# Frame of reference and the varying mass problem

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1. May 14, 2015

### HarikrishnanSB

Hi Friends ,

I had a discussion with a funny person last night.

He said the following argument :

Suppose there are 21 objects including the one object under consideration. Each of the 21 objects move with a velocity. Since there are 21 objects, for the object we are considering has 21 velocities with respect to 21 reference frames. We are fixing the reference frame as each object.

Now since mass = (Force x Time ) / Velocity. There are 21 different mass for the same object for 21 reference frame !!!

I argued him saying that mass is a constant and it cannot change. But he proved me there can be 21 different masses for 21 reference frames.

How can this be possible ? ? ?

Any replies will be appreciated.

2. May 14, 2015

### Staff: Mentor

I suspect that this is an attempt to invert Newton's 2nd law: F = ma → m = F/a. So what?

He proved no such thing.

Why do you think he proved anything?

3. May 14, 2015

### Staff: Mentor

If this is, as Doc Al suspects, an attempt to invert Newton's Second Law, it's an erroneous attempt. Acceleration is the derivative of velocity with respect to time, not the ratio of velocity to time.

Also, his argument up to that point has said nothing at all about force. What force is he talking about? And has he allowed for the fact that force also transforms between frames?

4. May 14, 2015

### DrStupid

This equation for the mass is misleading. In classical mechanics the full one-dimensional equation for bodies with constant mass is

$m = \frac{p}{v} = \frac{{m \cdot v_0 + \int {F \cdot dt} }}{v}$

and therefore

$m = \frac{{\int {F \cdot dt} }}{{v - v_0 }}$

With vo=0 and constant Force it turns into your equation. But this simplified equation applies to special frames of reference only and therefore must not be used for transformation. The Galilean transformation of the full equation results in

$m' = \frac{{\int {F' \cdot dt'} }}{{v' - v'_0 }} = \frac{{\int {F \cdot dt} }}{{\left( {v - u} \right) - \left( {v_0 - u} \right)}} = \frac{{\int {F \cdot dt} }}{{v - v_0 }} = m$