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nitzer219
2004-Mar-24, 04:58 AM
Thought you guys might enjoy this - I presented it as a thought experiment in a "Relativity" class (back when we used papyrus).

Anyway - I have forgotten the right answer, and hoping one of you math whizzes can refresh my sole-remaining neuron with it:

Assumption: The Space Shuttle can travel at lightspeed.

1. An electromagnetic video link-up exists between ground-control (gc), and the shuttle, such that the astronauts, and GC can see each other 'live' (almost instantaneous).

2. The shuttle takes off into orbit, then accelerates out of the solar system, first: near the speed of light; 2. Then ultimately achieves C^2.

Questions: At ground control...what happens to the video feed as the shuttle grows closer to c^2? It is time-delayed longer, the faster the shuttle goes, however...is there a formula to determine the geometric time dilation of the video feed?

It makes sense that AT C^2, the feed can never reach the earth again, or vice versa, correct? So...the issue is...what is the function mathematically for the time-delay?

Thanks in advance to those who reply! =D>

Lomitus
2004-Mar-24, 05:12 AM
Hey Mike,
I have no idea what the answer would be, but thats a really cool question! Mathmatically speaking, I can't even begin take a stab at it, but if Einstien's theories are correct and barring techilogical issues of the transmission (meaning that what ever the technology used, the feed would appear "live" on both ends), then one might assume that the video of the astronauts that ground control was seeing would appear in "slow motion" getting progressivly slower as the Shuttle aproached the speed of light. The same could be assumed of the reverse...that the astronauts would be seeing the people at mission control moving and speaking faster and faster and faster.

Ouch! Brain cramp! Guess thats why I'm a guitar player and not a mathmatician! LOL!!!

Bright Blessings,
Jim

Yumblie
2004-Mar-24, 06:15 AM
You mean just regular c and not c^2, right? Just regular c is the speed of light. Going c^2 is a whole different story.

The time dilation would use the good old factor gamma, sqrt(1-v^2/c^2). The time slowdown due to speed would be figured out by taking the "initial" time (in this case, the time at mission control) and multiplying it by gamma. It doesn't have much of an effect until you get very very close to c, then with each decimal place it gets exponentially higher. Going very close to c, the guys at mission control could end up waiting years for the guys board the shuttle to finish one word.

novaderrik
2004-Mar-24, 07:59 AM
if they had the same communication setup that Star Trek uses, they could hold normal conversations from halfway across the galaxy going warp 9.5...

Glom
2004-Mar-24, 11:38 AM
Hmm. Doppler shift: f' = sqrt((1-v²/c²)/(1+v²/c²)) × f

Assuming they were using UHF, which I believe is around E9 Hz, upon reaching v = c², f' = i Hz. So demodulating the signal would be a complex task.

AstroSmurf
2004-Mar-24, 12:06 PM
Glom, the relationship can be written on a simpler form:

f' = sqrt[(1-v/c)/(1+v/c)] x f

And you used v = c² ... :P

To return to the original question:

The shuttle would never reach lightspeed, no matter how powerful its engines are. To observers on the ground, its acceleration would decrease ever-so-slowly, and the downlink would also slow down ever-so-slowly, but never quite halt.

Observers on the shuttle would see the surrounding universe change shape, becoming ever "flatter" in the direction they're travelling. If they persist in accelerating, everything around them would coalesce into a thin disk. But they'll also be treated to a front-row view of the heat death of the universe...
Their uplink would behave in the same fashion as the downlink, becoming ever slower.

nitzer219
2004-Mar-24, 02:45 PM
You mean just regular c and not c^2, right? Just regular c is the speed of light. Going c^2 is a whole different story.

Yes Indeed! That is what I should have said correctly, Yumbie (I tried to warn you guys I'm at mathematical dolt...hence the post). :oops:

- Inferential Statistics...a bit rusty, but no problem, as it's required of my career. Anything above "Trig"...forget it. I stayed home those days.

Thanks very much for the catch!

I noticed too, that there is some dispute about the formula. I am just getting to know a prominent Theoretical Astrophysicist...so I'll see if he'll field it and reply....then the person with the closest answer can buy me a candy bar! :P

Thank you all for the responses - they're fascinating and you've collectively put my remaining neuron on a mini-treadmill!

AstroSmurf
2004-Mar-24, 03:03 PM
My formula and Glom's are equivalent, but mine is on a simpler form.

Nereid
2004-Mar-25, 02:18 AM
If the assumptions are inconsistent with GR (they are), then we're all free to write any answer we wish, and no one can gainsay us! :P

If we stick to known physics, then it's the same as communicating with Pioneer 10 and Pioneer 11, and Voyager 1 and Voyager 2, except that the frequency of the 'downlink' gets shifted to beyond what gc can detect on Earth ('ELF' - or is it 'ULF'? - radio gets absorbed inside the inner heliosphere).

JohnOwens
2004-Mar-25, 08:16 AM
Hmm. Doppler shift: f' = sqrt((1-v²/c²)/(1+v²/c²)) × f

Assuming they were using UHF, which I believe is around E9 Hz, upon reaching v = c², f' = i Hz. So demodulating the signal would be a complex task.

Glom, the relationship can be written on a simpler form:

f' = sqrt[(1-v/c)/(1+v/c)] x f
...
My formula and Glom's are equivalent, but mine is on a simpler form.
No, they are not at all equivalent. For a simple proof of this, assume v/c=1/2. Your formula will become
f' = sqrt[(1-1/2)/(1+1/2)] x f
f' = sqrt[(1/2)/(3/2)] x f
f' = sqrt(1/3) x f

Glom's will become
f' = sqrt[(1-v²/c²)/(1+v²/c²)] × f
f' = sqrt[(1-(1/2)^2)/(1+(1/2)^2)] x f
f' = sqrt[(1-1/4)/(1+1/4)] x f
f' = sqrt[(3/4)/(5/4)] x f
f' = sqrt(3/5) x f

Now just try and convince me that sqrt(1/3) = sqrt(3/5), I dare you. :wink:
This only proves that the two are not equivalent, not which one is correct, but as far as I can recall, I'm pretty sure Glom's is the right one.

And you used v = c² ... :P
OK, you've got this one right, certainly.

AstroSmurf
2004-Mar-25, 03:40 PM
Serves me right for doing sloppy math. Glom's formula is indeed not equivalent, but that's not my fault.

Derivation: The moving object is subject to time dilation, shifting its 'perceived transmission frequency' .
Further, it's moving away from us, adding straightforward Doppler shift to this.

So: (v counted positive away from observer)
f' = (dilation factor) × (doppler factor) × f
f' = sqrt(1 - v²/c²) × 1 / (1 + v / c) × f
f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f

Btw, if you run this experiment with e.g. a moving mirror and a laserbeam, the relativistic Doppler shift is applied twice.

daver
2004-Mar-25, 05:09 PM
f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f

Btw, if you run this experiment with e.g. a moving mirror and a laserbeam, the relativistic Doppler shift is applied twice.

AstroSmurf's equation matches what I remember and what the first site I googled posts.

JohnOwens
2004-Mar-25, 09:24 PM
So: (v counted positive away from observer)
f' = (dilation factor) × (doppler factor) × f
f' = sqrt(1 - v²/c²) × 1 / (1 + v / c) × f
f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f
Still not right here, assuming the third equation there was intended to be a simplification of the second, rather than an alternative possibility. (1 - v²/c²) != (1 - v/c). (1 - v²/c²) = (1 - (v/c)²), and in general (v/c)² != (v/c) (it is equal for v/c=0 or 1, of course). You can't simplify (1 - v²/c²) to (1 - v/c), only to (1 - (v/c)²).

Like I said before, though, I'm not real sure about which is the right one here. I know I remember a number of the formulae having v²/c² terms in them, but I'm not sure just which ones they were.

Added: Scratch that, I just found the right page in one of my notebooks. I was using beta in the equations, but unfortunately I was so comfortable with it at the time that I didn't bother to write down what it was, but I'm pretty sure it's beta = v/c. If that's correct, then what I have written is f' = f*sqrt((1 +/- beta) / (1 -/+ beta)), which would be equivalent to your f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f. I guess the squared beta was for other things. Unless I misremember and beta was v²/c², or something like that.

AstroSmurf
2004-Mar-26, 08:05 AM
So: (v counted positive away from observer)
f' = (dilation factor) × (doppler factor) × f
f' = sqrt(1 - v²/c²) × 1 / (1 + v / c) × f
f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f
Still not right here, assuming the third equation there was intended to be a simplification of the second, rather than an alternative possibility. (1 - v²/c²) != (1 - v/c). (1 - v²/c²) = (1 - (v/c)²), and in general (v/c)² != (v/c) (it is equal for v/c=0 or 1, of course). You can't simplify (1 - v²/c²) to (1 - v/c), only to (1 - (v/c)²).

No, but I can expand (1 - v²/c²) to (1 - v/c) × (1 + v/c) . The (1 + v/c) then cancel each other out :
sqrt(1 - v²/c²) × 1 / (1 + v / c) = sqrt[ (1 - v/c) × ( 1 + v/c) / (1 + v/c)² ] = sqrt[ (1 - v/c) / (1 + v/c) ]

Btw, interestingly enough, the Doppler shift factor for an object moving away is the inverse of the shift for one moving towards you with the same speed.

JohnOwens
2004-Mar-26, 08:30 AM
So: (v counted positive away from observer)
f' = (dilation factor) × (doppler factor) × f
f' = sqrt(1 - v²/c²) × 1 / (1 + v / c) × f
f' = sqrt[ (1 - v/c) / (1 + v/c) ] × f
Still not right here, assuming the third equation there was intended to be a simplification of the second, rather than an alternative possibility. (1 - v²/c²) != (1 - v/c). (1 - v²/c²) = (1 - (v/c)²), and in general (v/c)² != (v/c) (it is equal for v/c=0 or 1, of course). You can't simplify (1 - v²/c²) to (1 - v/c), only to (1 - (v/c)²).

No, but I can expand (1 - v²/c²) to (1 - v/c) × (1 + v/c) . The (1 + v/c) then cancel each other out :
sqrt(1 - v²/c²) × 1 / (1 + v / c) = sqrt[ (1 - v/c) × ( 1 + v/c) / (1 + v/c)² ] = sqrt[ (1 - v/c) / (1 + v/c) ]
OK, now I finally see where I messed up! :) :oops: I misparsed the sqrt in the second equation as being around the whole of [(1 - v²/c²) × 1 / (1 + v / c)], which, of course, it isn't; it only applies to (1 - v²/c²). Glad we finally got that worked out. =D>

Glom
2004-Mar-26, 12:54 PM
So who was right?

JohnOwens
2004-Mar-26, 01:15 PM
So who was right?
LOL, I think Astrosmurf was right, and you and I were wrong? I know I was misunderstanding what that one square root applied to, at least. :-k
Hmm, I'm not sure now, was my f' = f*sqrt((1 +/- beta) / (1 -/+ beta)) correct as well? I think so.... :o

nitzer219
2004-Mar-27, 12:58 AM
Just for another perspective - since I am not qualified to evaluate the mathematics of the thought experiment, as previously mentioned, I am going to send this to an astrophysicist.

Hopefully I will be able to post his proof in a day or two.

nitzer219
2004-Mar-31, 02:23 PM
Eroica posed a mathematical proof on another thread for this: "Relativistic Jerk" under Gen. Astronomy.

However, I was curious about the following, as I'm truly under-qualified to address the mathematics of the problem - maybe putting it in a physically plausible context would make it possible to solve?

Eroica -

Question:

Let's assume that neither the transmission (video feed to ground control (gc)), nor the ship can actually reach c, since as far as we know now, they can't.

Is this then, a correct assumption?

1. The shuttle, fractionally, exceeds the speed of the transmission.
2. The transmission would not equal 'c', as it's not in a complete vacuum (there is matter in space, so c is not going to be attained totally).
3. If the ship exceeds the speed of the transmission, would not the transmission fail to ever reach earth again, at the point the shuttle surpasses the speed of the transmission?

I'm still waiting on a formal reply from the guru on this...I have to believe this is solvable. Thanks for seeking out feedback for the answer.

Although I thought this nightmare up, I've never had the mathematical experience, practice, or knowledge to prove it...so I'm in my own kind of hell with this one...and what the heck..I figured...why not share my hell with others.
_________________
Mike
____________________
There is a valid theory to Doomsayers' claims: It's called, "The PT Barnum" theory.

Nylex
2004-Mar-31, 03:22 PM
So who was right?

Is that Josh?

milli360
2004-Mar-31, 05:21 PM
3. If the ship exceeds the speed of the transmission, would not the transmission fail to ever reach earth again, at the point the shuttle surpasses the speed of the transmission?
I don't see why it would. Why would it?

nitzer219
2004-Apr-01, 04:10 AM
[quote=nitzer219]

I don't see why it would. Why would it?

Let me be more specific. I think upon further thought, the following would be correct:

1. If the speed of the shuttle surpasses that of the speed of two-way transmission, then transmissions TO the shuttle would never again be reached.

2. Transmissions FROM the shuttle would still reach Earth, although, time dilation principles, I believe, would still apply.

milli360
2004-Apr-01, 11:26 AM
Let me be more specific. I think upon further thought, the following would be correct:

1. If the speed of the shuttle surpasses that of the speed of two-way transmission, then transmissions TO the shuttle would never again be reached.
At some point, you have to assume that the transmission is by radio/light, don't you?

2. Transmissions FROM the shuttle would still reach Earth, although, time dilation principles, I believe, would still apply.
That's what I was getting at, but doesn't the same apply TO also?

JohnOwens
2004-Apr-02, 12:57 AM
I'd guess that there might be something about the speed, relative to the observer, of the medium that reduces c, which might affect this one way or another. After all, well, let's call the speed of light in the interstellar medium c', and your velocity relative to the Earth (or Earth's velocity relative to you) v. At this point, c' &lt; v &lt; c. But the interstellar medium seems to be moving past you at v as well. I certainly won't swear by this, but it seems to me that this means the medium might not slow down the light as much in that direction, from your (the Shuttle's) point of view, so c' in that direction might be higher than v. On the other hand, the medium would also seem much denser due to dilation, but the apparent distance would be similarly reduced.