PDA

View Full Version : Relativistic gas mileage



violentquaker
2009-Nov-11, 05:12 AM
If I understand relativity correctly, a fast traveler between two points sharing an inertial frame will perceive his trip as taking less time than the time as measured by an observer in the rest frame.

Q1: Is this accurate?
Q2: If the traveler is a spacecraft between two stars, does this mean the spacecraft only needs enough fuel for the (shorter) time as perceived by the traveler?
Q2a: How does this fit with the first law of thermodynamics?
Q3: If A1 and A2 are yes, is there theoretically a point where additional acceleration actually reduces the total fuel expenditure due to time contraction (imagining that the acceleration comes from some sort of reaction mass like a rocket)?

Sorry if the questions are dumb; this relativity thing is a little counter-intuitive for me.:shifty:

Jens
2009-Nov-11, 05:20 AM
I don't understand it that well either! But this may be a possible solution. The observer will see the time decreasing, but the mass increasing, so it may be that the observer will see the rocket using a heavier mass of fuel but for a shorter time, leading to the same total?

WayneFrancis
2009-Nov-11, 06:56 AM
If I understand relativity correctly, a fast traveler between two points sharing an inertial frame will perceive his trip as taking less time than the time as measured by an observer in the rest frame.

Q1: Is this accurate?
Q2: If the traveler is a spacecraft between two stars, does this mean the spacecraft only needs enough fuel for the (shorter) time as perceived by the traveler?
Q2a: How does this fit with the first law of thermodynamics?
Q3: If A1 and A2 are yes, is there theoretically a point where additional acceleration actually reduces the total fuel expenditure due to time contraction (imagining that the acceleration comes from some sort of reaction mass like a rocket)?

Sorry if the questions are dumb; this relativity thing is a little counter-intuitive for me.:shifty:

Correct an observer travelling in a vehicle will perceive their trip shorter then an external observer. Now the problem we have is just because you get a certain mileage from your fuel in one frame does not mean that you will get the same mileage no matter what.

There are other problems. Such as travelling from place to place only requires energy expenditure for accelerating and decelerating. The fact that we have to dump even more energy to overcome friction is an artefact of the world we live in.

Basically the equations work just fine if you are in empty space without any friction. There is no paradox of "if I go faster and space & time dilate more will I save fuel" because the answer is no because you expend the energy getting to that frame not while you are in that frame.

I.E. If I want to make a trip to another star and say that I want to travel at .5c the fuel use is in the acceleration and then you have to use the calculations for fuel use that say the faster you are going the more energy you have to use to go faster per acceleration rate.

IE It takes more energy to go from .4c to .5c then it does .3c to .4c

Once at .5c you wouldn't be burning your rocket any more, because that would continue accelerating you and then the "an object in motion tends to stay in motion" comes into effect. So you coast along until you determine you need to start slowing down then you would expend the same amount of energy you used in accelerating to stop yourself.

So even in optimal situations you don't save fuel by going faster.

WayneFrancis
2009-Nov-11, 06:58 AM
I don't understand it that well either! But this may be a possible solution. The observer will see the time decreasing, but the mass increasing, so it may be that the observer will see the rocket using a heavier mass of fuel but for a shorter time, leading to the same total?

What will happen is the observer in the vehicle will think they are accelerating at rate of x but to an external observer the rate will slowly go down.

violentquaker
2009-Nov-11, 08:33 AM
Thanks for the response!


Correct an observer travelling in a vehicle will perceive their trip shorter then an external observer. Now the problem we have is just because you get a certain mileage from your fuel in one frame does not mean that you will get the same mileage no matter what.

I realize "mileage" is not a standard astronautical term, but technically wouldn't the "fuel expenditure per unit distance" be the same to observers in both frames while the "fuel expenditure per unit time" would be different? Once our spaceship decelerates back to its initial reference frame, I would imagine that both observers would agree on the mass of the spaceship and therefore the total mass expended.

Jens
2009-Nov-11, 08:40 AM
I realize "mileage" is not a standard astronautical term, but technically wouldn't the "fuel expenditure per unit distance" be the same to observers in both frames while the "fuel expenditure per unit time" would be different? Once our spaceship decelerates back to its initial reference frame, I would imagine that both observers would agree on the mass of the spaceship and therefore the total mass expended.

One thing to consider is that velocity has nothing to do with the frame of reference. It's only the acceleration. So it's when the spaceship is accelerating or decelerating that things will happen, not when it's going at a different velocity.

speedfreek
2009-Nov-11, 07:45 PM
How much fuel required largely depends on whether you want to slow down when you reach your destination, or not.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html has a very good explanation of how this all works.

Simply put, at relativistic speeds you need a whole lot more fuel to slow you down than you did to speed you up, due to the increase in your relativistic mass.

WayneFrancis
2009-Nov-11, 10:46 PM
I'll have the read that page speedfreek but the logic doesn't sound right.

While the rocket will have a higher relativistic mass, to an outside frame, and need to slow that down it had to put that energy in during its acceleration.

If the start and destination are both in the same frame then the same amount of energy should be used. This is because the 2 vectors should be exact opposites in a both the local frame and any frame not moving with respect to the start and end points. The total acceleration on both end, because slowing down is essentially an acceleration in the opposite direction, should be identical. To the rocket it never picks up any mass so relative to itself.

Woops even worse slowing down should take less fuel...as you've got less total relative mass at the end. IE you've burned half your fuel so your ship will be, in the local frame lighter thus easier to accelerate in the opposite direction.

speedfreek
2009-Nov-11, 11:16 PM
Yes I did put it badly. What I should have said was you need a whole lot more fuel if you want to come to rest at your destination than if you want to accelerate all the way there and sail past.

If you want to accelerate at 1g all the way to Alpha Centauri and sail past (at 0.77c!), you would only use approximately 1/4 of the fuel required to accelerate at 1g to the halfway point and then decelerate at 1g to come to rest when you arrived.

The article I linked explains it better than I can! :)

worzel
2009-Nov-12, 12:17 AM
I realize "mileage" is not a standard astronautical term, but technically wouldn't the "fuel expenditure per unit distance" be the same to observers in both frames while the "fuel expenditure per unit time" would be different?
As others have said, it is the acceleration that takes energy, not the coasting. In other words, you could travel 1 light year, 10 light years, or 100, and still use exactly the same fuel by first accelerating to .5c and then decelerating to 0c again when you get there. The only difference is how long you coast, which uses no fuel. So there is no meaningful fuel expenditure per unit distance or time.

mugaliens
2009-Nov-12, 08:27 PM
Q2: If the traveler is a spacecraft between two stars, does this mean the spacecraft only needs enough fuel for the (shorter) time as perceived by the traveler?

The decrease in perceived time is offset by the relativistic increase in mass.

worzel
2009-Nov-13, 06:30 AM
The decrease in perceived time is offset by the relativistic increase in mass.
But the traveller doesn't perceive any increase in their mass.

NorthernBoy
2009-Nov-13, 11:38 AM
While the rocket will have a higher relativistic mass, to an outside frame,

Relativistic mass seems to be creeping in to more and more discussions round here, so maybe this is a good point to remind everyone that it is not a concept that tends to be employed in physics nowadays, for several reasons. The best one of these is perhaps that we already have a very good word for this quantity (when, of course, we select the appropriate units for length and time), and that word is "Energy".

In the relativistic world in which I worked (particle accelerators), "mass" has meant "rest mass" for many years now.