# Finding inverse function

• Dec 6th 2009, 06:12 PM
acc100jt
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?
• Dec 6th 2009, 08:43 PM
Roam
Quote:

Originally Posted by acc100jt
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.
• Dec 6th 2009, 09:29 PM
werepurple
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.
• Dec 7th 2009, 02:53 AM
mr fantastic
Quote:

Originally Posted by acc100jt
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.
• Dec 7th 2009, 05:50 PM
acc100jt
I see, so it is my choice to choose the restriction. Thanks :)