When accelerating at high %c, can apparent velocites exceed c?by BitWiz Tags: accelerating, apparent, exceed, velocites 

#1
Nov2912, 04:25 PM

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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 



#2
Nov3012, 04:12 AM

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#3
Nov3012, 05:40 AM

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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 lightyears. Measurements are about as real as it gets in science. 



#4
Nov3012, 08:41 AM

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When accelerating at high %c, can apparent velocites exceed c?
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 Earthframe 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 oneyear 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 doubleintegrating 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 onboard clock measure the same duration to accelerate to 87%c (given constant acceleration per the onboard accelerometer) that an Earthobserver would measure? I really need to understand this, and again, thank you very much for your time. Chris 



#5
Nov3012, 05:50 PM

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#6
Nov3012, 06:44 PM

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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 



#7
Nov3012, 07:27 PM

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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. 



#8
Nov3012, 08:21 PM

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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 



#9
Nov3012, 11:44 PM

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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. 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. 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. 



#10
Dec112, 04:02 AM

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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. 



#11
Dec112, 09:41 AM

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That's an amazing link. Thank you. 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 



#12
Dec112, 10:44 AM

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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 



#13
Dec412, 04:15 AM

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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: "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 speedrelated quantites one can define in SR). [fixed] Intergalactic spaceflight: an uncommon way to relativistic kinematics and dynamics http://arxiv.org/abs/physics/0608040 



#14
Dec412, 09:21 AM

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A possibly helpful analogy:
In prerelativistic physics, if you are on a merrygoround, distant objects go faster in your frame the further away they are, without bound. However, this type of noninertial 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 spacetime. Continuous acceleration is then analogous to continuous rotation (in the xt plane, for example). Viewed this way, it is not surprising to see >> c 'frame' speed of objects in accelerated frames. 



#15
Dec512, 08:52 PM

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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 



#16
Dec512, 09:38 PM

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Thanks for the reply PAllen!
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 deltaDoppler, deltadistance and deltav. 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 



#17
Dec812, 07:11 AM

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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") 



#18
Dec812, 01:45 PM

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Hi, jartsa,
Thanks for helping out. I very much appreciate it. For now I want to eliminate the Observer on Earth and deal only with Observers on the Rocket and Target. Initially, Rocket has a fixed relative velocity toward Target. This has occurred long enough that Target, observing Rocket, sees the same velocity. The situation is symmetric as each observer sees the same thing  the other moving directly toward them at the same rate. There are no other objects (Earth, background stars, etc). I believe both objects can be said to share the same inertial frame. They will also see the same Doppler. Their accelerometers read zero. Their clocks remain synchonized as far as each's ability to read the other's. Distance and velocity measurements to the other using parallax, subtended angle, or visual magnitude are identical. In other words, I'm trying to zero out as many things as possible, including eliminating all other objects besides Rocket, Target, and their measuring apparatuses (apparati? ;) ). Rocket accelerates. This "breaks" the inertial frame into two(?). Rocket sees instantaneous changes in accelerometer, Doppler, and distance measurements. Target will see these later. There are a lot of relativistic effects  distance compression, angular compression, time dilation, optical abberation et al  and many of these effects cancel out with respect to Rocket at constant velocity or constant acceleration, and some do not. So here's my question: How does Rocket know that what it observes using whatever measuring devices it has at it disposal, that any differences in what it measures which are at odds with Newtonian predictions, are due to relativistic effects, and are NOT due to coincidental motion changes initiated by Target? [EDIT 20121208 2036 I'm not trying to discredit Relativity, but find where it is (and isn't) symmetric between Rocket and Target.] Thanks! Chris 


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