_(s(tau))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/%d_(s(tau))

(2.1)

Substituting for d_(s(tau))from (1.2) above

subs(%d_(s(tau)) = Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2), w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/%d_(s(tau)))

w[`~mu`] = Physics:-d_[nu](u[`~mu`], [X])*Physics:-d_(Physics:-SpaceTimeVector[`~nu`](X))/(Physics:-d_(tau)*(1-v(tau)^2/c^2)^(1/2))

(2.2)

Introducing now this definition (2.2)  into the system,

Define(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2)), quiet)

lhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))) = TensorArray(rhs(w[`~mu`] = Physics[d_][nu](u[`~mu`], [X])*Physics[d_](Physics[SpaceTimeVector][`~nu`](X))/(Physics[d_](tau)*(1-v(tau)^2/c^2)^(1/2))))

w[`~mu`] = Array(%id = 36893488148327765764)

(2.3)

Recalling that tau = c*t, we get

"PDETools:-dchange([tau=c*t],?,[t],params=c)"

w[`~mu`] = Array(%id = 36893488148324030572)

(2.4)

Introducing anew this definition (2.4)  into the system,

"Define(w[~mu]=rhs(?),redo,quiet):"

NULL

In the proper referential, the velocity of the particle vanishes and the tridimensional acceleration is directed along the positive x-axis, denote its value by `#msub(mi("w"),mn("0"))`

Hence, proceeding to the relevant substitutions and introducing the corresponding definition into the system, the four-acceleration in the proper referential reads

  "Define(`w__o`[~mu]= subs(v(t)=`w__0`, v(t)=0,rhs(?)),quiet):"

w__o[`~mu`] = TensorArray(w__o[`~mu`])

w__o[`~mu`] = Array(%id = 36893488148076604940)

(2.5)

NULL

The differential equation solving the problem

 

NULL``

Everything is now set up for us to establish the differential equation that will solve our problem. It is at this juncture that we make use of the invariant condition stated in the introduction.

The relativistic invariant condition of uniform acceleration must lie in the constancy of a 4-scalar coinciding with `w__&mu;`*w^mu  in the proper reference frame.

We simply write the stated invariance of the four scalar (d*u^mu*(1/(d*s)))^2 thus:

w[mu]^2 = w__o[mu]^2

w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`]

(3.1)

TensorArray(w[mu]*w[`~mu`] = w__o[mu]*w__o[`~mu`])

(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4

(3.2)

NULL

This gives us a first order differential equation for the velocity.

 

Solving the differential equation for the velocity and computation of the distance travelled

 

NULL

Assuming the proper reference frame is starting from rest, with its origin at that instant coinciding with the origin of the fixed reference frame, and travelling along the positive x-axis, we get successively,

NULL

dsolve({(diff(v(t), t))^2*c^2/(v(t)^2-c^2)^3 = -w__0^2/c^4, v(0) = 0})

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)

(4.1)

NULL

As just explained, the motion being along the positive x-axis, we take the first expression.

[v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2), v(t) = -t*c*w__0/(t^2*w__0^2+c^2)^(1/2)][1]

v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)

(4.2)

This can be rewritten thus

v(t) = w__0*t/sqrt(1+w__0^2*t^2/c^2)

v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)

(4.3)

It is interesting to note that the ultimate speed reached is the speed of light, as it should be.

`assuming`([limit(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = infinity)], [w__0 > 0, c > 0])

limit(v(t), t = infinity) = c

(4.4)

NULL

The space travelled is simply

x(t) = Int(rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2)), t = 0 .. t)

x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t)

(4.5)

`assuming`([value(x(t) = Int(w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t = 0 .. t))], [c > 0])

x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0

(4.6)

expand(x(t) = c*((t^2*w__0^2+c^2)^(1/2)-c)/w__0)

x(t) = c*(t^2*w__0^2+c^2)^(1/2)/w__0-c^2/w__0

(4.7)

This can be rewritten in the form

x(t) = c^2*(sqrt(1+w__0^2*t^2/c^2)-1)/w__0

x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0

(4.8)

NULL

The classical limit corresponds to an infinite velocity of light; this entails an instantaneous propagation of the interactions, as is conjectured in Newtonian mechanics.
The asymptotic development gives,

lhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0) = asympt(rhs(x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0), c, 4)

x(t) = (1/2)*w__0*t^2+O(1/c^2)

(4.9)

As for the velocity, we get

lhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)) = asympt(rhs(v(t) = t*c*w__0/(t^2*w__0^2+c^2)^(1/2)), c, 2)

v(t) = t*w__0+O(1/c^2)

(4.10)

Thus, the classical laws are recovered.

NULL

Proper time

 

NULL

This quantity is given by "t'= &int; dt sqrt(1-(v^(2))/(c^(2)))" the integral being  taken between the initial and final improper instants of time

Here the initial instant is the origin and we denote the final instant of time t.

NULL

`#mrow(mi("t"),mo("&prime;"))` = Int(sqrt(1-rhs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2))^2/c^2), t = 0 .. t)

`#mrow(mi("t"),mo("&prime;"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t)

(5.1)

Finally the proper time reads

`assuming`([value(`#mrow(mi("t"),mo("&prime;"))` = Int((1-w__0^2*t^2/((1+w__0^2*t^2/c^2)*c^2))^(1/2), t = 0 .. t))], [w__0 > 0, c > 0, t > 0])

`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0

(5.2)

When proc (t) options operator, arrow; infinity end proc, the proper time grows much more slowly than t according to the law

`assuming`([lhs(`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0) = asympt(rhs(`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0), t, 1)], [w__0 > 0, c > 0])

`#mrow(mi("t"),mo("&prime;"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2)

(5.3)

combine(`#mrow(mi("t"),mo("&prime;"))` = (ln(2*w__0/c)+ln(t))*c/w__0+O(1/t^2), ln, symbolic)

`#mrow(mi("t"),mo("&prime;"))` = ln(2*t*w__0/c)*c/w__0+O(1/t^2)

(5.4)

NULL

Evolution of the four-acceleration of the moving frame as observed from the fixed reference frame

 

NULL

To obtain the four-acceleration as a function of time, simply substitute for the 3-velocity (4.3)  in the 4-acceleration (2.4)

" simplify(subs(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2),?),symbolic)"

w[`~mu`] = Array(%id = 36893488148142539108)

(6.1)

" w[t->infinity]^(  mu)=map(limit,rhs(?),t=infinity) assuming `w__0`>0,c>0"

`#msubsup(mi("w"),mrow(mi("t"),mo("&rarr;"),mo("&infin;")),mrow(mo("&InvisibleTimes;"),mo("&InvisibleTimes;"),mi("&mu;",fontstyle = "normal")))` = Array(%id = 36893488148142506460)

(6.2)

We observe that the non-vanishing components of the four-acceleration of the accelerating reference frame get infinite while those components in the moving reference frame keep their constant values . (2.5)

NULL

Evolution of the three-acceleration as observed from the fixed reference frame

 

NULL

This quantity is obtained simply by differentiating the velocity v(t)given by  with respect to the time t.

 

simplify(diff(v(t) = w__0*t/(1+w__0^2*t^2/c^2)^(1/2), t), size)

diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)

(7.1)

Here also, it is interesting to note that the three-acceleration tends to zero. This fact was somewhat unexpected.

map(limit, diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2), t = infinity)

limit(diff(v(t), t), t = infinity) = 0

(7.2)

NULL

At the beginning of the motion, the acceleration should be w__0, as Newton's mechanics applies then

NULL

`assuming`([lhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)) = series(rhs(diff(v(t), t) = w__0/(1+w__0^2*t^2/c^2)^(3/2)), t = 0, 2)], [c > 0])

diff(v(t), t) = series(w__0+O(t^2),t,2)

(7.3)

NULL

Justification of the name hyperbolic motion

 

NULL

Recall the expressions for x and diff(t(x), x)and obtain a parametric description of a curve, with diff(t(x), x)as parameter. This curve will turn out to be a hyperbola.

subs(x(t) = x, x(t) = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0)

x = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0

(8.1)

`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0

`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0

(8.2)

The idea is to express the variables x and t in terms of diff(t(x), x).

 

isolate(`#mrow(mi("t"),mo("&prime;"))` = arcsinh(t*w__0/c)*c/w__0, t)

t = sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)*c/w__0

(8.3)

subs(t = sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)*c/w__0, x = c^2*((1+w__0^2*t^2/c^2)^(1/2)-1)/w__0)

x = c^2*((1+sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2)^(1/2)-1)/w__0

(8.4)

`assuming`([simplify(x = c^2*((1+sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2)^(1/2)-1)/w__0)], [positive])

x = c^2*(cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)-1)/w__0

(8.5)

We now show that the equations (8.3) and (8.5) are parametric equations of a hyperbola with parameter the proper time diff(t(x), x)

 

Recall the hyperbolic trigonometric identity

cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2 = 1

cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2 = 1

(8.6)

Then isolating the sinh and the cosh from equations (8.3) and (8.5),

NULL

isolate(t = sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)*c/w__0, sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c))

sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c) = t*w__0/c

(8.7)

isolate(x = c^2*(cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)-1)/w__0, cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c))

cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c) = x*w__0/c^2+1

(8.8)

and substituting these in (8.6) , we get the looked-for Cartesian equation

 

subs(sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c) = t*w__0/c, cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c) = x*w__0/c^2+1, cosh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2-sinh(`#mrow(mi("t"),mo("&prime;"))`*w__0/c)^2 = 1)

(x*w__0/c^2+1)^2-w__0^2*t^2/c^2 = 1

(8.9)

NULL

This is the Cartesian equation of a hyperbola, hence the name hyperbolic motion

NULL

Reference

 

[1] Landau, L.D., and Lifshitz, E.M. The Classical Theory of Fields, Course of Theoretical Physics Volume 2, fourth revised English edition. Elsevier, 1975.

NULL

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In solving the brachistochrone for a fine string of length L under the pull of g passing through the point (0,0) and (a,b) using the euler-lagange method,I stumble on this non-linear relation:

C*sinh(mu*g*a/C)=L

which I need to solve for C.

Maple give me the famous RootOf:

RootOf(A*exp(_Z)^2 - 2*_Z*L*exp(_Z) - A)

where A = mu*g*a

Can it be solve for C or am I force to use numeric method?

Thank you in advance for your help.

Mario

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