0

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
  • 201,751
Konacq
  • 1

2 Answers2

1

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
  • 35,459
  • 1
    Thank you for answering the OP's question without feeling compelled (as so many others do) to descend into a rant about how relativistic mass is a "deprecated" concept --- as if there were some official governing body that determines how we're allowed to group the terms in our equations. (In a better world, this thank-you would of course be unnecessary.) – WillO Dec 11 '18 at 16:04
  • Thanks for the answer. In my construct, I think a better description would be that Observer B is travelling relativistically with respect to Observer A. Therefore the rest mass of Probe A, when at rest on planet B, (as measured by Observer B) is the relativistic mass from Observer A’s perspective. Therefore, Probe A’s rest mass for Observer B is the same as the Probe A’s relativistic mass for the same Probe A. – Konacq Dec 11 '18 at 16:21
  • @Konacq The rest mass of Probe A will be the same when measured by observers A and B, since the rest mass is frame-independent. The relativistic mass is equal to the rest mass for Observer B, and is greater than the rest mass for Observer A. Therefore, it cannot be true that the rest mass of Probe A is equal to the relativistic mass of Probe A in Observer A's frame. – probably_someone Dec 11 '18 at 16:31
  • 1
    @Konacq This might make more sense once you realize that what we've been labeling as relativistic mass is, in fact, just the total energy of the object (divided by $c^2$, which is usually taken to be 1 in relativity). Probe A obviously has higher kinetic energy in Observer A's frame than in Observer B's frame, so it makes sense that the total energy of the probe is greater. This is one of the reasons people feel that relativistic mass is unintuitive, as they would rather just use an already-existing quantity for which we have a pretty solid intuition. – probably_someone Dec 11 '18 at 16:43
0

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
  • 7,391