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## Interstellar travel question....

One of the (probably many) things I don't understand about relativity is how examples can be shown from a travellers POV which include relatavistic effects. If I may borrow part of a reply which Grant Hutchinson made in another thread..

Relativity means that we can never reach lightspeed, but can get pretty close to it for a fairly short elapsed time aboard our spaceship. After one year accelerating at 1g, we've reached 0.75 lightspeed; at two years, 0.96; five years, 0.9999; ten years, 0.99999999. (Unfortunately, after ten years of acceleration, 9,000 years would have passed at home on Earth.)

How could these speeds apply to the traveller himself as he, his ship, engines, fuel etc are all in the same reference frame for the whole journey?

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Originally Posted by Jerryf
How could these speeds apply to the traveller himself as he, his ship, engines, fuel etc are all in the same reference frame for the whole journey?
In the same way that you can travel at 60mph while sitting stationary in the reference frame of your car. The motion is relative to your departure point: in this case, that includes the unaccelerated Universe outside the traveller's window.
After examining the view from his window, the traveller would be able to judge that he was moving away from his departure point at some speed.

Grant Hutchison

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<<After one year accelerating at 1g, we've reached 0.75 lightspeed>>

Ah but are the reference frames mixed here?

The difficulty is that, people watching the mission would see the acceleration of 1g reduce as the ship picked up speed. So relative to the unaccelarated people, it does indeed take about a year for the ship to get to 0.75c.

However on the ship, the acceleration is constant at 1g. Time goes slower for them, so they would perceive themselves to get to 0.75c more quickly than a year, in fact 265 days.

Is this correct ?

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Originally Posted by kzb
However on the ship, the acceleration is constant at 1g. Time goes slower for them, so they would perceive themselves to get to 0.75c more quickly than a year ...
Those aboard the ship might calculate the velocity they could achieve after so many days at 1g; but if they looked out the window, they would find that their velocity relative to the outside Universe was slower than their calculations suggested.
So it would take a year of on-board time before they'd be able to observe the outside stars shifting at 0.75c relative to their spaceship.

Grant Hutchison

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Originally Posted by grant hutchison
Those aboard the ship might calculate the velocity they could achieve after so many days at 1g; but if they looked out the window, they would find that their velocity relative to the outside Universe was slower than their calculations suggested.
So it would take a year of on-board time before they'd be able to observe the outside stars shifting at 0.75c relative to their spaceship.
Thank you Grant. This is pretty much what I'm asking - if there was a specific goal to reach a particular star 10 light years away from Earth why wouldn't distance contract in front of the ship if it's dilating behind?

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Originally Posted by Jerryf
This is pretty much what I'm asking - if there was a specific goal to reach a particular star 10 light years away from Earth why wouldn't distance contract in front of the ship if it's dilating behind?
Distance does contract for the traveller, both in front of and behind his ship: this means the accelerating traveller finds he is initially slow to get away from his starting point, but quick to approach his destination; when he starts to decelerate, however, he finds his starting point moves away quite quickly, but his destination approaches only slowly.

Grant Hutchison

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Originally Posted by grant hutchison
Distance does contract for the traveller, both in front of and behind his ship:
in that case relatavistic effects are not dependant on direction of motion relative to the observer...that's not right though is it?

[/QUOTE]this means the accelerating traveller finds he is initially slow to get away from his starting point, but quick to approach his destination; when he starts to decelerate, however, he finds his starting point moves away quite quickly, but his destination approaches only slowly.[/QUOTE]

Could I ask that you be more specific with your velocity terms here....if possible as specific as your original reply? Just trying to understand.....

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Originally Posted by Jerryf
in that case relatavistic effects are not dependant on direction of motion relative to the observer...that's not right though is it?
Length contraction is always observed in the direction of the velocity vector: so the traveller sees the Universe contracted along his direction of travel; and the stay-at-home observers see the traveller contracted along his direction of travel.

Originally Posted by Jerryf
Could I ask that you be more specific with your velocity terms here....if possible as specific as your original reply? Just trying to understand.....
Yes, sorry, I see how my mixed use of the word "slow" is muddying the waters.
Let's for the moment imagine that the traveller doesn't accelerate in the conventional way, but simply flips up to some significant fraction of the speed of light instantaneously. When he reaches his destination, he flips back down to local rest instantaneously. As soon as he gets up to near lightspeed, he'll see the distance he has to travel contracted ahead of him, and he'll cross it in a correspondingly short time. While he's doing the crossing, he'll appear to proceed in the customary way: the length of his journey doesn't change, and he'll run down the distance ahead in a very regular way.

Things get a bit tricky if he accelerates for half the journey and decelerates for the other half of the journey, though. In that case, the total length of his journey is changing as he travels, getting shorter during acceleration and then returning to "normal" during deceleration. During acceleration, the relativistic decrease in the journey length means that he runs down the distance ahead faster than he increases the distance astern. This is just a symptom of the fact the overall distance between departure point and destination is decreasing as he accelerates. The effects are reversed when he decelerates.
This has the distinctly odd effect that if you accelerate away from the Earth at 1g, you never measure yourself as being much more than a light-year from it: the increasing distant covered in the local rest frame is offset by the relativistic length contraction in the traveller's frame.

Notice I'm talking about what the traveller would deduce after he has allowed for the weird visual affects that near-lightspeed travel induces.

Grant Hutchison

9. Originally Posted by grant hutchison
Those aboard the ship might calculate the velocity they could achieve after so many days at 1g; but if they looked out the window, they would find that their velocity relative to the outside Universe was slower than their calculations suggested.
So it would take a year of on-board time before they'd be able to observe the outside stars shifting at 0.75c relative to their spaceship.
This is all included in Grant's explanation, but perhaps one thing confusing readers is that there are actually two different meanings of speed that must be distinguished when acceleration is also going on. Grant is using it in the standard measurable relative-velocity sense, the instantaneous rate of change of distance in the absence of acceleration. This is also what you'd infer from the blueshift of the star you are headed toward. However, another meaning for speed is the instantaneous rate that the distance you have yet to go is being reduced, and that is a larger number in this situation (and does not have to be less than c) because of the increasing amount of length contraction as you continue your acceleration. So this is the speed that others are talking about when they say that you are always increasing that speed at rate 1g, but Grant's point is that acceleration actually gets split into two pieces by relativity-- one piece is the rate of change of relative velocity, and the other is the rate of change of the length contraction. The latter requires that you maintain your acceleration, but the former is there no matter what you do.
This has the distinctly odd effect that if you accelerate away from the Earth at 1g, you never measure yourself as being much more than a light-year from it: the increasing distant covered in the local rest frame is offset by the relativistic length contraction in the traveller's frame.
That's cute. A way to think about this is that in the backward direction, these two speed components are working at odds to each other, as Grant said.

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I can feel a diagram coming on.
Here is a chart of a journey at 1g over a distance of 98 light years, with "turnaround" marking the midway point, at which our starship swings its engines around to the forward direction, so that it stops accelerating and starts decelerating.

Distance as measured by the travellers is on the vertical axis, shipboard time on the horizontal. Since we're looking at the ship's reference frame, the ship is always centred relative to the vertical axis of the graph, marking "zero distance".
At the start of the journey, "home" is where the ship is, and the destination is 98 light years away. As the ship accelerates, it pulls in the destination quite rapidly initially, but fails to shake off "home" astern: its motion is conspiring with length contraction to bring the destination quickly closer, but motion and length contraction are opposing each other astern, preventing home getting any farther away than one light year.
By the midpoint of the journey (4.5 years elapsed aboard ship), the ship is sitting halfway between home and destination, but measures their separation as only 2 light years.
Once the ship starts to decelerate, the waning of length contraction offsets the approach of the destination, so the ship creeps up on it very slowly. Meanwhile, as length contraction wanes astern, it becomes increasingly evident how far the ship has travelled, so that home recedes briskly into the distance.

Grant Hutchison

11. One thing I find noteworthy, and Rindler shows this in its own way, is that length contraction for the most part occurs *behind* you as you accelerate.

You can see this imagining watching a rigid rod (or spaceship) accelerating. The tail end has to accelerate faster than the nose in order to remain rigid in the accelerating rest frame. So watching that rod contract, it's the tail end that moves toward the nose, rather than the whole thing sort of contracting evenly about the center.

Another way to picture this, from the accelerating frame perspective, is, when you start accelerating, imagine your coordinate axis along the direction of acceleration as being pinned at infinity ahead of you. The contraction comes from behind, "compressing" that axis toward that fixed anchor at infinity.

-Richard

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I never really understood the Rindler horizon, Richard, but it seems like something interesting goes on in the acceleration/deceleration scenario in that regard.
As the ship accelerates away, "home" slides asymptotically towards the Rindler horizon (if I translate from SR to GR correctly). Turnaround comes when the destination is as far ahead as home is behind, so when deceleration begins, the destination has just managed to tuck itself inside the Rindler horizon for the decelerating spacecraft.

Also, if the ship were to continue to accelerate, it would pass the destination quite soon after the erstwhile "midpoint" of the journey. The destination would then slip aft and approach the Rindler horizon asymptotically. In fact, everything the ship passed during its acceleration would progressively end up pasted against this "barrier", a light year aft.

Grant Hutchison

13. The way I think of all this is, the Rindler coordinatization is a pretty arbitrary and not terribly meaningful one. Note that any coordinatization is allowed, because it's all nonlocal-- you aren't actually conferring with any local observers at the point in question what they think rulers and clocks are saying (and if you did, it would be an interesting puzzle to figure out what reference frame they'd need to be in to get the same answer as Rindler. Richard?). The only time the Rindler coordinatization corresponds to something physical is if you maintain constant acceleration indefinitely, and then the horizon is indeed the place that everything aft "gathers" to. But if you dart about and change your acceleration willy nilly, then the instantaneous Rindler metric loses any physical significance, in my view. It's rather like the simultaneity convention for things not causally connected to you, and how that can bounce all around if you start dodging and darting about, hopping forward and backward in time in a way that is not terribly physically meaningful but is just kind of an arbitrary means of bookkeeping (which is basically what nonlocal coordinatizations are).

14. Originally Posted by grant hutchison
I never really understood the Rindler horizon, Richard, but it seems like something interesting goes on in the acceleration/deceleration scenario in that regard.
As the ship accelerates away, "home" slides asymptotically towards the Rindler horizon (if I translate from SR to GR correctly). Turnaround comes when the destination is as far ahead as home is behind, so when deceleration begins, the destination has just managed to tuck itself inside the Rindler horizon for the decelerating spacecraft.

Also, if the ship were to continue to accelerate, it would pass the destination quite soon after the erstwhile "midpoint" of the journey. The destination would then slip aft and approach the Rindler horizon asymptotically. In fact, everything the ship passed during its acceleration would progressively end up pasted against this "barrier", a light year aft.

Grant Hutchison
Yes, when you do the turnaround, and change the direction of your proper acceleration, things get weird.

I've been thinking about this sort of off and on in the back of my mind ever since I got in the thorny bush of the backwards jump in simultaneity for things behind an observer who increases his velocity instaneously.

I realized there will be such a "jump" involved in *any instantaneous change in reference frame*. If we fire our rocket, while our velocity may change smoothly -- no kink/corner in the x(t) worldine -- there will be a kink/coner in the dv/dt curve. There's no avoiding step-function like behavior at some level of derivative.

So, when you jump into Rindler, there's going to be a jump of what things behind you are doing, most notably past the Rindler horizon.

From an inertial frame watching, you can visualize the Rindler horizon as the front of a light pulse fired at the horizon distance (- c^2/a) when the rocket is fired. That light pulse chases the accelerating observer but never quite makes it. The distance between them is his notion of the fixed distance to the horizon behind him. Any events that happen after that chasing light pulse passes never occur in the Rindler frame.

Now, to see what happened to stuff already behind the horizon when he started accelerated (something farther than -c^2/a), you just extend that light pulse backward in time, pretending it came in from infinity.

And that's the jump for the Rindler observer. Things take a backward jump in time to agree with that light pulse marking the horizon in a rather strange way. Things beyond the horizon jumped backwards and got stuck there *before he started accelerating*.

And what I realized is that these "jumps" do not necessarily and completely show up in the metric, they show up in the coordinate transforms themselves. For example, the metric of Lorentz observers remains all 1's before and after a jump in velocity. However, the coordinate transforms reflect that jump in matching up the origin of the observer. The Rindler observer has a jump in his metric, but he has other jumps in the coordinates themselves.

But those jumps never affect anything he will actually *see* from light information. The only light from the entire universe behind him he will see is that which was *in transit ahead of the c^2/g behind him* when he started. And the redshifting he's see of that will agree with the jump backwards and freezing against his horizon *in his past* with the light just getting there.

Now, when he turns around and fires his rocket the other way, you just do the jump again, except it now applies to the stuff that was in front of him. The stuff frozen on the horizon sort of "unwinds" and streches out, but the stuff in front of him now has a jump backwards.

That's a headful to think about, but it (I trust!) works out.

-Richard

15. Ken,

What I would say is the Rindler coordinates apply *locally*, right in the little Lorentz neighborhood of the guy who's accelerating. That's his clock and that's his ruler. So they tell us what he sees of stuff whizzing by him locally.

As far as what other reference frames would agree with that, well I don't know. I don't know if there would be any at all. Now, we can construct a familiy of observers with his exact same notion of distance. Those are just observers who would be mutually stationary in their own respective frames.

In Rindler, the force required to remain stationary decreases with distance ahead of the horizon. So that, from an inertial frame watching that would be a string of observers who started accelerating, but with decreasing proper acceleration down the line. They would appear to be getting closer together, with the tail end catching up faster than the fore end.

They would all be stationary in their own local frames, agreeing with constant Rindler distance. But their clock rates would be different.

And then you might construct some family of observers with constant Rindler clock rates, but they would be moving relative to each other.

-Richard

16. Something else sort of interesting about Rindler is the null geodesics, the path light appears to take. They are perfect semi-circular arcs. If a light pulse is fired from a point, the path will be along a circular arc whose tangent is parallel with the initial direction of the light beam in one spot and perpendicular to the horizon plane in another spot. The light follows that path to the horizon.

If the light is launched horizontally (we're accelerating in the z direction, so this is light launched in the z-y plane), it will follow the path of the circle made by taking a compass with one end at the horizon behind us, and the other end at the starting path. Sweep out the circle down back down to the horizon and that's the path light appears to take in our coordinates.

For light launched with an initial z component, you'd have to shift your radius so you were tangent at the start and perpendicular to the horizon at the end, but it would still be a circular path, just with a larger radius.

-Richard

17. Yes I agree-- the Rindler metric is physically meaningful in a local sense, meaning in terms of differentials, whereas the Rindler horizon, and backward time jumps, are nonlocal integrals of those differentials along some arbitrary constant extension of the local Rindler metric. So they really aren't physically meaningful unless the acceleration is maintained for so long that such nonlocal integrals really start tracing the history (and future) of the motion. I would have said that the way to make the Rindler metric globally meaningful is to find the family of observers who really measure space and time according to that metric. You found a family for space, and for time, but does this mean you can't find a family for both at once? If not, it would seem that the global extension has no physical meaning at all.

18. OK. I think I had better go to bed now before my head explodes!

Maybe I'll read it again tomorrow.

BTW, thanks guys. I spend time here to learn and it may be working!
Last edited by C18H27NO3; 2007-Jan-16 at 03:32 AM. Reason: Typos. Note to self: Proof read before hitting the submit button!

19. Originally Posted by Ken G
Yes I agree-- the Rindler metric is physically meaningful in a local sense, meaning in terms of differentials, whereas the Rindler horizon, and backward time jumps, are nonlocal integrals of those differentials along some arbitrary constant extension of the local Rindler metric. So they really aren't physically meaningful unless the acceleration is maintained for so long that such nonlocal integrals really start tracing the history (and future) of the motion. I would have said that the way to make the Rindler metric globally meaningful is to find the family of observers who really measure space and time according to that metric. You found a family for space, and for time, but does this mean you can't find a family for both at once? If not, it would seem that the global extension has no physical meaning at all.
Well, if some "global class" exists depends on just what you have. In Schwarzshild, any observers at the same 'r' share the same rulers and clocks.

And that is akin to one example here so trivial I didn't mention which is all observers in a plane all sharing the same proper acceleration normal to the plane. They would all see themselves in Rindler, just at different x and y coordinates. But their clock rates and notions of z distance would all agree. But, in z-t space, I don't think there is any other observer would share it.

In both those cases, that's because the metric depends on only 'z' or 'r'. The other directions aren't there, so any observer at different values of those coordinates shares the same local rules and clocks.

So, I guess we can consider a global class of observers stuck to that x-y plane, but is that really the kind of global extent we want or just sort of 2 dimensions worth?

-Richard

20. And, if you want some real fun, imagine two of those observers a large distance apart trying to synchronize their clocks. In general (general GR metrics) that isn't always possible. When light doesn't travel in a straight line, and it's coordinate speed varies along the way, the familiar synchronization procedure doesn't work very well.

If there is a unique light path between A and B that is constant with time, then you can do something, although it would be complicated. If there is not a unique path, or it varies with time along the way, then you're SOL.

-RIchard

21. Originally Posted by publius
But, in z-t space, I don't think there is any other observer would share it.
That's the issue, the plane of observers is kind of an ignorable extension, but the z extension is what really would make the horizon globally meaningful. You said you could do it if they were moving relative to each other, and that's fine, I just want their clocks to instantaneously measure time at the same rate as the Rindler metric, and their lengths to instantaneously do the same if laid end to end. Relative motion isn't a problem, that's like the relative motion in the expanding universe in true GR. I think what you are in effect doing is describing the analogy to the "comoving frame" coordinates in cosmology. Imagine this string of observers were actually galaxy clusters, all measuring spacetime the way the Rindler metric says, all stationary in their own frames, and cobbling together a coordinate system that one might call Rindler coordinates. That should be a perfectly allowable gravity-free universe, should it not? What would be the observables in such a universe? I think that universe is the closest thing to a physical meaning for the global Rindler metric.

22. Originally Posted by publius
If there is a unique light path between A and B that is constant with time, then you can do something, although it would be complicated. If there is not a unique path, or it varies with time along the way, then you're SOL.
Since synchronization is more a matter of convention than anything else, I would tend to just define "now" as the first time that a light beam I send out "now" gets to any other point. What's wrong with that? It's unambiguous. It's not reciprocal between me and others who are stationary with respect to me, but what's so special about them anyway? In GR, there isn't even any unique meaning to being nonlocally "stationary".

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Ken G wrote:
<<acceleration actually gets split into two pieces by relativity-- one piece is the rate of change of relative velocity, and the other is the rate of change of the length contraction.>>

But the Lorentz factor SQRT(1-(v/c)^2) is exactly the same for both time and distance. So do you get two doses of this factor or is it one factor divided up?

I've not seen this in popular science books.

24. Originally Posted by kzb
But the Lorentz factor SQRT(1-(v/c)^2) is exactly the same for both time and distance. So do you get two doses of this factor or is it one factor divided up?
It is the distance part that matters here, because we are talking about the rate of change of distance using the current time measure. You can think of the length contraction as being there because of the time dilation, but changing time dilation is not relevant to the rate that distance is changing, whereas changing the length contraction is directly relevant, as well as movement "through space". Those are the two components of the rate of change of distance to target.
I've not seen this in popular science books.
They're not as insightful as they could be.

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Originally Posted by Ken G
... but Grant's point is that acceleration actually gets split into two pieces by relativity-- one piece is the rate of change of relative velocity, and the other is the rate of change of the length contraction.
I've just been thinking about this. Shouldn't the latter component of the acceleration be the rate of change of rate of change of the length contraction; that is, the second derivative?

Grant Hutchison

26. Originally Posted by grant hutchison
I've just been thinking about this. Shouldn't the latter component of the acceleration be the rate of change of rate of change of the length contraction; that is, the second derivative?

Grant Hutchison
You would think that, but it's tricky. Think of "his ruler and clock", the moving observer's proper time (4-)vector, call it T, and his proper ruler vector, call it X (expressed in our coordinates). Both of those are constant with velocity. The only change if his velocity changes.

So (in our frame), we can think of dT/dt and dX/dt, which are two vector derivatives that are non-zero only if dv/dt is non-zero. dX/dt has to do with the rate of change of length contraction. The existence of a first time derivative of that requires acceleration.

And the final step is taking both of those notions into the proper time of the accelerating observer himself, dT/dtau and dX/dtau. The former(times c, actually I think) is actually the proper 4-acceleration defined as the dV/dtau, the rate of change of 4-velocity per unit proper time.

There is a helpful way to get a mental picture of this. Consider a fixed curve in the x-y plane, and imagine a particle moving along it at constant speed (here we're completely Netwonian, Galilean, and Euclidean ). Its (vector) velocity at any time can be written:

V = v*T, where T is the (unit) tangent vector of the curve, and v is the constant speed. Take the derivative:

dV/dt = v* dT/dt. Since the magnitude of v is constant. That's the acceleration. And since T dot T = 1, dT*T = 0, changes in the tangent vector are always perpendicular to the tangent. This is the "normal component of acceleration", and is a property of entirely of the curve geometry itself, not how fast the particle is moving along it. In the general case, where v itself was not constant, you'd get an addition T*dv/dt term, which would be the tangential component of acceleration.

And finally, noting that ds = v*dt -- > dt = ds/v, we get:

dV/dt = v^2 * dT/ds. Look familiar? A velocity squared time 1/length (T is defined as dimensionless, you let the dimension ride with the scalars). Just like constant speed circular motion, with the magnitude of dT/ds = 1/r. Constant circular motion is just one example of constant speed along a curve.

Indeed, dT/ds actually can define a simple notion of "curavature" applicable to simple curves in any number of dimensions. That is the change of tangent vector per unit arc length along the curve.

Now, in relativity, those curves are world lines, and we're constrained to move along them at 'c'. It is the same thing as constant speed motion along curves, except now we're letting time itself be a dimension, and have a non-Euclidean Minkowksi geometry.

Proper acceleration in SR can be written exactly in the above form,

'a' = c^2 * dT/ds = c* dT/dtau.

-Richard

27. Grant,

And to continue this tangent vector rambling (again back in Newton and Galileo):

a = T*dv/dt + v^2*dT/ds in the general case. That's the tangential component (change in scalar speed) plus a normal component due to the curvature of the trajectory. I just think that's a neat way to break it down. You can think that as a "regular" component along your line of motion, plus a "centrifugal" component perpendicular to your line of motion.

And just as you can piece together a curve by differential straight line segments, you can also piece it together with differential pieces of circles. At each point 's' along the curve, you are on a piece of a circle with radius given by
r = 1/|dT/ds|

That just strikes me as just plain cool, somehow. I'll leave it to any bright and bushy tailed young mathematicians to locate the center of that circle.

In Minkowski space, v = c = constant above, and all acceleration is "centrifugal". But because of the non-Euclidean way that mess works, it's not circles we have, **but hyperbolas**.

In Euclidean space, constant curvature (curves) are circles. In Minkowski, constant curvature are hyberbolas. And that's what constant "straight line" acceleration is. In SR, that is the equivalent of a constant circular motion. And again, I'll let the young bushy tailed budding mathematicians play with that to prove it.

-Richard

28. That is a very impressive and higher-level mathematical way of thinking about all this, Richard. I have no doubt you are correct, and I see the elegance of that approach. It isn't the way my physical intuition is currently configured, however, so I'll try to address Grant's question from a different angle, because I need to correct my poor wording. In the absence of acceleration, we are saying that a stationary observer thinks the traveler has a distance D yet to travel, so at speed V he will take a time D/V. The traveler thinks that distance is D/gamma, so the time will be D/V/gamma in his own frame. The stationary observer calls that time dilation, the traveler calls it length contraction. Now if we say the traveler has a fixed acceleration A, which means that he would perceive any outside object that is coasting with him as acquiring in a small proper time dTau a speed A*dTau, then by the velocity addition formula we find the change in the approaching speed of the target is dV = A*dTau/gamma^2. The stationary observer thinks he speeds up by A*dt/gamma^3 because they agree on the speed but not on how long it took to achieve-- the stationary observer uses dt not dTau, so the acceleration looks like A/gamma^3 and is not being maintained as gamma grows.

All that's the usual kind of stuff, but the accelerating traveler doesn't really care about any of it, he wants to know how much closer he's getting to his goal, and how long it will take (and how much longer will he hear "are we there yet") and none of these numbers tell that story if he jams on the accelerator pedal. If D is the distance he has left to go from the Earth's reference frame, then D/gamma is how far he thinks he has left to go, and the negative rate of change of D/gamma with respect to Tau is what we might call his "relevant" velocity. When there's acceleration, that's what has a second term beyond just v (and that second term contains the rate of change of 1/gamma in it), and was what I was really talking about. What I meant to say was, when there's acceleration, the relevant closing speed gets split into two terms by relativity, not the acceleration itself. You might also define a "relevant acceleration" in the same way, and then yes, you do get funky second derivatives of 1/gamma in there, and it all gets pretty complicated (and not terribly "insightful"). I'm sure Richard's way is far more elegant, and therefore easier to catch errors, if you like Minkowski geometry.

29. Ken,

Yeah, it's whatever floats your intuitive boat, I suppose.

Me, my fancy is just plain tickled by the view that we're all moving along our worldines at constant speed, c. That "direction" (in Minkowski) we're moving defines our proper time unit 4-vector. ds/dtau = cT, where T is the unit tangent 4-vector. And cT is V, the 4 velocity, and the magnitude of the 4-velocity is always c.

Acceleration is just a change in the "direction" of our tangent vector, not its magnitude.

And that picture is just what floats my boat.

-Richard

30. Originally Posted by publius
Me, my fancy is just plain tickled by the view that we're all moving along our worldines at constant speed, c.
Don't get me wrong, I think that's the deepest possible insight into relativity. I just don't visualize physical problems that way.

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