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Energy required to accelerate, and frames of reference

  1. Jun 19, 2012 #1
    It is my understanding that the faster an object moves, the more energy is required to accelerate it. As an object approaches the speed of light, an infinite amount of energy is required to further accelerate it, which is why no object can travel faster than the speed of light. But movement is relative to a frame of reference, which is where I get confused.

    Say, for example, the only two objects in the universe are two spaceships, A and B. A stays stationary (relative to B), while B accelerates away from from A. As B's acceleration increases, it needs more and more energy to increase it's acceleration. Let's say that A then starts moving and catches up to B. Now they are both accelerating through space, but they are neck and neck with each other and have no other objects to reference in order to gauge their speed. In fact, if they stopped accelerating and feeling whatever g forces they were, they could both claim that they are stationary. They then start accelerating again. Would they require more energy than they previously did to accelerate further, or less because they were technically stationary before they resumed accelerating?

    I hope that clear. Obviously I'm a bit confounded by this. Thanks for your help.
     
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  3. Jun 19, 2012 #2

    PeterDonis

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    The key point here is that "energy" is relative to a frame of reference, just like movement is. This may help in clearing up your confusion.

    What this means in terms of acceleration is that how much "energy" it takes to accelerate an object depends on how you do the accelerating. For example, particle accelerators routinely accelerate subatomic particles to speeds very close to the speed of light; when they do, they find that the energy needed, in terms of the power they have to draw from the power grid, say, goes up, as you say.

    However, suppose we have a rocket out in space, far away from everything else. Its rocket engine accelerates it. From the viewpoint of observers on the rocket, the energy needed to produce a given acceleration (meaning what the observers feel, or measure with an accelerometer) is constant; the rocket burns the same amount of fuel to produce the same acceleration, even as the rocket speeds up relative to the rest of the universe. (Actually, over time, the mass of the rocket decreases because of the exhaust ejected out the back, so the energy required for a constant acceleration goes down. But this is really a separate effect from what you're asking about.)

    From the viewpoint of us watching the rocket, of course, the "energy" required to accelerate it goes up as its speed increases, just as for the particles in the particle accelerator. But we're not supplying the energy, the rocket is, so the increased "energy" in this case doesn't show up as anything we observe directly.

    Keeping all of the above in mind may help with the scenario you pose:

    As you state it, this can't possibly be true; A can't remain stationary relative to B if B is accelerating away from A. I think what you meant to say is that A remains stationary relative to the original frame in which both A and B are at rest, while B accelerates relative to that frame. I assume you would also specify that A does not accelerate at this point--he feels no force and is weightless.

    From A's viewpoint, yes. Things may look different from B's viewpoint; see above.

    This means A has to accelerate *harder* than B is; otherwise he won't catch up. What happens when he "catches up", though, depends on the details of how he accelerates. See below.

    No, but they have each other. However, this scenario isn't as simple as you appear to think it is. I would suggest reading up on Rindler observers and the Bell spaceship paradox; these two links are good places to start:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken]

    http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

    You will find that the meaning of "constant acceleration" is ambiguous; if both A and B feel the same acceleration, then the distance between them, as seen by either of them, will increase, not stay the same. If the distance between them, as seen by either of them, stays the same, then the one in front (B in this case) will feel *less* acceleration than the one behind (A in this case).

    (Note: the above assumes that the separation between A and B is in the same direction as they are accelerating. If they are really "neck and neck", right beside each other, separated only perpendicular to the direction they are accelerating, then they can both feel the same acceleration and stay the same distance apart. But A would have had to accelerate harder for some period to catch up, since B started first.)

    Yes, but whether they were actually stationary, relative to each other, at this point, would depend on how exactly they stopped.

    It depends on how they are accelerated, and from which frame you are judging "energy". See above.
     
    Last edited by a moderator: May 6, 2017
  4. Jun 19, 2012 #3
    Thank you so much for your prompt and thorough response. These forums truly are a treasure.

    Let's see if I understand this.

    The rocket in question's fuel source is on board, and therefore the crew of the rocket doesn't notice any increase in the amount of energy needed to accelerate the craft, but mission control back on Earth sees them using up more and more energy to create less and less acceleration.

    If the fuel source was back on Earth somehow (say we have a really long hose connected from Earth to the spaceship's fuel tank), we would notice the same thing as above. This is because regardless of where the fuel it, the observers on Earth are still in a different frame of reference than the crew on the spaceship.

    Correct?

    Does this relativity of energy required occur because of time dilation and length contraction? In other words, do we observe this phenomenon because to the observer on Earth, the spaceship is accelerating less and less, whereas on the spaceship, they observe themselves accelerating differently (faster)?
     
  5. Jun 19, 2012 #4

    PeterDonis

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    Yes, mission control will "see" this, but it won't correspond to any actual increased effort on the part of mission control. The "energy" being used up is not coming from mission control, so they don't really care how much "energy" it takes from their viewpoint to produce a given acceleration.

    No, because the fuel has to move to where the spaceship is before it can be burned and produce thrust. That changes things: now there has to be a pumping station or something that is imparting energy to the fuel to get it to the spaceship. That station will have to impart more and more energy to the fuel just to get it to the ship, as the ship accelerates. So now mission control *does* care about the situation, in a way they didn't before.

    Note that this doesn't necessarily mean that the additional energy that mission control has to expend in this case to get the fuel to the ship, is exactly equal to the additional "energy" they saw the ship expending in the first case, above. The two cases are different, and the analogy between them is kind of thin.

    Not really. It's related, because time dilation and length contraction are effects of relative motion, and relativity of energy is too. But time dilation and length contraction do not "cause" relativity of energy. They are all effects of something more fundamental, the nature of spacetime.

    Not really. The "acceleration" of the ship with respect to the Earth is not really a good thing to focus on, because it's frame dependent; an astronaut orbiting the Earth, say, would see the spaceship to have a different acceleration in this sense than an observer at rest on the Earth. The only real invariant quantity is the acceleration the crew members of the spaceship *feel*, or directly measure with accelerometers. In relativity it's usually a good idea to focus on invariant quantities like this rather than quantities that depend on your frame of reference.
     
  6. Jun 19, 2012 #5
    The Earth observers do not see them using more energy in the form of increased thrust or mass ejection.They do see a decreased acceleration rate with increased velocity. Decreasing by a factor of 1/gamma 3
    If the Earth frame measures the rocket energy by measuring the amount of ejected mass over time then this measurement would also decrease as the coordinate distance and time for a given increase in relative velocity increased in this frame.
    Eventually as v ----->c the acceleration would be virtually undetectable (requiring such long spatial distances for measurement) and likewise the ejected mass
     
  7. Jun 20, 2012 #6
    So the energy output of the spaceship doesn't increase from either the crew or Earth's perspective, but from Earth's perspective, the acceleration that the ship gains from using that same amount of fuel decreases, which means that the ship needs to use more fuel in order to continue to accelerate?
     
  8. Jun 20, 2012 #7

    PeterDonis

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    I may have caused some confusion by not taking my own advice to focus on invariants. So let me try to further clarify some things.

    First of all, a useful quick reference for this topic is the Usenet Physics FAQ page on the relativistic rocket:

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

    This gives simple mathematical formulas for many of the things we are talking about.

    Second, a clarification about "energy". Suppose we have a rocket sitting on Earth, with its tanks full of fuel. It blasts off, leaves Earth and keeps on accelerating indefinitely, all without any external interaction--it just uses the fuel in its tank, burns it, and sends exhaust out the back. Strictly speaking, from the standpoint of the observers on Earth, the rocket sitting on the launch pad already has all the energy it will ever have--the total energy of the rocket "system" (rocket + exhaust) never changes! It's just that the energy is stored in the fuel instead of being "visible" as kinetic energy of the rocket and its exhaust. (Btw, to work the math properly, you have to keep track of the the rocket exhaust as well as the rocket itself. We've been focusing on the rocket alone, since that's the part that's interesting to mission control and to the crew. :wink: But the exhaust is part of the physics and we can't completely ignore it, as we'll see directly.)

    Furthermore, if we consider both the rocket and its exhaust, the statement I just made, that the total energy of the "rocket system"--rocket plus exhaust--never changes, is frame independent. The rocket sitting on the launch pad is at rest in some particular frame, its own center of mass frame, which is also the "Earth" frame in this problem. (We're ignoring Earth's gravity in all of this; assume the rocket launches from a space station in free space that happens to be close to Earth so the crew can take shore leave in between trips. :wink:) It has a total energy E in that frame. But we can rephrase that as follows: the rocket "system" has a total 4-momentum vector with a norm of E, that "points" in a particular direction in spacetime. Now the rocket launches; it starts burning fuel, ejecting exhaust, and accelerating. But by conservation of energy-momentum, the total 4-momentum of the system--rocket + exhaust--stays the same; it has to, since there is no external interaction; the system is self-contained. And since 4-momentum is a 4-vector, we can view it from any frame we like, and it will still have a norm of E (its length will not change), and its components in any frame will stay the same (because the 4-vector as a whole is conserved).

    The above should make it clear what is different about the scenario where the fuel is pumped through a long hose to the rocket: the system is no longer self-contained. That means the whole analysis is different (and I haven't worked that version of the problem in detail).

    Of course, as I said above, the "energy" we are really concerned with is the energy of the rocket itself, not the rocket + exhaust. If we just look at the rocket itself, two things are happening:

    (1) The rocket is burning fuel and ejecting exhaust, so it is losing energy from that standpoint, both in its own instantaneous rest frame and in the Earth frame;

    (2) The rocket is accelerating, so it is gaining energy in the Earth frame. The acceleration means it is also gaining energy in its instantaneous rest frame; we can construct a "rest frame" for the rocket in which it is not gaining energy, but this frame will not be an inertial frame and will have some different properties. I'd rather not go into that since it brings in other complications.

    At this point I want to comment on some things Austin0 said:

    Yes; in fact the observed thrust and mass ejection will decrease, even in the rocket's instantaneous rest frame, because the rocket's own mass, in its instantaneous rest frame, is decreasing as it burns fuel and ejects exhaust. See below.

    This is correct; the observed acceleration rate does decrease by a factor of 1/gamma^3. This can be thought of as increased time dilation of the accelerated rocket, relative to the Earth frame.

    There is a time dilation effect here as well, but there's also another effect. The rocket thrust felt by the crew of the rocket, or measured by their accelerometers, is constant. Since the rocket's mass, as seen in its instantaneous rest frame, is decreasing (as it burns fuel and ejects exhaust), the amount of fuel it needs to burn and exhaust that it needs to eject to maintain a constant thrust (and acceleration as measured by its own accelerometers) also decreases, as seen in the rocket's instantaneous rest frame. As seen from the Earth frame, this decrease in rate of fuel consumption and exhaust ejection is compounded by the time dilation effect.

    It's also important to distinguish the "rate of acceleration" from the "rate of energy gain" of the rocket, as seen in the Earth frame. Here's one way of looking at that: relative to the Earth frame, as the rocket's velocity gets closer and closer to c, more and more of the (constant) applied rocket thrust goes into increasing the rocket's mass (as seen in the Earth frame) instead of increasing its velocity. So the rocket gains total energy, as seen from the Earth frame, at a faster rate than it gains velocity, so to speak. (This also means the rate of observed fuel consumption and exhaust ejection, as see from Earth, does not decrease as fast as the observed acceleration as seen from Earth does.) Here "mass" no longer means "rest mass", but what is called "relativistic mass". There are some pitfalls in viewing things this way, but it's one way of looking at why the rocket's acceleration, as viewed from the Earth frame, gets smaller and smaller even though the acceleration as measured on board the rocket stays constant.

    Hopefully I haven't confused things further with this rather long post, but I think it's important to have some idea of the issues involved even with what seems like a simple scenario.
     
    Last edited by a moderator: May 6, 2017
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