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

Why are time dilatation & length contraction infinite at light speed?

  1. Aug 1, 2014 #1
    Considering that speed of light is constant and finite, then why are the time dilatation and length contraction infinite to a frame of reference moving at the speed of light?
    We know that a moving frame of reference experiments time dilatation and length contraction from the point of view of a “static” observer. And, at the limit, if the frame of reference reached the speed of light, time would completely stop (what I called here “infinite dilatation”) and length would completely contract to zero (what I called here “infinite contraction”).
    I know that it’s not possible for a body to reach that speed (although I really don’t understand why), but let’s think about the practical applications of this. For example, I heard that currently they are successful at accelerating particles at 99,999% the speed of light at the Large Hadron Collider.
    So what does it means to have a particle that needs 1 second to move 299 792 458 meters? It looks like a normal speed to me, or at least as natural as any other.
    I agree and understand the concept of electromagnetic waves traveling at this particular number, and that this measurement does not depend of the speed of the observer. But why does this creates a physical limit to movement? Why can’t we reach and actually beat that speed with enough acceleration in a particle?
    And why a particle moving at this speed seems to cease to exist? When I say it ceases to exist I mean it has no dimension at all and it is not affected by time.
    What can an observer say that happens to a particle in the LHC at this speed?
     
  2. jcsd
  3. Aug 1, 2014 #2

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    You have to be careful when you talk about the "point of view" of an observer. Usually that term means the Inertial Reference Frame in which the observer is at rest. Then we can calculate the Length Contraction and Time Dilation of objects moving according to that IRF but the observer himself cannot directly observe those effects. He has to do a lot of work to figure out what they are.

    We can then transform the coordinates of events in one IRF to another IRF moving with respect to the first one using the Lorentz Transformation process but the process does not allow going all the way to the speed of light so forget about that.

    Speeds are normal in the sense that once you have defined an IRF, you can assign a speed to any particle, short of c, transform to its rest IRF, define what it looks like, then transform back to the original IRF to determine its Length Contraction and Time Dilation. Since we can't transform to c, that limits the speed of particles.

    When you accelerate an object in its rest frame, its speed increases by some amount. When you transform to its new rest frame you can repeat the process as many times as you like. When you transform back to the original IRF, you will see that its speed approaches c but never reaches it.
     
  4. Aug 1, 2014 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    And what this is telling you is that a frame of reference "reaching the speed of light" is not possible: the Lorentz transformation is undefined (because you end up having to divide by zero).

    You just gave the reason why. To expand on it a little: there is a factor called ##\gamma## in the Lorentz transformation, where ##\gamma = 1 / \sqrt{1 - v^2 / c^2}##. If ##v = c##, ##\gamma## is undefined (because you would have to divide by zero to evaluate it); but as ##v \rightarrow c##, ##\gamma## increases without bound. Physically, what this means is that, as you add more and more energy to an object (which is what increasing ##\gamma## corresponds to, physically), its speed gets closer and closer to ##c## but never reaches it; there is no way to actually reach ##c## this way (because there is no well-defined value of ##\gamma## for that speed).
     
  5. Aug 1, 2014 #4
    Ok, I understand that the domain of the γ function does not include the c value as it will force to divide by zero, and for that matter, any number larger than c will involve imaginary numbers. But this for me seems a to be a problem about mathematical definition.

    However, it looks logical to me that c/2 is in the domain of the γ function. So, if that’s the case, let’s suppose the following scenario:
    Three observers in a train station. One of them remains in the platform. The other two observers get on a train each going in opposite directions at c/2.
    For the observer in the platform, he will see that the other two are getting away at c/2. Also, for both observers on the trains, the person on the platform is getting away at c/2.

    But what would the observers on the trains see when looking at each other?
     
  6. Aug 1, 2014 #5

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

  7. Aug 1, 2014 #6

    phinds

    User Avatar
    Gold Member
    2016 Award

    No, this is one of those cases where the math models reality quite well, so it is NOT a problem about the math it's a description of an actual physical limitation.
     
  8. Aug 1, 2014 #7
    Ok, with this formula for velocity addition it has become clear that there is no way to achieve c by adding relative movement. That settles one of my doubts quite well.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why are time dilatation & length contraction infinite at light speed?
Loading...