Average Speed Challenge

Assuming:

• the star is on a parallel course to our sun,
• and it is perpendicular to the suns trajectory, and
• Ignoring time dilation.

It takes 4 times longer to get back than to get there for a total of 5 time periods. The average is: (.8C + .2C + .2C + .2C + .2C)/5 = .32C Wherever you calculate it it will be the same. Wherever you experience it it will be different.

3.2C

3.2C?

Ouch!! I hope Guest above meant 0.32c, which may perhaps be right...

SL

Hi Jorrie,

A good question!

At first, I was going to congratulate techno for his stripping the solution done to its basic elements ..... and he is certainly correct within the framework he established, in particular: ignoring time dilation.

However, you don't say that time dilation is ignored.

If I understand the recent "solution" of Einstein's Twin Paradox correctly, it would be impossible to state ahead of time what the average speed as calculated here on earth would be, since we can't calculate from anything you said, how much time dilation would have occurred here. That assumes that the "solution" is correct also. In which case Techno would be wrong when he says "where you calculate it stays the same".

Regards, Greg

Hi Greg, you wrote: "If I understand the recent "solution" of Einstein's Twin Paradox correctly, it would be impossible to state ahead of time what the average speed as calculated here on earth would be, since we can't calculate from anything you said, how much time dilation would have occurred here."

I intended all speeds and time dilations to be relative to us here on Earth, hence relativistically, we cannot speak of time dilation here! What is troublesome though, is where will Earth be at the time of return of the spaceship, as pointed out by Techno in his assumptions - and does it make any difference?

Regards, Jorrie

Hi Jorrie - I assumed for the purpose of the question the Earth and the distant star both treated as fixed points, otherwise we'd need to know velocities. I agree with various postings on 0.32c calculated here on Earth.

From point of view of the spaceship, I get time of outward trip down by factor 0.6 and return trip by ~ 0.98, (compared with Earth-based measurement) giving average speed 0.354c. Is this correct? I'm not too sure about it.

Hi Codemaster, given your assumptions, the answer is right. The only thing is, how good are those assumptions?

You also asked: "From point of view of the spaceship, I get time of outward trip down by factor 0.6 and return trip by ~ 0.98, (compared with Earth-based measurement) giving average speed 0.354c. Is this correct?"

No, I do not agree with that answer, unless you can prove it. Remember that not only the times, but also the distances are different in the ship's frame...

Regards, Jorrie

PS: If the same simplifying assumptions are made than for the earth frame, the answer in the ship's frame is in fact calculable - give it a bash!

Hi Jorrie, I assumed that velocity in spaceship frame would be Earth-frame divided by factors below, giving result below.

(I've just worked out how to insert images here, but unless I improve my technique, it's about as much trouble as it's worth)

If I assume both time and distance change, both so as to increase velocity, I get

But this is pretty much a guess, it's a long time since I was involved in special relativity, and I didn't understand it too well then!

But I believe that in spaceship frame, as v → c time → 0, so in principle can travel any distance in practically no time. This relates to the Twin Paradox, which I've seen a few refs to recently (ref to explaining it away). Am I right in thinking it's not a paradox at all, the postulated results are what would actually happen?

Also you haven't said yet whether everybody's right who said 0.32c for the original challenge, are you saving that for a bit?

Cheers...Codey

Hi again Jorrie, ref last bit of my post #14, just noticed you said 0.32c is correct in first line of #10. Apologies.

The change of time in the E=MC² as you approach the speed of light is based on the assumption that the speed of light is constant. What if it is not?

We see in the doppler effect that frequency changes dependant on speed of an object. With a frequency change the time period between cycles have changed.

If the doppler effect on light was great enough it could shift the frequency (and time between waves right out of our visible light spectrum. Dark matter. Then we would only be able to see matter going at a specific speed relative to our current speed.

Mabee it is not time that changes, only relative speeds and the perception of time.

Hi Codey, you said: "Also you haven't said yet whether everybody's right who said 0.32c for the original challenge, are you saving that for a bit?"

I think I did say in my last reply to you that, given the simplifying assumptions, the 0.32c is right... I also implied that part of the challenge is to know how valid the assumptions are...

In any case, the first set of formulae that you used above is right - you've got the time dilation factor right. The second set that you used is wrong - the process is a bit more complex than that. I will post a solution to this (which was not part of the challenge, actually) at some stage.

Answering your question: "This relates to the Twin Paradox, which I've seen a few refs to recently (ref to explaining it away). Am I right in thinking it's not a paradox at all, the postulated results are what would actually happen?"

The twins have never quite been a paradox - Einstein solved it around a century ago. The only problem is that us engineers (and some scientists, or shall I be bold and call them pseudo-scientists) sometimes find it difficult to wrap our heads around it! This is the way it turns out in experiment after experiment!

Regards, Jorrie

Hi Codey, here's a solution to the riddle from the spaceship's view, which I spoke about last time: "I will post a solution to this (which was not part of the challenge, actually) at some stage"

The time dilation factors of 0.6 and ~0.98 that you worked out is correct. For the ideal case, where the Earth remains motionless during the mission, they are applied in as follows:

v = (0.98+0.6)v₁v₂ / (0.98v₁+0.6v₂) = 0.28c

where v₁ = 0.8, the outbound speed relative to Earth and v₂ = 0.2.

The factor 0.98 + 0.6 = 1.58 represents the Lorentz contracted distance and the rest is the time dilation factor. A brute force calculation, taking the distance to the star as 1 ly, total distance 2 ly, gives a spaceship distance travelled of 1.58 ly.

The time dilated spaceship time for the trip is 5.65 years. Dividing the distance by time gives 0.28c. This answer is, obviously, distance independent.

Regards, Jorrie

Hi Jorrie,

"For the ideal case, where the Earth remains motionless during the mission, ..."

Now I'm lost. You didn't say that in the question.

I am not claiming expertise here. The last time I actually dealt with some of this (besides reading of course) was in college, and those problems were carefully crafted and somewhat simplified, as many college physics problems are (an extension of the "frictionless pulley" thing). I am sure we addressed the flaws in the "paradox", but I'll be damned if I can remember the details, other than involving a third observer I think. I do know it wasn't explained in the same way as the "solution" I read.

My understanding of the "solution" to the Twins Paradox, which you say was never really a paradox but a misunderstanding, is that you cannot measure time dilation by simply ascribing the separation velocity of two objects (earth and spaceship) as the "actual" velocity of one of them. Rather that to determine relative time dilation you must take into account the relative velocities of both objects in relation to the universe as a whole, or a statistical representation of at least the local portion of the universe.

What if the sun (and the earth along with it) is moving at .5X velocity away from the center of the universe, and the ship travels at .5X velocity toward the center of the universe, yielding a separation velocity of X? "Who" experiences "what" time dilation from each perspective? (I realize it is in some ways an oversimplification.)

Please explain the paradox "solution" a bit and where I am either correct or wrong.

Regards, Greg

Hi Greg, you wrote: "For the ideal case, where the Earth remains motionless during the mission, ..."

Now I'm lost. You didn't say that in the question."

Yes, but the context of my reply was that the 0.32c is correct if the simplifying assumption is made, with nothing said about the validity of the assumptions, I think…

For your questions about the twin paradox, I suggest you first read the following Blog entries:

http://cr4.globalspec.com/blogentry/1211/Rocket-Challenge-Solution#coments

where Masu pointed me to some article (I think it was reply #7) and my responses to that.

I have done a three-part mini-series on the twin paradox in my Blog here:

http://cr4.globalspec.com/blogentry/696/Paradoxes-of-Relativity-Part-2A-The-Twin-Paradox

I'll be more than happy to discuss it further, but it may be better if you read those refs and question the contents.

Regards, Jorrie

Hi Jorrie - first 2 paras of your post #20 - I spotted that after posting #14, see my #17.

Thanks for the calculation, but I'm even more puzzled now! I would have assumed the velocity measured by the spaceship would be greater than Earth measurement, approaching ∞ as v → c (as my #14) but your answer 0.28c is less.

I must get round to reading up your relativity publications sometime.

Cheers... Codey

Hi Codey, you wrote: "first 2 paras of your post #20 - I spotted that after posting #14, see my #17"

This happens a lot when posts follow shortly after one another - often, while I'm busy typing a reply, someone replies and corrects or solves the issue, while I'm busy to deal with the original reply (being interrupted a few times by work, spouse, children, friends...)

Your "I would have assumed the velocity measured by the spaceship would be greater than Earth measurement, approaching ∞ as v → c" is a common misconception. If only time was influenced by movement, this would have been so.

However, distances are equally influenced and hence the spaceship measures exactly the same constant speed as what we measure on Earth. Put differently, during the outbound trip, the spaceship would perceive Earth as moving away from it at the same 0.8c, just with an opposite sign.

Regards, Jorrie

PS: I'll post the 'official solution' in my Blog on Monday.

Hello Jorrie, thanks for that, interesting.

But am I right thinking that if an astronaut does a journey eg 10 lightyear, and back sufficiently close to c, time measured on his clock, and his biological processes, could be say 1 year? While his brother left on Earth would measure just over 20 years, hence the twin paradox.

Cheers....Codey

Hi Codey, you're perfectly right.

It will require a speed of ~0.999c, for which the Lorentz factor is ~20. The astronaut would reckon that the distance to the star as ~10/20 = 0.5 light-year and it took him ~1 year for the round trip.

Regards, Jorrie

It would make a difference since my assumptions assume that our current velocity (momentum) in the direction of the suns path is maintained (perfect vacuum) as we know there is many things in space that could affect our angular velocity.

If the straight line path was altered in any way, the distance would increase for that leg of the journey. We know that the return path would not be the same route because of the current speed of our solar system. Therefore the answer is only a best guess. Since the return path takes longer, it potentially would have more things affecting the trajectory than the outgoing trip. this would cause the return trip to be a longer distance. This adds more time periods at the lower speed and therfore the avereage speed would be less than the calculated answer of .32C

Hi techno, you wrote: "It would make a difference since my assumptions assume that our current velocity (momentum) in the direction of the suns path is maintained (perfect vacuum) as we know there is many things in space that could affect our angular velocity."

Good simplifying assumption, but difficult to connect with reality. What about we, here on Earth, beam a carrier frequency towards the spacecraft for the whole duration of the trip. Now let the spacecraft measure its redshift/blueshift relative to this signal and keep on adjusting it's speed to maintain a constant 0.8c or -0.2c by means of a spectrometer or some other receiver gadget.

Would this not be almost equivalent to the original simplifying assumptions, despite the fact that Earth's position and velocity might change?

Regards, Jorrie

True,

I was treating the vehicle as iif it was at a constant speed.

A real challenge would be to add the calculation of the doppler effect if the star was not perpendicular and parallel to our sun's trajectory. In this case the distance would have and the speeds of both solar systems would have to be known.

Ouch! my head hurts

I make it 0.32c assuming constant units of time and the distance is the same out and back.

Without thinking too much about it, I'd say that the average speed is the harmonic mean of the two speeds.

0.368C?

Cowering anonymously, LOL. This is not my area at all.

Since speed is a function of distance and time it makes sense "to me" to account for the change in time as perceived on Earth. (assuming the speeds were measured on the spaceship)

The fact that everyone is reading so deeply into this, makes me question if I'm interpreting it correctly. But you asked for the average speed/velocity ignoring distance, time dialation, relativity, etc. Off the top of my head....

[(Velocity2 - Velocity1)/2] + Velocity1

.5c

oops, I had those labeled wrong. To clarify...

(.8-.2)/2 + .2 =.5

Hi Guest, you wrote: "(.8-.2)/2 + .2 =.5"

I'm afraid that won't please Newton!

As many have concluded, in the simplified case, 0.32c is the correct value.

However, none ventured quantitatively into the practical case to any depth...

Regards, Jorrie