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I understand relativistic mass and the equations underpinning it. My question deals with how to calculate relativistic mass when an object is viewed from different frames of reference.

Consider a space probe launched from a planet orbiting Star A (we will call it Probe A). Probe A launches into space, away from its planet at 90% of the speed of light relative to Star A. The relativistic mass will be 5.3 times the rest mass.

Now consider an observer on a planet orbiting a distant Star B (we will call him Observer B). Star B is hurling through space at the exact same velocity (speed and direction) as Probe A, but tangential to the path of Probe A, so there is no danger of a collision between Star A and Star B. Observer B is not aware his star system is moving – to him, of course, his star is stationary. When Observer B views Probe A, he measures a velocity of zero, and therefore the relativistic mass of Probe A equals the rest mass when viewed by Observer B.

How can the same object (Probe A) have different relativistic masses based on the observer?

Qmechanic
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Konacq
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2 Answers2

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In a word, because that's how relativistic mass is defined. It isn't frame-independent because it isn't supposed to be frame-independent.

The momentum of an object moving with speed $\frac{v}{c}=\beta$, rest mass $m$, and Lorentz factor $\gamma=\frac{1}{\sqrt{1-\beta^2}}$ is:

$$p=\gamma\beta mc$$

In order to gain some intuition on momentum, we have to answer the following question: "What do we want to do with the $\gamma$?" There are two ways to answer this question:

  • Combine the $\gamma$ and the rest mass $m$ into a new quantity which we define as the relativistic mass $m_{rel}=\gamma m$. Then we have that $p=m_{rel}\beta c=m_{rel}v$, in an analogy with classical mechanics, but we also have a new quantity which itself may not behave intuitively. For example, its value depends on the frame from which it is measured.

  • Leave the $\gamma$ in the expression as-is. This means there is no possibly-unintuitive quantity to work with, but it also means that we can't make the same analogy with classical mechanics. We have to accept that, in special relativity, momentum and velocity have a nonlinear relationship.

Which one is chosen is entirely a matter of convention, and the same physics arises from each choice. Nowadays, it seems that the concept of relativistic mass is falling out of favor (the analogy with classical mechanics that it's supposed to preserve ends up falling apart when accelerations are involved, anyway), and so if it seems like an unintuitive concept, it may be comforting to know that it's not strictly necessary to even define.

probably_someone
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Relativistic mass is actually the time component of the relativistic four-momentum (I'm using $c = 1$ to simplify things, as is usually done in relativity): \begin{equation}\tag{1} p^0 \equiv E = \gamma \, m_0 \equiv m. \end{equation} Thus relativistic mass transforms as shown below, when you change reference frame (here, $\vec{u}$ is the relative velocity between the two inertial frames and $\Gamma = 1/\sqrt{1 - u^2}$. Of course $\vec{p} = \gamma \, m_0 \vec{v}$ is the particle's momentum in the first frame): \begin{align} \tilde{p}^0 \equiv \tilde{E} = \tilde{\gamma} \, m_0 &= \Gamma \, (E - \vec{p} \cdot{u}) \\[12pt] &= \Gamma \, \gamma \, (1 - \vec{v} \cdot \vec{u}) \, m_0, \tag{2} \end{align} which implies \begin{equation}\tag{3} \tilde{\gamma} = \Gamma \, \gamma \, (1 - \vec{u} \cdot \vec{v}), \end{equation} or if you prefer: \begin{equation}\tag{4} \tilde{m} = \frac{1 - \vec{u} \cdot \vec{v}}{\sqrt{1 - u^2}} \, m, \end{equation} This is the transformation law of the "relativistic mass" (or energy, which is a much better wording !). Notice that $\tilde{m} = m_0$ if $\vec{u} = \vec{v}$ (rest-frame of the particle, where $E = m_0$).

Cham
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