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Relativity: Mass with respect to velocity

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data
    In the theory of relativity, mass of a particle is

    m= m[itex]_{0}[/itex]/ [itex]\sqrt{1-v^{2}/c^{2}}[/itex]

    m is the mass when the particle moves at v relative to the observer, and m[itex]_{0}[/itex] is the rest mass.

    Sketch the graph of m with respect to v.


    2. Relevant equations
    E = mc[itex]^{2}[/itex]


    3. The attempt at a solution
    No idea where to start :(
     
    Last edited by a moderator: Mar 23, 2013
  2. jcsd
  3. Mar 23, 2013 #2

    Curious3141

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    Don't you know how to sketch a curve? Think of what the vertical intercept is, and whether there's an asymptote.
     
  4. Mar 23, 2013 #3

    LCKurtz

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    Would you know where to start if the problem was to graph$$
    y = \frac 1 {\sqrt{1-\frac{x^2} 9}}$$?
     
  5. Mar 25, 2013 #4
    Theory of special relativity graph

    1. The problem statement, all variables and given/known data

    In the theory of special relativity, mass of a particle is

    [itex]m = \frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}[/itex]

    m is the mass when the particle moves at v relative to the observer, and m0 is the rest mass.

    Sketch the graph of m with respect to v.


    2. Relevant equations

    [itex]m = \frac{m_{0}}{\sqrt{1-v^{2}/c^{2}}}[/itex]



    3. The attempt at a solution

    In order to sketch the graph, I had to first (before finding the derivative and second derivative):

    - Define the domain
    - Find the x and y intercepts
    - Find where there would be asymptotes (both vertical and horizontal)
    - Check for symmetry

    DOMAIN

    When the x-axis is the velocity of a particle, the domain is 0 [itex]\leq x [/itex][itex]\leq c[/itex]

    INTERCEPTS

    [itex]f(x) = y = m [/itex]

    [itex]x = v [/itex]

    therefore,

    [itex] f(x) = \frac{m_{0}}{\sqrt{1-x^{2}/c^{2}}} [/itex]
    I didn't find any x-intercepts

    Found a y-intercept at [itex] y = m_{0} [/itex]

    ASYMPTOTES
    since domain has been restricted to be between 0 and C, there is no horizontal asymptote.

    Vertical asymptote:
    Took the just the denominator and made it = to zero, got v=c. Since v is x. It is [itex] x = c[/itex], the vertical asymptote.


    SYMMETRY

    There is Even symmetry, [itex] f(x) = f(-x) [/itex]

    Ok, so now the 1st and 2nd derivatives

    1st derivative

    [itex]m^{\prime} = \frac{-mv}{c^{2}-v^{2}}[/itex]

    x-intercept at 0, therefore there is an interval of increase between 0 and c.

    2nd derivative

    [itex] m^{\prime \prime} = \frac{-(m + v) + 2m^{\prime}v}{c^{2} - v^{2}} [/itex]

    Here is when i am really stuck, how do we find the zeros for the second derivative?

    BTW, pic of question and answer is attached. (number 53)
     

    Attached Files:

    Last edited by a moderator: Mar 26, 2013
  6. Mar 26, 2013 #5

    Simon Bridge

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    You have an equation for m and m' to substitute in, m''=0, so solve for v.
    Shortcut - what would m''=0 tell you?
    What would that mean for relativity?
    Is that reasonable physically?

    http://www.weburbia.com/physics/mass.html
     
  7. Mar 28, 2013 #6
    Thanks, that helped alot. Have a great day.
     
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