Freddy Baudine

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Special Theory of Relativity: Uniformly ...

Posted: Freddy Baudine 70 Product: Maple 2022

September 13 2022

5 0

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., where 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.

let's introduce the coordinate system, with

(1)

(2)

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.

The four velocity can therefore be computing using

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

(1.1)

As to the interval , it is easily obtained from (2) . See Equation (4.1.5) here with for in the moving reference frame we have that .

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

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,

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

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

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

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

(1.6)

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.

Applying the definition just given,

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

(2.1)

Substituting for from (1.2) above

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,

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

(2.3)

Recalling that , we get

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

(2.4)

Introducing anew this definition (2.4) into the system,

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

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

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

(2.5)

The differential equation solving the problem

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 in the proper reference frame.

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

(3.1)

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

(3.2)

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

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

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,

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

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)

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)

(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.

(4.4)

The space travelled is simply

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

(4.5)

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

(4.6)

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)

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,

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

(4.9)

As for the velocity, we get

(4.10)

Thus, the classical laws are recovered.

Proper time

This quantity is given by "t'= ∫ 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 .

`#mrow(mi("t"),mo("′"))` = 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("′"))` = 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

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

(5.2)

When , the proper time grows much more slowly than according to the law

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

(5.3)

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

(5.4)

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

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

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

(6.1)

(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)

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

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

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.

(7.2)

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

(7.3)

Justification of the name hyperbolic motion

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

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

(8.1)

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

(8.2)

The idea is to express the variables and in terms of .

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

(8.3)

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

(8.4)

x = c^2*(cosh(`#mrow(mi("t"),mo("′"))`*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

Recall the hyperbolic trigonometric identity

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

(8.6)

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

(8.7)

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

(8.8)

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

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

(8.9)

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

	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.

Download Uniformly_accelerated_motion.mw

What to take care of when entering a tet...

Posted: Freddy Baudine 70 Product: Maple 2020

October 27 2020

6 1

In the study of the Gödel spacetime model, a tetrad was suggested in the literature [1]. Alas, upon entering the tetrad in question, Maple's Tetrad's package complained that that matrix was not a tetrad! What went wrong? After an exchange with Edgardo S. Cheb-Terrab, Edgardo provided us with awfully useful comments regarding the use of the package and suggested that the problem together with its solution be presented in a post, as others may find it of some use for their work as well.

The Gödel spacetime solution to Einsten's equations is as follows.

(1)

((`Defined as spacetime tensors representing the NP null vectors of the tetrad formalism `*`see <a href='http://www.maplesoft.com/support/help/search.aspx?term=Physics,tetrads`*`,' target='_new'>?Physics,tetrads`*`,</a> `*l[mu]*`, `)*n[mu]*`, `*m[mu]*`, `)*conjugate(m[mu])

(2)

Working with Cartesian coordinates,

(3)

the Gödel line element is

(4)

Setting the metric

Physics:-g_[mu, nu] = Matrix(%id = 18446744078354506566)

$[metric = {(1, 1) = -1, (2, 2) = -1, (3, 3) = (1/2)*exp(2*q*y), (3, 4) = exp(q*y), (4, 4) = 1}, spaceindices = lowercaselatin_is]$

(5)

The problem appeared upon entering the matrix M below supposedly representing the alleged tetrad.

Matrix(%id = 18446744078162949534)

(6)

Each of the rows of this matrix is supposed to be one of the null vectors . Before setting this alleged tetrad, Maple was asked to settle the nature of it, and the answer was that M was not a tetrad! With the Physics Updates v.857, a more detailed message was issued:

`Warning, the given components form a`*null*`tetrad, `*`with a contravariant spacetime index`*`, only if you change the signature from `*`- - - +`*` to `*`+ - - -`*`.
You can do that by entering (copy and paste): `*Setup(signature = "+ - - -")

(7)

So there were actually three problems:

1.	The entered entity was a null tetrad, while the default of the Physics package is an orthonormal tetrad. This can be seen in the form of the tetrad metric, or using the library commands:

Physics:-Tetrads:-eta_[a, b] = Matrix(%id = 18446744078354552462)

(8)

(9)

(10)

The matrix M would only be a tetrad if the spacetime index is contravariant. On the other hand, the command IsTetrad will return true only when M represents a tetrad with both indices covariant. For instance, if the command IsTetrad is issued about the tetrad automatically computed by Maple, but is passed the matrix corresponding to with the spacetime index contravariant, false is returned:

Physics:-Tetrads:-e_[a, `~μ`] = Matrix(%id = 18446744078297840926)

(11)

(12)

3.	The matrix M corresponds to a tetrad with different signature, (+---), instead of Maple's default (---+). Although these two signatures represent the same physics, they differ in the ordering of rows and columns: the timelike component is respectively in positions 1 and 4.

The issue, then, became how to correct the matrix M to be a valid tetrad: either change the setup, or change the matrix M. Below the two courses of action are provided.

First the simplest: change the settings. According to the message (7), setting the tetrad to be null, changing the signature to be (+---) and indicating that M represents a tetrad with its spacetime index contravariant would suffice:

(13)

The null tetrad metric is now as in the reference used.

Physics:-Tetrads:-eta_[a, b] = Matrix(%id = 18446744078298386174)

(14)

Checking now with the spacetime index contravariant

Physics:-Tetrads:-e_[a, `~μ`] = Matrix(%id = 18446744078162949534)

(15)

At this point, the command IsTetrad provided with the equation (15), where the left-hand side has the information that the spacetime index is contravariant

(16)

Great! one can now set the tetrad M exactly as entered, without changing anything else. In the next line it will only be necessary to indicate that the spacetime index, , is contravariant.

$[tetrad = {(1, 1) = -(1/2)*2^(1/2), (1, 3) = (1/2)*2^(1/2)*exp(q*y), (1, 4) = (1/2)*2^(1/2), (2, 1) = (1/2)*2^(1/2), (2, 3) = (1/2)*2^(1/2)*exp(q*y), (2, 4) = (1/2)*2^(1/2), (3, 2) = -(1/2)*2^(1/2), (3, 3) = ((1/2)*I)*exp(q*y), (3, 4) = 0, (4, 2) = -(1/2)*2^(1/2), (4, 3) = -((1/2)*I)*exp(q*y), (4, 4) = 0}]$

(17)

The tetrad is now the matrix M. In addition to checking this tetrad making use of the IsTetrad command, it is also possible to check the definitions of tetrads and null vectors using TensorArray.

(18)

Matrix(%id = 18446744078353048270)

(19)

For the null vectors:

(20)

[0 = 0, 1 = 1, 0 = 0, 0 = 0, Matrix(%id = 18446744078414241910)]

(21)

From its Weyl scalars, this tetrad is already in the canonical form for a spacetime of Petrov type "D": only

(22)

(23)

Attempting to transform it into canonicalform returns the tetrad (17) itself

Matrix(%id = 18446744078396685478)

(24)

Let's now obtain the correct tetrad without changing the signature as done in (13).

Start by changing the signature back to

(25)

So again, M is not a tetrad, even if the spacetime index is specified as contravariant.

(26)

By construction, the tetrad M has its rows formed by the null vectors with the ordering . To understand what needs to be changed in M, define those vectors, independent of the null vectors (with underscore) that come with the Tetrads package.

and set their components using the matrix M taking into account that its spacetime index is contravariant, and equating the rows of M using the ordering :

[l[`~μ`] = Vector[row](%id = 18446744078368885086), n[`~μ`] = Vector[row](%id = 18446744078368885206), m[`~μ`] = Vector[row](%id = 18446744078368885326), mb[`~μ`] = Vector[row](%id = 18446744078368885446)]

(27)

${Physics:-D_[mu], Physics:-Dgamma[mu], Physics:-Psigma[mu], Physics:-Ricci[mu, nu], Physics:-Riemann[mu, nu, alpha, beta], Physics:-Weyl[mu, nu, alpha, beta], Physics:-d_[mu], Physics:-Tetrads:-e_[a, mu], Physics:-Tetrads:-eta_[a, b], Physics:-g_[mu, nu], Physics:-gamma_[i, j], Physics:-Tetrads:-gamma_[a, b, c], l[mu], Physics:-Tetrads:-l_[mu], Physics:-Tetrads:-lambda_[a, b, c], m[mu], Physics:-Tetrads:-m_[mu], mb[mu], Physics:-Tetrads:-mb_[mu], n[mu], Physics:-Tetrads:-n_[mu], Physics:-Christoffel[mu, nu, alpha], Physics:-Einstein[mu, nu], Physics:-LeviCivita[alpha, beta, mu, nu], Physics:-SpaceTimeVector[mu](X)}$

(28)

Check the covariant components of these vectors towards comparing them with the lines of the Maple's tetrad

l[mu] = Array(%id = 18446744078298368710), n[mu] = Array(%id = 18446744078298365214), m[mu] = Array(%id = 18446744078298359558), mb[mu] = Array(%id = 18446744078298341734)

(29)

This shows the null vectors (with underscore) that come with Tetrads package

Physics:-Tetrads:-l_[mu] = Vector[row](%id = 18446744078354520414), Physics:-Tetrads:-n_[mu] = Vector[row](%id = 18446744078354520534), Physics:-Tetrads:-m_[mu] = Vector[row](%id = 18446744078354520654), Physics:-Tetrads:-mb_[mu] = Vector[row](%id = 18446744078354520774)

(30)

So (29) computed from M is the same as (30) computed from Maple's tetrad.

But, from (30) and the form of Maple's tetrad

Physics:-Tetrads:-e_[a, mu] = Matrix(%id = 18446744078297844182)

(31)

for the current signature

(32)

we see the ordering of the null vectors is , not used in [1] with the signature (+ - - -). So the adjustment required in M, resulting in , consists of reordering M's rows to be