General equation for hyperbolic motion in a 2D plane -- Help needed please

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Need a general equation for relativistic hyperbolic motion in a 2D plane for arbitrary initial parallel and transverse velocities, to complete a real time simulation I am coding for acceleration of extended 2D objects in a plane.
Hi. I'm working on coding a 2+1D simulation. The simulator moves 2D objects in a plane in real time, demonstrating Wigner rotation and other relativistic scenarios. So far I have created the algorithms for active and passive Lorentz transformation of extended 2D objects and for relativistic velocity addition in any direction in the plane. Now I need an algorithm for the general case of Born rigid acceleration in a 2D plane. Wikipedia gives the special case of acceleration in the x direction with a non-zero initial velocity in the parallel x direction, which can be written like this:

$$ X(T)~ = ~\frac{ 1 }{\alpha } \left( \sqrt{ 1+\left( V_X ~ G_{X} + \alpha ~ T \right)^2 } - G_{X} \right), ~~~~~~~~_{(Eq1a)}$$
$$ V(T) ~=~ \frac{ V_X ~ G_X + \alpha ~ T }{ \sqrt{ 1+\left( V_X ~ G_X + \alpha ~ T \right)^2 } } , ~~~~~~~~_{(Eq1b)}$$

where [itex] \alpha[/itex] is the constant proper acceleration in the x direction, [itex] V_X [/itex] is the initial velocity in the x direction, [itex]G_{X} [/itex] is the gamma factor of the initial parallel velocity and using units such that the speed of light is unity.

For motion in the 2D plane, my program needs to cope with initial transverse velocities. After some trial and error I came up with this equation to handle initial transverse velocity:

$$ X(T) ~=~ \frac{ 1 }{\alpha } \left( \sqrt{ 1+\left( \alpha ~T~/~G_Y \right)^2 } - 1 \right), ~~~~~~~~_{(Eq2)}$$
$$ V(T) ~=~ \frac{ \alpha ~ T~/~G_Y }{ \sqrt{ 1+\left( \alpha ~ T~ / ~G_Y \right)^2 }}~*~\frac{1}{G_Y} , ~~~~~~~~_{(Eq2b)}$$

where [itex]G_{Y} [/itex] is the gamma factor of the initial transverse velocity in the Y direction. This equation works well if the initial X velocity is zero.

Now if the equations are combined like this:

$$ X (T)~=~ \frac{ 1 }{\alpha } \left( \sqrt{ 1+\left( V_X ~ G_{X} + ~ \alpha ~ T~/~G_Y \right)^2 } - G_{X} \right), ~~~~~~~~_{(Eq3a)}$$
$$ V(T) ~=~ \frac{ V_X ~ G_X + \alpha ~ T ~/~G_Y }{ \sqrt{ 1+\left( V_X ~ G_X + \alpha ~ T~/~G_Y \right)^2 } } ~*~\frac{1}{G_Y}, ~~~~~~~~_{(Eq3b)}$$

I do not get the desired result. The final equations will only work correctly if either the initial parallel x velocity or the initial transverse y velocity are zero. Can anyone derive or locate the the fully general equation for hyperbolic motion in a 2D plane, so I can continue with my coding project? I have come up with some ad hock equations and fudge factors for limited special cases, but the whole thing is becoming messy and inaccurate. I have spent weeks on coding this simulator, but I seem to have reached a dead end. This is a shame as I think it would be very useful to anyone interested in relativity. I would very much appreciate any help with this.

I would be happy with an equation that is limited to acceleration only in the x direction (It is always possible to make sure this is the case by rotating the whole frame before the calculation and rotating everything back before displaying on the screen).
 
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  • #2
While I don't know the answer to this, I can point you at some references which may be of some help. See https://en.wikipedia.org/w/index.php?title=Born_rigidity&oldid=1145936959 and it's discussion of the Herglotz-Noether theorem.

I think what you are looking for would be classified as "irrotational" Born rigid motion, classified as class A in the wiki article. It's likely that the reference by Moeller or LeMaitre in this article would be helpful to you. I don't feel that the Wiki article itself answers your questions, unfortunately. But encouraginly, it does address the issues of existence and uniqueness, and my interpretation is that the sort of motion you're looking for does exist and is unique.

wiki said:
In general, it was shown that irrotational Born motion corresponds to those Fermi congruences of which any worldline can be used as baseline (homogeneous Fermi congruence).[19]

Unfortunately, the answers given in the Wiki give a metric for an observer located at this "fixed" point, rather than as a set (congruence) of worldlines in some particular inertial frame with Minkowskii coordinates. I don't have access to the cited articles, and I can't really guarantee that they'll directly answer your question. But the wiki article might give you a place to start. It looks like it'll be a somwhat involved project to me, though - good luck.
 
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  • #3
Possibly also useful and probably also not actually the answer you're looking for is the definition of Born rigid motion in terms of a congruence of worldlines, given in https://arxiv.org/abs/0802.4345. The pdf version of this is https://arxiv.org/pdf/0802.4345.pdf.

A congruence of worldlines mathematically is simply a 4-vector field, that associates a 4-velocity to every point in the space-time. This 4-velocity is physically interpretable as the 4-velocity of a point on the rigid object in question.

The motion of a point on the rigid body would then be the integral curves of the congruence. Using a fluid flow analogy, the motion of a point on the rigid body is a streamline of the velocity field given by the 4-velocity u. So the trajectory followed by a point on the rigid congruence would be a streamline of the velocity field u which defines the congruence.

The mathematics gives us mathematical condition for the congruence, the vector field u, to be a Born rigid congruence.

An inertial frame of reference will have a Minkowskii metric g. We will use coordinates such that the components of the metric have a +--- signature to be compatible with the reference. Then the metric g has components forming a diagonal of (c^2, -1, -1, -1). We can then express the mathematical conditions for the congruence as defined by the vector field u to be rigid by $3.3 from the previously mentioned arxiv reference.

$$\script{L}_u c^{-2} u^a u^b -g = 0$$

Where ##\script{L}_u## is the Lie derivative, https://en.wikipedia.org/wiki/Lie_derivative

I personally use a -+++ signature, and use geometric units so that c=1. I believe that this would make the required condition for the congruence to be Born rigid as:

$$\script{L}_u u^a u^b + g = 0$$

I don't use Lie derivatives enough to express the Lie derivative in terms of covariant derivatives, which would then be ordinary partial derivatives given the Minkowskii metric. Such an expression is a bit backwards, as the Lie derivative exists independent of the metric. But knowing this relatinonship would allow us to write a partial differential equation for the vector field u.

Then solving your problem would be solving the partial differential equations with some boundary conditions to get the vector field u, and then plotting the resulting streamlines of said vector field.

It is very likely that I am making this more complicated than it needs to be, possibly someone else has a simpler way.
 
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  • #4
Thanks for the pointers, pervect. :smile: I'm afraid that route may be above my ability and access level. However, the Wikipedia page on acceleration in Special relativity seems to hint that their is a simpler way. Referring to the section on Curved worldlines, Its says the basic equations (6a) for hyperbolic motion are obtained by integration of the equations for proper acceleration (3a), but I have not figured out how that works yet, or if they are even referring to the right equation.

In the bibliography of that page it states that Max Born "noticed that his transformation can be used to transform into a 'hyperbolically accelerated reference system'" and von Laue derived the same formulas as in 1911 for the transformation of rest acceleration and hyperbolic motion, and the hyperbolic reference frame, which sounds useful in this context.
 
  • #5
I assume that this is a pointlike object that passes through event ##(t_0,x_0,y_0)## with velocity ##(u_x,u_y)## eternally accelerating in the ##+x## direction with proper acceleration ##\alpha##.

According to Wikipedia a general Lorentz boost is$$
B(v)=\left(
\begin{array}{ccc}
\gamma&-\gamma v_x&-\gamma v_y\\
-\gamma v_x&1+(\gamma-1)\frac{v_x^2}{v^2}&(\gamma-1)\frac{v_xv_y}{v^2}\\
-\gamma v_y&(\gamma-1)\frac{v_xv_y}{v^2}&1+(\gamma-1)\frac{v_y^2}{v^2}
\end{array}
\right)$$where ##v^2=v_x^2+v_y^2## and ##\gamma=(1-v^2)^{-1/2}##.

Take your initial position ##X=(t_0,x_0,y_0)^T## and compute ##X'=B(-u)X##. (Sanity check: take your initial four velocity, ##U=(\gamma_u,\gamma_uu_x,\gamma_uu_y)^T## and confirm that ##B(-u)U## is ##(1,0,0)^T##.) You now have a point on the worldline of your object in a frame where it is instantaneously at rest. You can easily write down its trajectory in this frame - it's just ##(t',X'_x+\sqrt{(t'-X'_t)^2+\alpha^{-2}}-\left|\alpha^{-1}\right|,X'_y)^T##. Then you apply ##B(u)## to this to get the trajectory in the original frame. You also need to apply ##B(-u)## to the general position ##(t,x,y)^T## in order to get an expression for ##t'## in terms of the coordinates you are using - then you can eliminate it from the above. The result will be a mess that you'll need to rearrange into ##x(t)## somehow.

Caveat: I typed this on my phone. Typos may exist.
 
  • #6
This might be an easier way.

You have a four-velocity ##U=(\gamma, \gamma u_x, \gamma u_y)##. The associated four-acceleration is$$
\begin{eqnarray*}
A &=& \frac{d}{d\tau}U\\
&=&\frac{dt}{d\tau}\frac{d}{dt}U\\
&=&\gamma\frac{d}{dt}U\\
&=&\left(\gamma^4u_x\frac{du_x}{dt},
\gamma^4(1-u_y^2)\frac{du_x}{dt},
\gamma^4u_xu_y\frac{du_x}{dt}\right)
\end{eqnarray*}$$where, I've used ##\frac{d}{dt}u_y=0## to impose acceleration parallel to the x-axis, and the identities ##\frac{d}{dt}\gamma=\gamma^3u_x\frac{d}{dt}u_x## and ##\gamma^4u_x^2+\gamma^2=(1-u_y^2)\gamma^4## in getting the last line.

Now, the modulus-squared of the four acceleration is ##-a^2##, where ##a## is the acceleration measured by an onboard accelerometer. Remembering that the modulus-squared of a four-vector ##(V_t,V_x,V_y)## is ##V_t^2-V_x^2-V_y^2##, we can set ##|A|^2=-a^2## and get a bit of a mess that simplifies to$$
\left(\frac{du_x}{dt}\right)^2=\frac{a^2}{1-u_y^2}\left(1-u_x^2-u_y^2\right)^3
$$which is separable, giving you$$
\frac{\sqrt{1-u_y^2}du_x}{\sqrt{1-u_x^2-u_y^2}^3}=adt
$$Integrating the right hand side is trivial; the left needs a trigonometric substitution. You end up with$$
\left[\frac{u_x}{\sqrt{1-u_y^2}\sqrt{1-u_y^2-u_x^2}}\right]_{u_x=u_{xi}}^{u_{xf}}=a(t-t_i)
$$where ##u_{xi}## and ##t_i## are initial conditions and ##u_{xf}## is the x component of velocity at time ##t## that you want.

I'm out of time now, but I think that if you plug in the limits you end up with a fourth-order polynomial that is actually a quadratic in ##u_{xf}^2##. Whether that can be integrated to get ##x(t)## I don't know.

Same caveat as above applies.
 
  • #7
Ibix said:
where, I've used ##\frac{d}{dt}u_y=0## to impose acceleration parallel to the x-axis,
Ok, I can live with acceleration constrained to the x-axis. It does seem to simplify the calculations greatly.
Ibix said:
Now, the modulus-squared of the four acceleration is ##-a^2##, where ##a## is the acceleration measured by an onboard accelerometer.
So we are talking about proper acceleration which is what is required.
Ibix said:
$$\frac{\sqrt{1-u_y^2}du_x}{\sqrt{1-u_x^2-u_y^2}^3}=adt$$
Integrating the right hand side is trivial; the left needs a trigonometric substitution. You end up with$$
\left[\frac{u_x}{\sqrt{1-u_y^2}\sqrt{1-u_y^2-u_x^2}}\right]_{u_x=u_{xi}}^{u_{xf}}=a(t-t_i)$$
This is the part that has me puzzled. In my limited understanding of Born rigid motion, the proper acceleration is constant, whereas you appear to be talking about acceleration that is a function of time...?

The Wikipedia page on hyperbolic motion mentions that "Solving for the equation of motion gives the desired formulas". If someone could show me how the hyperbolic equations are derived from the equations of motion for the simple case of zero initial velocity, then I might be able to extend to the general case myself.
 
  • #8
KDP said:
the proper acceleration is constant, whereas you appear to be talking about acceleration that is a function of time...?
No, his equations are for constant proper acceleration. The RHS of the last equation you post is the product of the proper acceleration with the difference in coordinate times over the range of integration.
 
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  • #9
KDP said:
In my limited understanding of Born rigid motion, the proper acceleration is constant, whereas you appear to be talking about acceleration that is a function of time...?
The proper acceleration is a constant ##a## - the RHS of the expression you quoted is ##a\times(t-t_i)##, not ##a## as a function of ##t##. This does, of course, mean that ##du_x/dt\neq\mathrm{const}##.
KDP said:
If someone could show me how the hyperbolic equations are derived from the equations of motion for the simple case of zero initial velocity, then I might be able to extend to the general case myself.
Take my last expression, set ##u_y=u_{xi}=t_i=0##, solve for ##u_{xf}## (which is ##dx/dt##) and integrate to get ##x(t)##.
 
  • #10
Let's do it from scratch in a covariant way. Hyperbolic motion can be introduced as the motion of a charged point particle in a homogeneous electric field. Let's make the field ##\vec{E}=E \vec{e}_1## and we restrict ourselves to the motion along the 1-axis. Then the equations of motion read
$$u^{\mu}=\dot{x}^{\mu}, \quad \begin{pmatrix}\dot{u}^0 \\ dot{u}^1 \end{pmatrix}=\frac{q E}{m c} \begin{pmatrix}u^1 \\ u^0 \end{pmatrix}.$$
The dot is the derivative wrt. proper time, and thus we have the additional constraint
$$u^{\mu} u_{\mu}=(u^0)^2-(u^1)^2=c^2.$$
To solve the equations of motion, we derive the equation for ##u^1## with respect to ##\tau## and use the equation of motion for ##u^0##:
$$\ddot{u}^1=\frac{q E}{m c} \dot{u}^0 = \left (\frac{q E}{mc} \right) u^1.$$
The general solution is
$$u^1(\tau)=A \sinh(\alpha \tau)+B \cosh(\alpha \tau), \quad \alpha=\frac{q E}{mc}.$$
From the original equation of motion for ##u^1## we get
$$u^0(\tau)=\frac{1}{\alpha} \dot{u}^1=A \cosh(\alpha \tau) + B \sinh(\alpha \tau).$$
This can be written in matrix form as
$$\begin{pmatrix}u^0(\tau) \\ u^1(\tau) \end{pmatrix} = \begin{pmatrix} \cosh(\alpha \tau) & \sinh(\alpha \tau) \\ \sinh(\alpha \tau) & \cosh(\alpha \tau) \end{pmatrix} \begin{pmatrix}A \\ B \end{pmatrix}.$$
This is a Lorentz transformation with rapidity ##\eta=\alpha \tau## of the vector ##(A,B)##. Thus
$$(u^0)^2-(u^1)^2=A^2-B^2 \stackrel{!}{=} c^2.$$
Let's now start from rest, i.e., ##u^1(0)=0##. Then ##B=0## and ##A=+c##, because we must have ##u^0>0##:
$$\begin{pmatrix}u^0(\tau) \\ u^1(\tau)\end{pmatrix}=\begin{pmatrix}c \cosh(\alpha \tau) \\ c \sinh(\alpha \tau) \end{pmatrix}.$$
This we can integrate once more to get the world-line. We make ##t=0## for ##\tau=0## and ##x^1(0)=x_0^1##:
$$\begin{pmatrix} x^0(\tau) \\ x^1(\tau) \end{pmatrix}=\begin{pmatrix}\frac{c}{\alpha} \sinh(\alpha \tau) \\ x_0^1 + \frac{c}{\alpha} [\cosh(\alpha \tau)-1] \end{pmatrix}.$$
It's now also easy to express this result in terms of the coordinate time ##t## since
$$t=x^0/c=\frac{1}{\alpha} \sinh(\alpha \tau)$$
and thus
$$x^1(t)=x_0^1 + \frac{c}{\alpha} [\sqrt{1+\alpha^2 t^2}-1].$$
 
  • #11
Ibix said:
The proper acceleration is a constant ##a## - the RHS of the expression you quoted is ##a\times(t-t_i)##, not ##a## as a function of ##t##. This does, of course, mean that ##du_x/dt\neq\mathrm{const}##.
PeterDonis said:
No, his equations are for constant proper acceleration. The RHS of the last equation you post is the product of the proper acceleration with the difference in coordinate times over the range of integration.
Thanks Peter and Ibix for clearing up my misunderstanding. I think I can move forward now. :smile:

vanhees71 said:
and thus
$$x^1(t)=x_0^1 + \frac{c}{\alpha} [\sqrt{1+\alpha^2 t^2}-1].$$
Wow Vanhees, thanks very much for the detailed derivation. I think you have definitely given me something to sink my teeth into.
Once again, thanks you all (including pervect of course) for taking the time to point me in the right direction ..
 
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  • #12
I've been busy with other things, but I've occasionally thought about the question in this thread. My thoughts are a bit decoupled from the way the thread has progressed, however.

First in my thoughts is an example of a fairly well-known rigid motion, the sliding block. We've had some interesting threads on it in the past where we've talked about this. It's a fairly simple idea, one has an inertial frame of reference with a flat floor, and one makes this floor accelerate upwards. This is a rigid sort of motion, most easily described in Rindler coordinates. We imagine a block sliding across the floor, and the block's motion is also rigid. There's nothing too surprising here - so far.

Now, what is interesting and (IMO) somewhat surprising thing about the sliding block in that it's a rotational motion, not an irrotational motion, in terms of the wiki classification of rigid motion into these two categories.

If one is familiar with "Thomas precession", a gyroscope mounted on such a sliding block would precess, as discussed in https://arxiv.org/abs/0708.2490v1, pdf version https://arxiv.org/pdf/0708.2490v1.pdf, which is a simple argument for why I say the motion would be classed as "rotational".

A consequence of the fact that the sliding block motion is rotational is that, according to Born's original defintion of rigidity, one can't change the blocks sliding speed in a rigid manner, as this changes its state of rotation. Of course, one might get around this by considering Born's revised notion of rigidity, but the scope of my thoughts is about Born's original notion, not the revised one.

So, the interesting question is - what would the irrotational equivalent of the sliding block motion be, and how would we compute it?

This is a particular case of the general problem.

My thinking is that a logical way to do would be to start with the curve of an accelerating reference point at the center of the block. The motion of this point would describe a curve through space-time, and differentiating it would give us a vector field at every point along the curve, the tangent of the curve.

We'd like to extend this vector field to describe the rigid motion of an extended body. Rather than try and extend it through all of space, we might take the approach of extending it to a nearby "world tube", for points "close" to the reference point. Exactly how to do this is unclear - we'd need a transport law similar to Fermi-Walker transport. I'm only used to seeing Fermi-Walker transport discussed for transport of a vector along time-like curves, however - what we want here is something similar, but along a space-like curve.
 
  • #13
pervect said:
Born's revised notion of rigidity
Which notion are you referring to?
 
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  • #14
pervect said:
what would the irrotational equivalent of the sliding block motion be
I don't think there is one; I believe the Herglotz-Noether theorem rules it out. That theorem classifies possible rigid non-inertial motions in flat spacetime, and IIRC the only irrotational one is linear constant acceleration in a single direction with no sideways "sliding" at all, i.e., the standard Rindler congruence.
 
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  • #15
Hi Pervect,
it might interest you to know I now have a solution to this problem, but its a work-around algorithm, rather than a simple equation. It works like this. In any given inertial frame (S1) where our test object has initial constant motion before the acceleration phase, we can find another inertial reference frame (S0) where the initial motion is zero. This simplifies things a lot as all the motion is in a straight line in this alternative frame rather than travelling in a curve in the original frame. In the spirit of keeping things simple my extended 2D objects are simple squares and I only move the 4 vertices. Now I accelerate the object in the new frame using the basic hyperbolic motion equations and then Lorentz transform the 4 points back to the original frame moment by moment. Before the transform the 4 vertices can all have different accelerations and velocities at any given instant in S0 and after the transformation things get more complicated as all 4 corners can have different time coordinates in frame S1. Correcting for this requires knowing the history and future of each point which requires knowing exactly how the object accelerates in S1, bringing us full circle. This is circumvented by starting at a given time point in S1 and transforming to 4 different time coordinates in S0. Compute where each vertex was in S0 at those times and then transform back to S1 where they all arrive simultaneously as prearranged. A further simplification for pragmatic reasons is to rotate everything in S0 so that all motion is calculated parallel to the x-axis and then rotate everything back before transforming everything back to S1.

The fact that it works is confirmed by checking that all events in S0 like a light signal arriving at vertex when a thruster turns on or off in S0 has a corresponding event in S1.

It works, but it would be nice to have a single equation, which would be less computationally intensive as all calculations have to be done 50 times a second to keep the animation smooth. This could become a problem as additional objects are added to the simulations.

pervect said:
Now, what is interesting and (IMO) somewhat surprising thing about the sliding block in that it's a rotational motion, not an irrotational motion, in terms of the wiki classification of rigid motion into these two categories.

If one is familiar with "Thomas precession", a gyroscope mounted on such a sliding block would precess, as discussed in https://arxiv.org/abs/0708.2490v1, pdf version https://arxiv.org/pdf/0708.2490v1.pdf, which is a simple argument for why I say the motion would be classed as "rotational".

From what I gathered from my studies of Wigner rotation, is that when an object is transported in a closed loop on a 2D plane, in a born rigid manner, with orthogonal acceleration, then when the object arrives back at its starting point it is rotated in a real way relative to its starting orientation, but any gyroscopes mounted on the object, (including a Sagnac device) will not detect any rotation at all. Irrotational motion? They would of course notice distant stars rotating around them.

An initial observation from my simulation is that Wigner rotation is always in the opposite sense to the path of the object. For example if you follow a path where you are always turning to the right you follow a clockwise path and the Wigner rotation would then be anti-clockwise. This is a worthwhile observation, as the Wigner formula for the rotation only gives the magnitude but not the direction of the rotation. I have of course checked that after objects are accelerated instant by instant in my simulation, that the object final orientation of the object matches the theoretical Wigner angle exactly and that the object maintains its exact square shape and area.

A further cross check in the simulation is to draw some lines in S0 passing through the 4 vertices and remaining parallel to the x-axis. These lines represent a physical object such gun barrel. The idea is that if the square represents a projectile in a gun barrel then if the projectile always remains inside the barrel in S0 and its sides are always parallel to the gun barrel in S0, then the same must be true at all instants in S1 during the acceleration, despite the complex curved paths and rotations that occur in S1.
 
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