1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove Speed Any Particle with Mass >0 is Always < c

  1. Jun 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Use:
    ##\beta =\dfrac {\overline {u}} {c}=\dfrac {\overline {p}c} {E}\&\&E^{2}=\left( pc^{2}\right) +\left( mc^{2}\right) ^{2}##


    2. Relevant equations



    3. The attempt at a solution
    Solving the second equation for E, I get:
    ##\left( \dfrac {E u} {c}\right) ^{2}+m^{2}c^{4}>0##
    ##E^{2}u^{2}>-m^{2}c^{6}##

    ##\left( p\dfrac {c^{2}} {u}\right) ^{2}u^{2}>-m^{2}c^{6}##
    p^2 > -m^2 c^2

    ##\gamma ^{2}m^{2}v^{2}## > -m^2 c^2
    v^2 >-c^2 +v^2
    v^2<c^2 -v^2
    √2 v<c
    v < (c) / √2, and because √2 << c
    v<c

    Is this correct? I don't think it is, but I cannot think of any other way.
     
    Last edited: Jun 15, 2013
  2. jcsd
  3. Jun 15, 2013 #2
    You din't proved v<c!. You proved sqrt(2)*v<c, which is certainly not true.

    There are many arguments that prove this. The best i remember is the Force method.

    Hint: What is the equation of Force?
     
  4. Jun 15, 2013 #3
    I am required to use these equations:
     
  5. Jun 15, 2013 #4
    As i told you there are many methods. These equations might get used in some or other. But you dont need these equations in force method. First derive the equation of Force. Use F = dp/dt.
     
  6. Jun 15, 2013 #5
    I don't know how to proceed :/
    dp= gamma (m*v_2 - m*v_1 )
    dt= gamma(d t_0), by time dilation formula

    F=ma=dp/dt
    m du/dt = gamma( (m*v_2 - m*v_1 ) )/dt
     
    Last edited: Jun 15, 2013
  7. Jun 15, 2013 #6

    Ibix

    User Avatar
    Science Advisor

    Your first approach isn't far off. Try not dumping terms just because it gets you an inequality. See where it takes you.
     
  8. Jun 15, 2013 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your second equation as written is incorrect. It should be
    $$E^2 = (pc)^2 + (mc^2)^2.$$ Just use this equation to eliminate E in the first equation and show that ##1/\beta > 1##.
     
  9. Jun 15, 2013 #8
    Ok, thanks. I will follow what vela and lbix suggested.

    Vela which second equation are you referring to?
     
  10. Jun 16, 2013 #9
    These are not correct.
    dp=d(gamma*mv).
    dt =dt.(you are at rest not moving. Time dilation is for moving frame)

    Now F= dp/dt, remember gamma is a function of v, not a constant, write the whole function and then diffrentiate dp/dt.

    You can go with your first approach. Remeber the constraint given in the problem. It says something about the mass the object!


    As i said to you 'the best i remember is force method' because only the force method gives you physical understanding of why the speed of any object can't exceed 'c'. The method you are using is all about solving equations mathematically, it won't give you physical understanding.
     
  11. Jun 16, 2013 #10

    Ibix

    User Avatar
    Science Advisor

    Vela was observing a missing square in [itex]E^2=p^2c^2+m^2c^4[/itex] in your first post. I ignored it because the rest of your post implies the correct form.
     
  12. Jun 16, 2013 #11
    These are not correct.
    dp=d(gamma*mv).
    dt =dt.(you are at rest not moving. Time dilation is for moving frame)

    Now F= dp/dt, remember gamma is a function of v, not a constant, write the whole function and then diffrentiate dp/dt.

    You can go with your first approach. Remeber the constraint given in the problem. It says something about the mass the object!


    As i said to you 'the best i remember is force method' because only the force method gives you physical understanding of why the speed of any object can't exceed 'c'. The method you are using is all about solving equations mathematically, it won't give you physical understanding.
     
  13. Jun 17, 2013 #12
    I think I found the proof.
    ## E^{2}-E^{2}\dfrac {u^{2}} {c^{2}}=\left( mc^{2}\right) ^{2}##

    ## E^{2}\left(1-\dfrac{u^{2}}{c^{2}}\right)=m^{2}c^{4} ##

    ##\dfrac{E^{2}}{c^{6}}\left(c^{2}-u^{2}\right)=m^{2} ##

    ##\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} =m ##

    If m>0, then u<c.

    Is this a correct proof?
     
  14. Jun 17, 2013 #13
    That is correct. Good work!
     
  15. Jun 17, 2013 #14
    How do I make this a robust proof?
     
  16. Jun 17, 2013 #15
    I don't know what you mean by robust proof. But if you are looking for the proof which gives some physical understanding, then derive the equation for force as i told you earlier.
     
  17. Jun 17, 2013 #16
    I don't know the exact definition of a robust proof either, but I have heard it so many times in textbooks that I know there's a difference between a weak proof and a robust proof.

    I will try you approach, darkxponent.

    Can anyone tell me, if this proof is robust?

    Thank you all for your help. :)
     
  18. Jun 17, 2013 #17
    I can see a much more simple proof, which might seem more 'robust' (and which has been mentioned before).

    Look at your second equation. If m>0 then E > (pc). Now your first equation is v/c = (pc)/E.

    What does that tell you about v/c ?
     
  19. Jun 17, 2013 #18
    If E>pc, then pc/E <1.
    Then, the ratio of v/c <1 .

    For v/c <1, c>v or v<c.

    :>
    I am not sure what robust means, but I am ok with this one too.
     
  20. Jun 17, 2013 #19

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It wouldn't be a very good proof, even if it were correct. Ideally, you end up with an inequality that is what you're trying to show, i.e. ##|u|<c##. Instead, you have a relatively big leap of logic. How exactly does m>0 imply u<c? You should consider that when you take square root, there's another possible solution:
    $$\dfrac{E}{c^{3}}\left(\left(c-u\right)\left(c+u\right)\right)^{\dfrac{1}{2}} = -m.$$ This ruins your (already flawed) argument. According to your logic, since the righthand side is negative, u isn't necessarily less than c, right? Can you even justify taking the square root? By assuming you can take the square root of the lefthand side, you've implicitly assumed what you're trying to prove.

    Go back to the second line in your proof. If you can somehow justify that ##E \ne 0##, you can divide both sides by ##E^2##. The righthand side will be ##\frac{m^2c^4}{E^2} = \left(\frac{mc^2}{E}\right)^2##. What can you conclude about the sign of that term?
     
  21. Jun 17, 2013 #20
    Good to see. So proofs are raining from everywhere. As i told there are many arguments. You allready got two of them. I dont know how many of these proofs are. Both of these are more or less similar(solving inequalities). If you are looking for the so called robust proof, try the Force method. I can show it to you(as you allready have done yours), but i would like you to do it yourself.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Prove Speed Any Particle with Mass >0 is Always < c
Loading...