# Special theory of relativity: Variation of relativistic mass with velocity

There is a topic called variation of mass with velocity. In which to get the equation of mass in relativistic case we consider collision of two particles. We take steady frame S1 and moving frame S2. When we consider moving frame S2 we assume collision of two masses, each of having same value m moving in the opposite direction to each other with the same speed. using such information we get value of velocities of the particles for corresponding steady frame S1 using velocity transformation formula. When we write equation of conservation of the momentum for the masses in steady frame S1 we consider masses to be different i.e. m1, m2. In moving frame masses are considered identical but for steady frame they are considered different. Why?

The simple answer to your question is that since (presumably) we're considering mass to be a function of speed, since the bodies have the same speed in S2 they will have the same mass, assuming that they are identical bodies.

I feel bound to point out, though, that it's becoming less and less common among physicists to talk about mass varying with speed. The argument concerns how to interpret the formula for the momentum, $\vec{p}$, of a body moving at velocity $\vec{u}$, speed u, namely$$\vec{p}=\gamma m_0 \vec{u}\ \ \ \text{in which}\ \ \ \gamma=\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}.$$ Old interpretation

$m_0$, a constant for the body independent of its motion, was called the rest mass of the body. $γm_0$, which was sometimes denoted by m, was called the body's relativistic mass, and is speed-dependent. The idea of calling $γm_0$ 'relativistic mass' was that you could continue to use the Newtonian formula, $\vec{p}=m \vec{u}$, provided that you used so-called 'relativistic mass' instead of rest mass.

One of the reasons why this interpretation has fallen out of favour is that putting $γm_0$ instead of $m_0$ doesn't make other Newtonian formulas into relativistic ones. For example, it won't make $KE=\frac{1}{2}m_0 u^2$ into the relativistic formula $$KE=(\gamma-1)m_0 c^2.$$ Modern interpretation

Instead of regarding $γm_0$ as a speed-dependent mass that, when multiplied by $\vec{u}$, gives us $\vec{p}$, it's at least as logical to regard $\vec{p}=\gamma m_0 \vec{u}\$ as $\ \vec{p}= m_0 (\gamma \vec{u})$, that is as $m_0$, a constant for the body, multiplied by $\gamma \vec{u}$, a kinematic quantity (called proper velocity) that replaces $\vec{u}$ in the ordinary Newtonian expression.

This means that we don't call $γm_0$ 'relativistic mass'. If we want to call it anything, it's the mass equivalent of the body's total energy (KE + rest energy), $\gamma m c^2$. Remember that $c^2$ is merely a constant!

In that case there's no need to call $m_0$ 'rest mass' – it's the only sort of mass we talk about. Nor is there any need for the zero subscript! So the two formulas we quoted earlier are usually written$$\vec{p}=\gamma m \vec{u}$$and$$KE=(\gamma-1)m c^2.$$

• The first sentence of your answer says that mass changes with speed. But actually this is the result of this topic. How can we use the result before getting result ? – ADR Feb 27 '18 at 13:48
• I thought you were asking about why masses are treated as the same in one frame but not in the other. My first paragraph answered this question. I assumed you were talking about so-called 'relativistic masses', as the question wouldn't make sense if you were talking about rest masses. As I went on to say, $relativistic mass$, that changes with a body's speed (or according to our frame of reference) is falling out of favour as a concept. Some would say that it fell out of favour forty years ago! – Philip Wood Feb 27 '18 at 14:01
• Just to clarify things: I'd say (with modern use of 'mass') that the bodies' masses are the same in both S1 and S2, and, if the bodies are identical, the same as each other – Philip Wood Feb 27 '18 at 15:00
• Yes i am talking about relativistic masses. You are right the topic in the book starts with the assumption that mass is dependent on the velocity. That means based on the assumption we are getting the result, but what was the reason to consider such assumption. Had it been experimentally proved before? – ADR Feb 27 '18 at 15:14
• @AmritDas "How can we use the result before getting result ?" You may be missing the way you can choose a frame that you understand things in, work out the problem in that frame and then transform the kinematics only and ask what that requires of dynamics. If you know how positions, times, and velocities are affected by changing frames you can use that knowledge to learn how dynamics (force, energy, momentum, or—if you really must—"relativistic mass") transform, as well. The frame where they have the same speed must be simple. – dmckee Feb 27 '18 at 15:31

The relativistic mass concept historically was defined to extend the conservation of momentum law to SR (special relativity) as well. What is assumed is that the mass of a particle is function of the velocity in a given reference frame, that is $m = m(u)$ with the Newtoniam limit $m(u = 0) = m_0$ where $m_0$ is the mass of the particle in a reference frame where the particle is at rest.

The logical steps of the demonstration are:
1. It is considered an elastic collision between two identical bodies, that is having the same rest mass, in two reference frames in relative motion. In the frame where the velocities of the bodies are the same in absolute value the masses are assumed to be the same, otherwise they are labelled differently.
2. The velocity Lorentz transformation is applied to translate the velocities from one reference frame to the other.
3. It is required that the conservation of momentum law holds in SR as well.

As a consequence you get the relation
$m = \gamma m_0$
where:
$\gamma = 1 / \sqrt{1 - u^2/c^2}$

The demonstration is consistent and it is not correct to say that the outcome was already embedded in the assumption. The demonstration started from assuming a freedom in the definition of mass of a moving particle and then requiring it to comply with the conservation of momentum in SR. The relativistic momentum was validated by the experimental evidence, even if the $\gamma$ factor should be read together with the velocity and leave the rest mass as the mass of the particle.