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F=ma at light speed

  1. May 10, 2008 #1
    Im just an amateur, and I read that Newton's Equation F=ma doesnt work when approaching light speeds because of a relative increase in mass at higher velocicites. What is the reason for this, if any, and is there a ratio between veolcity and relative mass I could use to make my calculations more accurate when dealing with light speed? Thanks!
     
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  3. May 10, 2008 #2

    Hootenanny

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    [itex]F = ma[/itex] may not work, but the full version of Newton's Second law does,

    [tex]\mathbf{F} = \frac{d\mathbf{p}}{dt}[/tex]

    even at velocities close to the speed of light. However, you should note that unlike the classical case, the acceleration is not generally in the same direction as the force.
     
    Last edited: May 10, 2008
  4. May 10, 2008 #3
    Thanks, but can you tell me what the variables d, p, and t are? Im new here and not yet familiar with most full formulas
     
  5. May 10, 2008 #4
    In special relativity it is also F = m0A. And this forumla always works!

    But the difference is that A is a 4-vector. A is the 4-acceleration vector which is basically the derivative of the 4-velocity vector with respect to proper time. And the 4-velocity vector is basically the change in direction in spacetime with respect to proper time. But note that we have four, not three, directions in spacetime!
    In addition there is the 4-energy-momentum vector which is mass times the 4-velocity vector.

    Four vectors are simple and elegant.

    As you can see it is pretty straightforward, we simply have to add the extra direction in relativity. :smile:
     
  6. May 10, 2008 #5

    Hootenanny

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    p is the momentum of the particle and,

    [tex]\frac{d}{dt}[/tex]

    Is the first derivative with respect to time. I'm guessing that you haven't done any calculus if you haven't seen this notation before. If that is the case, a more appropriate formulation would be,

    [tex]m\mathbf{a} = \mathbf{F} - \frac{\left(\mathbf{F}\mathbf{\cdot}\mathbf{v}\right)\mathbf{v}}{c^2}[/tex]

    You should note that the product [itex]\left(\mathbf{F}\mathbf{\cdot}\mathbf{v}\right)[/itex] isn't simply a scalar multiplication, rather it's the dot product between the two vectors (force and velocity).
     
  7. May 10, 2008 #6

    malawi_glenn

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    the d's mean that you take the derivative of momentum (p), with respect to time (t)

    Veolcity = [tex] \frac{d\vec{r}}{dt} [/tex]

    Acceleration = [tex] \frac{d\vec{v}}{dt} [/tex]

    Momentum = [tex] \vec{p} = \gamma m \vec{v} [/tex]

    Where: [tex] \gamma = \frac{1}{\sqrt{1-(v/c)^2}} [/tex]
     
  8. May 10, 2008 #7
    You lost me. Could you explain that in terms that an 8th grader who's only in Geometry could understand?
     
  9. May 10, 2008 #8

    Hootenanny

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    8th grade, so you're about fourteen right? Without wanting to sound too harsh, I'm afraid that you're going to have a difficult time understanding this. Perhaps in a few years when you've done some more advanced mathematics (and basic physics) courses you might be able to have another go, but at the moment relativity is a little above your level. Which begs the question why are you studying SR? Is it for a project, or are you just generally interested?
     
  10. May 10, 2008 #9
    I am interested, and I wanted to see around how much U-235 you would have to fission (at 100% efficiency) to propel the space shuttle to light speed (i know its not that simple, but i wanted a round number to see how much you would need) and at the moment all I can do is use F=ma because thats all I know, but I was hoping that there would be some sort of ratio to how fast you go to how much mass you gain, because then I could simply tack that on to my equation and plug in the numbers and solve again, with a more accurate answer.
     
  11. May 10, 2008 #10
    I think it is great that someone who is 14 years old is asking questions about relativity. I think a 14 year old should already understand derivatives. Myself I was 12 years old when I became familiar and understood special relativity.

    Why anyone would discourage, or worse put down, the young and enthusiastic people is beyond me.
     
  12. May 10, 2008 #11

    HallsofIvy

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    Now that's a challenge! You know of us had to spend years and years in classes just to beign to understand this stuff!

    The "derivative" of a quantity is essentially its instantaneous rate of change as opposed to an "average" rate of change. The "a" (acceleration) in F= ma is the instantaneous rate of change of velocity. That could be written [itex]F= m dv/dt[/itex]. That is true as long as the mass (m) is constant. If it is changing itself it is better to say "Force = rate of change of momentum" and momentum, of course, is "mass times velocity" so "F= rate of change of mass times velocity" or F= d(mv)/dt.

    By the way, you titled this "F= ma at light speed". Since nothing with mass can move at light speed, the question is vacuous.
     
  13. May 10, 2008 #12
    So I take it no one has ever found the ratio between an object's velocity and its relative gained mass, am I correct in assuming this?
     
  14. May 10, 2008 #13

    robphy

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    Could you elaborate on what "understood" means in this sentence?
    I was about 13 when I was introduced to relativity [via a video on PBS]... but understanding it took a while.

    ...and use the Minkowski metric.
     
    Last edited: May 10, 2008
  15. May 10, 2008 #14

    Hootenanny

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    It's good that your interested, but there are a couple of points to mention here. Firstly, no matter how much energy you have your space shuttle will never reach the speed of light, it can approach arbitrarily close to the speed of light, but it will never actually get there. Secondly as I have said previously, the acceleration produced by a given force isn't necessarily in the same direction as the force. One can use the concept of relativistic mass, as described by MG, but you simply can't substitute this force the mass term in the classical expression.
    It is not an insult to tell someone that they haven't the sufficient knowlage to understand a concept. It is far worse to lead someone to believe that they do understand something when they don't.
     
  16. May 10, 2008 #15

    Hootenanny

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    Pretty standard 8th grade material :rolleyes:
     
  17. May 10, 2008 #16

    robphy

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    I was just making the point that
    "relativity" is more about adding a dimension to "space" and dealing with a higher-dimensional space. Indeed, most of the trouble folks have with relativity is not about its dimensionality. Just consider every example with "motion in the x-direction".

    There are ways to handle the non-euclidean-signature metric without explicitly invoking the metric or the Lorentz transformation.... but most textbooks don't do them (or seem to know about them).
     
    Last edited by a moderator: May 10, 2008
  18. May 10, 2008 #17
    In my generation the concept of derivatives was taught at schools when the student was between 13 and 14 years old. Is it now different?
     
  19. May 10, 2008 #18

    Hootenanny

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    I would assume so since the OP states that he hasn't met them before. From what I know of the US education system calculus is usually introduced in the 11th grade (16-17).

    Anyway, we're getting off topic. Are you still around rocky14159?
     
  20. May 10, 2008 #19

    HallsofIvy

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    Certainly not. An object with rest mass m0, moving at speed v, will have relativistic mass of
    [tex]\frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}[/tex]

    If you know about mass increasing I'm suprised you hadn't see that formula before.
     
  21. May 10, 2008 #20

    malawi_glenn

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    I learned that stuff in kindergarden :biggrin: I had P.A.M Dirac as teacher ;)
     
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