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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?

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marked as duplicate by Kyle Kanos, garyp, Bill N, DilithiumMatrix, Cort Ammon Mar 6 '17 at 21:52

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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 It was used in my 1st relativity book but I was not at ease with it and all of my teachers rejected it as an old and outdated "notation".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 $\vec{p}=m\vec{v}$.

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  • $\begingroup$ Could you please explain that for different observerS relativistic mass will be different . So they will write down different equation. So is that right $\endgroup$ – Shashaank Mar 10 '17 at 11:25
  • $\begingroup$ Of course relativistic mass is dependent on the frame. For example, the relativistic mass of a particle in its rest frame is the same as its invariant mass $m_{rel}=m_0$. While if I observe the same particle from another reference frame where the particle has a certain speed $v$, then FOR ME, the particle has a relativistic mass $m_{rel}>m_0$. $\endgroup$ – Luthien Mar 10 '17 at 21:44
  • $\begingroup$ Ok thanks for the explanation . Could you please explain just one thing more. Like length contraction and time dilation are for both the observerS. Like A sees B going by. Then for A B's clock go slow and for B A's clock go slow. Similarly for A B's mass increases. For B , A's mass increases. But we evade such things in length contraction by saying both are Measuring different things. How can we evade it here. Whose mass increases. $\endgroup$ – Shashaank Mar 10 '17 at 21:49
  • $\begingroup$ As I said in my answer, I would avoid using relativistic mass, it's an old notation and it can lead to misunderstandings. Please not that the "real" mass of the particle, the so-called invariant mass or rest mass, stays the same in every reference frame. Every observer will say that the rest mass of the particle is $m_0$. $\endgroup$ – Luthien Mar 10 '17 at 21:56
  • $\begingroup$ Plus, in length contraction for example, we measure the SAME thing. I'm on a ultrarelativistic train and there is a bar of lenght $L_0$ (proper length) on the train with me. A guy from outside the train, who sees the train moving with the speed $v$, will see the SAME bar contracted and he will say that the bar is $L=L_0/\gamma$. Similarly, with relativistic mass, if we have two observers moving with relative speed $v$, than every one of them will see the other one's relativistic mass being $m_{rel}>m_0$. But really, I'd recommend to avoid this notation. $\endgroup$ – Luthien Mar 10 '17 at 22:00
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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.

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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.

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