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

This is probably just a coincidence, but

  1. Jan 13, 2006 #1
    [tex]EK = mv^2[/tex]
    energy kinetic = mass times velocity squared

    So... Isn't one way to interpret time dilation that everything moves through time and space with a sum velocity of c (somehow), and that as velocity in space increases velocity in time therefore decreases? I've read this in several places, anyway.

    If this is the case, then the permanent space-time velocity of all objects is c... Which means that their kinetic energy would be:

    [tex]EK = mv^2[/tex]

    [tex]EK = mc^2[/tex]

    So er... [tex]E=mc^2[/tex]
     
  2. jcsd
  3. Jan 14, 2006 #2
    The equation [itex]E=mc^2[/itex] only applies when [itex]v=0[/itex]. The general equation for any [itex]v[/itex] is
    [tex]E=\sqrt{(mc^2)^2+c^2p^2}[/tex]
    where [itex]p[/itex] is momentum.
    So indeed just a coincidence.
     
  4. Jan 14, 2006 #3

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    1. Since when is kinetic energy equal to mv^2? What happened to the 1/2?

    2. What exactly are "velocity in space" and "velocity in time"?

    3. What is "space-time velocity"?

    4. This statement is puzzling: "..... Isn't one way to interpret time dilation that everything moves through time and space with a sum velocity of c (somehow),... " Everything does NOT move with a "sum velocity of c (somehow)".

    Zz.
     
  5. Jan 14, 2006 #4
    Of course. I completeley overlooked this most obvious argument!
     
  6. Jan 15, 2006 #5
    No, its [tex] E_K = \frac{1}{2} m v^2[/tex]


    [tex]E=mc^2[/tex] only applies to a an object when [tex] v= 0[/tex] as another pointed out. It has nothing to do with kinetic energy.
     
  7. Jan 15, 2006 #6
    The length of the velocity four-vector of any particle is 1 in geometric units, which corresponds to c in SI units. A particle in it's rest frame is moving through time at a second a second. To compare lengths with times, we multiply by c, to find that it is moving c metres a second.
     
  8. Jan 15, 2006 #7

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    But really, do you honestly think the OP knew about this and that this is what he/she is describing by making that erroneous statement? I highly doubt it.

    Zz.
     
  9. Jan 15, 2006 #8
    I believe one of Greene's recent popular science texts (with some grandiose title) expresses this idea (and I assume this is where the OP got the idea from); whether or not the OP fully understood it or not is, of course, another matter.
     
  10. Jan 15, 2006 #9

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I'm not so sure... Greene cannot make a silly mistake of equating KE with mv^2 and missing out that 1/2, which is the OP starting premise.

    Zz.
     
  11. Jan 15, 2006 #10
    I comprehend the idea, but the important thing is that I apparently don't comprehend simple Newtonian physics. Forgive me for the incorrect formula and the idiotic post... I just saw something that sort-of corresponded and then decided to post it, and consequently wound up with a bunch of gibberish.

    Again, sorry. I don't know what I thought when I was writing EK=mv^2. Thanks for replying to this, anyway.
     
  12. Jan 15, 2006 #11
    Sorry, I meant the notion of having a velocity of c at all times, as opposed to the definition of non-relativistic classical kinetic energy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: This is probably just a coincidence, but
  1. Just wondering (Replies: 8)

  2. Just a question (Replies: 4)

  3. This is just wierd (Replies: 5)

Loading...