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Velocity addition formula

  1. Feb 9, 2015 #1
    Imagine two space ships each with thrusters that can accelerate them .1c or .3c almost instantly. The velocity addition formula indicates that if one space ship applied the .1c thruster 3 times then if c is estimated to be 300,000,000 m/s the velocity of the ship after the accelerations the ship will be approximately 87,669,903 m/s. Which is less than the 90,000,000 m/s which would have been achieved using the .3c thruster. So does this mean that the energy needed to go .1c in space < (energy needed to go .3c in space)/3?

    Also given the formula (u + v) / (1 + ((u*v)/(c*c)))

    Is it saying it is more expensive to try to go .1c then .4c or .4c then .1c than go .2c then .3c or .3c then .2c (because u + v = .5c in both cases, but the u * v = .04c with the .1c and the .4c, but .06c with the .2c and the .3c) but you would achieve a higher velocity?
     
    Last edited: Feb 9, 2015
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  3. Feb 9, 2015 #2

    DEvens

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    I really do not know if it is saying that because I cannot understand what you said.

    What it is asking is, is the energy required to accelerate to 0.1C less than one third the energy required to accelerate to 0.3C?

    It's kind of a poor question because you can't really get it from the information provided. In order to compare energy for the different things you would need to calculate energy. And you can't get that from the velocity composition formula.
     
  4. Feb 9, 2015 #3
    The question in the text previous to the text you quoted was about whether it required less energy to go accelerate to 0.1c was less than one third the energy required to accelerate 0.3c. The text you quoted was about whether it required less energy to try to accelerate 0.4c and then 0.1c _or_ 0.1c then 0.4c vs. trying to accelerate 0.2c and then 0.3c _or_ 0.3c then 0.2c. I didn't realise the question was so poor, I thought there might be a general principle like it requiring more energy to achieve a greater velocity. So you could look at the velocity achieved by using the equation and then make statement about whether there was a lower energy route to achieving the lower velocity than the lowest energy route for achieving the higher velocity. Was I mistaken?
     
  5. Feb 9, 2015 #4

    DEvens

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    The velocity composition formula does not tell you anything about the energy required to achieve those velocities.

    Also, the question asks about the relative energies of velocities 0.1 and 0.3. But the information given refers to velocities 0.1 and 0.292 (the speed just less than 90 million m/s). So it's not the right speeds. And it does not tell you what energy is required of the engines to achieve those speeds.

    And in order to properly understand the energy required to achieve a given speed you need to be doing the full calculation. Even at non-relativistic velocity there are surprises. Example: If you have a rocket motor that lets you add 1 m/s to your speed, and you accelerate from 0 m/s to 1 m/s, and your ship masses 100 kg, then you added 1/2 mv^2 = 50 Joules of kinetic energy. But if you start at 1 m/s, and do the same thing going from 1 m/s to 2 m/s, you have gone from 50 Joules at 1 m/s,
    to 200 Joules at 2 m/s. And so you have added 150 Joules to your kinetic energy. How can that be? How can your rocket motor give you three times as much kinetic energy doing the same thing? Once you figure that one out, then try the relativistic question again.
     
  6. Feb 9, 2015 #5

    PeterDonis

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    Since the energy per unit rest mass can be computed from the velocity, knowing the velocities does tell you something about the energy required. The problem is that the OP is making an incorrect assumption about the relationship between velocity and energy. See below.

    Note that for this to be true, the 0.1c thruster, when it fires the second and third times, must change its ship's speed by 0.1c relative to its current rest frame, not relative to the original rest frame. (When it fires the first time, the original rest frame and the ship's current rest frame are the same.)

    Not really. The statement just quoted is true, but not for the reason you think.

    Energy is not a linear function of velocity. The energy of an object with rest mass ##m## moving at speed ##v## is ##m c^2 / \sqrt{1 - v^2 / c^2}##. The kinetic energy (what you have to add to move the object from rest to speed ##v##) is just that total energy minus the rest energy ##m c^2##. If you calculate the kinetic energy (or better, the kinetic energy per unit rest mass) for ##v = 0.3c##, you will see that it is a lot more than three times the kinetic energy for ##v = 0.1c##.

    In fact, if you calculate the kinetic energy per unit rest mass for the speed the ship with the 0.1c thruster will be at after the thruster fires three times (87,669,903 m/s, or about 0.292c), you will find that that is also a lot more than three times the kinetic energy per unit rest mass for a speed of 0.1c. (This is just the relativistic version of what DEvens was telling you at the end of post #4.) So clearly the relationship between speed and the energy required to reach that speed is more complicated than you have been assuming. You are leaving out a key factor. Rather than give it explicitly here, I'll suggest that you google "relativistic rocket equation".
     
    Last edited: Feb 9, 2015
  7. Feb 10, 2015 #6
    I'm now just trying to check whether I'm understanding it correctly or whether I'm making any any mathematical mistakes I'm making or logical errors.

    Imagine spaceships A, B, and C have acceleration switches, which in their rest frame will increase their speed by either 0.1c or 0.2c or 0.3c or 0.4c depending on the switch. Each takes the ship from 0 to the velocity it says on the button from the perspective of the rest frame it was in before it pressed the button. What I'm interested in is the velocity achieved from spaceship A's perspective. The energy to accelerate the each ship after an acceleration button has been pressed can be imagined to come from a ship behind which is follows it through each acceleration, and be measured by each ship to be the same each time. I'm not concerned with the fuel on the ship behind.

    T = 0 : B applies 0.1c acceleration switch, C applies 0.3c acceleration switch.
    T = m : B applies 0.1c acceleration switch again.
    T = m + n : B applies 0.1c acceleration switch again.

    I'm assuming that from B's perspective the same amount of energy received is the same each time it presses the button, because I'm assuming the theory allows B to consider itself to be at rest prior to pressing the button, and that the same energy would be required for the same effect each time from its perspective. And I'm assuming from B's perspective each press of the button takes it to 0.1c (30,000,000m/s) faster than prior to when it pressed the button, and at the end of three presses it has increased in velocity 0.1c each time, and so i at 90,000,000 m/s at the end of 3 presses of the "0.1c" button. From A's perspective the 3 presses took B to 87,669,903 m/s rather than the 90,000,000 m/s that would have been expected if each "0.1c" fuel consumption had produced a 30,000,000 m/s increase. Presumably A will agree with B about the energy received for each acceleration. So from A's perspective the amount of energy required to increase velocity has increased. I'll just run through that again, because maybe the way I write things isn't always clear. At any point prior to an acceleration B could have considered itself the rest frame (assuming rest frame history doesn't matter), and so the effect of pressing the button will appear the same each time, including the amount of energy it then receives to do it. Presumably A can agree with B about the energy consumed to accelerate it to its state, it would just disagree about the increase in velocity achieved per press of the button. If so then from A's perspective the amount of fuel required to increase velocity has increased, as those three same percentages being used up don't achieve three times the velocity increase. So a press of the ".3c" acceleration switch would require more energy as 90,000,000 m/s is a higher velocity than the 87,669,903 m/s that results from the consumption of 3 ".1c" units. I'm assuming the spaceships are efficient.

    If my understanding is ok, then given the formula
    (u + v) / (1 + ((u*v)/(c*c)))
    it is clear that
    (0.4c + 0.1c)/(1 + ((0.4c * 0.1c) / (c * c))) > (0.2c + 0.3c) / (1 + ((0.2c * 0.3c) / (c * c)))
    from both the numerators being the same and the denominator of the smaller being bigger when the equation is considered as a numerator expression and a denominator expression.

    The equation seems to be saying that if Observer A wanted to achieve a higher velocity and could only press each button once, then it would be better for it to press the "0.4c" button followed by the "0.1c" button, rather than the "0.2c" button followed by the "0.3c" button. The extra energy in a "0.4c" unit compared to 4 "0.1c" units outweighs the extra energy of a "0.2c" unit and a "0.3c" unit when compared to a "0.1c" unit. Have I misunderstood?
     
  8. Feb 10, 2015 #7

    PeterDonis

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    No, he won't--at least not for any acceleration after the first. See below.

    That's true anyway, because, as already pointed out, energy is not a linear function of velocity. The amount of energy A sees B receiving in the second acceleration is larger (quite a bit larger) than the amount of energy A sees B receiving in the first acceleration, even though A sees B's speed increase less in the second acceleration than in the first.

    However, in each case, B sees himself receiving the same amount of energy (relative to his instantaneous rest frame before each acceleration starts). So for the first acceleration, A and B will agree on the energy B receives; but for the second, and even more for the third, A will see B receiving more energy than B sees himself receiving. And this is true even though A sees B's speed increasing less for the second and third accelerations than B sees his own speed increasing.
     
  9. Feb 10, 2015 #8
    I had said "Presumably A will agree with B about the energy received for each acceleration." to which you replied
    I had also said: "So from A's perspective the amount of energy required to increase velocity has increased." To which you replied:
    If the fuel had been on-board space ship B, are you saying that A would be disagreeing what % had been used up in the 3 thrusts?
     
  10. Feb 10, 2015 #9

    PeterDonis

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    No. The fuel used up is a direct observable (at least, it is if you use an appropriate invariant measure of how much fuel is used, such as how much rest mass of fuel is used). But the fuel used up is not the same as "the energy added to spaceship B" without qualification, because the latter is frame-dependent and the former is not.

    The direct relationship will be between the rest mass of fuel used up for a given thrust, and the energy added to spaceship B, in the instantaneous rest frame of B when the thrust starts. (This is assuming the thrust is over a short enough time.) But as noted, that will not be the same as the energy added to spaceship B, in spaceship A's rest frame.
     
  11. Feb 10, 2015 #10
    So they'll agree on the amount of fuel used up, but disagree on the amount of energy that using that fuel would provide the system/spaceship?
     
  12. Feb 10, 2015 #11

    PeterDonis

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    Yes. Again, energy is frame-dependent.
     
  13. Feb 10, 2015 #12
    I hadn't realised that the energy released from chemical reactions was said to change with velocity. But have used the energy equation you provided and can see it on a spread sheet, so thanks :)
     
  14. Feb 10, 2015 #13
    How is it tested btw?
     
  15. Feb 10, 2015 #14

    PeterDonis

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    The energy of anything depends on the coordinates you are using. The chemical reactions themselves don't change.
     
  16. Feb 10, 2015 #15

    PeterDonis

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    How is what tested? If you mean, have we run chemical reactions on a spaceship moving at relativistic speed relative to Earth, and tried to measure their energies from Earth, of course not; that's well beyond our current or near-future technology. Chemical reaction energies are measured in the rest frame of the reaction vessel. To figure out what they would be in a frame in which the reaction vessel was moving at relativistic speed, you would use the equations of SR to calculate it.

    If you mean, how are the laws of special relativity tested, they've been tested thousands of times in particle accelerators and other devices all over the world, not to mention plenty of experiments specifically testing key postulates. See here for a comprehensive review:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
     
  17. Feb 10, 2015 #16
    I meant how can they tell that it wasn't that the energy given off was the same but the resultant increase in velocity was different, and that actually it was a different resultant velocity + the energy given off was different.
     
  18. Feb 10, 2015 #17

    PeterDonis

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    I don't understand what this means. Once you have chosen a frame, the relationship between energy and velocity is fixed by the equations of SR, which have been thoroughly tested. So testing the equations of SR is testing that relationship.
     
  19. Feb 10, 2015 #18

    Vanadium 50

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    Name123, before you go too far down that rabbit hole, you do realize that energy is frame-dependent even in Newtonian mechanics, right?
     
  20. Feb 11, 2015 #19
    No I didn't, I had assumed that energy wasn't relative. That there was a certain amount of energy "in play" so to speak in the universe, and that the amount was thought to be objective.
     
  21. Feb 11, 2015 #20

    A.T.

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    Classical kinetic energy has velocity right in the formula, and that velocity is obviously frame dependent. When you accelerate your car from 0 to V in the ground frame, it's velocity changes from U to U+V in some other frame. The change in kinetic energy is different in each frame, but the fuel burned is the same.
     
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