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I Velocity addition: Physical meaning of non-singularity at c?

  1. Jun 21, 2016 #1
    Hi.

    In the derivation of the relativistic formula for adding velocities, the Lorentz factor drops out. Mathematically, the formula works for inertial frames with relative velocity c and even gives an answer to Einstein's famous question about what happens if you drive at the speed of light and turn your headlights on.

    However, is there a physical meaning to this? Massive light sources can't reach c because the required energy would be infinite. But is there some framework (QFT maybe?) where a photon decays into multiple photons such that one of them can be regarded as "headlights" and another as "light from the headlights"?
     
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  3. Jun 21, 2016 #2

    vanhees71

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    No, in free space a photon cannot decay into two photons because it's forbidden by the conservation laws together with the on-shell conditions (just check it out!).

    The meaning is that no matter, how the light source moves the light has always (phase) velocity ##c##. That's what the Lorentz transformation tells you in terms of the "velocity addition rule" taken in the limit where one velocity has magnitude ##c##.
     
  4. Jun 21, 2016 #3

    Nugatory

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    The velocity addition formula is best understood as transforming the velocity (of something) measured in one inertial frame into a the velocity of the same something as measured in another inertial frame. Thus, I can use it to answer the question "something is moving at the speed of light relative to me. What is its speed ##w## as measured in an inertialframe in which you are rest, given that you are moving at speed ##u## relative to me?": ##w=(u+c)/(1+\frac{uc}{c^2})=c##. That has natural physical significance; it's telling us that the speed of light is the same in all inertial frames.

    However, if we set ##u=c## as well, the question becomes: "something is moving at the speed of light relative to me. What is its speed ##w## as measured in an inertial frame in which you are rest, given that you are moving at speed ##c## relative to me?". That question is meaningless because there is no such inertial frame. The calculation comes out to ##c##, but it has no more physical significance than if it had come out to be infinity (as happens with many of the other physically meaningless computations you can do if you make the bogus assumption that something can be at rest in one inertial frame yet moving at the speed of light in another).
     
  5. Jun 21, 2016 #4
    I think that since the Lorentz factor is infinite, i.e. undefined, for frame relative velocity c, the final formula for relativistic addition of velocities is not valid for this case, even though the Lorentz factor drops out.
     
    Last edited: Jun 21, 2016
  6. Jun 21, 2016 #5
    Actually, what is the exact reason that there is no such inertial frame? Can this be derived directly from the postulates?

    Earlier, I might had argued that there is no such inertial frame since the Lorentz transformation for space and time coordinates to such a frame becomes mathematically undefined if the relative velocity is c. But this is no the case for the transformation of velocities. So I don't find this mathematical argument very convincing anymore. After all we know about the existence of coordinate singularities in different areas of physics, why should the singularity of the Lorentz factor at c be physically meaningful and not just be an artifact of the particular representation of the Lorentz transform I'm used to?
     
  7. Jun 21, 2016 #6

    PeterDonis

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    Why not? The velocity transformation alone cannot be used to go from one inertial frame to another. You need the full Lorentz transformation. So if the latter is singular, you can't have an inertial frame.

    Because the singularity of the Lorentz factor at c is not a coordinate singularity. It's a manifestation of the physical fact that timelike objects and lightlike objects are fundamentally different and you can't transform one into the other.

    An alternate way of seeing this is to consider an ordinary non-singular Lorentz transformation, in one spatial dimension for simplicity, and ask how it acts on the two types of vectors--timelike and lightlike. For timelike vectors, a Lorentz transformation acts like a hyperbolic rotation: if you think of a vector with its tail at the origin and its tip on a unit hyperbola to the future of the origin, a Lorentz transformation moves the tip of the vector along the hyperbola to point in a different direction. For for lightlike vectors, a Lorentz transformation acts like a dilation; it doesn't change the direction of the vector at all (it is still lightlike and still points along the same 45 degree line in the same direction), but it changes its length (the tip is now either closer to or further from the origin, depending on the direction of the Lorentz transformation).

    Now suppose we could somehow define a valid Lorentz transformation with relative velocity c. How would it act on the two types of vectors? The problem is that there is no possible valid action it could have. For a timelike vector, it would have to move the tip an infinite distance along the hyperbola--but that doesn't result in a valid vector. (It might seem like it is changing the timelike vector to a lightlike one, pointing along the 45 degree line, but this "lightlike vector" isn't a valid vector because its tip would have to be at infinity.) For a lightlike vector, it would have to either move the tip out to infinity along the 45 degree line, or collapse the tip to the origin so the vector arrow became a single point--but neither of those results in a valid vector either. So the concept of "Lorentz transformation with relative velocity c" simply doesn't make sense--it doesn't take one valid 4-vector to another valid 4-vector.
     
    Last edited: Jun 21, 2016
  8. Jun 21, 2016 #7
    As I stated above, since the formula for transformation of velocities makes use of the Lorentz factor and that factor is undefined at v=c, so is the velocity formula similarly undefined. The fact that the Lorentz factors drop out during the derivation of the velocity formula does not change that fact.
     
  9. Jun 21, 2016 #8
    Are there other meaningful quantities whose transformation formulae remain well-defined when the relative velocity approaches c? I noticed that
    $$\frac{E}{p}=\frac{\gamma m_0 c^2}{\gamma m_0 v}=\frac{c^2}{v}$$
    turns into ##E/p=c## known for photons, but this isn't particularly interesting.
     
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