Results 1 to 5 of 5

Math Help - Finding inverse function

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    184
    Thanks
    1

    Finding inverse function

    In the theory of relativity, the mass of a particle with velocity v is

    m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}

    where m_{0} is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.

    I've tried to make v the subject and I got this,

    \displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)

    When I square root both sides to get v, should I choose the +ve or the -ve square root?

    The v here is just the speed and no direction is involved here, am I right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Apr 2008
    Posts
    191
    Quote Originally Posted by acc100jt View Post
    In the theory of relativity, the mass of a particle with velocity v is

    m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}

    where m_{0} is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.

    I've tried to make v the subject and I got this,

    \displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)

    When I square root both sides to get v, should I choose the +ve or the -ve square root?

    The v here is just the speed and no direction is involved here, am I right?
    This is the formula for mass increase:

    m=f(v)=\frac{m_{0}}{\sqrt{1-(\frac{v}{c}})^2}
    m= mass when obeject is moving at speed v
    m_0= mass when object is not moving

    So, now you're solving for velocity. The velocity CAN be negative since it is a vector. Given that the direction can be 180 degrees away from the reference, it can be considered negative. It really depends on your problem and the value you are interested in.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Dec 2009
    Posts
    17
    I'm not really good at physics or anything, but I'm making a guess.

    In physics, the values are highly directional. To really know the eventual signs of the values, one must first understand the equation.

    As we also know, the rest mass of a particle is, as far as I have learnt, positive; consequently, the mass of a particle with velocity v is always positive.

    Let the rest mass of the particle be M.

    And, (m) = (+)/(+) = (+)

    More importantly, you have to know the inequality shared between M & m.

    Firstly,

    let g(v) = [1 - (M^2)/(m^2)]^(1/2), and the range: (0,1)

    The range is in consideration that no particle can attain the speed of light.

    Hence, let [1 - (M^2)/(m^2)]^(1/2) be k.

    => m = M/k

    when k tends to 1, m ≈ M > M
    when k tends to 0, m ≈ ∞ > M

    Hence, the inequality is this: m > M

    Hence, as we both know, physics is very particular about the nature and direction of all matters, or whatever in the universe, we choose signs; in other words, this is not a matter of mathematics where we show a full range of values, instead, in physics, we show the full range of possible, and desired range.

    Following your method, by square rooting the expression, I will arrive at this:

    v = [(c)^(1/2)][1-(M^2)/(m^2)]^(1/2)

    Using what we know about the values of M & m as positive values, and the fact that c is merely speed, and thus 'positive'.

    Since M < m
    (M^2)/(m^2) is always < 1, hence,
    1-(M^2)/(m^2) is always positive.

    (v) = (+)(+) = (+), hence v is always positive.

    There you have it, you should most probably choose the +ve square root, that is if you always define it in the direction it goes, or it moves in a linear fashion you defined it positive.
    Last edited by werepurple; December 6th 2009 at 09:49 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by acc100jt View Post
    In the theory of relativity, the mass of a particle with velocity v is

    m=f(v)=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}

    where m_{0} is the rest mass of the particle and c is the speed of light in vacuum. Find the inverse function of f.

    I've tried to make v the subject and I got this,

    \displaystyle v^{2}=c^{2}\left(1-\frac{m^{2}_{0}}{m^{2}}\right)

    When I square root both sides to get v, should I choose the +ve or the -ve square root?

    The v here is just the speed and no direction is involved here, am I right?
    Since you want an inverse function, f has to be restricted so that it's 1-to-1. The restriction used will determine whether you want the +ve or -ve square root in your inverse function. The restriction v > 0 => +ve root.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2006
    Posts
    184
    Thanks
    1
    I see, so it is my choice to choose the restriction. Thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need help finding an inverse function
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 3rd 2011, 03:57 PM
  2. Finding a Function's Inverse
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 6th 2010, 01:48 PM
  3. Finding inverse of this function?
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: February 2nd 2010, 09:53 PM
  4. finding the inverse of a function
    Posted in the Calculus Forum
    Replies: 8
    Last Post: October 15th 2009, 08:54 PM
  5. Finding the inverse of a function
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 11th 2009, 05:28 PM

Search Tags


/mathhelpforum @mathhelpforum