Your photonic rocket reaches planet A, 10 light-years away in 10 years.

In summary, you will reach planetary system A in 10.0000... years, as measured by your clock or calendar, but not in 10 years as measured by an external observer.
  • #1
danR
352
4
I've been posing this question to physicist folk for 10 years, mySpace 6 years ago, and here (don't know when, when I had an older account), and I see that others bring it up on occasion. I will refine it to meet current objections.

You have a photonic drive spaceship. It has a (classical) mass ratio (fueled to empty weight) sufficient for it to classically reach or exceed its exhaust velocity.

Claim: You (as far as you're concerned, not an external observer) will reach planetary system A, 10 light-years away, in 10.0000... years (or even 9.9 years).

So, when your onboard calendar/clock, not some external observer's, reaches 10 years on the dot, will you be at planetary star system A?

[I have stripped the usual baggage from this question. It's not asking about ultimate speed, but arrival time, a distinct physical and temporal terminus ad quem. It's not about mass increase in antimatter fuel, hence getting back the energy investment dividends to re-expend in infinity-busting thrust. No, it's not about that. It's not about external observers. I don't care whether you look flatter, or your clock is slow, or your mass is on steroids. And just to be clear, you turn your ray-gun on all external beings in the universe. So you have none to bring into this gedankenexperiment. This universe has only one person, you. In your spaceship.]

And if there's any flaw in my exposition, but not the argument, I trust someone will repair it, then satisfactorily refute the corrected argument.
 
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  • #2
For that matter, I could almost have posted this in 'Classical' Physics section, though it would probably be move here.
 
  • #3
Sure, I'll bite. :wink:

Your terminology is ambiguous (and in so far as there is any "problem" involved in what you're stating, that's where it's coming from), so I'll clarify how I'm interpreting things as I go.

danR said:
You have a photonic drive spaceship. It has a (classical) mass ratio (fueled to empty weight) sufficient for it to classically reach or exceed its exhaust velocity.

By "photonic drive", I'm assuming you mean a drive with an "exhaust velocity" of c, the speed of light; in other words, the rocket emits photons out the back, and thereby drives itself forward. Since nothing can go faster than light, the rocket can't exceed its exhaust velocity; and since the rocket is (presumably, since it's carrying humans) made of matter, not radiation, it can't reach c either, so it can't *reach* its exhaust velocity. So the bit in your exposition about "sufficient to reach or exceed its exhaust velocity" is impossible as stated, but it doesn't matter; the problem can be analyzed by just removing that statement, so I'll consider it removed.

A-wal said:
Claim: You (as far as you're concerned, not an external observer) will reach planetary system A, 10 light-years away, in 10.0000... years (or even 9.9 years).

By "10 light-years away" I assume you mean "10 light-years away in the mutual rest frame of the Earth and planet A" (for simplicity we'll assume that Earth and planet A are in fact at rest relative to one another). However, in order to make your claim true, I *cannot* also interpret "in 10.0000... years (or even 9.9 years)" as "in 10 years (or 9.9 years) as measured in the rest frame of Earth and planet A", because, as I suspect you are already aware, if I interpret it that way the claim is false. As noted above, the rocket can't reach the speed of light, so to cover 10 light-years in the rest frame of Earth and planet A, the rocket must take *more* than 10 years as measured in that frame. So basically, you're using distance as measured in one frame and time as measured in a different frame, which is of course going to give you meaningless results.

A-wal said:
So, when your onboard calendar/clock, not some external observer's, reaches 10 years on the dot, will you be at planetary star system A?

Yes, of course, since (given the interpretation of the claim above that I described) the time is being measured in your frame, aboard the spacecraft , *not* in the Earth/planet A frame. But to link this time meaningfully to a distance, you have to measure distance in the same frame (yours, aboard the rocket) as you're measuring time. In your frame, aboard the rocket, the distance between Earth and planet A is not 10 light-years; it's shorter, because of length contraction. So you, in the rocket, observe planet A flying towards you at some high speed, such that it takes less than 10 years for it to reach you--but it's also covering less than 10 light-years to do it, so planet A never reaches the speed of light in your frame, just as you never reach the speed of light in the Earth/planet A frame.
 
  • #4
danR said:
I've been posing this question to physicist folk for 10 years, mySpace 6 years ago, and here (don't know when, when I had an older account), and I see that others bring it up on occasion. I will refine it to meet current objections.

You have a photonic drive spaceship. It has a (classical) mass ratio (fueled to empty weight) sufficient for it to classically reach or exceed its exhaust velocity.

Claim: You (as far as you're concerned, not an external observer) will reach planetary system A, 10 light-years away, in 10.0000... years (or even 9.9 years).

So, when your onboard calendar/clock, not some external observer's, reaches 10 years on the dot, will you be at planetary star system A?

[I have stripped the usual baggage from this question. It's not asking about ultimate speed, but arrival time, a distinct physical and temporal terminus ad quem. It's not about mass increase in antimatter fuel, hence getting back the energy investment dividends to re-expend in infinity-busting thrust. No, it's not about that. It's not about external observers. I don't care whether you look flatter, or your clock is slow, or your mass is on steroids. And just to be clear, you turn your ray-gun on all external beings in the universe. So you have none to bring into this gedankenexperiment. This universe has only one person, you. In your spaceship.]

And if there's any flaw in my exposition, but not the argument, I trust someone will repair it, then satisfactorily refute the corrected argument.

If the 10 light years to the planetary system is measured by observers at rest with respect to the planetary system, you will reach it way before your own clock shows 10 years.

A separate issue is that for a photonic rocket, nothing will change the fact that the rocket will never move at speed greater than c as observed by any inertial observer.
 
  • #5
In all this time, you've never run across the "relativistic rocket equations?

http://www.desy.de/user/projects/Physics/Relativity/SR/rocket.html, for example.

It's got a number of worked examples. For instance you could reach the center of the galaxy, 30,000 light years away, flipping over at midpoint so you arrive at a low velocity, in about 20 years of proper time (which is equivalent to your onboard calendar or watch) if you could accelerate continuously with a proper acceleration of 1g.

With ideal efficiency this would take 955,000 tonnes of matter/antimatter fuel for a 1kg payload.

You could cut the time roughly in half if you just want to do a flyby, and reduce the fuel requirements as well, to only 62 tonnes / kg of payload.
 
  • #6
danR said:
I've been posing this question to physicist folk for 10 years, mySpace 6 years ago, and here (don't know when, when I had an older account), and I see that others bring it up on occasion. I will refine it to meet current objections.

You have a photonic drive spaceship. It has a (classical) mass ratio (fueled to empty weight) sufficient for it to classically reach or exceed its exhaust velocity.

Claim: You (as far as you're concerned, not an external observer) will reach planetary system A, 10 light-years away, in 10.0000... years (or even 9.9 years).

So, when your onboard calendar/clock, not some external observer's, reaches 10 years on the dot, will you be at planetary star system A?

[I have stripped the usual baggage from this question. It's not asking about ultimate speed, but arrival time, a distinct physical and temporal terminus ad quem. It's not about mass increase in antimatter fuel, hence getting back the energy investment dividends to re-expend in infinity-busting thrust. No, it's not about that. It's not about external observers. I don't care whether you look flatter, or your clock is slow, or your mass is on steroids. And just to be clear, you turn your ray-gun on all external beings in the universe. So you have none to bring into this gedankenexperiment. This universe has only one person, you. In your spaceship.]

And if there's any flaw in my exposition, but not the argument, I trust someone will repair it, then satisfactorily refute the corrected argument.
Your question can have only one meaningful answer: Yes.

We have to assume that when you specify the distance to the planetary system, you have not yet turned on your thrusters to get there, in other words, you start out at rest with respect to your destination. This being the case, and assuming that you have all the technology available to carry out your mission, you can select any time you want to take to get there. If you want to get there in exactly 10 years or 9.9 years or any other time, there is an acceleration profile and a cruising velocity that will get you there in that specified length of time according to "your onboard calendar/clock, not some external observer's", as you specified.

This should be an obvious answer because if you accelerated instantly to near the speed of light, you would get there in a time approaching zero. If instead you accelerated just a tiny bit so that you were going incredibly slow, you could take a time approaching infinity. Therefore, somewhere between these two extremes is a speed that will get you there in any time you desire.

This is true no matter what the distance is that you have to travel. You can make any trip to any place take any amount of time you want.

I took a first cut at a formula to tell you what speed you have to travel at to go a distance d (in light-years) in a time t (in years) or any other compatible units you want:

1 / √[1 + (t/d)²]

I haven't checked this but it looks right.
 
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  • #7
ghwellsjr said:
I took a first cut at a formula to tell you what speed you have to travel at to go a distance d (in light-years) in a time t (in years) or any other compatible units you want:

1 / √[1 + (t/d)²]

I haven't checked this but it looks right.
I have confirmed that this formula is correct but I should have mentioned that the units for speed are light-years/year or a fraction of the speed of light (the same as beta).

So for your example, you want to travel 10 light-years in 10 years. Here's the calculation of 1 / √[1 + (t/d)²] with t=10 and d=10:

1 / √[1 + (10/10)²]
1 / √[1 + (1)²]
1 / √[1 + 1]
1 / √[2]
1 / 1.414
0.707

So at a speed (β) of 70.7% of the speed of light we can calculate gamma as follows:

1 / √[1 - β²]
1 / √[1 - 0.707²]
1 / √[1 - 0.5]
1 / √[0.5]
1 / 0.707
1.414

This means that in the original frame where you were at rest and your destination was 10 light-years away and you were traveling toward it at 0.707c it will take:

t = d/v
t = 10/0.707
t = 14.14 years

But since your time is dilated by the factor of 1/gamma, it will take you:

t = 14.14 * [1/1.414]
t = 10 years

You can also see by the formula that for very short times, the speed approaches 1 and for very long times the speed approaches d/t.
 
  • #8
Very thorough discussions. Thanks all.

I used to ask a more paradox question, as the wording may suggest, but that one seems to have been brought up and answered several times on PP; something like this:

If you're investing more energy as mass in your matter/antimatter fuel, you should get an energy return for your drive (as, say, more energetic exhaust photons). The implication being that external observers also see you get promptly to your destination, even at velocity c (or faster, even if it does entail disappearing as burst of tachyons or going backward in time, or into a spacio-temporal worm-hole, or whatever)
 
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  • #9
I don't know why you think there is an implication that external observers will see you getting promptly to your destination. If you are talking about an external observer that was present with you at the start of your trip, then he can assume that it takes you 14.14 years to arrive at your target 10 light-years away because your speed is 0.707c. I calculated this result for you as the second to last calculation in post #7. However, he won't see you arrive for another 10 years since you are 10 light-years away. So he won't see you arrive until 24.14 years after you leave but you will experience only 10 years.

I hope you aren't getting confused into thinking that since you have traveled 10 light-years in 10 years, you are traveling at the speed of light. Or if you got there in 9.9 years you would be traveling faster than the speed of light. In the first case, as I said in post #7, you would only be traveling at 0.707 of the speed of light. If you were going a little faster, 0.709 c, you could make it in 9.9 years.

Keep in mind that I'm assuming that you are accelerating at a very high rate just at the start of your trip to get your speed up to what my formula calculates and then you coast the rest of the trip. If you want to accelerate over a longer period of time at a lower rate then you will have to go faster to get there in the desired time.
 
  • #10
ghwellsjr said:
I don't know why you think there is an implication that external observers will see you getting promptly to your destination. If you are talking about an external observer that was present with you at the start of your trip, then he can assume that it takes you 14.14 years to arrive at your target 10 light-years away because your speed is 0.707c. I calculated this result for you as the second to last calculation in post #7. However, he won't see you arrive for another 10 years since you are 10 light-years away. So he won't see you arrive until 24.14 years after you leave but you will experience only 10 years.

I hope you aren't getting confused into thinking that since you have traveled 10 light-years in 10 years, you are traveling at the speed of light. Or if you got there in 9.9 years you would be traveling faster than the speed of light. In the first case, as I said in post #7, you would only be traveling at 0.707 of the speed of light. If you were going a little faster, 0.709 c, you could make it in 9.9 years.

Keep in mind that I'm assuming that you are accelerating at a very high rate just at the start of your trip to get your speed up to what my formula calculates and then you coast the rest of the trip. If you want to accelerate over a longer period of time at a lower rate then you will have to go faster to get there in the desired time.

This is good, yes, and I take it these calculations hold regardless of whether I have a drive that can cheat the relativistic parasitic mass/energy entailed by going faster, to an observer, than 0c?

The way the older question was put was: why can't we actually exploit the 'parasitic' relativistic mass increase of the fuel as more energetic photons? This is not a chemical rocket after all, or an electron accelerated by external magnetic fields in a linear accelerator. Rather, they are imparting more and more momentum (as far as an external observer can see, or as long as he can survive the stray photon leakage that is approaching Planck's length as the pilot approaches c) as we go faster and faster to our infinitanium drive-plate mirrors (a mere engineering problem).

The current explanation is that 'mass' increase has gone out of favour for 'energy' increase. Presumably the hydrogen/antihydrogen yield remains the same electron volts to both pilot and observer.
 
  • #11
danR said:
The way the older question was put was: why can't we actually exploit the 'parasitic' relativistic mass increase of the fuel as more energetic photons? This is not a chemical rocket after all, or an electron accelerated by external magnetic fields in a linear accelerator. Rather, they are imparting more and more momentum (as far as an external observer can see, or as long as he can survive the stray photon leakage that is approaching Planck's length as the pilot approaches c) as we go faster and faster to our infinitanium drive-plate mirrors (a mere engineering problem).

I'm not sure why you think the "mass increase" makes any difference. In the rest frame of Earth/planet A, the rocket's "mass" does increase (I put "mass" in quotes because, as you note, referring to this increased observed energy as "mass" is not recommended because it causes confusion), but that just means the observed acceleration of the rocket (the actual change in velocity for a given amount of thrust) goes down, approaching zero as the velocity approaches c (so the velocity never actually reaches c, no matter how much thrust you expend). In the rest frame of the rocket, there is *no* mass increase, and your engine has to "burn" the same amount of fuel to make the same thrust during the entire trip; the "mass increase" of the fuel in the rest frame of Earth/planet A doesn't make any difference to the thrust, because the thrust is generated in the rest frame of the rocket, so the fuel mass/energy in that frame is what matters.
 
  • #12
PeterDonis said:
In the rest frame of Earth/planet A, the rocket's "mass" does increase...

No it doesn't. In any rocket that carries its fuel with it, all the kinetic energy the rocket will ever have already exists as chemical (or matter/antimatter) energy in the fuel before takeoff. During flight, some of this fuel is exhausted out the back, meaning that the total energy content of the rocket (="Relativistic Mass") decreases with time.

Reason #89 why "Relativistic Mass" is the enemy of pedagogy.
 
  • #13
The point is, that it won't matter what frame you work the problem in, as long as you work it out correctly.

So, changing your viewpoint is NOT going to make the rocket perform any better - or any worse. It will do what it will do. You can't "beat the system" by working the problem out in a different frame and getting a "better answer".

One can either spend a lot of time fooling with frame-dependent quantities, like "relativistic mass", and hope one doesn't make any mistakes in converting between frames. OR, one can, instead, work the problem in frame-independent terms, as much as possible, and not have to worry about making sure that one has accounted for the manner in which frame-dependent things change, and have a much greater chance of getting the right answer. And, with less effort to boot, once one becomes familiar with the frame-independent techniques.
 
  • #14
ZikZak said:
No it doesn't. In any rocket that carries its fuel with it, all the kinetic energy the rocket will ever have already exists as chemical (or matter/antimatter) energy in the fuel before takeoff. During flight, some of this fuel is exhausted out the back, meaning that the total energy content of the rocket (="Relativistic Mass") decreases with time.

Reason #89 why "Relativistic Mass" is the enemy of pedagogy.

Oops, good point! (And true even of a "photonic rocket", for which technically is isn't the "fuel" that's exhausted out the back, since the energy in the photons comes from the fuel.)
 
  • #15
ZikZak said:
No it doesn't. In any rocket that carries its fuel with it, all the kinetic energy the rocket will ever have already exists as chemical (or matter/antimatter) energy in the fuel before takeoff. During flight, some of this fuel is exhausted out the back, meaning that the total energy content of the rocket (="Relativistic Mass") decreases with time.

Reason #89 why "Relativistic Mass" is the enemy of pedagogy.

Could we briefly say then that to an external (earth, planet A, midpoint) observer, there is no real interest, simple or compound, on the matter/antimatter dollar, only relativistic inflation?

And we would say that mid-point observer seeing the ship go past at fuel exhaustion will not see any higher energy stray-leakage photons than the Earth observer saw when the ship blasted off?

Of course, I've always felt that otherwise, there would be conservation of energy problem.
 

1. How is it possible for the photonic rocket to travel 10 light-years in only 10 years?

The photonic rocket utilizes the properties of light, specifically the speed of light, to travel at incredibly high speeds. This allows it to cover large distances in a relatively short amount of time.

2. What is a photonic rocket made of?

A photonic rocket is made of lightweight materials that are able to withstand the intense heat and pressure generated by the propulsion system. It also contains highly reflective mirrors that help to direct and focus the photons in the desired direction.

3. How is the photonic rocket powered?

The photonic rocket is powered by a laser or other high-energy light source. This light source is directed at the reflective mirrors, which reflect and amplify the photons to create a powerful beam of light that propels the rocket forward.

4. What makes the photonic rocket a more efficient form of space travel?

Unlike traditional rocket engines that require fuel, the photonic rocket does not need to carry any extra weight, making it much more efficient. Additionally, since it travels at the speed of light, it can cover large distances in a shorter amount of time, reducing travel time and increasing efficiency.

5. How does the photonic rocket navigate and steer towards its destination?

The photonic rocket is equipped with advanced navigation and guidance systems that use sensors and computer algorithms to calculate the trajectory and make adjustments as needed. The reflective mirrors also play a crucial role in steering the rocket towards its destination by directing the photons in the desired direction.

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