Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativity and the universal speed limit

  1. Aug 7, 2014 #1
    Hey guys, this is a question that has been bothering me since I finished my special relativity course last year.

    I was told that nothing can travel faster than the speed of light. Thinking relativistically, I take this to mean no one thing in the universe can travel faster than any other thing in the universe by a difference 'c'. This seems like a difficult condition to satisfy-- with so many objects moving around in different directions, some of them spinning really fast, like neutron stars, they would have to be configured just so... I guess it's just hard for me to imagine why, if I am in a space ship, I can't just keep accelerating in one direction. Will my constant thrust start producing diminishing acceleration after a while? Put another way, if some object is hurtling towards me at c/2, will it be easier for me to accelerate away from the object than towards it?
     
  2. jcsd
  3. Aug 7, 2014 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    Yes, Relativity has shown that ANY acceleration still produces diminishing additional velocity as you approach c asymptotically.
     
  4. Aug 7, 2014 #3

    Nugatory

    User Avatar

    Staff: Mentor

    Another form of this question that sometimes throws people: I have a gun that fires a bullet with a muzzle velocity of .5c, point it in one direction and shoot, then point it in the other direction and shoot. Don't I end up with two bullets moving at speed ##c## relative to one another? Or suppose I mount this gun on a vehicle traveling at .75c and then fire it straight ahead? Don't I end up with a bullet moving at ##1.25c##?

    I don't, and the reason is to be found in the rule for "relativistic addition of velocity" which google will find for you pretty quickly. The diminishing acceleration effect that you suspected might exist and phinds confirmed can be derived from this rule.
     
  5. Aug 7, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The reason why "ANY acceleration still produces diminishing additional velocity" is that as the spacecraft's speed increases (relative to some reference frame) so does its mass. The closer the speed of the spacecraft gets to the speed of light, the greater the mass gets. that's why "constant thrust" will "start producing diminishing acceleration"- a= f/m.
     
  6. Aug 8, 2014 #5

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Using relativistic mass? A little outdated?

    The explanation in terms of 4-velocity and 4-acceleration, translated to English description, suggests that time dilation is the key concept not change of mass. Thus any finite acceleration experienced in the momentarily comoving frame of the accelerating body, when translated back to a frame where the object is moving near c, causes the experienced acceleration (proper acceleration) to be divided by gamma (due to dt versus d tau). As a result, any finite acceleration pattern experienced by a body, when referred to a give inertial frame, always has speed < c, and coordinate acceleration decreasing as speed approaches c.
     
  7. Aug 8, 2014 #6

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    Actually, there's a bit more to the calculation than this, and the factor is [itex]\gamma^3[/itex], not [itex]\gamma[/itex].
     
  8. Aug 8, 2014 #7

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    ... and it actually depends on direction of acceleration in relation to current current velocity in the inertial frame (γ3 is only for collinear). I thought of this right after posting, but didn't want to complicate my main point that the better (IMO) way to look at the barrier is the transformation of acceleration (mass not involved).
     
  9. Aug 8, 2014 #8

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    To really think relativistically, it might be good to look at the spacetime viewpoint.
    A nonzero-mass particle has a unit 4-velocity, whose tip _always_ lies on the future-timelike unbounded unit-hyperboloid that is asymptotic to the light-cone [in the tangent space].

    Since the 4-acceleration is orthogonal to the unit-4-velocity, the 4-acceleration is tangent to the hyperboloid. Thus, the 4-acceleration effectively moves the tip of the unit-4-velocity to another point on the hyperboloid, possibly parametrized by angles [rapidities="arc-lengths on the unit-hyperboloid"]. No matter what is done, the unit-4-velocity is always inside the future-light-cone. The relative-speed (regarded as the ratio of the spacelike and timelike components of the unit-4-velocity wrt the lab-frame's 4-velocity) is bounded by 1 and won't change very much when the relative-rapidity is large (as it rides up near the light-cone in the lab-frame).

    In the Galilean case [hindsighted from Minkowski spacetime], the unit-4-velocity would have its tips on the unbounded spacelike-plane where its timelike-component is always 1. In this case, the relative-speed is unbounded (and is proportional to the Galilean analogue of the rapidity). Since our everyday intuition is Galilean, we find it hard to see the SR-case, especially if we rely too heavily on purely Galilean methods.
     
  10. Aug 9, 2014 #9
    Quantum and relativity are so confusing to me. I need help.
     
  11. Aug 13, 2014 #10
    We can help you, but you have to pick one or the other. You can not possibly understand both at once because they are incompatible ;)
     
  12. Aug 13, 2014 #11

    Dale

    Staff: Mentor

    QM and special relativity are compatible. And QM is even compatible with curved spacetime, so I wouldn't say they are incompatible without some qualifications.
     
  13. Aug 13, 2014 #12
    This is true.

    This is true. However if I observe one object travelling away from me at 0.75c and another object travelling away from me in the opposite direction at 0.75c, I measure their relative velocity as 1.5c.

    Yes, in the sense that the rate of change of your relative velocity as measured by you will be higher if you accelerate away from the object. The rate of change of your velocity measured by an independent observer will be the same either way.

    As measured by you, yes.
     
  14. Aug 13, 2014 #13

    jtbell

    User Avatar

    Staff: Mentor

    The separation speed of the two bullets, as measured by you in your reference frame, is c. That is, the distance between the two bullets increases at the rate c in your reference frame.

    The relative speed of the two bullets is 0.8c, as per the "relativistic velocity addition" formula. This is the speed of one bullet in the reference frame in which the other one is stationary, i.e. the speed of one bullet as "observed" by the other one.

    http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html#c2

    (be careful of + and - signs for the different directions)
     
    Last edited: Aug 13, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook