Relativistic mass and velocity

Question

I have read somewhere that when an object is approaching the speed of light then its relative mass is increasing by the formula $m=\frac{m_0}{\sqrt{1-(\frac{u}{c})^2}}$. Is this true? If yes then why this happens? Does an object really increase its mass when it has bigger velocity or it has to do with the mommentum? I am confused.If here on earth we measure 1 kg and then accelerate it at $u=c/2$ then its mass will be bigger?

Answer 1

It's just another way of expressing things. I think that nowadays teachers tend to avoid the concept of "relativistic mass", because it may lead to misunderstandings. It was used in my 1st relativity book but I was not at ease with it and all of my teachers rejected it as an old and outdated "notation". As far as I'm concerned, it's better to stick with invariant mass $m_0$.

Relativistic mass is good to give you a sort of "intuition". As you approach the speed of light $c$ your relativistic mass becomes infinity (your inertia is incredibly high), so it would take an infinite force to change your state of motion accelerating. This means that you can't reach a velocity higher than $c$.

But, except for this "intuition aspect" that relativistic mass can provide you, it's just really a matter of how you want to express things. For example, if you want to express Energy you may write $E=\gamma m_0c^2$ as well as $E=mc^2$, where $m$ is your relativistic mass.

Let's take a look at the relativistic spatial momentum \begin{equation} \vec{p}=\gamma m_0 \vec{v} \end{equation} If you stick your $\gamma$ to your velocity you get the spatial part of the so-called four-velocity, namely $\vec{u}=\gamma \vec{v}$, so you can express your momentum as $\vec{p}=m_0\vec{u}$. Either way you can choose relativistic mass and express your momentum as $\vec{p}=m\vec{v}$.

Answer 2

Using relativistic mass, we have

\begin{align*} \gamma &= \frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}} \\ m &= \gamma \, m_{0} \\ \mathbf{p} &= mv \end{align*}

the relativistic Newton's second law like this:

\begin{align*} \mathbf{F} &= \frac{d\mathbf{p}}{dt} \\ &= m\frac{d\mathbf{v}}{dt}+\frac{dm}{dt}\mathbf{v} \\ &= m_0\left( \gamma \, \mathbf{a}+\gamma^3 \frac{\mathbf{a}\cdot \mathbf{v}}{c^2} \mathbf{v} \right) \end{align*}

Relativistic mass somehow is a bridge for beginners to transit Newtonian mechanics to Special Relativity. If one who is familiar with $4$-vectors and tensors, then the dynamics will be

\begin{align*} d\tau &= \frac{dt}{\gamma} \tag{proper time differential} \\ X^{\mu} &= (ct,\mathbf{r}) \tag{4-position vector} \\ U^{\mu} &= \frac{dX^{\mu}}{d\tau} \tag{4-velocity} \\ A^{\mu} &= \frac{dU^{\mu}}{d\tau} \tag{4-acceleration} \\ P^{\mu} &= m_{0} U^{\mu} \tag{4-momentum} \\ F^{\mu} &= \frac{dP^{\mu}}{d\tau} \tag{Minkowski Force} \\ &= m_{0} A^{\mu} \tag{relativistic Newton's second law} \\ \end{align*}

which still looks similar to Newtonian mechanics.

However, in General Theory of Relativity, the gravitational mass (which is the same as inertial mass according to the Principle of Equivalence) is not equal to relativistic mass. The rest mass is intrinsic property of matter.

Answer 3

Mass is an intrinsic property . It means that the mass of a particle is like its identity (in particle physics). So , no: the mass of the particle is invariant (it is the same in all reference frames in the world).

Now, if you accelerate a body , it will gain momentum which causes the particle to "look" heavier (because $p = \gamma \times m \times v$, some people call $\gamma\times m$ the "relativistic mass", which is larger than the rest mass). But $m$ is $m$ , it is always the rest mass which is invariant.

If you want to use "relativistic mass" , the momentum will be : $P= m_{\hbox{(relativistic)}}\times v$ .

I took a course in Special Relativity , and we treated the thing of "relativistic mass" as old terminology.