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

Why can light travel at the speed c?

  1. Jul 2, 2011 #1
    hey guys, i'm new to this forum so i dunno if this has been already asked before

    i saw this equation on special relativity explaining why objects with mass cannot travel at the speed of light

    E = mc^2/(1-(v^2/c^2))^0.5
    ( hard to write on a computer. view the attachment for the equation or link here: http://upload.wikimedia.org/math/2/4/0/24090a520815f2d76b2be996acc6c9e2.png)

    this makes sense right; if an object travels at the speed of light, the denominator becomes 0 and if the object has mass (say for example one arbitrary unit) you get a large number divided by zero - in other words an undefined energy to make this happen - impossible
    but what confuses me it that if you take a photon which can travel at c, you still get an undefined energy. my question is: how are photons able to travel at c in a vacuum? do they carry undefined amounts of energy? or does the equation not work for light?
    thanks
     

    Attached Files:

  2. jcsd
  3. Jul 2, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That's it. You need to use this more general equation, which works for all particles, including massless photons:

    [tex] E^2 = mc^4 + p^2c^2 [/tex]

    (p is the momentum of the particle.)
     
  4. Jul 2, 2011 #3

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Actually, a large number divided by zero is not undefined, it's infinite.
    In this case you are dividing zero by zero which is indeterminate but can be an exact number, you just have to solve the problem some other way.
     
  5. Jul 2, 2011 #4

    Drakkith

    User Avatar

    Staff: Mentor

    I thought ALL numbers divided by zero are undefined.
     
  6. Jul 2, 2011 #5

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If 0/0=x, then x should be a solution of 0=0x. All values of x are solutions, so any real number is a possible value of 0/0. That's why it's called an indeterminate form.

    It's misleading to say that 1/0 and 0/0 are both undefined and just leave it at that, because 1/0 is infinite, but 0/0 is an indeterminate form.
     
  7. Jul 2, 2011 #6

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Consider A/B=C.

    Then A=B*C, correct?

    Now if A=0 and B=0, we have:

    0=0*C, correct?

    This equation is true for every value of C, correct?

    Therefore, 0/0 can be any number you want.
     
    Last edited: Jul 2, 2011
  8. Jul 2, 2011 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, that is why it is called "indeterminant". But "infinite" is included in "undefined".
     
  9. Jul 2, 2011 #8

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not saying it's wrong to call it undefined, just that it's misleading. It hides an important distinction.
     
  10. Jul 2, 2011 #9

    Pengwuino

    User Avatar
    Gold Member

    The past few posts have given me a little confusion because when you're talking about a photon, you're separately taking the limit of [itex]m \to 0[/itex] and [itex]v \to c[/itex] but one is not dependent on the other. If I could decide I simply want to take m to 0, you should find the first term drops out and end of story, it's 0, not infinite.
     
  11. Jul 2, 2011 #10

    atyy

    User Avatar
    Science Advisor

    Can you explain more? I have never understood this point. It was always explained to me that the relativistic mass is a good heuristic for showing that only massless particles can travel at the speed of light. I have never found the explanation that the true equation is E=rest mass+momentum2 to be any more rigourous, since momentum is not defined in the same way as for massive particles anyway. The only way I know to make it "correct" is to invoke quantum theory and say that mass is ultimately a dispersion relation of a wave. It'd be cool if a beloved 0/0 heuristic has more to it than I learnt.
     
  12. Jul 2, 2011 #11

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    I'll reiterate my favorite solution, which I haven't seen any book. I makes no use of division by zero, no funny definition of momentum.

    For any particle with mass, it is straightforward to derive that:

    sqrt(1 - v^2/c^2) = 1 - KE / (KE + E0)

    If you want to have a particle with E0=0 and KE > 0, and have it follow as many of the same rules as normal particles as possible, then v=c is immediately implied. No limit, no zerodivide, no alternate momentum definition. THEN, having established this, you use :

    E2=mc^4+p^2c^2

    to arrive at p=E/c, where we know that for massless particles all E is kinetic, and m=0.
     
  13. Jul 2, 2011 #12
    If 1/0 = infinite then 1/infinite - 1/0 = 0 but 1/infinite is undefined so it could not be true.
     
  14. Jul 2, 2011 #13

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    The limit as x approaches 0 for 1/x doesn't even exist since the limit approaches plus and minus infinity depending on which way you go, so how can you define 1/0 as infinity? Defining it as negative infinity would also be equally valid.

    Defining 1/0 to be infinity causes all sorts of problems. It would define, then, infinity*0=1. But infinity*0 is an indeterminate form.

    Also, defining 1/0 to be infinity would first require the definition of infinity to be one of the integers. This definition would ruin the fundamental theorem of arithmetic since infinity could not be uniquely factored into primes.
     
  15. Jul 3, 2011 #14

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    The OP was asking about an equation for the energy of a moving particle. He stated that a massive particle has an undefined energy when moving at the speed of light and said that was impossible. I've never heard of anyone stating that the energy of a massive particle becomes undefined when traveling at the speed of light. I've always heard that it takes an infinite amount of energy to accelerate a massive particle to the speed of light.

    Then he stated that a massless particle (a photon) also has an undefined energy and he wondered if photons therefore have undefined amounts of energy or if the equation doesn't work for light. But the equation doesn't yield an undefined energy for light in the same way that it does for a massive particle, rather 0/0 is indeterminate, it can have any value, you just have to determine that value some other way. It's not that the equation doesn't work for light in the sense that it is no longer true, but rather it just doesn't tell you which of the many possible answers is the correct one.
     
  16. Jul 3, 2011 #15

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    I'm not following this at all. If 1/0 is infinite then 1/infinite = 0. Where did you get - 1/0? And why do you say that 1/infinite is undefined?
     
  17. Jul 3, 2011 #16
    how would this equation work in terms of particles like the tachyon which moves at superluminal speeds (assuming it exists - which it probably doesn't)
     
  18. Jul 3, 2011 #17

    Doc Al

    User Avatar

    Staff: Mentor

    It still works for tachyons, but E, p, and m are all imaginary numbers. :bugeye:
     
  19. Jul 3, 2011 #18
    awesome, thanks again for the quick reply. i am somewhat enlightened.
    although i'm sure that energy cannot be imaginary.
     
  20. Jul 3, 2011 #19

    Doc Al

    User Avatar

    Staff: Mentor

    But you're OK with imaginary momentum and mass? :wink:
     
  21. Jul 3, 2011 #20
    i guess you're right. its all crazy!!
    i suppose negative or imaginary mass would sum up to gravity having a repulsive effect or something. i guess that's feasible. if mass is imaginary then using any momentum equation:
    p=mv or planck's constant/lambda; i suppose the momentum also is imaginary.
    but the energy works out to be imaginary too (according to your equation) and that surely cant be right?
    if you slot the numbers into the original equation i had before, you get a complex number divided by a complex number - a real number. how does that work??
    surely tachyons cause the equations to lose their meaning - my main reason for thinking they don't exist
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why can light travel at the speed c?
Loading...