When accelerating at high %c, can apparent velocites exceed c?

BitWiz

If I travel toward a star 100 ly distance (measured when at "rest") at a low % of c, the star will continue to appear about 100 ly away.

If within 1 year, I accelerate to γ=2 (about 87% c), time dilation will halve my experience of the trip's duration, and to compensate, distance compression (in the direction of motion) will prevent me from observing that *I* am moving faster than light, i.e. that I'm managing to travel 100 LY in less than 100 experienced years. Instead, my target will (appear to) be about 50 ly away. If I have this correct, then:

Question 1: The target star has "moved" from 100 ly to 50 ly away in our mutual frame. Is this an optical artifact or is the object really closer?

Question 2: The target star has (appeared to) move 50 ly toward me in the space of a single year. How?

Question 3: Various video simulations on the web show the optical effects of accelerating toward an object at a high percentrages of c, and the initial movement of the target object appears (visually?) to be *away* from the observer during acceleration. How does this reconcile with the above, i.e. if the remaining distance decreases proportional to gamma as you accelerate, why does the target appear to be receding?

Question 4: If Question 3 above is due to an optical artifact, then where is my target "really?"

Thanks for your time,
Chris

pervect

The distance of 50 light years is the distance you calculate after compensating for all light-speed delays. So to give a non-metaphysical answer to a metaphysical quesiton, it's the "real" distance. Or at least the one you measure.


The target didn't move - it's just that distance is not a property that's independent of the oberver in SR.

I'm not aware of any video simulations that show an object appear to receede as you accelerate towards it. The effects that you see visually are generally covered under the mathematics called "Terrel rotation",

ZikZak

No it won't. You do not suffer time dilation. It does not "halve your experience" of the trip's duration. Time dilation is something you observe about other things, not something that ever happens to you yourself, the observer. In this case, you will observe clocks on the Earth and destination star to be dilated, i.e., to run slow, because they are traveling at 0.87c.

But you don't travel 100LY in less than 100 years. You travel a distance of 50 light years, because that is the distance between the Earth and the destination in your reference frame. Or, rather, you are at rest and the destination travels 50LY to you.

You are still stuck thinking that the Earth/star is an absolute reference frame and there is something more "really real" about the distance being 100LY rather than 50. In the ship's reference frame, the distance measured is 50 light-years. Measurements are about as real as it gets in science.

BitWiz

Thanks for the replies pervect and ZikZak. I'm afraid I may not have "framed" my initial question properly. Let me try again:

Given: a star 100 ly from Earth. A rocket leaves Earth toward that star, and over an Earth-frame duration of one year, the rocket accelerates to 87%c (gamma=2). An Earth observer then sees the rocket travel approx 100 yrs / 87% = ~113 yrs until it reaches the star.

Now take Earth out of the frame. The observer is now on the rocket. The frame now includes only the observer and the star.

After the approx one-year acceleration phase:

1) What is the distance to the star that the observer sees (approx)?

2) What will an onboard clock say is the duration of the total trip (approx)?

The rocket has an onboard accelerometer. During the acceleration phase:

3) A decrease in the distance to the star can be measured (Newtonian) by double-integrating the accelerometer readings down to distance traveled using the onboard clock. Will these numbers agree with observed -- and actual -- distances to the star? If not, by how much?

4) Will an on-board clock measure the same duration to accelerate to 87%c (given constant acceleration per the onboard accelerometer) that an Earth-observer would measure?

I really need to understand this, and again, thank you very much for your time.

Chris

jartsa

Actual distance according to the observer is about 50 lys

About 57 yrs

First integration will give the actual speed. Second integration gives a distance traveled, which has nothing to do with the actual distance traveled, or actual distance left, because these distances depend also on the velocity.

The duration is shorter. The duration of the trip is shorter, and the durations of all parts of the trip are shorter.

BitWiz

Thank you, jartsa! Then I have this question for you:

If I, an observer on the rocket, see the star at 100 ly distant at the time of launch, and then one experienced year later, see the star at 50 ly distant, have I observed that the star has moved (relatively) toward me at approx 50 times the speed of light?

Thanks,
Chris

jartsa

The distance shrunk by 50 lys in one year. Average shrinking pace was 50 * 300000 km/s.

It does not sound terribly wrong to me to say that the star "moved" towards you.

Let's say you are traveling at speed 0.87 c towards a star, and the star makes a short spurt at speed 0.87 c, which eliminates the optical effects for a while. Then for a short while you can see a star that is quite close and quite large.

BitWiz

i.e. 50 times faster-than-light (FTL), and if acceleration is constant, the pace is quite a bit faster than 50c at the end of the acceleration period.

Agreed. It's all "relative." It also implies that an observer on the star will see the ship moving toward the observer FTL. I would also think this "gamma" component of the motion is free of Doppler(?).

Toward or away from the observer?

Away from the observer, the optical effects disappear, and the distance between them will revert to gamma=1, the distance widening over the duration of the spurt's acceleration.

Toward the observer, there will be Doppler increase equivalent to 87%c, but I believe the relative effects will cause gamma to become 4 (multiplicative?) or about 97%c. Observers at both the star and rocket will observe additional FTL increases of 50%.

If I, an observer on a rocket ship, can accelerate to 99.99999%c in say one meter and then decelerate, I could presumably take a picture of the star from less than 0.05 ly away without leaving my driveway. It makes me wonder what distances photons see to the objects in their path -- zero? -- and whether they are responsive at all to time in a vacuum.

But I digress. The idea that I can observe an accelerating (relative) mass object moving FTL is interesting to say the least. It's an outcome that I have been counting on to solve a different problem, but I've had trouble finding anything about this on the web. That of course makes me suspicious. I'm hoping you and others on the forum can help me sort this out.

Thanks for your time,
Chris

pervect

Use the formulae for the relativistic rocket, http://math.ucr.edu/home/baez/physic...SR/rocket.html

Let c= 1, and let time be measured in years, distance in light years, and accelerations in light years/ years^2

Then, if gamma =2 at the end of one Earth year, we must have

sqrt(1+a) = 2, or a=3. So you had to accelerate about 3 light years/years^2 to reach a gamma of 2 in one earth year, that's approx 3g.

At the end of 1 year at this acceleration, the in the Earth frame distance from the earth to the rocket is only 1/3 of a light year. Thus the distance in the earth frame from the rocket to the star is 100- 1/3) = 99.66 ly. The distance in the rocket frame (what you asked for) is 99.66 / gamma = 49.83 ly.

It takes the(1/3)*arcsinh(3) - .6 years of proper time (i.e. rocket time) for the rocket to stop accelerating and start coasting.

It then takes 49.83 / .866 = 57.54 years of proper time for the rocket to reach the star. So the total proper time (rocket time) is 58.14 years.

No, doing a Newtonian integral will not give the correct relativistic results. Using the Newtonian formula, you'd predict that you moved .5 a t^2, with a=3 and t=1, or 1.5 light years towars the star in the acceleration phase, but you actually moved .33 light years according to the relativistic calculation.

Nope, in Earth frame the rocket stopped accelerating at 1y, the proper time on the rocket was .6 years.

I hped going through the numbers helps more than I think it will. Basically, distance and duration are not indpendent of the observer. What is independent of the observer is the so-called Lorentz interval, and what you need to study to understand it is the Lorentz transform.

The Lorentz transform is a simple linear equation. The coefficients in the equation can be understood as length contraction, time dilation, and the relativity of simultaneity.

jartsa

Indeed the "motion" is free of Doppler. And the "motion" is not relative.

Surely not. They disagreed about the distance when their velocities were different. When their velocities become the same, they agree about the distance. The one who changes his velocity changes his mind.

That's all wrong too.

Why did you decelerate? Deceleration cancels acceleration. You only gained two meters.

EDIT: Oh yes, the picture was taken after the acceleration and before the deceleration. Well I don't know, but I guess the picture would not be so spectacular.

BitWiz

Hi, pervect,

That's an amazing link. Thank you.
Is there something wrong here? 3g (~30m/s^2) over one year (~3E7 seconds) gives us a velocity of 9E8 m/s or ~3c. If we integrate acceleration over that year for mass effects, do we really come up with gamma=2? If so, for which observer? Rocket or independent?

Oh, ye of little faith. ;-) I think I understand Lorentz somewhat, though perhaps not yet well enough. Regardless, you have been very helpful. My goal though is to understand the "experience" of relativity in a kind of reverse-empiracle way from the math. If I was a photon and had direct relativisitic experience from birth, I'm sure this would all make perfect, intuitive sense. Regardless, I think we tend to get buried in the abstract and forget to ask "what does it all mean?"

My initial question remains: If I accelerate toward an object, the distance to that object diminishes. Relativistically, with only myself and the object in the frame, I'm conjecturing that my acceleraton toward the object is indistinguishable from that object accelerating toward me, and the observed -- and measurable -- reductions in distance between me and that object can be thought of as me moving toward it, it moving toward me, or any combination of the two, and they are all indistinguishable.

The interesting part is not that I observe apparent velocity toward the object or vice versa, but rather what happens when that velocity changes. Time dilation, space compression, Lorentz seem to indicate that with certain combinations of acceleration and apparent initial velocity (not speed), the objects can approach each other at velocities much greater than c within their shared frame, and further, it does not matter which object is actually experiencing the g's. Observers on either object will see the other approaching at the same rate. Is that how you see it?

Again, thanks very much for your time.
Chris

BitWiz

I'm not surprised. ;-) I inadvertantly let "exuberence" trump my reserve at times.

You would just need an anamorphic lens. ;-) One of the issues seems to be that, though you would be "closer," I think the angle of incidence would be (much) smaller than if you were at rest at the apparent distance. So though you were closer, you would be looking at a smaller object, and there would be no increase in resolution. However you could video the object's history up to within 0.05 years. I'm speculating at all of this.

It might also imply that I can substitute a data emitter for the camera, and transmit a signal much closer to the star than from Earth. The Doppler might fry the recipient, but the data would get there faster. Encoded neutrinos?

Thanks again for your time.
Chris

pervect

You might want to double check my math, I'm not infallible. The results I mention below are in the original post:

earth time: 1 year
proper time (rocketship time) .6 year
final velocity: .877 c
final gamma: 2
acceleation 3 light years/year^2
distance travelled: 1/3 ly

These come from the equations in the link I mentioned:

I'm not convinced it's safe to assume this. I'm not convinced it's wrong , either. I think you need to do the calculations, from the webpage you sholud have the tools to do that.

No. I see that the proper velocity, also known as celerity. can increase well above "c", but the ordinary velocity cannot increase above c. It's important not to confuse the two concepts.

"Intergalactic spaceflight: an uncommon way to relativistic
kinematics and dynamics" http://arxiv.org/abs/physics/0608040

has a brief discussion on this isssue (ordinary velocity, proper velocity, celerity, and rapidity - some of the different speed-related quantites one can define in SR).

[fixed]
Intergalactic spaceflight: an uncommon way to relativistic kinematics and dynamics
http://arxiv.org/abs/physics/0608040

PAllen

A possibly helpful analogy:

In pre-relativistic physics, if you are on a merry-go-round, distant objects go faster in your frame the further away they are, without bound. However, this type of non-inertial frame motion is not associated with momentum or any other physical quantity in the same simple way it is for an inertial frame. In SR, a rotating frame still has these general effects, and apparent (frame) speed >>c is not a problem because it isn't a relative velocity. If you expressed momentum in a a rotating frame, it would have have complex formula rather than be proportional to frame velocity.

In relativistic physics (SR for this purpose), a change in relative velocity is a 'boost' which has a lot in common with rotation in space-time. Continuous acceleration is then analogous to continuous rotation (in the x-t plane, for example). Viewed this way, it is not surprising to see >> c 'frame' speed of objects in accelerated frames.

BitWiz

Hi, pervect.

Another great link. Thank you. I'll respond on a couple of other things later, but wanted to take a minute to let you know how much I appreciate your help.

Thanks,
Chris

BitWiz

Thanks for the reply PAllen!
It's an interesting way to look at it. Using your analogy, are you're saying that lengthening the spokes has no more effect than my turning in place at the hub, that it's a perspective change rather than real target movement in a frame? If so, it seems that the length unit dimension of c has no meaning since it can assume any value > 0. What am I missing ...
I'm trying to get a handle on the four types of velocity in pervect's link (just above), and "rapidity" seems to be a good candidate for what you're describing. I'll come back to this after I've rested some overloaded synapses. Even so, maybe you can help me with this:

I'm troubled by the apparent lack of symmetry implied by a couple of replies in the case where a single observer accelerates toward a target. The Observer can measure four things: proper local time, apparent distance to the target, proper local acceleration, and Target Doppler.

When in the (single?) frame that includes *only* the Observer, the measuring devices, and the Target, as Observer accelerates he measures the Target's delta-Doppler, delta-distance and delta-v. Apparently none of these will agree with readings and integrations from his accelerometer and clock.(?) How does Observer distinguish between Lorentz effects and actual changes in acceleration by Target toward the Observer?

Thanks very much for wading in,
Chris

jartsa

Let's consider a spaceship with constant proper acceleration.

Accelerometers and clocks don't care anything about "motion" caused by Lorentz contraction.

Same is true for a device that measures relativistic Doppler shift.

Also an observer that was left standing on the launch pad does not observe Lorentz contraction.

These three things have the same opinion about the velocity of the spacesip.

(if we ask the guy standing on the launch pad: "what velocity does a person on the spaceship obtain, using his clock", he will answer: "on the spaceship the motion of the clock hand slowed down, while on the launch pad the motion of the hand of the velocity meter measuring the speed of the spaceship slowed down, so the velocity obtained using the onboard clock happens to be the real velocity, the same velocity that I am observing")