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Reasons why lightspeed travel is impossible

  1. Sep 17, 2010 #1
    So, as something with rest mass gets closer to the speed of light, it gains more mass, making it require greater force to accelerate it making it gain even more mass and so on.
    So the reason we use to say nothing can travel at or faster than light is because it gains more mass, and requires more and more energy to accelerate at the same rate the closer its speed is to light. (That's the only reason I've seen for why faster than light travel is impossible)

    Suppose a rocket is propelling itself by burning fuel. The rate of energy released from burning the fuel is constant from the rocket's point of view, and therefore has a constant force applied to it from its point of view. But from an observer's view, the rate of energy released from burning the fuel, and the force resulting from it, is less than that viewed by the rocket because time for the rocket is slower than time for the observer.

    So...
    1. Can the rocket detect its own change in mass from near lightspeed travel?

    2. If it can't, then isn't its acceleration from its view still constant because the force applied by the fuel is also constant from its view?

    3. Even though the rocket perceives itself as having a constant acceleration, from the observer's view it's acceleration is decreasing right?

    4. As the rocket gets faster, time for it slows, and its acceleration due to the fuel burning will decrease from the observer's view right?

    So my question is really that can't you say that lightspeed travel is impossible not (or also?) because it gains mass and requires an infinite amount of energy, but because time slows down by an exponentially increasing factor the closer it gets to the speed of light, which would make it require an infinite amount of time?

    I've looked but I haven't seen anywhere that says requiring an infinite amount of time is a reason that faster than light travel is impossible

    (Please correct me if I'm wrong about anything which I probably am, I may have misused the word force...)
     
    Last edited: Sep 17, 2010
  2. jcsd
  3. Sep 17, 2010 #2
    I see this argument a lot and I have reservations about this argument.

    How could one argue that the increase of relativistic mass, from a different frame nevertheless, has an impact on the rocket's ability to accelerate?

    If a rocket accelerates with 1g for a long time it will actually lose rest mass instead of gain it as it needs to burn up a lot of fuel to keep accelerating. So in fact the rocket will accelerate easier in a later stage because it has converted mass to energy and that energy is dispatched in the opposite direction (in some amount, as seen from the observer's frame, related to the Doppler factor if I am not mistaken).

    Perhaps when a rocket is pushed the argument is valid.

    Who can shine some light on this?
     
    Last edited: Sep 17, 2010
  4. Sep 17, 2010 #3

    Dale

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    No.
    Yes.
    Yes.
    Yes.
     
  5. Sep 17, 2010 #4
    If [tex]a[/tex] is the proper acceleration measured by the rocket, then, its speed wrt to an observer is known to be :

    [tex]v=\frac{at}{\sqrt{1+(at/c)^2}}[/tex]

    The above explains why the rocket can never reach [tex]v=c[/tex]
     
  6. Sep 17, 2010 #5

    Janus

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    Another explanation use the addition of velocities theorem.

    Consider point 2.

    From the Rocket's point of view its acceleration is constant (or even increasing due to the changing fuel to payload ratio.)

    Now assume that the rocket is dropping off space buoys as it travels, with each buoy retaining the velocity of the rocket at the moment it is released. Furthermore, let's say that these buoys are dropped off as the ship attains a fixed velocity difference from the last buoy as measured from the ship. If the ship drops the first buoy when it achieves a velocity of 0.1 c relative to its starting point it drops a buoy, and then when it achieves a velocity of 0.1c relative to that buoy, it drops another, etc.

    The formula for velocity addition is

    [tex]w= \frac{u+v}{1+\frac{uv}{c^2}}[/tex]

    Which means if u is the velocity of a buoy with respect to the starting point, and v is the velocity of the ship with respect to the buoy (as measured by the ship), then w is the velocity of the ship with respect to the starting point.

    If the first buoy is dropped off at 0.1c and then we calculate the ship's velocity when it it moving at 0.1c relative to the buoy (and ready to drop the second buoy), we get an answer of:
    0.198c
    Which is also the velocity of the second buoy with respect to the starting point.

    If the ship drops off a third buoy when it it moving at 0.1c with respect to the second buoy it and the third buoy will now have a velocity with respect to the staring point of:
    0.292c

    And here is a list of the velocities of the next 12 buoys.
    0.381c
    0.463c
    0.538c
    0.606c
    0.666c
    0.718c
    0.763c
    0.802c
    0.835c
    0.863c
    0.886c
    0.906c

    Note that even though the ship, by just adding up its changes of velocity from buoy to buoy would calculate that it has accelerated by a total of 1.4c, it is actually only moving at 0.906c relative to where it started.

    Furthermore, if you look at the velocity addition formula again:

    [tex]w= \frac{u+v}{1+\frac{uv}{c^2}}[/tex]

    You will note that no matter how many buoys the ship drops off and accelerates from, it will never achieve a velocity of c relative to its starting point. As long as u and v are smaller than c, w will come out to be smaller than c, and since each subsequent calculation uses the answer from the previous one for u, you always get an answer smaller than c.
     
  7. Sep 17, 2010 #6

    bcrowell

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    This seems fine to me, except for the misuse of the term "exponentially" -- time only occurs in the base of the relevant expressions, not the exponent.

    To the observer in the earth's frame, the rocket would need infinite energy in order to reach c, and if the rocket had a source of energy that allowed it to keep blasting its rockets with constant thrust (constant as measured in the rocket's frame), its velocity would approach c, but would never reach it. (I.e., it would take infinite time to reach c.)

    To the observer in the rocket's accelerating frame, the rocket's kinetic energy is always zero. To this observer, the universe is permeated by a gravitational field (the equivalence principle at work), and this gravitational field is accelerating the earth. To hover in this gravitational field, the rocket engine is continually engaged in the business of spewing out high-energy exhaust gases. The earth's velocity approaches c, but never reaches it. (I.e., it would take infinite time to reach c.)
     
  8. Sep 17, 2010 #7
    What do you mean, I can't follow what you say.
     
  9. Sep 17, 2010 #8
    Okay, so then requiring an infinite amount of time is another reason for why an object with rest mass can't travel at or faster than the speed of light right? I looked around some and I haven't seen that supplied as a reason. Does anyone know why it isn't normally suppplied as a reason?

    Thanks for giving another explanation. I've never heard of the addition of velocities theorem before, but I like it as an explanation. It easily shows mathematically why it's impossible.

    And sorry for misusing 'exponentially' I thought it would be suitable since the rate at which the factor changes increases the closer the speed is to c. So I guess would asymptotically be more appropriate then?

    Also, its amazing how this post got 6 replies within a few hours while another post I made yesterday in this same forum has no replies. I guess some questions are more interesting or have more people that have an answer for it?
     
  10. Sep 17, 2010 #9

    DaveC426913

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    It is important for you to understand that these are all nothing more than looking at the same thing from different points of view.

    Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass.
     
  11. Sep 18, 2010 #10
    From the viewpoint of a spaceship with a clock and an accelerometer it can accelerate to speeds that are faster than light as measured by integrating the readings from the two instruments, but space-time as manifested by the view of the stars and a timing clock pulse from Earth does weird things, from the ship's perspective. But it depends on what we measure. A ship that continuously accelerates from rest to then stop at a distant destination will find that a clock signal from it's origin has counted out just a couple of years of flight, even if it travels millions of light-years (as measured by other means.) If the distance is instead measured by a series of radar pulses from the ship, then it'll get a quite significantly different measurement. From it's point of view the space-time displacement between it's origin and it's destination shrank compared to measurements taken by a stationary observer.
     
  12. Sep 18, 2010 #11

    bcrowell

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    You can integrate [itex]\int a d\tau[/itex], where a is the proper acceleration and [itex]\tau[/itex] is the proper time. However, that doesn't give your actual velocity relative to the stars as you pass by them. That's because an integral is basically a Riemann sum, but velocity increments don't add linearly.
     
  13. Sep 18, 2010 #12

    Janus

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    Classically, the rocket equation is:

    [tex]V_f = V_e \ln \left ( \frac {M_i}{M_f} \right )[/tex]

    With Vf being the final velocity of the rocket,
    Ve the exhaust velocity,
    Mf the final mass of the rocket
    and
    Mi the intial mass of the rocket.

    Factoring in Relativity gives us:

    [tex]V_f = c \tanh \left (\frac{Ve}{c} \ln \left(\frac{M_i}{M_f}\right) \right )[/tex]

    Which gives us the result that any ratio of initial mass to final mass less than infinite results in a final velocity of less than c.

    A rocket accelerates forward by throwing mass backwards, the efficiency of this process depends on the exhaust velocity.

    So if I'm in the "stationary" frame watching the rocket, the efficiency of the rocket will be determined by the difference between the rocket's velocity and the exhaust velocity.

    Assume that the rocket is already moving at 0.9c relative to me and the exhaust velocity relative to the ship is 0.1c. Relative to me, the exhaust velocity is

    [tex]\frac{0.9c-0.1c}{1- \frac{0.9c(0.1c)}{c^2}} =[/tex] 0.879c

    and difference between rocket and exhaust velocity is 0.021c

    When the rocket reaches 0.95c, the exhaust velocity with relative to me will be 0.939c for a difference of 0.011c , almost half that of the difference at 0.9c.

    At a rocket velocity of 0.99c the difference drops off to 0.002c

    From my frame, the rocket's efficiency drops off as it approaches c and the rocket's acceleration decreases the nearer it gets to c.
     
  14. Sep 18, 2010 #13

    JesseM

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    I don't think it's quite the same. If you ignore dynamical issues like energy and just think in terms of kinematics, you can write down an equation describing the worldline of an object that reaches c in finite time (for example, x(t) = (a/2)*t^2, where x and t are coordinates of an inertial frame and a is some constant coordinate acceleration), and if you calculated proper acceleration as and proper time using the usual kinematical equations of SR, I believe you would find that at the moment the object reaches c the proper acceleration has gone to infinity but the proper time is still at some finite value.
     
  15. Sep 18, 2010 #14
    I don't think so, see post #4. DaveC426913 is correct.
     
  16. Sep 18, 2010 #15

    JesseM

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    What specific statement in my post are you disagreeing with? Again remember that I am separating dynamics from kinematics--of course it is impossible dynamically to accelerate to c in finite time since this would require infinite energy and force, but we can consider the possibility from a purely kinematical point of view, and applying the usual kinematical equations of SR the proper time would not go to infinity for the worldline I mentioned (which reaches c in finite coordinate time, the proper acceleration going to infinity at that time).
     
  17. Sep 18, 2010 #16
    The fact that you disagree with Dave's initial claim that finite proper acceleration for an infinite time produces the same effect as infinite proper acceleration for a finite time. His claim is correct, you can prove it trivially from math in post 4.


    You don't need to, the formula [tex]v=\frac{at}{\sqrt{1+(at/c)^2}}[/tex] encapsulates both the kinematics and the dynamics.
     
  18. Sep 18, 2010 #17

    bcrowell

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    Well, I can think of a few purely kinematical objections:

    (1) If the object reached v=c at t=tc, instantaneously comoving Minkowski frames would exist for all t<tc, but not for [itex]t \ge t_c[/itex].

    (2) There is causality violation at t>tc.

    (3) If the object has finite size, and its size as measured by comoving observers is constant (i.e., it's Born-rigid), then its longitudinal size as measured by inertial observers becomes zero at t=tc. But if its front and back coincide at t=tc, and its front and back both have v=c at t=tc, then it's hard to imagine how any description of the time-evolution of the system could make its front and back *not* coincide at t>tc; if the initial conditions are the same, how can the final conditions be different?
     
  19. Sep 18, 2010 #18

    JesseM

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    How is that a kinematical objection? You wouldn't object to the fact that light moves at c on kinematical grounds, in spite of the fact that light has no comoving Minkowski frame.
    Again, I don't understand how you are defining "kinematical objection". Tachyons are consistent with the axioms of SR, we can imagine a logically possible set of laws of physics where tachyons exist and causality is violated but the basic kinematical laws work the same as in SR. If someone said they could rule something out on purely kinematical grounds, I would normally interpret that to mean that they could show it was impossible using only the kinematical rules with no additional assumptions about the detailed laws of physics (for example, if the kinematical rules are those of SR, on purely kinematical grounds we can rule out the possibility that there would be some non-inertial path a clock could take between two events A and B such that it would have elapsed more time than a clock which moves inertially from event A to event B)
    Is it even possible for an object to remain Born rigid if each point on the object doesn't have uniform proper acceleration? (not a rhetorical question, I'm really not sure) Anyway I see no reason to impose Born rigidity as a condition if we're only talking about kinematics, not what is realistic according to dynamical laws dealing with solid objects.
     
  20. Sep 18, 2010 #19

    JesseM

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    Uh, I never disagreed with his "initial claim", I only objected to the claim of his I quoted, where he said "Saying a rocket would require an infinite acceleration to reach c - is the same thing as saying it would take an infinite amount of time - is the same thing as saying it would acquire infinite mass."
    I think you need to look into the distinction between kinematics and dynamics more carefully, that equation is purely kinematical since it can be derived from the definitions of velocity and proper acceleration and the coordinate transformation of SR, without any consideration of the dynamical laws which predict what trajectory a given object will actually take in a specific physical situation (and the equation only deals with velocity as a function of time for special the case of constant proper acceleration, whereas I was talking about a kinematical analysis of a hypothetical object moving with ever-increasing proper acceleration)
     
  21. Sep 18, 2010 #20

    bcrowell

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    Point taken, but I think it's important to keep a clear distinction between observers and objects that can't be observers. When we talk about a rocket, we implicitly assume that it's possible for an observer to be aboard. When we talk about a ray of light, we implicitly assume that the ray of light isn't an observer. In your example, the object suddenly makes the transition from being in the same class as the rocket (observer-compatible) to being the same class as the ray of light (observer-incompatible). This seems to me like a purely kinematic issue: can you tie the origin of a coordinate system to a particular world-line?

    Yes, but again we have two classes of objects: those that move at less than c and those that move at more than c. A tachyon can't slow below c any more than a normal object can accelerate above c. I think you run into real kinematic difficulties when you try to posit objects that switch between the two classes.

    Actually I think if it's going to remain Born-rigid then the proper accelerations have to be *un*-equal. This is basically the Bell spaceship paradox.

    Well, Born-rigidity is a purely kinematical condition: all it says is that the radar distance between nearby points remains constant. And when, e.g., we find that a Born-rigid disk can't have an angular acceleration, that is a purely kinematical prohibition, not a dynamical one; the problem is that if you wanted to have an angular acceleration, accelerations would have to occur simultaneously at all points around a circumference, but that simultaneity is kinematically impossible.

    If you don't require Born-rigidity, then I think you have a problem, because without some sort of requirement of this type, you could have the object get infinitely distorted according to a comoving observer. I think it's hard to maintain the idea of an "object" when the object is distorted by a factor of *infinity* in its own comoving frame.
     
    Last edited: Sep 18, 2010
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