What is the constant velocity curve of an accelerating spaceship?

[pervect]

I've got a rather messy unsolved (AFAIK) problem I've been thinking about a bit, and I thought I'd share it with the group.

The basic idea is fairly simply stated. Suppose we have an accelerating spaceship which has a constant proper acceleration g.

The question is, what is the curve (pick any coordinates that make the problem simple) that has a constant velocity 'v' with respect to any observer who is "stationary" with respect to the accelerating ship.

The notion of "stationary wrt to the ship" invites one to choose a coordinate system in which the metric is static (not a function of time) and in which the location of the spaceship is the origin. Then objects which are stationary in this coordinate system are also stationary with respect to the ship. Other more complex formulations of this notion are possible, but I think it's well defined and coordinate independent. (If anyone doesn't agree, we can argue about it more :-).)

The velocity 'v' is measured by the local clocks and local ruler of said "stationary observer", when the observer and the moving object are at the same place at the same time.

I'm particularly interested in the magnitude of the 4-acceleration of v, i.e. the "gravity" an observer on such a trajectory would experience.

 

[robphy]

Well... if I understood the question correctly, the trivial v=0 case would be the concentric hyperbolas with worldline curvature g. The nontrivial case would be... um... [let me get back to you on that one]. Is there a nontrivial case?

 

[pervect]

Well... if I understood the question correctly, the trivial v=0 case would be the concentric hyperbolas with worldline curvature g. The nontrivial case would be... um... [let me get back to you on that one]. Is there a nontrivial case?

That's not quite the result I'm getting. Here's my argument

If we use Minkowski coordinates, an accelerated observer with constant proper accelration g can be represented by the curve

ta = cosh(g*tau)/g
xa = sinh(g*tau)/g

We can use the idea of "radar coordinates" as a convenient way to define the worldline of a stationary observer. Somone who stays a constant "radar distance" away from our accelerating observer will have "v=0".

This is the same idea as saying that someone could use radar to hover a constant distance above the Earth, or any other gravitating body, in the accelerating frame we would argue that the light signals always took the "same path".

Let the radar distance be $\chi$. Then a signal emitted from our accelerating observer at time $\tau$ must hit our "stationary observer", and return at time $\tau + 2 \chi$.

We can actually trace two null geodesics, one starting at time $\tau$ and proceding in the +x direction, another starting at time $\tau + 2 \chi$ and proceeding in the -x direction. Where these two null geodesics meet will be the worldline of our "stationary" observer.

We can then write a whole bunch of equations

t1 = cosh(g*tau)/g + k1
x1 = sinh(g*tau)/g + k1
t2 = cosh(g*tau+2*chi)/g + k2
x2 = sinh(g*tau+2*chi)/g - k2

and we look for the k1 and k2 that satisfy t1=t2, x1=x2, which is where the two null geodesic meet.

By considering t1-t2+x1-x2 = 0 and t1-t2+x2-x1 = 0, we can solve for k1 and k2 fairly easily. After a lot of manipulation I get the worldline for the stationary observer, (ts, xs).

ts = t1 = t2 = (exp(g*tau+2*chi) + exp(-g*tau))/2g
xs = x1 = x2 = (exp(g*tau+2*chi) - exp(-g*tau))/2g

This worldline hasn't been normalized yet, i.e. (dts/dtau)^2 - (dxs/dtau)^2 != -1. I won't go through the details, but after a change of variables, I get a proper acceleration for the "stationary observer" of

g*exp(-chi)

which means that the proper acceleration is lower for more distant observers. This is what we expect considering the "Bell spaceship" paradox - if the lead ship stays a constant distance away from the tail ship, it must accelerate at a lower proper acceleration.

This was the "easy" part, now we want to find the worldline which has a constant velocity relative to the above 'stationary' worldlines for v != 0.

Note that we don't want to mesure the velocity with the radar - we want the stationary observer to personally measure the velocity "on site". I just used radar as a convenient device to define what a "stationary" obsever was.

This should yield the same idea of "stationariness" that applies to an observer at a constant Scwarzschild R coordinate relative to a black hole - the metric in the radar coordinates should be static - unless I'm making some conceptual error.

 

[George Jones]

After a lot of manipulation I get the worldline for the stationary observer, (ts, xs).

ts = t1 = t2 = (exp(g*tau+2*chi) + exp(-g*tau))/2g
xs = x1 = x2 = (exp(g*tau+2*chi) - exp(-g*tau))/2g

I agree with this for the constant distance/zero relative speed case. Looking at (ts - xs)(ts + xs) gives something fairly nice for the lines of consant distance. I also made a similar http://groups.google.ca/group/sci.physics/msg/91d8704c115d8899?dmode=source&hl=en" on sci.physics.

Regards,
George

 

[George Jones]

Well... if I understood the question correctly, the trivial v=0 case would be the concentric hyperbolas with worldline curvature g.

This doesn't quite work because the line of simultaneity for a comoving observer is not always horizontal, and thus lines of simultaneity connect events at which the acclerated observers have non-zero relative speed.

Regards,
George

 

[George Jones]

Let $\left( t,x\right)$ be global inertial coordinates in which the coordinates of the accelerated ship are $\left( g^{-1}\sinh\left( g\tau\right) ,g^{-1}\cosh\left( g\tau\right) \right)$. An observer in this ship sets up (as pervect did, and as I did in a link given previously) a coodinate system $\left( t^{\prime},x^{\prime}\right)$. In global inertial coodinates, curves of constant $x^{\prime}$ are hyperbolae defined implicitly by

$$t^{2}-x^{2}=-g^{-2}e^{2gx^{\prime}}.$$

The hyperbolae all have asymptotes $t=x$ and $t=-x$. For $x>0$, these hyperbolae successively become less sharply curved as $x$ increases.

Lines of constant $t^{\prime}$ are straight lines

$$t=\frac{e^{2gt^{\prime}}-1}{e^{2gt^{\prime}}+1}x$$

that pass through the origin. As $t^{\prime}\rightarrow\infty$, the lines approach $t=x$; as $t^{\prime}\rightarrow-\infty$, the lines approach $t=-x$.

These equations implicitly define the transformation between the inertial coordinate system $\left( t,x\right)$ and the accelerated coordinate system $\left( t^{\prime},x^{\prime}\right)$. The accelerated coordinate system $\left( t^{\prime},x^{\prime}\right)$ only covers the wedge $x>\left|t\right|$ of the global inertial coordinate system, with the halflines $t=x$ and $t=-x$ playing the role of horizons.

The above gives

$$\begin{align*} t\left( e^{2gt^{\prime}}+1\right) & =x\left( e^{2gt^{\prime}}-1\right) \\ \left( t-x\right) e^{2gt^{\prime}} & = - \left( t+x \right). \end{align*}$$

After transforming to lightcone coordinates $x^{+}=t+x$ and $x^{-}=t-x$, the transformation equations become

$$x^{+} x^{-}=-g^{-2}e^{2gx^{\prime}}$$

and

$$e^{2gt^{\prime}}= - \frac{x^+}{x^-}.$$

Now let another ship move through the accelerated coordinate system with speed $w$, so that $x^{\prime}=wt^{\prime}+\alpha$ for this second ship. Using this gives

$$\begin{align} x^{+} x^{-} & =-g^{-2}e^{2g\left( wt^{\prime}+\alpha\right) }\\ & =-g^{-2}\left( - \frac{x^+}{x^-}\right) ^{w}e^{2g\alpha}, \end{align}$$

so that

$$\left( x^{+} \right)^{1-w} \left( x^{-} \right)^{1+w}= \left( -1 \right)^{1+w} g^{-2}e^{2g\alpha}.$$

This implicitly defines the path of the second ship in the global inertial coordinate system.

It might now be prudent to chose some specific values for $w$, maybe $1/2$, and $\alpha$. Also, $g$ has a nice value when time is measured in years and distance is measured in lightyears.

Still working on it.

Regards,
George

 

[robphy]

[The trivial case I suggested above was probably an inaccurate way to describe the observer coincident with the given observer with acceleration g... the coincident observer has relative velocity v=0 with the given observer. If this isn't right, I am probably misunderstanding the whole problem.]

Let me try to rephrase your question [to get it straight in my mind].

You have a uniformly accelerating observer with acceleration g; its worldline is a hyperbola asymptotic to the lightcone of event (call it) O. Worldlines at rest according to this observer [determined by radar methods] are also hyperbolas asymptotic to this lightcone, and thus are also uniformly accelerating observers with acceleration inversely proportional to distance r from O.

Are you seeking a family of worldlines that intersect the above family of uniformly-accelerated worldlines such that the dot-products of their [unit] tangent vectors at these intersection events are constant?

[Hopefully I haven't totally misunderstood the question.]

 

[George Jones]

Here's my take on the situation. Ship 1 has constant 4-acceleration, and ship 2 moves with constant relative speed with respect to 1. I think this means the following.

Let A and B be arbitrary event on 1's worldline.

Draw a line simultaneity through A for an obsrever comoving with 1 at A. This line intersects 2's worldline at event A', say. Compute the relative speed by taking the inner product of the 4-velocities at A and A'.

Now do the same for B and B'. The relative speeds should be the same.

I guess I didn't understand "concentric hyperbolas with worldline curvature g."

Sorry if I misinterpreted what you, robphy.

It still quite possible that we're all thinking of completely different things!

Regards,
George

 

[pervect]

I agree with this for the constant distance/zero relative speed case. Looking at (ts - xs)(ts + xs) gives something fairly nice for the lines of consant distance. I also made a similar http://groups.google.ca/group/sci.physics/msg/91d8704c115d8899?dmode=source&hl=en" on sci.physics.

Regards,
George

Glad we agree - afater reading your usenet post I realized it would have been better if I would have made the radar signals emit at $$\tau-\chi$$ and received at $$\tau+\chi$$ rather than the choice I originally made. This makes both $$\tau$$ and $$\chi$$ radar coordinates, and yields the more pleasing equivalent expressions for the worldlines of "stationary observers" of

$$ts = e^{\chi} \, \mathrm{cosh}(g \tau)/g$$
$$xs = e^{\chi} \, \mathrm{sinh}(g \tau)/g$$

 

[George Jones]

I have found a mistake in my sci.physics post. There I wrote

$$t = \frac{e^{2at'} + 1}{e^{2at'} - 1}x,$$

when it should be

$$t = \frac{e^{2at'} - 1}{e^{2at'} + 1}x.$$

I have edited post #6 in this thread to reflect this change.

I have not calculated the acceleration for the second ship, but I am now able to plot worldlines of both ships. I might attach some .pdf plots in a later post.

Regards,
George

 

[Hurkyl]

I guess I don't understad the difficulty in this. If we take two observers whose worldlines are given by:

((1/g) sinh (g tau), (1/g) cosh (g tau))

and

((1/h) sinh (h tau), (1/h) cosh (h tau))

then the lines of simultaneity for both observers are exactly the lines passing through the origin.

Furthermore, the proper distance along the line of simultaneity between the two observers is a constant.


I was originally interested in the case of a uniformly accelerating spaceship, with the stipulation that the ship didn't appear to deform to the occupants: my two observers were at the front and rear of the ship.

Since the result extends to any number of observers, I took this to mean that an occupant at any point in the ship would not perceive any sort of deformation there.


Wouldn't it be reasonable to also interpret this as being that the observers are stationary wrt each other?

 

[pervect]

Let me try to rephrase your question [to get it straight in my mind].

You have a uniformly accelerating observer with acceleration g; its worldline is a hyperbola asymptotic to the lightcone of event (call it) O. Worldlines at rest according to this observer [determined by radar methods] are also hyperbolas asymptotic to this lightcone, and thus are also uniformly accelerating observers with acceleration inversely proportional to distance r from O.

I think this works now - I wasn't aware till you mentioned it that all these "stationary" hyperbolic observers were asymptotic to the same event O, but after some thought, I think you're right about that, and it's a very helpful observation. I still need to do some algebraic manipulation to see how the algebra works out.

Are you seeking a family of worldlines that intersect the above family of uniformly-accelerated worldlines such that the dot-products of their [unit] tangent vectors at these intersection events are constant?

[Hopefully I haven't totally misunderstood the question.]

That's a nice simple way of putting the problem - I was going through a lot of mental contortions to describe the relative velocity, but requiring that the dot product of the unit tangent vectors be constant should do the job just fine.

 

[pervect]

I guess I don't understad the difficulty in this. If we take two observers whose worldlines are given by:

((1/g) sinh (g tau), (1/g) cosh (g tau))

and

((1/h) sinh (h tau), (1/h) cosh (h tau))

then the lines of simultaneity for both observers are exactly the lines passing through the origin.

Furthermore, the proper distance along the line of simultaneity between the two observers is a constant.

I think this is essentially the same observation as robphy made, that all the observers we are calling "stationary" are hyperbolic and asymptotic to the same event O. I have to go through some alegebra to see how this result could formally be equivalent to the one George and I got, but I think it must be the same result. I've normalized our result, but haven't yet showed that an affine transformation takes it to the above form (though I think one should exist).

 

[pervect]

OK, if we take

$$ts = e^{\chi} \, \mathrm{cosh}(g \tau)/g$$
$$xs = e^{\chi} \, \mathrm{sinh}(g \tau)/g$$

and make the substitutions

$$g = h e^{\chi}$$
$$\tau = \lambda e^{-\chi}$$

we get

$$ts = \mathrm{cosh}(h \lambda)/h$$
$$xs = \mathrm{sinh}(h \lambda)/h$$

therefore my/George's result is equivalent to robphy/Hurkyl's result.

 

[robphy]

To follow up on my [hopefully correct] formulation of the problem...

one set of curves that satisfies the "constant dot-product" condition is the set of spacelike rays from the vertex, which, as pointed out by Hurkyl, are the lines of simultaneity for those uniformly accelerating observers. Those rays, of course, don't count as worldlines... However, worldlines with unit tangents orthogonal to those spacelike rays should satisfy condition as well... but that's just the trivial case of the coincident observers again.

So, along these lines, one should tighten up the condition to read that the dot-products of their unit tangents be constant and positive (with the +--- signature convention) and that the dot-product of their spatial components doesn't change sign (to ensure that the velocities don't change direction... that is, same speed is not sufficient).

I'm beginning to wonder: is there a nontrivial solution? [Or is my formulation of the original problem incorrect?]

In spite of this, let me stick my neck out and make a guess at the answer... I don't have a lot of free time to check it out... but here it goes... maybe someone can prove or disprove it:

My conjecture:
The sought after worldlines are a spacetime-translation of the set of worldlines of our uniformly accelerating observers... probably with some restriction on the displacement between the vertices of their asymptotic lightcones. (The trivial case, of course, is zero displacement.)

 

[George Jones]

Suppose the second ship crosses the (inertial) x-axis halfway bewteen the origin of the inertial coodinates and the place where the constant-acceleration ship crosses the x-axis, i.e, with inertial coodinates $\left( 0,\frac{1}{2g}\right)$. Suppose further that the second ship moves through the accelerated reference frame $\left( t' , x' \right)$ such

$$x' = \frac{1}{2}t'+\alpha,$$

where

$$\alpha = \frac{1}{2g} \ln \left( \frac{1}{4} \right).$$

Denote the proper time of the second ship by $T$, so as not confuse it with the proper time $\tau$ of the first ship.

After the some calculation I get that the worldline of the second ship is given by

$$t \left( T \right) = \frac{2\sqrt{3}}{9}g^{2}T^{3}-\frac{\sqrt{3}}{8g^{2}}\frac{1}{T}$$

$$x \left( T \right) = \frac{2\sqrt{3}}{9}g^{2}T^{3}+\frac{\sqrt{3}}{8g^{2}}\frac{1}{T}.$$

for $T>0$.

This leads to (after calculation):

$-4/T^2$ for the square of the magnitude of the 4-acceleration;

$2/\sqrt{3}$ for the constant inner product of the appropriate 4-velocities (as it should be).

Regards,
George

 

[pervect]

I'm getting that the general solution is a rather messy differential equation.

I'll be looking at George's results in more detail soon.

The way I derived the equation is as follows:

As long as we put the asymptotic point of the accelerating observer at the origin, we know that a line passing through the origin to (t,x) will be perpendicular to the 4-velocity of any observer "stationary" with respect to the accelerating obsever.

The unit vector perpendicular to (t,x) is just (x,t). Therfore that is the 4-velocity (before normalization) of the stationary observer.

If we let the desired path be defined by a single function, x(t), the 4-velocity of the desired "constant velocity" observer before normalization is just (1,dx/dt).

This leads to the equation

(1,dx/dt) dotproduct (x,t) = k

where we must normalize the two vectors before taking the dot product.

Putting all the normalization factors in with the appropriate signs gives us

$$\left( x \left( t \right) -t \, \frac{dx}{dt} \right) ^{2}={k}^{2} \left( \left( x \left( t \right) \right) ^{2}-{t}^{2} \right) \left( 1-{{ \left( \frac{dx}{dt} \right) }}^{2} \right)$$

where k is the constant value of the dot product, a number greater than 1 and equal to the "gamma factor" of the constant velocity.

Solving the quadratic for dx/dt, taking the plus sign of the quadratic (which implies positive velocities) we get the following diffeq

$$\frac{dx}{dt} = {\frac {t \, x \left( t \right) + \left( \left( x \left( t \right) \right) ^{2}-{t}^{2} \right) k\sqrt {-1+{k}^{2}}}{{k}^{2} \left( \left( x \left( t \right) \right) ^{2}-{t}^{2} \right) +{t}^{2}}}$$

Finding the 4-accleration is straightforwards but long, the result I get for the magnitude of the 4-acceleration is

$$\frac{\frac{d^2 x}{dt^2}}{\left( 1-\left( \frac{dx}{dt} \right) ^2 \right)^{\frac{3}{2}}}$$

The above formula gives sensible results for x = sqrt(1+t^2), which corresponds to x=cosh(tau), t=sinh(tau), x^2-t^2 = 1

 

[George Jones]

As long as we put the asymptotic point of the accelerating observer at the origin, we know that a line passing through the origin to (t,x) will be perpendicular to the 4-velocity of any observer "stationary" with respect to the accelerating obsever.

The unit vector perpendicular to (t,x) is just (x,t). Therfore that is the 4-velocity (before normalization) of the stationary observer.

If we let the desired path be defined by a single function, x(t), the 4-velocity of the desired "constant velocity" observer before normalization is just (1,dx/dt).

This leads to the equation

(1,dx/dt) dotproduct (x,t) = k

where we must normalize the two vectors before taking the dot product.

Putting all the normalization factors in with the appropriate signs gives us

$$\left( x \left( t \right) -t \, \frac{dx}{dt} \right) ^{2}={k}^{2} \left( \left( x \left( t \right) \right) ^{2}-{t}^{2} \right) \left( 1-{{ \left( \frac{dx}{dt} \right) }}^{2} \right)$$

where k is the constant value of the dot product, a number greater than 1 and equal to the "gamma factor" of the constant velocity.

Again, I agree with all this.

Solving the quadratic for dx/dt ...

I have been too lazy to check this.

Finding the 4-accleration is straightforwards but long, the result I get for the magnitude of the 4-acceleration is

$$\frac{\frac{d^2 x}{dt^2}}{\left( 1-\left( \frac{dx}{dt} \right) ^2 \right)^{\frac{3}{2}}}$$

This looks interesting.

I tried parametrizing my result by coodinate time t, instead of proper time T, and I ran up against a quartic whose solution filled a couple of screens.

I intend to fill in some of the steps missing from last post. Hopefully this will happen tomorrow, but I'm not going to miss tomorrow's hockey game https://www.physicsforums.com/showpost.php?p=913597&postcount=16", pervect .

I have attached an image of a plot of my results.

navy - ship accelerating at g

red - ship with constant speed 1/2 relative to navy

navy, green, blue - lines of constant distance in the acclerated frame

horizontal black, purple - lines of constant time (i.e., simultaneity) in the accelerated frame

Note that the tangent to navy at R is parallel to the tangent to green at Q, so what I a have said ("Let A and B be arbitrary event on navy's worldline. Draw a line simultaneity through A for an obsrever comoving with 1 at A. This line intersects 2's worldline at event A', say. Compute the relative speed by taking the inner product of the 4-velocities at A and A'. Now do the same for B and B'. The relative speeds should be the same.") is the same as what pervect and robphy have said ("Are you seeking a family of worldlines that intersect the above family of uniformly-accelerated worldlines such that the dot-products of their [unit] tangent vectors at these intersection events are constant?")

It seems we all agree on what constant speed means. Constant speed also means moving at constant (accelerated) coordinate speed. That's why I looked at the example x' = t'/2 + alpha.

On my diagram, it appears plausible that red moves with speed 1/2 both at P (blue and black time and space axes), and at Q (green an purple time and space axes).

Regards,
George

 

[pervect]

After the some calculation I get that the worldline of the second ship is given by

$$t \left( T \right) = \frac{2\sqrt{3}}{9}g^{2}T^{3}-\frac{\sqrt{3}}{8g^{2}}\frac{1}{T}$$

$$x \left( T \right) = \frac{2\sqrt{3}}{9}g^{2}T^{3}+\frac{\sqrt{3}}{8g^{2}}\frac{1}{T}.$$

for $T>0$.

I confirm that this is a solution, the dot-product is constant, i.e.

normalize(x,t) dotproduct (dt/dT, dx/dT) = constant

$$\frac{x \,dt/dT - t \, dx/dT}{\sqrt{x^2-t^2}} = \sqrt{3}/2$$

and (dt/dT)^2 - (dx/dT)^2 = 1

and the magnitude of the 4-acceleration is 2/T

Sorry about your hockey game :-)

 

[George Jones]

Canada lost 2-0 to Switzerland today - I should have worked on physics.

First a note about the diagram attached to post #18 - units on the time axis are years and on the space axis are lightyears, so g = 1.03/year.

This post extends post \#6 and fills in some of the details left out of post \#16.

Ship 1 moves with constant 4-acceleration $g$ and ship 2 moves with constant speed $w$ in the coordinate system $\left( t^{\prime},x^{\prime}\right)$ for an observer in ship 1, and consequently $x^{\prime}=wt^{\prime}+\alpha$ is the worldline of ship 2.To pin down $\alpha$, suppose the second ship crosses the (inertial) x-axis at event P halfway (in the inertial frame) between the origin of the inertial coodinates and the event where the constant-acceleration ship crosses the x-axis, i.e, P has inertial coordinates $\left( 0,1/\left( 2g\right) \right)$. The inertial coordinates of P, when used in

$$t=\frac{e^{2gt^{\prime}}-1}{e^{2gt^{\prime}}+1}x,$$

imply $$t^{\prime}=0$$, so that the accelerated-frame coordinates of P are $\left( 0,\alpha\right)$. Then, $t^{2}-x^{2}=-g^{-2} e^{2gx^{\prime}}$ applied to P implies

$$\begin{align*} 0^{2}-\left( \frac{1}{2g}\right) ^{2} & =-g^{-2}e^{2g\alpha}\\ \alpha & =\frac{1}{2g}\ln\left( \frac{1}{4}\right) . \end{align*}$$

As a concrete example, take the accelerated-frame speed of ship 2 to be
$w=1/2.$ Using these values in the post #6 result $\left( x^{+}\right)^{1-w}\left( x^{-}\right) ^{1+w}=\left( -1\right) ^{1+w}g^{-2}e^{2g\alpha}$ gives

$$\begin{align*} \left( x^{+}\right) ^{\frac{1}{2}}\left( x^{-}\right) ^{\frac{3}{2}} & =\left( -1\right) ^{\frac{3}{2}}g^{-2}e^{2g\alpha}\\ x^{+}\left( x^{-}\right) ^{3} & =\left( -1\right) ^{3}g^{-4}\frac{1} {16}\\ x^{+} & =-\frac{1}{16g^{4}\left( x^{-}\right) ^{3}}. \end{align*}$$

Inverting $x^{+}=t+x and x^{-}=t-x$, and using the above equation, results in

$$\begin{align*} t & =\frac{1}{2}\left( x^{+}+x^{-}\right) \\ & =\frac{1}{2}\left( -\frac{1}{16g^{4}\left( x^{-}\right) ^{3}} +x^{-}\right) \end{align*}$$

and

$$\begin{align*} x & =\frac{1}{2}\left( x^{+}-x^{-}\right) \\ & =-\frac{1}{2}\left( \frac{1}{16g^{4}\left( x^{-}\right) ^{3}} +x^{-}\right) . \end{align*}$$

$x^{-}$ is a valid parameter of the worldline of ship 2, so rename $x^{-}$ as $s$, and call the proper time of the second ship $T$, as $\tau$ is already used as the proper time of the first ship. The worldline of ship 2 is given parametrically by

$$\begin{align*} t\left( s\right) & =\frac{1}{2}\left( -\frac{1}{16g^{4}s^{3}}+s\right) \\ x\left( s\right) & =\frac{1}{2}\left( -\frac{1}{16g^{4}s^{3}}-s\right) . \end{align*}$$

This parametrization can be used to produce a plot of the motion of ship 2, but this is not how I produced the plot in post #18.

The unit magnitude of the proper velocity of the ship 2 gives

$$\begin{align*} 1 & =\left( \frac{dt}{dT}\right) ^{2}-\left( \frac{dx}{dT}\right) ^{2}\\ & =\left[ \left( \frac{dt}{ds}\right) ^{2}-\left( \frac{dx}{ds}\right) ^{2}\right] \left( \frac{ds}{dT}\right) ^{2}\\ & =\frac{3}{16g^{4}s^{4}}\left( \frac{ds}{dT}\right) ^{2}, \end{align*}$$

so

$$\frac{ds}{dT}=\frac{4g^{2}s^{2}}{\sqrt{3}}.$$

Conseqently,

$$\begin{align*} T-T_{0} & =\frac{\sqrt{3}}{4g^{2}}\int_{s_{0}}^{s}\frac{1}{x^{2}}dx\\ & =\frac{\sqrt{3}}{4g^{2}}\left( \frac{1}{s_{0}}-\frac{1}{s}\right) , \end{align*}$$

and choosing

$$-T_{0}=\frac{\sqrt{3}}{4g^{2}}\frac{1}{s_{0}}$$

gives

$$\begin{align*} T & =-\frac{\sqrt{3}}{4g^{2}}\frac{1}{s}\\ s & =-\frac{\sqrt{3}}{4g^{2}}\frac{1}{T}. \end{align*}$$

Thus, the worlline inertial coordinates of ship 2 as functions of proper time
are

$$\begin{align*} t\left( T\right) & =\frac{2\sqrt{3}}{9}g^{2}T^{3}-\frac{\sqrt{3}}{8g^{2} }\frac{1}{T}\\ x(T) & =\frac{2\sqrt{3}}{9}g^{2}T^{3}+\frac{\sqrt{3}}{8g^{2}}\frac{1}{T}. \end{align*}$$

As a check, I calculated $\left( dt/dT\right) ^{2}-\left( dx/dT\right)$ to make sure that it equalled 1. Another straightforward calculation shows that the magnitude of the 4-acceleration of the second ship is given by

$$\begin{align*} a^{2} & =\left( \frac{d^{2}t}{dT^{2}}\right) ^{2}-\left( \frac{d^{2} x}{dT^{2}}\right) ^{2}\\ & =-\frac{4}{T^{2}}. \end{align*}$$

Finally, the constant velocity condition can be verified either as pervect has done, or by ''dotting'' with the 4-velocity of ship 1 at an appropriate event on its worldline.

Regards,
George

 

[pervect]

Ship 1 moves with constant 4-acceleration $g$ and ship 2 moves with constant speed $w$ in the coordinate system $\left( t^{\prime},x^{\prime}\right)$ for an observer in ship 1, and consequently $x^{\prime}=wt^{\prime}+\alpha$ is the worldline of ship 2.

This is actually a rather interesting result in its own right (the linear relationship between distance and time). I don't believe that it holds true in all coordinate systems.

Would you hapen to know what the metric was in terms of the radar coordinates t' and x'?

 

[George Jones]

Would you hapen to know what the metric was in terms of the radar coordinates t' and x'?

The relationship between radar and inertial coordiantes is

$$\left( t,x\right) =\left( g^{-1}e^{gx^{\prime}}\sinh\left( gt^{\prime }\right) ,g^{-1}e^{gx^{\prime}}\cosh\left( gt^{\prime}\right) \right).$$

(Note: this probably offers a more familiar to the proper time aramtrization of the worldline of ship 2.)

This leads to

$$\begin{align*} \frac{\partial}{\partial t^{\prime}} & =\frac{\partial t}{\partial t^{\prime }}\frac{\partial}{\partial t}+\frac{\partial x}{\partial t^{\prime}} \frac{\partial}{\partial x}\\ & =e^{gx^{\prime}}\cosh\left( gt^{\prime}\right) \frac{\partial}{\partial t}+e^{gx^{\prime}}\sinh\left( gt^{\prime}\right) \frac{\partial}{\partial x}\\ & =g\left( x\frac{\partial}{\partial t}+t\frac{\partial}{\partial x}\right). \end{align*}$$

Similarly,

$$\frac{\partial}{\partial x^{\prime}}=g\left( t\frac{\partial}{\partial t}+x\frac{\partial}{\partial x}\right) .$$

The component of the metric with respect to radar coordinates are

$$\begin{align*} g_{0'0'} = \mathbf{g}\left( \frac{\partial}{\partial t^{\prime}},\frac{\partial }{\partial t^{\prime}}\right) & =g\left( x-t\right) \\ g_{1'1'} = \mathbf{g}\left( \frac{\partial}{\partial x^{\prime}},\frac{\partial }{\partial x^{\prime}}\right) & =g\left( t-x\right) \\ g_{0'1'} = \mathbf{g}\left( \frac{\partial}{\partial t^{\prime}},\frac{\partial }{\partial x^{\prime}}\right) & =0. \end{align*}$$

Eliminating $x^{+}$ from

$$\begin{align*} x^{+}x^{-} & =-g^{-2}e^{2gx^{\prime}}\\ \frac{x^{+}}{x^{-}} & =-g^{-2}e^{2gt^{\prime}} \end{align*}$$

gives these metric components in terms of radar coordinates.

I think that you'll find that $\partial/\partial t^{\prime}$ (or something very related) is a timelike killing vector field.

This should yield the same idea of ''stationariness'' that applies to an observer at a constant Scwarzschild R coordinate relative to a black hole - the metric in the radar coordinates should be static - unless I'm making some conceptual error.''

This comment goes to the heart of the Unruh effect, and to the heart of the analogy between Hawking and Unruh radiation.

Regards,
George

 

[Chronos]

I think you guys have basically got it right. Albeit the path seems a bit torturous. Isn't this easier if you plot spacetime positions using light cones?

 

[pervect]

I think you guys have basically got it right. Albeit the path seems a bit torturous. Isn't this easier if you plot spacetime positions using light cones?

I'm not quite sure what you mean. You'd have to sketch out your proposed simpler solution in a lot more detail for me to see what your'e getting at, I'm afraid :-(. I'm happy to have a solution of any form, even if it seems complicated. Speaking of which

I've got a few more interesting results:

A rather general form of the solution in inertial coordinates is

$$t = a \lambda^n - \frac{b}{\lambda}$$
$$x = a \lambda^n + \frac{b}{\lambda}$$

[add]It turns out, on further analysis, to be convenient to make a*b=1, this doesn't affect the generality of the solution at all, and removes a parameter.
[end add]

The solution hasn't been normalized yet, but is simple to write down in its unnormalized form. It satisfied the 'constant dot product' criterion. The constant dot product is

$$\frac{n+1}{2 \sqrt{n}}$$

[add]Another way of putting this is that n = (1+v)/(1-v), where v is the constant velocity as measured by the accelerated observer. So for v=.9c, n=19.
[end add]

The "hyperbolic" point of the accelerating observers must be at the origin of the coordiante system.

The reason that George's choice of coordinates perserves velocities is that they distort time and space equally in the x and t directions, i.e because g_00 = -g_11. Local clocks tick at a rate of $\sqrt{|g_{00}|}$ faster than coordinate clocks, and local rulers in the x direction read larger by a factor of $\sqrt{|g_{11}|}$ than coordinate rulers. Because |g_00| = |g_11|, there is no net effect on velocities in the x-direction. Because this is a 1d problem, we don't need to consider velocities in the y or z directions.