How Does Relativistic Deceleration Affect Velocity Calculations in Space Travel?

[Al68]

If I'm on a rocket accelerating away from earth, my velocity relative to Earth at any given time will be: $$v = \frac{at}{\sqrt{1+(\frac{at}{c})^2}}$$

This makes sense, since my velocity relative to Earth will be less than the Newtonian $$\Delta v = a \Delta t$$, due to relativistic effects.

But how about later, if I decelerate relative to earth? Obviously, I can't use the same equation, since my change in velocity relative to Earth will now be greater than $$\Delta v = a \Delta t$$.

Is there a similar equation I can use to calculate my velocity relative to Earth at any given time during deceleration?

Thanks,
Alan

 

[Meir Achuz]

If I'm on a rocket accelerating away from earth, my velocity relative to Earth at any given time will be: $$v = \frac{at}{\sqrt{1+(\frac{at}{c})^2}}$$

This makes sense, since my velocity relative to Earth will be less than the Newtonian $$\Delta v = a \Delta t$$, due to relativistic effects.

But how about later, if I decelerate relative to earth? Obviously, I can't use the same equation, since my change in velocity relative to Earth will now be greater than $$\Delta v = a \Delta t$$.

Is there a similar equation I can use to calculate my velocity relative to Earth at any given time during deceleration?

Thanks,
Alan

Your velocity will be
$$v = v_0-\frac{at}{\sqrt{1+(\frac{at}{c})^2}}$$,
where v_0 is the velocity at the start of the deceleration, and a is the magnitude of the acceleration. There is an a^2 in the denom, so that sign doesn't matter.

 

[pervect]

If I'm on a rocket accelerating away from earth, my velocity relative to Earth at any given time will be: $$v = \frac{at}{\sqrt{1+(\frac{at}{c})^2}}$$

This makes sense, since my velocity relative to Earth will be less than the Newtonian $$\Delta v = a \Delta t$$, due to relativistic effects.

But how about later, if I decelerate relative to earth? Obviously, I can't use the same equation, since my change in velocity relative to Earth will now be greater than $$\Delta v = a \Delta t$$.

Is there a similar equation I can use to calculate my velocity relative to Earth at any given time during deceleration?

Thanks,
Alan

The easiest approach, assuming a constant deacceleration, is to pick the point at which you will stop as a reference.

Let (tstop, xstop) be the coordinates where you come to a stop.

Then (correction) v(t) = -a(t-tstop)/sqrt(1+(a^2(t-tstop)^2)

Thus at t=tstop, v=0
and at t < tstop, v > 0

You are basically time reversing the problem, and accelerating away from your stopping point.

There are a few things about unformly accelerated notion that are extremely useful to know:

1) all such motion is hyperbolic, i.e for the correct choice of origin x^2 - t^2 = constant
(x^2 - c^2t^2 = constant if c is not equal to 1).

for example: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

(d + c^2/a) = (c^2/a) cosh(aT/c)
t = (c/a) sinh(aT/c)

thus (d+c^2/a)^2 - (ct)^2 = (c^2/a)^2

as cosh^2 - sinh^2 =1

[add]
See also, for instance
http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity )

2) the "lines of simultaneity" for a uniformly accelerated observer all pass through one point (the origin of the coordinate system, if the choice of the origin is made as in #1)

i.e. in the example above, all liens of simultaneity run through the point
x = -c^2/a; t=0

[add] Here is a proof of the later remark: assume c=1, then

x^2 - t^2 = constant

By taking the derivative, we get
2*x*dx - 2*t*dt = 0

therfore v = dx/dt = t/x. The slope of the line of simultaneity is 1/v, i.e. x/t. Thus a line from (0,0) to (x,t) is the line of simultaneity at (x,t) because it passes through (x,t) and it has the correct slope: dx/dt = x/t.

 

[pervect]

I've corrected one typo and added some proofs and more references to my previous post, the added content is significant enough that I thought I'd make a note of it.

 

[Al68]

What I was really looking for here was an equation for the ship's deceleration relative to Earth's apparent position. This would obviously not be the same as the ship's deceleration relative to its destination.

Thanks,
Alan

 

[pervect]

The deacceleration relative to Earth will be the same as the deacceleration relative to its destination, as long as inertial coordinate systems are used.

It's a simple space translation.

Suppose the destination is a distance L away from the Earth.

Then the coordiantes relative to the Earth will be (t,x) and the coordinates relative to the destination will be (t,x-L).

Thus when x=L, the ship is at its destination. The acceleration of the ship (d^2x/dt^2) will be the same relative to the Earth as it will be to the destination, because

d^2(x-L)/dt^2 = d^2x/dt^2

given that L=constant.

This simple "space translation" works the same in SR as it does in non-relativistic mechanics.

Time translation also works in the same way.

Note that we can also formulate the equations of motion using the principle of hyperbolic motion that I mentioned.

Let the destinaton be at (tstop, xstop), and let c=1.

then for x < (xstop/2), the hyperbolic point is at (t=0,x=-1/a) and

(x+1/a)^2 - t^2 = 1/a^2

and for x > xstop/2, the hyperbolic point is at (t=tstop, x=xstop+1/a)

(x-(xstop + 1/a))^2 - (t-tstop)^2 = 1/a^2

We can re-write the second expression to solve for x (using Maple). There are two solutions, one of which is:

$${\frac {{\it xstop}\,a+1-\sqrt {1+{a}^{2}{t}^{2}-2\,{a}^{2}t{\it tstop }+{a}^{2}{{\it tstop}}^{2}}}{a}}$$

the derivative of this expression is (agian using Maple)

$${\frac {a \left( t-{\it tstop} \right) }{\sqrt {1+{a}^{2}{t}^{2}-2\,{a }^{2}t{\it tstop}+{a}^{2}{{\it tstop}}^{2}}}}$$

the same as the expresssion I posted earlier.

See the Wikipedia link for the hyperbolic equation (you may have to manually add in the trailing ')' to the URL, for some reason the board omits this symbol and wiki doesn't find the page.

http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity )

You can confirm that the velocity at t=tstop/2 is the same in both expressions:

 

[Al68]

pervect,

If the distance L refers to the distance between Earth and the destination, according to the ship's frame, then L will not remain constant. It will get larger as the ship decelerates. Just like L got smaller as the ship accelerated away from earth. My understanding is that the relativistic rocket equation takes this into account, which is why the final velocity relative to Earth (after acceleration) is less than it would be if we didn't take length contraction into account, assuming a specified acceleration and time. In other words, velocity is less than a*t.

So, is there an equation that relates to a ship's deceleration relative to Earth's apparent position, that takes this length expansion (or de-contraction) into account?

Thanks,
Alan

 

[pervect]

The coordinates (x,t) are the inertial coordinates of the ship in the Earth frame.

The destination is a constant distance away in the Earth coordiantes.

Look for instance at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

(I assume this is where you initally got your equations?)

The proper time as measured by the crew of the rocket (i.e. how much they age) will be denoted by T, and the time as measured in the non-accelerating frame of reference in which they started (e.g. Earth) will be denoted by t. We assume that the stars are essentially at rest in this frame. The distance covered as measured in this frame of reference will be denoted by d and the final speed v. The time dilation or length contraction factor at any instant is the gamma factor γ.

Thus d and t are in the Earth coordinate system. T is in the rocket's coordinate system (proper time). The distance to the destination according to the rocket is not given on the webpage.

 

[JesseM]

If the distance L refers to the distance between Earth and the destination, according to the ship's frame, then L will not remain constant. It will get larger as the ship decelerates. Just like L got smaller as the ship accelerated away from earth.

The distance L is measured in an inertial frame, the frame where the rocket was at rest until it began to accelerate (the rest frame of the earth). You seem to be thinking it's the distance as seen in some non-inertial frame of the ship, or in the ship's instantaneous co-moving inertial rest frame, but that's not correct, the relativistic rocket equation describes the velocity as seen in a single inertial frame.

My understanding is that the relativistic rocket equation takes this into account, which is why the final velocity relative to Earth (after acceleration) is less than it would be if we didn't take length contraction into account, assuming a specified acceleration and time. In other words, velocity is less than a*t.

No, the reason it's less as seen in an inertial frame is because "a" refers to the acceleration experienced by passengers on the ship--constant "a" means that they feel a constant G-force, which means that their acceleration in their instantaneous co-moving inertial rest frame is the same from one moment to another. Because of the way velocities transform in relativity, this would mean that the rate their coordinate velocity increases (their coordinate acceleration, which is different from 'a') as seen in a single inertial frame is constantly decreasing.

So, is there an equation that relates to a ship's deceleration relative to Earth's apparent position, that takes this length expansion (or de-contraction) into account?

What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?

 

[Al68]

OK, then how would the ship's crew calculate their change in velocity? I assume they would have to take into account that as they accelerate, gamma would increase, and maybe we could picture the change in Earth's apparent position as a kind of velocity, in the same direction of the ship's velocity. Since the distance from earth, as measured by the ship is increasing due to acceleration, but increasing less due to length contraction.

So, is there an equation that relates to a ship's deceleration relative to Earth's apparent position, that takes this length expansion (or de-contraction) into account?

What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?

Yes, this is what I meant. I'm looking for how to calculate everything from the ship's point of view. And I know the ship's crew cannot observe anything in real time, but they should be able to figure out their velocity, (coordinate) acceleration, and (apparent) distance from Earth at any given time. Including during their deceleration. And I'm looking for how they would calculate this, not how they would observe it directly.

Thanks,
Alan

 

[JesseM]

OK, then how would the ship's crew calculate their change in velocity?

"Velocity" in what coordinate system? If you want to look at a non-inertial coordinate system where they are at rest, then of course their velocity is always zero in this system, although you could calculate the way that the Earth's velocity is changing in this coordinate system. But as I've said before, it's not like there's a single non-inertial coordinate system that counts as the "frame" of a non-inertial observer, you could construct such a coordinate system in a variety of ways, I don't think there's any strong physical motivation for preferring one over all the others.

What do you mean by "apparent"? Do you mean the distance as seen in the ship's instantaneous co-moving inertial frame?

Yes, this is what I meant. I'm looking for how to calculate everything from the ship's point of view.

Again, I think you're confusing the issue by using commonsense words like "observe" and "point of view" and "apparent distance" for what is measured in a particular non-inertial coordinate system. Do you think there is any reason to say that this particular type of non-inertial coordinate system--one where the ship's definitions of distance and simultaneity at any given moment match those of their instantaneous co-moving inertial rest frame--represent's the ship's "point of view" in a way that some other non-inertial coordinate system with different definitions of simultaneity and distance does not?

And I know the ship's crew cannot observe anything in real time, but they should be able to figure out their velocity, (coordinate) acceleration, and (apparent) distance from Earth at any given time. Including during their deceleration. And I'm looking for how they would calculate this, not how they would observe it directly.

You only put (coordinate) before acceleration, but you really should put it before velocity and distance from Earth as well; what you are really asking is for the value of these things in a particular non-inertial coordinate system (one whose definition of distance and simultaneity at each moment matches that of the ship's instantaneous co-moving inertial rest frame), but there is no reason to say that the coordinate distance of the Earth in this non-inertial frame has any more right to be called the "apparent distance" than the coordinate distance in some other frame.

Anyway, like I said earlier, in the coordinate system that I assume you're asking about, their own coordinate velocity and acceleration would be zero, because they're at rest at all moments in this coordinate system. But you can certainly ask about the coordinate velocity, acceleration and position of the Earth in this coordinate system. If you can figure out the coordinate position of the Earth as a function of time in this system, you could obtain the others in the normal way, by taking derivatives with respect to coordinate time. The calculation of distance as a function of time in this coordinate system seems like it'd be a bit tricky, although the formulas on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html help...I'll think about it and get back to you.

 

[pervect]

To compute or measure in the rockets coordinate system the same velocity that the Earth observer measures, the rocket needs to measure its velocity relative to an object that is stationary with respect to the Earth, but at the same location as the rocket.

"Stationary with respect to the Earth" is really defined in the Earth's frame of referece, not the rockets frame, however. There is a way to do this from the rocket's coordiante system, but it is complicated. It also requires making use of special features of the geomtery. In another post, I think I gave a longish quote about why there is no general notion of "relative velocity" of a distant object, so it should not be a surprise that finding the relative velocity of the rocket to the Earth requires us to make use of some "special features" of the geometry (namely, the fact that space-time is flat in this example).

The advanced way of doing this from the rocket's coordinate system: the rocket "parallel transports" the velocity vector of the Earth along some path to the rocket's current location. The rocket then finds the magnitude of the resulting "parallel transported" velocity to answer the question.

In general, this procedure of parallel transporting would give a result that depended on the exact path over which the vector was transported. But, becuase space-time in this example is flat, this process of parallel transport will give the same results, regardless of the path taken.

To avoid the process of parallel transport, the rocket could make use of (for example) an "Earth outpost" that knows it is stationary with respect to the Earth because the outpost has been exchanging light signals with the Earth for a long time, without noticing any change in the "propagation delay".

This "Earth outpost" appraoch is much simpler, though I don't know if it will satisfy Al68.

If the rocket measures its velocity V with respect to the Earths' outpost, it will measure the same velocity V that the Earth (and the Earth's outpost) measures for the rocket.

A full treatment of parallel transport requires some knowledge of tensors, which I do not believe that AL68 has.

A somewhat popular reference:

http://math.ucr.edu/home/baez/gr/parallel.transport.html

Note that when you parallel transport a vector around a closed path on the Earth's surface, as in the above example, it changes direction.

It is only when the geometry is fundamentally flat that parallel transporting a vector around a closed path does not change it's direction.

Because parallel transporting a vector around a closed path on a flat plane does not change its direction, the result of parallel transporting a vector along a path cannot depend on the path.

Schild's ladder would be an ideal way of explaining parallel transport, but the only reference I have which talks about it is MTW's "Gravitation", and I haven't found much on the WWW, either.

"Gravitation" is a book that is written at the graduate level, but has largish bits that may be comprehensible to someone who is willing to ignore the rest of the book (which won't be comprehensible without graduate level physics).

 

[Al68]

Jesse,

I think your last paragraph states my question more clearly than I did. I wanted to figure out the coordinate position of Earth as a function of time in the (non inertial) system in which the ship is at rest. And I agree with you that it would be tricky. That's why I was asking for help.

pervect,

You are correct about my knowledge of tensors. I can barely spell tensors. And I worded my question poorly, Jesse worded it better in his last paragraph.

Thanks,
Alan

 

[JesseM]

Alan, if I'm understanding you right you want to know how things would look in a particular noninertial coordinate system (call it NI) which has the following properties:

1. If a given tick of the ship's clock is simultaneous with a distant event A in the co-moving inertial rest frame of the ship at the moment of the tick, then A should also be simultaneous with that tick in NI. Also, if D is the distance between the ship at the moment of the tick and A in the co-moving inertial rest frame of the ship at that moment, then the coordinate distance between A and the ship at the moment of the tick should also be D in NI.

2. The time-coordinate assigned to an event on the ship's worldine by NI should be identical to the readings of clocks on the ship at the moment of the event (ie the proper time along the ship's worldine between that event and the event of leaving earth).

I think these conditions should be enough to uniquely specify the NI coordinate system. Note, though, that this will not be a well-behaved coordinate system throughout spacetime--as the lines of simultaneity change angles at different points along the ship's worldline, then there will be points where different lines of simultaneity cross each other, and events at or beyond the crossings will be assigned two or more sets of coordinates by NI. But as long as you consider only the regions of spacetime near enough to the ship's worldline that events within those regions are assigned unique coordinates, it should be OK (although I don't know how to calculate the borders of this region).

Nevertheless, I would say that you are incorrect to refer to this as "the (non inertial) system in which the ship is at rest"--as I keep emphasizing, over and over, and you seem to want to ignore, there is no single standard way to define the coordinate system of a non-inertial observer, you could certainly come up with other non-inertial coordinate systems in which the ship is at rest that do not have the properties I described for NI above, and there would be no compelling reason to treat NI as having more claim to represent the ship's "point of view" than any of these others. Please tell me, do you understand and agree with this? (and if so, will you quit using phrases like 'apparent distance' and 'the ship's point of view' to describe how things work in the NI coordinate system?)

Anyway, to figure out the Earth's distance D as a function of time T within the coordinate system NI, I think the first thing to do would be to figure out D as a function of d and v, the distance and velocity of the ship in the Earth's frame. You might think that D would just be $$d * \sqrt{1 - v^2/c^2}$$ due to Lorentz contraction, but that isn't right, the distance isn't like the length of a rigid object, it's constantly changing so you have to worry about simultaneity issues. (edit: I was wrong about that, see pervect's next post and my response...the distance is like the length of a rigid object, just imagine that there was a rod attached to the Earth of length d in the Earth's rest frame, and when the ship reaches the end of it, both frames agree that at that moment the length of the rod in their frame is equal to the ship's distance from the Earth in their frame.) So let's consider a simple inertial case where the ship is moving away from the Earth at constant velocity v instead of accelerating. In the Earth's frame, the ship will be at distance d at time d/v, and since in this frame the ship's clocks are running slow by a factor of $$\sqrt{1 - v^2/c^2}$$, that means the ship's clock will only read $$(d/v)*\sqrt{1 - v^2/c^2}$$ at this moment. But in the ship's frame, at the moment its own clock reads this time of $$(d/v)*\sqrt{1 - v^2/c^2}$$, it is the Earth's clock that reads less than this because it has been slow by a factor of $$\sqrt{1 - v^2/c^2}$$ since they departed, so according to the definition of simultaneity in the ship's frame the Earth's clock reads $$(d/v)*(1 - v^2/c^2)$$ at this moment. And since in the Earth's frame the Earth is also moving away at the same speed of v (the speeds with which two inertial observers see the other moving are always reciprocal), the Earth's distance at this moment can be found by multiplying velocity and time, giving $$v*(d/v)*(1 - v^2/c^2)$$, which gives $$d*(1 - v^2/c^2)$$. So, at the point on the ship's worldline where the distance in the Earth's frame is d, the distance in the ship's frame is $$d*(1 - v^2/c^2)$$.

Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the Earth's frame (they must have departed Earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the Earth in this instantaneous frame.

So, we know that at the point on the ship's worldline that corresponds to time t in the Earth's frame, when the distance in the Earth's frame is d, the NI system should say the distance of the Earth at that same point on the ship's worldline is $$d*(1 - v^2/c^2)$$. And from the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html we have an expression for d as a function of a and t: $$d = (c^2/a)*(\sqrt{1 + (at/c)^2} - 1)$$. So this means that the distance D in NI as a function of time t in the Earth's frame is $$D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (at/c)^2} - 1)$$. And the page also gives us a function for the time t as measured in the Earth's frame as a function of the time T as measured by clocks on the ship: $$t = (c/a)*sh(aT/c)$$, where the function sh x is the hyperbolic sine function (also called sinh -- you can find it on most calculators). So, substituting this into the formula for D as a function of t gives D as a function of T:

$$D = (1 - v^2/c^2)*(c^2/a)*(\sqrt{1 + (a/c)^2 * ((c/a)*sh(aT/c))^2} - 1)$$

simplifying a bit,

$$D = ((c^2 - v^2)/a)*(\sqrt{1 + (sh(aT/c))^2} - 1)$$

and this page mentions there's a hyperbolic function identity that says (ch(x))^2 = 1 + (sh(x))^2, so this simplifies to:

$$D = ((c^2 - v^2)/a)*(ch(aT/c) - 1)$$

There's a good chance I made an error somewhere along the line, but if not, this should be the formula for the coordinate distance of the Earth D as a function of coordinate time T in the NI coordinate system. To find the coordinate velocity of Earth you'd take the first derivative of D with respect to T, while to find the coordinate acceleration of Earth you'd take the second derivative. The page with the hyperbolic function identities above says that d sh(x) / dx = ch(x) and d ch(x) / dx = sh(x). So, taking these derivatives is just a matter of using the chain rule of calculus:

$$dD/dT = ((c^2 - v^2)/a)*(sh(aT/c) * (a/c))$$

or

$$dD/dT = ((c^2 - v^2)/c)*sh(aT/c)$$

Taking another derivative gives:

$$d^2 D /dT^2 = ((c^2 - v^2)/c)*(ch(aT/c) * (a/c))$$

or

$$d^2 D /dT^2 = (a*(c^2 - v^2)/c^2)*ch(aT/c)$$

Again, buyer beware, but if I haven't made any errors these should be the formulas for the coordinate velocity and coordinate acceleration of the Earth in the NI coordinate system.

 

[pervect]

I'm think that the distance in the ship's coordinate system should be just sqrt(1-v^2/c^2) times the distance in the Earth's coordinate system.

Basically, the angle betweent the line of simultaneity and the horizontal 'x' axis is just the rapidity, theta, therefore the length in the Earth frame divided by the length in the instantaneous ship frame is just cosh(theta).

Another way of putting this: letting c=1

The slope of the "line of simultaneity" is 1/v , thus delta-t = v*delta-x

If the Earth is at the origin of the coordinate system and the spaceship is at (t,d) in the Earth coordinate system

The point on the Earth's trajectory (t=*, x=0) simultaneous with (t,x) in the ship frame is just

(t-v*d,0)

The lorentz interval betweent these points is d^2 - (v*d)^2

The distance is the square root of the Lorentz interval, i.e d*sqrt(1-v^2)

The following diagram might have helped, if it came out :-(. (I was trying gnuplot this time around). I'll keep it, in case I can get it to work.

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\[
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[JesseM]

I'm think that the distance in the ship's coordinate system should be just sqrt(1-v^2/c^2) times the distance in the Earth's coordinate system.

Yes, I think you're correct. I confused myself by thinking about what the Earth's clock would read in the ship's frame at the moment that the ship's clock reads $$(d/v)*\sqrt{1 - v^2/c^2}$$, but what's important is the time in the ship's own frame at this moment--since in its frame the Earth has been moving at velocity v for a time $$(d/v)*\sqrt{1 - v^2/c^2}$$, at this time the Earth must be a distance $$v*(d/v)*\sqrt{1 - v^2/c^2}$$ in this frame, or $$d*\sqrt{1 - v^2/c^2}$$. So a corrected version of my argument after that point would look like this:

So, we know that at the point on the ship's worldline that corresponds to time t in the Earth's frame, when the distance in the Earth's frame is d, the NI system should say the distance of the Earth at that same point on the ship's worldline is $$d*\sqrt{1 - v^2/c^2}$$. And from the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html we have an expression for d as a function of a and t: $$d = (c^2/a)*(\sqrt{1 + (at/c)^2} - 1)$$. So this means that the distance D in NI as a function of time t in the Earth's frame is $$D =\sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (at/c)^2} - 1)$$. And the page also gives us a function for the time t as measured in the Earth's frame as a function of the time T as measured by clocks on the ship: $$t = (c/a)*sh(aT/c)$$, where the function sh x is the hyperbolic sine function (also called sinh -- you can find it on most calculators). So, substituting this into the formula for D as a function of t gives D as a function of T:

$$D = \sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (a/c)^2 * ((c/a)*sh(aT/c))^2} - 1)$$

simplifying a bit,

$$D = \sqrt{1 - v^2/c^2}*(c^2/a)*(\sqrt{1 + (sh(aT/c))^2} - 1)$$

and this page mentions there's a hyperbolic function identity that says (ch(x))^2 = 1 + (sh(x))^2, so this simplifies to:

$$D = \sqrt{1 - v^2/c^2}*(c^2/a)*(ch(aT/c) - 1)$$

There's a good chance I made an error somewhere along the line, but if not, this should be the formula for the coordinate distance of the Earth D as a function of coordinate time T in the NI coordinate system. To find the coordinate velocity of Earth you'd take the first derivative of D with respect to T, while to find the coordinate acceleration of Earth you'd take the second derivative. The page with the hyperbolic function identities above says that d sh(x) / dx = ch(x) and d ch(x) / dx = sh(x). So, taking these derivatives is just a matter of using the chain rule of calculus:

$$dD/dT = \sqrt{1 - v^2/c^2}*(c^2/a)*(sh(aT/c) * (a/c))$$

or

$$dD/dT = (c*\sqrt{1 - v^2/c^2})*sh(aT/c)$$

Taking another derivative gives:

$$d^2 D /dT^2 = (c*\sqrt{1 - v^2/c^2})*(ch(aT/c) * (a/c))$$

or

$$d^2 D /dT^2 = (a*\sqrt{1 - v^2/c^2})*ch(aT/c)$$

Again, buyer beware, but if I haven't made any errors these should be the formulas for the coordinate velocity and coordinate acceleration of the Earth in the NI coordinate system.

 

[Al68]

Jesse,

Thanks for all your work. I have to say, I liked your first answer. It fit in with what I was thinking. But I'd better reread and absorb everything better before I ask any more questions.

pervect, I appreciate your help, too. I just don't have the math skills to use some of it.

Thanks,
Alan

 

[Al68]

So, we know that at the point on the ship's worldline that corresponds to time t in the Earth's frame, when the distance in the Earth's frame is d, the NI system should say the distance of the Earth at that same point on the ship's worldline is $$d*\sqrt{1 - v^2/c^2}$$.

This makes sense to figure out D as a function of d at any moment. But if we were to have an event, like a signal sent out by the ship's crew, this signal would not be simultaneous in both frames. Is it safe to say, that if we wanted to figure out D, d, T, and t for this event, that your earlier post would be correct? At this event, the Earth's clock would read $$(d/v)*(1 - v^2/c^2)$$, and the rest of your earlier post would apply?

Thanks,
Alan

 

[JesseM]

This makes sense to figure out D as a function of d at any moment. But if we were to have an event, like a signal sent out by the ship's crew, this signal would not be simultaneous in both frames.

Simultaneous with what? It only makes sense to ask if two events are simultaneous within a single frame, you can't really ask whether a single event is simultaneous in two different frames (you can ask if it occurs at the same time-coordinate in two coordinate systems, but this is different, and it depends on the arbitrary choice of where you put the origin of each coordinate system, not just their relative velocity).

Is it safe to say, that if we wanted to figure out D, d, T, and t for this event, that your earlier post would be correct? At this event, the Earth's clock would read $$(d/v)*(1 - v^2/c^2)$$, and the rest of your earlier post would apply?

The Earth's clock would read $$(d/v)*(1 - v^2/c^2)$$ at the time of this event in the ship's frame (but you shouldn't say 'At this event, the Earth's clock would read...' because an 'event' in relativity refers only to a single point in spacetime, so anything happening elsewhere like on Earth would not be 'at this event'). But my calculation of the distance D in the ship's frame was wrong, the distance would be the time of the event according to the ship's clock (ie $$(d/v)*\sqrt{1 - v^2/c^2}$$) multiplied by v, since in the ship's frame the ship's own clock is ticking at the normal rate while the Earth is moving away at velocity v. Thus the distance D would be $$v*(d/v)*\sqrt{1 - v^2/c^2}$$, or $$d*\sqrt{1 - v^2/c^2}$$.

To make this more clear, think in terms of the example I mentioned in the edit to that earlier post--suppose there was a rod attached to the earth, of length d in the Earth's frame. If we look at the event of the ship passing next to the far end of the rod, then obviously in the Earth's frame, the distance between that event and an event happening "simultaneously" on Earth (like the event of the Earth's clock ticking d/v) would be d, the length of the rod. So if you now look at the same event of the ship passing the end of the rod in the ship's frame, and you want to calculate the distance between this event and an event on Earth which is simultaneous with it in the ship's frame (in this case, the event of the Earth's clock ticking $$(d/v)*(1 - v^2/c^2)$$), then the distance would just be equal to the length of the rod in the ship's frame, or $$d*\sqrt{1 - v^2/c^2}$$ due to Lorentz contraction. This would be the distance of the Earth at the moment the ship passes the end of the rod in the ship's frame.

 

[Al68]

OK, if the event of the ship sending out the signal would be simultaneous with the Earth's clock ticking d/v, in Earth's frame, wouldn't this signal also be simultaneous with the event of a rod attached to the ship (length d in Earth's frame) reaching earth? And the length of this rod would be $$d/\sqrt{1 - v^2/c^2}$$ in the ship's frame? It looks to me like we can't consider distance to be like a rigid rod, because it is constantly changing, which is what you said in your earlier post, before the edit. I have to say again that your earlier post matches up with what I was thinking. Maybe I'm biased because your first conclusion matched my own thoughts, but it still looks right to me.

Thanks,
Alan

 

[JesseM]

OK, if the event of the ship sending out the signal would be simultaneous with the Earth's clock ticking d/v, in Earth's frame, wouldn't this signal also be simultaneous with the event of a rod attached to the ship (length d in Earth's frame) reaching earth?

In the Earth's frame, yes. But these events would not be simultaneous in the ship's frame.

And the length of this rod would be $$d/\sqrt{1 - v^2/c^2}$$ in the ship's frame?

That's correct, but again, in the ship's frame the event of the ship reaching the end of the Earth's rod and sending out a signal would not be simultaneous with the event of the far end of its own rod passing earth. After all, you agree that in the ship's frame the Earth's rod only has length $$d*\sqrt{1 - v^2/c^2}$$, right? So as seen in the ship's frame, it must be a lot shorter than the ship's own rod, so at the moment that the ends of the two rods closest to the ship are next to each other, the far ends can't possibly be both next to Earth (and we know that the end of the Earth's rod is actually attached to the earth, so naturally when the ship passes next to the far end it will say the other end is next to earth, since that end is always next to Earth at a given moment in any frame).

If this is still unclear, you might try trying to identify the coordinates of the event of the ship passing next to the end of the Earth's rod, and the event of the Earth passing next to the end of the ship's rod, as seen in the Earth's coordinate system, then use the Lorentz transform to see when these events happen in the ship's coordinate system (and if you have trouble with this I could help out).

 

[Al68]

OK, but if the ship is a distance D from Earth (in the ship's frame) when the ship's crew sends the signal, can't we just as easily say that $$d = D*\sqrt{1 - v^2/c^2}$$ at the time of the signal (in Earth's frame), since this distance would be length contracted in Earth's frame? What if we specified that the signal would be sent when the ship was a specific distance (D) from Earth in the ship's frame. We could have the signal simultaneous with the end of the ship-rod passing Earth in the ship's frame. Wouldn't this distance be $$d = D*\sqrt{1 - v^2/c^2}$$ in Earth's frame due to length contraction? And now the signal would be sent when the ship reached the end of an earth-rod length D in the ship's frame, which would be $$D/\sqrt{1 - v^2/c^2}$$ in the Earth's frame, since the length of this rod would be length contracted in the ship's frame. It seems contradictory for distance contraction to be reciprical.

Can we really treat distance between events like the length of a rigid rod, since length contraction for rigid bodies is reciprical between frames?

Thanks,
Alan

 

[JesseM]

OK, but if the ship is a distance D from Earth (in the ship's frame) when the ship's crew sends the signal, can't we just as easily say that $$d = D*\sqrt{1 - v^2/c^2}$$ at the time of the signal (in Earth's frame), since this distance would be length contracted in Earth's frame?

No, because if the ship had a rod of length D attached to it, and sent a signal when the far end of its rod passed the Earth in its own frame, in the Earth's frame the event of the far end of the rod passing it and the event of the ship sending the signal wouldn't be simultaneous. So although it's true that the Earth will see the length of the ship's rod as $$D*\sqrt{1 - v^2/c^2}$$, that doesn't mean it will judge this to be the distance the ship was when it sent the signal.

If you're worried about reciprocality, it's reciprocal in this sense: if they both want to figure out the distance between the Earth and the ship at the moment of an event on the ship's worldline (like the ship sending a signal), then the distance will be shorter in the ship's frame. But if they both want to figure out the distance between the Earth and the ship at the moment of an event on the earth's worldline, then the distance will be shorter in the earth's frame. To see this, suppose we have a rod of length D attached to the ship, and the Earth sends out a signal at the moment it passes the far end of this rod. Both frames agree that the ship was at the other end of the rod at the moment the far end passed the earth, so in each frame the distance between the Earth and the ship at the moment of this event will just be the length of the rod in that frame, which means in the Earth's frame the distance must be $$D*\sqrt{1 - v^2/c^2}$$.

Another way of thinking about it: if they both want to know the distance between the Earth and the ship "at the same moment" as an event on the ship's worldline, their two rest frames will give different answers to which event on the Earth's worldline is simultaneous with that event; but if they both want to know the distance between the Earth and the ship "at the same moment" as an event on the Earth's worldline, their two rest frames will give different answers to which event on the ship's worldline is simultaneous with that event.

 

[Al68]

If you're worried about reciprocality, it's reciprocal in this sense: if they both want to figure out the distance between the Earth and the ship at the moment of an event on the ship's worldline (like the ship sending a signal), then the distance will be shorter in the ship's frame. But if they both want to figure out the distance between the Earth and the ship at the moment of an event on the earth's worldline, then the distance will be shorter in the earth's frame. To see this, suppose we have a rod of length D attached to the ship, and the Earth sends out a signal at the moment it passes the far end of this rod. Both frames agree that the ship was at the other end of the rod at the moment the far end passed the earth, so in each frame the distance between the Earth and the ship at the moment of this event will just be the length of the rod in that frame, which means in the Earth's frame the distance must be $$D*\sqrt{1 - v^2/c^2}$$.

OK, and for this example, the Earth's signal would be simultaneous (in the ship's frame) with the ship reaching the end of an earth-rod with a length of D in the ship's frame. This earth-rod would be $$D/\sqrt{1 - v^2/c^2}$$ in Earth's frame, but the event of the ship reaching the end of the earth-rod would not be simultaneous with Earth's signal in Earth's frame.

So if the ship were to send out a signal when it reached the end of the Earth rod, and when the end of the ship-rod reached earth, in the ship's frame, this signal would not be simultaneous with the ship reaching the end of the earth-rod in Earth's frame. In Earth's frame, the ship's signal would occur prior to the ship reaching the end of the earth-rod (length $$D/\sqrt{1 - v^2/c^2}$$) in Earth's frame. Is this correct?

Thanks,
Alan

 

[JesseM]

OK, and for this example, the Earth's signal would be simultaneous (in the ship's frame) with the ship reaching the end of an earth-rod with a length of D in the ship's frame. This earth-rod would be $$D/\sqrt{1 - v^2/c^2}$$ in Earth's frame, but the event of the ship reaching the end of the earth-rod would not be simultaneous with Earth's signal in Earth's frame.

Yup, exactly.

So if the ship were to send out a signal when it reached the end of the Earth rod, and when the end of the ship-rod reached earth, in the ship's frame, this signal would not be simultaneous with the ship reaching the end of the earth-rod in Earth's frame.

In the ship's frame, both rods have length D, so the event of the ship reaching the end of the Earth's rod and the event of the Earth reaching the end of the ship's rod would be simultaneous. Your question is unclear, are asking whether the signal from the ship in the ship's frame would be simultaneous with the signal from the Earth in the Earth's frame? Asking about simultaneity of events in different frames makes no sense, as I explained in an earlier post:

It only makes sense to ask if two events are simultaneous within a single frame, you can't really ask whether a single event is simultaneous in two different frames (you can ask if it occurs at the same time-coordinate in two coordinate systems, but this is different, and it depends on the arbitrary choice of where you put the origin of each coordinate system, not just their relative velocity).

In Earth's frame, the ship's signal would occur prior to the ship reaching the end of the earth-rod (length $$D/\sqrt{1 - v^2/c^2}$$) in Earth's frame. Is this correct?

My understanding of your scenario was that the ship sends the signal at the time and the place where it reaches the end of the Earth's rod--in that case, all frames must agree these two events have the same coordinates, if two events coincided in one frame but not in another that would represent contradictory physical predictions. Imagine the end of the Earth's rod was set to explode at the moment the ship passed it--it isn't possible for some frames to say to ship managed to send a signal before it was blown up while others say it was blown up before it could send the signal, that would be an obvious physical contradiction. Similarly, although different frames can disagree on the ages of the twins in the twins paradox as long as they apart, if they reunite at a single time and place all frames must agree on what their clocks read at that point in spacetime.

 

[Al68]

OK, and for this example, the Earth's signal would be simultaneous (in the ship's frame) with the ship reaching the end of an earth-rod with a length of D in the ship's frame. This earth-rod would be be $$D/\sqrt{1 - v^2/c^2}$$ in Earth's frame, but the event of the ship reaching the end of the earth-rod would not be simultaneous with Earth's signal in Earth's frame.

Yup, exactly.

So if the ship were to send out a signal when it reached the end of the Earth rod, and when the end of the ship-rod reached earth, in the ship's frame, this signal would not be simultaneous with the ship reaching the end of the earth-rod in Earth's frame.

In the ship's frame, both rods have length D, so the event of the ship reaching the end of the Earth's rod and the event of the Earth reaching the end of the ship's rod would be simultaneous. Your question is unclear, are asking whether the signal from the ship in the ship's frame would be simultaneous with the signal from the Earth in the Earth's frame?

No, I was saying that this ship's signal (when it reached the end of the earth-rod) would be received by earth, but not simultaneously with the ship reaching the end of the earth-rod in Earth's frame. In Earth's frame, the event of this ship's signal and the event of the ship reaching the end of the earth-rod are not simultaneous. Since the event of the ship reaching the end of the Earth rod and the event of the end of the ship-rod reaching Earth are simultaneous in the ship's frame, these events cannot be simultaneous in Earth's frame.

In Earth's frame, the ship's signal would occur prior to the ship reaching the end of the earth-rod (length $$D/\sqrt{1 - v^2/c^2}$$) in Earth's frame. Is this correct?

My understanding of your scenario was that the ship sends the signal at the time and the place where it reaches the end of the Earth's rod--in that case, all frames must agree these two events have the same coordinates, if two events coincided in one frame but not in another that would represent contradictory physical predictions. Imagine the end of the Earth's rod was set to explode at the moment the ship passed it--it isn't possible for some frames to say to ship managed to send a signal before it was blown up while others say it was blown up before it could send the signal, that would be an obvious physical contradiction. Similarly, although different frames can disagree on the ages of the twins in the twins paradox as long as they apart, if they reunite at a single time and place all frames must agree on what their clocks read at that point in spacetime.

Why would all frames have to agree that these two events (the ship's signal and the ship reaching the end of the Earth rod) would have the same coordinates? I believe you stated earlier that if the event of the Earth's clock ticking d/v and the event of a rod attached to the ship (length d in Earth's frame) reaching Earth were simultaneous in Earth's frame, these events would not be simultaneous in the ship's frame. If this is correct, then, in my scenario, the event of the ship's clock reading D/v (and sending a signal) and the event of the ship reaching the end of the earth-rod would be simultaneous in the ship's frame, but not in Earth's frame.

I agree this will result in contradictory physical predictions, or paradox so to speak. That's why I'm asking so many questions. I would also note that if we really had rigid rods, Earth could observe the ship's position in real time, by observing the ship-rod, and this would clearly violate the laws of physics. This was pointed out to me in an earlier topic. This is why I thought you were right when you said we could not treat distance like the length of a rigid rod.

(edit) Just wanted to add something here. What if, in the Twins Paradox, the destination star explodes when the ship reaches it (in the ship's frame), and destroys the ship. The event of the star's explosion and the event of the ship reaching the star would not be simultaneous in Earth's frame. Would the ship be destroyed in one frame but not the other?

Thanks,
Alan

 

[JesseM]

No, I was saying that this ship's signal (when it reached the end of the earth-rod) would be received by earth, but not simultaneously with the ship reaching the end of the earth-rod in Earth's frame.

Oh, I didn't realize you were talking about when the signal would be received, you'd never brought that up before.

In Earth's frame, the event of this ship's signal

Please, if you're going to talk about both the event of the signal being sent and the separate event of the signal being received, you can't use phrases like "the event of this ship's signal" or I won't know which one you're talking about. An event is something that happens at a single point in space and time, so transmission and reception are clearly separate events.

and the event of the ship reaching the end of the earth-rod are not simultaneous.

In this case, were you talking about the event of the signal being received on Earth not being simultaneous with the event of the signal being sent by the ship (which is at the same point in space and time that it reaches the end of the earth-rod)? Since the signal can't travel instantaneously fast but only at the finite speed of c, all frames must agree it took some time to get from the end of the rod with the ship to the other end where the Earth is, so all frames will agree the event of reception happened at a later time than the event of transmission.

Since the event of the ship reaching the end of the Earth rod and the event of the end of the ship-rod reaching Earth are simultaneous in the ship's frame, these events cannot be simultaneous in Earth's frame.

That has nothing to do with the fact that the event of the Earth receiving the signal (if that is indeed the event you were referring to above) happens later than the event of the ship sending the signal. Even if the ship's rod were a different length so that the event of the ship reaching the end of the Earth's rod did happen at the same moment as the event of the Earth reaching the end of the ship's rod in the Earth's frame, it would still be true that the event of the Earth receiving the ship's signal would happen at a later time than the event of the ship sending it.

Why would all frames have to agree that these two events (the ship's signal and the ship reaching the end of the Earth rod) would have the same coordinates?

Events that happen at the same time and place in one frame must happen at the same time and place in all frames, this is just one of the basic assumptions you need to make when constructing coordinate systems in relativity, otherwise you'd get different physical predictions in different coordinate systems as I mentioned. I think it definitely helps if you think in terms of spacetime diagrams. As an analogy, suppose we're drawing line segments on an ordinary piece of paper, and then using two different sets of x and y axes, with one set's axes rotated with respect to another's, to assign x and y coordinates to the endpoints of different segments. If we have two line segments that meet at a single point, isn't it obvious that no matter which set of x and y axes we use, those two endpoints will be assigned the same x and y coordinates? It's just the same with coordinate systems in relativity, you're just using differently-oriented space and time axes to describe the same pattern of worldlines in spacetime. If each worldine has dots drawn on it to represent successive ticks of a clock moving on that worldline (or any other set of events along it), then if two worldlines cross each other in such a way that two dots coincide, the events represented by those dots must coincide in all coordinate systems, just like dots drawn at the endpoints of the two line segments I desribed would have to coincide in all coordinate systems if they coincided in one.

You might also want to check out a thread I started a long time ago called An illustration of relativity with rulers and clocks, where I drew some diagrams of what would happen if you had two rulers moving at relativistic speed with clocks studded at regular intervals along their length--I made a point there of showing how even though they disagree about things like whose clocks are running slower or whose clocks are in sync, they always agree about the readings on a given pair of clocks from each ruler at the moment they pass next to each other.

I believe you stated earlier that if the event of the Earth's clock ticking d/v and the event of a rod attached to the ship (length d in Earth's frame) reaching Earth were simultaneous in Earth's frame, these events would not be simultaneous in the ship's frame.

Never, you must have misunderstood--that would lead to totally contradictory physical predictions in the two frames.

I agree this will result in contradictory physical predictions, or paradox so to speak. That's why I'm asking so many questions. I would also note that if we really had rigid rods, Earth could observe the ship's position in real time, by observing the ship-rod, and this would clearly violate the laws of physics.

When I said "rigid rods", I didn't mean they would remain rigid if the ship accelerated, as that would indeed violate the laws of physics. But as long as the ship and the Earth are moving inertially, the rods attached to them will appear "rigid" in the sense that their length is constant in each frame. That was all I meant by "rigid rods".

(edit) Just wanted to add something here. What if, in the Twins Paradox, the destination star explodes when the ship reaches it (in the ship's frame), and destroys the ship. The event of the star's explosion and the event of the ship reaching the star would not be simultaneous in Earth's frame.

Yes, they certainly would! And less violently, if there is a clock at the star, then both frames will predict the same thing about what the star's clock will read and what the traveling twin's clock will read at the moment he reaches the star. To understand this you have to take into account different frames' definitions of simultaneity, as well as their different opinions on the rate that different clocks tick. For example, say the star's clock is synchronized with the Earth's clock in their mutual rest frame. Then in this frame, the star's clock read zero at the moment the traveling twin left earth, when his clock also read zero; since his clock is running slow in this frame, they'll predict his clock will be behind the star's clock when they meet. But in the traveling twin's frame, the star's clock was not synchronized with the Earth's clock at the moment he left, instead it was significantly ahead of the Earth's clock. Thus, even though in his frame the star's clock is ticking slower than his, he'll agree in the prediction that the star's clock will be ahead of his when they meet, because it had that "head start". You might want to try actually plugging some numbers into a scenario like this, using the Lorentz transform to convince yourself that both frames make precisely the same prediction; if you like I could help you out with this.

 

[Al68]

Jesse,

You're right, I misread your earlier post completely. That caused a chain reaction that led me to several mistakes before I realized it. I still have questions, but I'd better take my time with them before I post.

Thanks,
Alan

 

[Al68]

If we say that the distance from the Earth to the ship (in earth’s frame) is equal to the length of the ship-rod (in earth’s frame) when the end of the ship-rod reaches Earth (in earth’s frame), can we possibly be correct? Since Earth can measure the length of the ship-rod before the ship leaves earth, mark the rod along it’s length, and then just watch the ship-rod go by? Then Earth could know the ship’s location and velocity in real time. And, of course the ship is not moving inertially in this case, it is accelerating.

Thanks,
Alan

 

[JesseM]

If we say that the distance from the Earth to the ship (in earth’s frame) is equal to the length of the ship-rod (in earth’s frame) when the end of the ship-rod reaches Earth (in earth’s frame), can we possibly be correct? Since Earth can measure the length of the ship-rod before the ship leaves earth, mark the rod along it’s length, and then just watch the ship-rod go by? Then Earth could know the ship’s location and velocity in real time. And, of course the ship is not moving inertially in this case, it is accelerating.

I was assuming a purely inertial situation, like a ship that moves inertially past the Earth rather than one that takes off from Earth after originally being at rest with respect to it. If the ship accelerates, the rod can't stay rigid, so in the Earth frame the back end of the rod will not begin to accelerate with the ship until well after the ship and the front end begin to accelerate (this is also true in every other inertial frame)--the back end will only start to accelerate once a signal moving at the speed of sound in the rod has reached it from the front end, starting at the moment the front end began to accelerate. Only after all parts of the rod have stopped accelerating and the rod is moving inertially again will its length be constant in the Earth's frame, and its new length will be shorter than its length before acceleration, because of Lorentz contraction. But like I said, it's easier just to think in terms of a ship that moves past the Earth inertially without ever accelerating.

 

[Al68]

I was assuming a purely inertial situation, like a ship that moves inertially past the Earth rather than one that takes off from Earth after originally being at rest with respect to it. If the ship accelerates, the rod can't stay rigid, so in the Earth frame the back end of the rod will not begin to accelerate with the ship until well after the ship and the front end begin to accelerate--the back end will only start to accelerate once a signal moving at the speed of sound in the rod has reached it from the front end, starting at the moment the front end began to accelerate. Only after all parts of the rod have stopped accelerating and the rod is moving inertially again will its length be constant in the Earth's frame, and its new length will be shorter than its length before acceleration, because of Lorentz contraction. But like I said, it's easier just to think in terms of a ship that moves past the Earth inertially without ever accelerating.

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct?

Thanks,
Alan

 

[JesseM]

It almost looks like, with an accelerating ship, you were right the first time, when you said that distance isn't like the length of a rigid object. Would this be correct?

Once again, there is no such thing as "the" distance for an accelerating ship, because there is no single standard choice of coordinate systems for accelerating objects. If you use the specific type of coordinate system that I labeled "NI" in that long post from page 1, where the accelerating observer's definition of distance at any moment is defined to match that of their instantaneous inertial rest frame, then the argument I made in that post shows why we can figure out distance at any moment in the NI coordinate system by considering the distance in the rest frame of a passing inertial observer:

Now, this conclusion shouldn't be altered if we are talking about the instantaneous co-moving inertial rest frame of an accelerating observer rather than the constant inertial rest frame of an inertial ship. After all, we could imagine that we have both an inertial ship and an accelerating ship moving away from the earth, and that they cross each other's paths at the moment that they are at a distance d in the Earth's frame (they must have departed Earth at different times for this to work); at the point in spacetime where they cross, they both have the same instantaneous inertial rest frame, so they would both measure the same distance to the Earth in this instantaneous frame.

Of course, the distance in the NI system would not match that of a physical ruler carried by the accelerating observer, because that ruler would be constantly deformed due to the acceleration and thus wouldn't be a good measure of distance at all. But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

 

[pervect]

But this is why I keep trying to emphasize that there is no standard way for an accelerating observer to define things like distance, because there's no single obvious physical way to measure them like for the inertial observer, so your choice of which non-inertial coordinate system to use is fairly arbitrary with no reason to say one is a better representation of the accelerating observer's "point of view" than any other.

I'm not sure how much this will help the discussion, but I can talk a little bit about the reasons the coordinate system commonly used for an accelerated observer is the coordinate system of an instantaneously co-moving observer.

Let's start with an object at a point. Through that point, there are an infinite number of vectors in 4-d space-time that could be consider the direction of "time".

When we settle on the velocity of an observer at that point, though, that picks out a unique vector that represents that particular's observer's concept of "time".

There will be a 3-dimensional space orthogonal to that time vector. This constitutes that particular observer's notion of "space".

The basis vectors that represent "space" at that point can be picked arbitrarily if one doesn't mind a rotating coordinate system. Rotating coordinage systems are sometimes annoying. To avoid this annoyance, there are procedures for picking out a "non-rotating" space vector. I won't go into the details at this point, I'll just mention that it's something that can be done with enough effort. Of course our choice is not really unique, we have a rotational degree of freedom in picking out our coordinate axes.

We've now set up a purely local coordinate system at that point. We now have to discusss how we extend that coordinate system to cover significant distances, rather than infinitesimal distances.

Relativity has a concept that's roughly similar to a "straight line" - it is called a geodesic.

The previous efforts have given us the information needed to construct the various time-like and space-like geodesics ("straight lines") passing through our initial point. These geodesics form the coordinate axes of our coordinate system. While we don't have to use relativity's notion of a "straight line" (i.e. a geodesic) to define how these coordinates propagate, it is a convenient and naturally choice, to make our axes the best possible equivalent to a "straight line".

The next issue comes up - how do we make "tic marks" on these coordinate axes, to represent equal intervals? We can use the concept of the Lorentz interval along the geodesic path to define where our "tic-marks" should be, such that distance between tic marks, along our geodesic paths, is uniform. This is also a rather convenient choice. (The technically minded might appreciate that this particular choice of tic-mark spacing makes the Christoffel symbol $\Gamma^x_{xx}$ zero.

If we carry out this procedure in flat space-time, the above procedure will necessarily define as our coordinate system the coordinate system of an inertial observer at that point.

The above procedure is a bit more general than this, though, in that it can be carried out in curved space-times (where inertial frames do not really exist) as well as flat space-times (where inertial observers do exist).

So we can see that using the instantaneous inertial coordinates for the coordinate system of an accelerated observer is a rather "natural" choice. But it does have a limitation. Eventually, the different geodesic lines cross. In Euclidean geoemtry, parallel lines never cross, but unfortunately in the more general case, they do, and in particular, for any accelerated observer, a simple diagram shows that the timelike geodesics through different points in the trajectory of the accelerated observer (which we which to specify as the "origin" of our coordinate system) must intersect.

When this happens, when the geodesics ("straight lines") intersect, the coordinate system is no longer a 1:1 mapping. The result is that this general procedure for creating a coordinate system of an accelerated observer has a limited "range" - it is only a "local" coordinate system. The range limit is equal to c^2/a, where a is the accleration of the observer.

 

[JesseM]

I realize I was simplifying things a bit for the sake of the argument, in practice there may be many cases in general relativity where one particular coordinate system may be seen as the most "elegant" choice, or the easiest to work with mathematically (aside from the case of an accelerating observer, physicists might also say that the most elegant coordinate system for dealing with the spacetime outside a black hole is Schwarzschild coordinates, or that the most elegant coordinate system for dealing with cosmology is the one where the distribution of matter and energy appears most uniform in space). But the theory of special relativity was originally designed around the idea of inertial reference frames, and they have a very simple physical interpretation in terms of local readings on a network of inertial rulers and clocks; in general relativity there isn't anything like this, GR was actually designed around the idea that you should be able to use any coordinate system you want (this can be true in SR too if you formulate the laws in tensor form, but this wasn't how they were originally formulated or how they'd be described in undergraduate texts). It would actually be incorrect to apply the non-tensor form of SR to anything but an inertial frame, while the laws of GR could be applied in any non-inertial coordinate system you like. And even if the type of coordinate system you describe for an accelerating observer may be the most "natural" in some sense, I don't think there's any simple interpretation of the coordinates assigned to distant events in terms of local readings on measuring-devices like the inertial rulers and clocks of an inertial coordinate system.

 

[Al68]

Jesse,

OK, I almost forgot the reason I started this thread. I wanted to figure out the coordinate position of the Earth as a function of time in system NI. It looks like you've done that. But I also wanted to figure out how to do this for the ship's deceleration and coming to rest relative to earth. Is there a way to figure out the coordinate position of the Earth as a function of time in a decelerating system NI? That is in motion relative to Earth and starts decelerating at a specified distance D from Earth at a specified time T in the ship's frame, and eventually comes to rest relative to earth. In other words, similar to the ship coming to rest at the star system in the Twins Paradox.

Thanks,
Alan