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Relativity and infinities

  1. Jul 26, 2004 #1
    Is it impossible to accelerate a rest mass to the speed
    of light or does special relativity have an infinity that
    should be removed.For example, what if

    rest mass / (1 - v^2 / c^2)^1/2

    should really be rest mass / (1 - v^2 / c^2 + small constant)^1/2

    The small constant would also stop lengths contracting to zero.
     
  2. jcsd
  3. Jul 26, 2004 #2
    I've never been sure how to respond to a question like this where someone asks if nature should be a different way than it appears to be.

    Its all pretty straight forward. If you want to know what "really" is then you start with the postulates and then go from there. The postulates of SR, coupled with the other laws of nature (i.e. energy conservation, Maxwell's equations etc.). With that as a start Einstein showed that the amount of work, W, done to accelerate a particle of proper mass m0 to a speed v is given by the relation

    [tex] W = m_{0}c^{2} \left ( \frac{1}{\sqrt{1-v^2/c^2}} - 1 \right )[/tex]

    The momentum, p, of a particle of proper mass m0 moving at speed v is given by

    [tex]\bold p = \frac{m_{0}\bold v}{\sqrt{1-v^2/c^2}} [/tex]

    Since both W and p are both infinite when v = c it follows that there would be a violation of the principles of the conservation of energy and conservation of momentum for the particle to move that fast since if the particle started out at rest that energy and momentum had to come from somewhere.

    Pete
     
    Last edited: Jul 26, 2004
  4. Jul 26, 2004 #3
    Surely all particles - even photons- must be accelerated up to the speed they have -
    afterall some photons are made of two rest masses - positrons and electrons fpr example.
     
  5. Jul 26, 2004 #4
    Photons can't be accelerted in an inertial frame. But what does this have to do with your question? Thanks

    Pete
     
  6. Jul 26, 2004 #5
    The point I'm making is that it might take a large but finite energy to accelerate a particle with rest mass to the speed of light.Infinity is not accepted in quantum field theory why is it accepted in special relativity - infinity is not a number.
     
  7. Jul 26, 2004 #6
    I understand your point. You seem to have the impression that infinity is a number. It is not a number. You also seem to think that infinite energy/momentum is something allowable in SR. It is not allowable.When a physicists poses a question then he uses the laws of physics, expressed in the form of equations, to answer them. When the result of such an answer is "infinite mass/momentum etc" then the way to interpret that is that one or more of the starting assumptions are wrong. The starting assumption that is flawed is assuming that a particle can have the speed v. So when the answer is infinite then it means "Can't be done".

    The answer to your question "should really be rest mass / (1 - v^2 / c^2 + small constant)^1/2" the answer is no.

    Pete
     
  8. Jul 26, 2004 #7
    The small constant would also stop lengths contracting to zero.

    Wouldn't that prevent the formation of a singularity in black holes
    and at the beginning of the universe?
    And ignoring the fact that you don't agree with having the small constant in the first place,if it did exist,what effects would it have on the theory of general relativity?
     
  9. Aug 4, 2004 #8
    good pt.

    I am no pro at any of this, but I have always just taken speed of light and never thought too much about it but,

    if you have a flashlight and you turn it on, i can't see how instintaneously the photons could be moving the speed of light, i know i have been taight they always move at the speed of light which i can accept but, logically thinking for any thing to get from point a to point b in a certain amount of time from rest they must start going and if they want to speed up to get to a certain limit such as c you would get a vel then an acceleation by def of the next derivative.

    I know DW can explain this well hes brilliant, how can you go from rest in flashlight to C without first a increasing vel causing acc. only way would be to travel linearly to the speed of light which would take hellalong time, well even that would have acc but it would be constant.

    DW i await your elegant solution.
     
  10. Aug 4, 2004 #9

    DW

    User Avatar

    The solution is simple. You are making the assumption that before turning on the light there are photons sitting at rest with respect to the flashlight which then speed up in the process of emission. That is a false assumption. The photons in the beam didn't exist prior to emission. They are created with an initial velocity of c, so there was no acceleration process at all. Nothing massless can travel at any other speed than c.
     
  11. Aug 4, 2004 #10
    interesting, I like it, but how was that proved, or how is it now proved, is itjust because all massless particles have V at C at all pts in time, and if so how is that prooved.
     
  12. Aug 4, 2004 #11
    EDITED VERSION (thx@DW 4 help):
    ----------------------------------

    Dispersion relation for free particles:
    [tex] E(p) = \sqrt{ p^2c^2 + m^2c^4} [/tex] (1)


    Wave function of a free particle:
    [tex] \Psi = C \cdot \exp \left[ \frac{-i}{\hbar} \left( Et - px \right) \right] [/tex] (2)

    Renaming
    [tex] \frac{E}{\hbar} =: \omega, \, \frac{p}{\hbar} =: k [/tex] (*)
    leads to:
    [tex] \Psi = C \cdot \exp \left[ -i \left( \omega t - kx \right) \right][/tex] (2*)
    Which is the classical wave-equation in the variables usually used.

    Velocity of a wave is:
    [tex] v = \frac{d\omega}{dk} \stackrel{(*)}{=} \frac{dE}{dp} [/tex] (3)

    Plugging in the dispersion relation (1) in (3) results in:
    [tex] v = \frac{dE}{dp} = \frac{d}{dp}\sqrt{ p^2c^2 + m^2c^4} = \frac{pc}{E(p)}c [/tex] (4)

    From (4) using (1) one can see that v=c <=> m=0 and v<c <=> m>0 (possible problems at m=p=0 not considered)

    That would be my guess for a proof. There are, of course, several points where further investigation would be nessecary to make it a complete proof (what happenes when I take solutions of the Dirac-Equation instead of the Klein-Gordon-eq, for example) but I like it as it is.
     
    Last edited: Aug 5, 2004
  13. Aug 4, 2004 #12
    [tex] E(p) = \sqrt{ p^2c^2 + m^2c^4} [/tex] (1)
    So for photons with m=0:
    [tex] E = cp [/tex] (1*)

    Wave function of a free particle:
    [tex] \Psi = C \cdot \exp \left[ \frac{i}{\hbar} \left( Et - px \right) \right] = C \cdot \exp \left[ i \left( \omega t - kx \right) \right][/tex] (2)

    Velocity of a wave is:
    [tex] v = \frac{\omega}{k} = \frac{E}{p} [/tex] (3)
    note that 2nd "=" comes from (2).

    (1*) in (3) => v = c.

    Ok wow this is over me but I'm gonna try cause I asked for it right. so first thanx athiest alot for this, your getting me started.

    Ok I follwed the beginning till you got to wave function and I didin't see how that fit in. Also the velocity of wave I didn't see how that came about and how it was equal to E/P, i believe you and that it is, but still a lil confused,
    and last I know no one will want to do this but can someone tell me what some of those variables are, like in EM right now every variable is used for different things, seems like every chapter has a v but for different things, but if someone can tell me what i , k, x, exp are.

    thanx
     
  14. Aug 4, 2004 #13
    i : Imaginary unit (-> imaginary numbers, complex numbers)
    k: k and omega are p and E divided by h-bar respectively => omega/k = E/p, because h-bar cancels out.
    x : Position in space.
    exp: Exponential function.

    > How the wave-function fit in:
    All particles are described by wave-functions.

    > Where the formula v = omega/k comes from:
    I remebered having learned something like that during my first semesters of studying, so I looked up the formula in book (dunno the english term for those types of books: "equation collection" would be the direct translation from german) and found it there. No explanation given there, though.

    EDIT: Flaw i complained about is probably found.
     
    Last edited: Aug 5, 2004
  15. Aug 4, 2004 #14
    has is there an imaginary in the wave function, does not the wave function show the probability of a particle to be someone at a certain time, how could the root of a negative number survive in that.

    Still not understnad the correlation of the h bar and wave function to the e/p sorry, i try to followyou and can't see it.
     
  16. Aug 4, 2004 #15

    DW

    User Avatar

    It is correct, it is just that for massive particles v is not [tex]\frac{\omega }{k}[/tex]. It is [tex]v = \frac{d\omega }{dk}[/tex] or in terms of a monochromatic wave state one can define the quantum frequency such that [tex]v = \frac{c^{2}k}{\omega }[/tex].
     
  17. Aug 4, 2004 #16
    if v = w/k and for massive objects we take dw/dk would that not be dv with respect to somehting not sure to me thats an acceleration am i wrong?? probably as always

    what does just plain K stand for

    and what is a monochromatic wave.

    thanx
     
  18. Aug 4, 2004 #17

    DW

    User Avatar

    v does not equal w/k for massive objects. v is dw/dk.

    k unbold is 2*pi divided by the wavelength.

    A wave comprised of a single wavelength.
     
  19. Aug 4, 2004 #18
    ok so is v = w/k for small and then derivative of that for massive objects.
     
  20. Aug 4, 2004 #19

    DW

    User Avatar

    [tex]v = \frac{\omega }{k}[/tex] is only the "special case" for light speed waves. In all cases in general it is [tex]v = \frac{d\omega }{dk }[/tex]. It just happens that at v = c we get [tex]\frac{d\omega }{dk } = \frac{\omega }{k}[/tex]. Here is one way to go about it. Start by defining a quantum frequency in terms of the wave number by [tex]\omega ^{2} = c^{2}k^{2} + constant[/tex] The constant will turn out to be proportional to the square of the mass, but for now let it be arbitrary. From that equation, find [tex]\frac{d\omega }{dk}[/tex]. You will get:
    [tex]\frac{d\omega }{dk} = \frac{c^{2}k}{\omega}[/tex]
    For a light speed particle [tex]v = \frac{\omega }{k} = c[/tex] and inserted into the above results in
    [tex]\frac{d\omega }{dk} = \frac{c^{2}k}{\omega} = c[/tex]
    For all particles light speed or not, the numerator is proportional to the momentum and the denominator is proportional to the energy resulting in
    [tex]\frac{d\omega }{dk} = \frac{pc^{2}}{E}[/tex]
    Again, for light like particles E = pc and that yields v = c, otherwise its results are consistent with
    [tex]p = \gamma mv[/tex] and [tex]E = \gamma mc^{2}[/tex].
     
    Last edited: Aug 5, 2004
  21. Aug 5, 2004 #20
    Thx 4 help

    That sounds nice and seems to give good results. Thx a lot! I´ll edit my first post, then.
    In case you know that off your head: Could you give a quick explanation why v = dw/dk ? I just remembered having done smth like that some years ago (also remembered there was one velocity with v=w/k and one with v=dw/dk, but I made a mistake when trying dw/dk -forgot factor 0.5 when taking derivativce of sqrt- and came to senseless a result) and just looked that up, but an explanation would be nice.
     
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