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I Subbing Planck length into length contraction equation?

  1. Aug 11, 2016 #1
    I was wondering if it is possible to work out the maximum amount of energy an object with mass can have using the length contraction equation (i.e. "actual" length divided by Lorentz factor).

    The way I thought of doing this was by rearranging e = mc^2 to get c^2 = e/m. Then, substitute e/m into the length contraction equation. Once you've done that, just rearrange the equation to get energy and substitute Plank length for the contracted length.

    Now assuming you have the values for the rest of the variables (i.e. mass, "actual" length, speed), you can calculate the maximum amount of energy that an object with that mass can have. My reasoning behind this is that the Planck length is the shortest distance in the universe (it may well not be but that's an assumption I'm making in this case) and so the length can't be contracted any further than this value. Thus, the output value of the equation is the maximum energy that the object can have.

    This is probably wrong, but I'm curious to know about the mistakes I may have made.
    [tex]
    \frac{mv^{2}}{\left ( \frac{l}{l\bullet } \right )^{2}-1} = e
    [/tex]
     
  2. jcsd
  3. Aug 11, 2016 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    There is no such thing in classical relativity. The energy of an object with nonzero rest mass, relative to a particular frame, is unbounded.

    This assumption is incompatible with classical relativity, which assumes that spacetime is a continuum, with no "shortest" distance (or time).

    There are speculative theories which attempt to quantize spacetime, but they're just speculative at this point. There are also speculative theories (Google "doubly special relativity") which attempt to modify the Lorentz transformation so that there is a shortest distance (and time) while still building the theory on a non-quantized spacetime; these are also just speculative. None of these speculative theories will use the equations of standard relativity as you are attempting to use them.
     
  4. Aug 11, 2016 #3
    Ah yes, I understand- thanks for the clarification.
     
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