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Electron with infinite mass

  1. Jan 14, 2005 #1
    According to Einstein, any thing moves or tends to move with or nearly equaly to the velocity of light will have its mass equal to infinity.
    Therefore my question is that:
    what is the reason that electrons being matter and moving with velocity nearly equal to light doesnot have mass equal to light?
     
  2. jcsd
  3. Jan 14, 2005 #2
    Most of the time electrons move slowly when compared to the speed of light.
    I think a case in which the high velocity has forced physicists to use relativistic expressions has been that of cosmic rays.
    Of course also in the case of particle accelerators you might have electrons moving "fast". (just my guess, I don't know enough about particle physics)
     
  4. Jan 14, 2005 #3

    jtbell

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    The Stanford Linear Accelerator (SLAC) accelerates electrons to an energy of 50 GeV which corresponds to a speed of 0.99999999995c. (Assuming I counted the decimal places correctly! :uhh:)
     
  5. Jan 14, 2005 #4
    That's it's effective kinematic mass, not it's gravitational mass. All it means is that the faster something goes, the harder it is to accelerate further, and becoming infinite at the speed of light, this makes this an unreachable limit (you would have to push infinitely hard).

    The mass is defined by

    m=gamma * restmass

    where gamme = 1/sqrt(1-(v/c)^2)

    where v is the velocity of the electron and c is the speed of light.

    Hope this helps,
    Chris.
     
  6. Jan 14, 2005 #5
    Aren't always gravitational and kinematic mass the same as far as we know?
    I think the only thing you can say remains the same is the rest mass. The relativistic mass will increase as speed increases.
    Am I wrong to say that the electrons in SLAC are attracted much more strongly to the earth than a non-relativistic electron?
     
  7. Jan 14, 2005 #6
    With the speed given by jtbell (v=0.9...95*c) I obtained gamma=31623. So the relativistic mass of these electrons is not very big (roughly a Fluorine nucleus at rest).
     
  8. Jan 14, 2005 #7

    dextercioby

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    That's a really poor analogy... :yuck: A fluorine nucleus has a rest mass of approximately 20GeV (actually under 19),which is waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay below 50GeV...

    Learn to manipulate numbers and physical constants...

    Daniel.
     
  9. Jan 14, 2005 #8

    dextercioby

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    Why did u bring gravity into discussion...??I'm sure u're well aware that the gravity effects (wrt relativistic mass) inside an accelerator of (fundamental) particles are much,very much less,than the ones SR predicts...

    Daniel.

    PS.Therefore it makes no point into bringing into discussion relativistic mass in gravity fields...
     
  10. Jan 14, 2005 #9
    GeV is used for energy. I calculated a MASS (Kg)!
     
  11. Jan 14, 2005 #10

    dextercioby

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    If i'm not mistaking,the gravity mass is
    [tex] m_{rel.gravity}=\frac{m_{0}}{\sqrt{1+\Phi/c^{2}-\beta^{2}}} [/tex]

    So if u're defining
    [tex] m_{rel.kinetic}=:\frac{m_{0}}{\sqrt{1-\beta^{2}}} [/tex]

    ,i think you can draw your own conclusions.

    Daniel.
     
  12. Jan 14, 2005 #11
    Hi Daniel,

    I ask you to solve this problem:

    "An electron is moving with constant velocity so that its relativistic (kinetic) mass is equal with the mass of Fluorine nucleus (at rest). Calculate the velocity of this electron."

    BTW......what's so wrong with this analogy? Maybe the numbers are not so accurate but the message is very clear: even at the speed presented by jtbell, kinetic masses are still small.
     
    Last edited: Jan 14, 2005
  13. Jan 14, 2005 #12

    dextercioby

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    You didn't get the point,did u...?The analogy is bad,because it specifies a wrong nuclei.You need a nuclei with A>=50 (titanium,vanadium,that area of periodic table) to get the analogy correct.
    Numbers are important in physics.

    What do you mean "small"??50 Gev cf.0.5MeV i'd say it's pretty big,wouldn't u say so??

    Daniel.
     
  14. Jan 14, 2005 #13
    Daniel, can you answer my question (the first one)?
     
    Last edited: Jan 14, 2005
  15. Jan 14, 2005 #14

    dextercioby

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    Are u making fun of of me???
    [tex] \gamma\sim\frac{20GeV}{0.5MeV}=40,000. [/tex](1)

    [tex] \beta=\frac{\sqrt{\gamma^{2}-1}}{\gamma}\sim \frac{\sqrt{1,599,999,999}}{40,000} [/tex] (2)

    [tex] v=\beta\cdot c\sim 7.5\cdot 10^{3}\sqrt{1,599,999,999}ms^{-1} [/tex]


    Daniel.
     
  16. Jan 14, 2005 #15
    Thanks Daniel!

    Now you see that the answer is in fact the velocity given by jtbell. And the relativistic mass (Kg) of an an electron with that velocity is equal with the rest mass (Kg) of a Fluorine nucleus!!!
     
  17. Jan 14, 2005 #16

    Hans de Vries

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    0.99999999995c ==> gamma = 100,000
    0.9999999995c ==> gamma = 31,622

    The calculation is ok but the input is wrong :rolleyes: So many 9's.....

    Regards, Hans
     
  18. Jan 15, 2005 #17

    dextercioby

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    He,he,i knew there was something fishy with his calculations.Too many 9-s indeed.That's why i didn't extract the sq root and make the division through gamma.

    Daniel.
     
  19. Jan 15, 2005 #18

    jtbell

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    OK, here's how I calculated it... I started with

    [tex]\frac {v}{c} = \frac {pc}{E}[/tex]

    which I used a lot when I was in particle physics once upon a time...

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

    Applying the binomial approximation

    [tex](1-x)^n \cong 1-nx[/tex]

    I got

    [tex]\frac {v}{c} \cong 1 - \frac {1}{2} \left(\frac {mc^2} {E} \right)^2[/tex]

    Plugging in numbers,

    [tex]\frac{v}{c} \cong 1 - \frac {1}{2} \left( \frac {0.0005 GeV}{50 GeV} \right)^2 = 1 - \frac {1}{2} 10^{-10} = 1 - 0.00000000005 = 0.99999999995[/tex]

    (with ten 9's)
     
  20. Jan 15, 2005 #19
    Yes, indeed there are ten 9's. The "equivalent" nucleus would be a "little" heavier, as Daniel told me before. :grumpy:
     
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