# Thread:

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?

2. 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.

3. 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.

4. 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.

5. I see, so it is my choice to choose the restriction. Thanks
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