# Imaginary Velocity?

1. Jun 20, 2010

### NegativeGPA

So, according to my understanding,

m= m_o/√(1-(v^2/c^2 ))

gives the mass of an object in respect to the object's original mass and its velocity. I wondered what happened if the mass of an object became lower than the rest mass? [I have no idea how this would happen, but it was a, what if it did? kind of question]

I made the substitution m=m_o-a for some arbitrary amount lower than the rest mass. After solving the equation for velocity, i got

v=c√(1-(m_o^2/(m_o-a)^2 )

in the equation, you can see that,
(m_o-a)^2 < m_o^2

thus,
(m_o^2/(m_o-a)^2 > 1

so we would end up with the square root of a negative number, giving the object with an imaginary velocity.

This matched my predictions because the reason I wondered this was because I wanted to know what was so special about the rest mass of an object. Why is that amount of energy in that amount of space a particle, and why does additional energy cause what we call velocity? So i assumed that a lower amount of energy would do somehow the opposite of velocity, but what is that?

2. Jun 20, 2010

### Kevin_Axion

When the rest mass of an object becomes imaginary this is the description of a Tachyon: http://en.wikipedia.org/wiki/Tachyon
The Tachyon's velocity would hence be super-luminal.

3. Jun 20, 2010

### NegativeGPA

Haha thanks, but I'm talking about an imaginary velocity, not an imaginary mass.

I'm asking what would happen if an object's mass became lower than its rest mass, and apparently it has an imaginary velocity. I'm asking if anyone has any idea what that means

4. Jun 20, 2010

### Kevin_Axion

5. Sep 29, 2012

### natoasto

I suppose the 'opposite of velocity' would make it so that the object required velocity to be added to it in order to be brought to a state of rest.

6. Sep 30, 2012

### kurros

Well, the rest mass is special because it is frame invariant (this may seem slightly circular but the rest frame has unique properties, so we are not being arbitrary at least), so we may be justified in thinking it a property of the object rather than something just due to the way we look at it.

I think the fact that you are getting imaginary numbers out really just tells you that it is not possible to go to some reference frame where your object has a lower mass than in the rest frame. All measurable observables are real numbers in physics.

7. Oct 1, 2012

### Staff: Mentor

Please, do not use this relativistic mass. Apart from some old books, nobody uses it any more. You mean the energy of the particle:
E= m c^2 /√(1-(v^2/c^2 ))

At rest, the energy is E = m c^2. This is the lowest possible energy of the particle. If it moves, you add kinetic energy and the total energy has to increase. There is no way to get a lower energy, and therefore you should not expect a meaningful result in your calculation.