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Suppose a system (eg: rocket) consists of $N$ atoms. It starts moving away from the origin of an inertial frame at speed 0.9c. Will $N$ changes and if it changes where does this change come (if increases) or goes (if decreases)?

Update: Let's suppose there is no fuel in the rocket and it attains this speed through a sequence (can be a large number) of gravitational slingshots, and the mass 0 we are talking about is calculated after receiving the first slingshot

Follow-up Question: Since $N$ will not change and Total Mass = Sum(mass of all atoms in the system), and according to the equation

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$there is △m increase in Total mass. I want to understand how does this △m came into the system, does the particle mass increase or something else
ancad ancad
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2 Answers2

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No there will not be any change. Particle number is Lorentz invariant i.e. the number of Atoms that make up a particular object is the same no matter from which frame of reference you measure it.

SPHerical
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A rocket uses fuel to move so the fuel particles escape the system; if you roughly know $\Delta m$ and the mass of the fuel particles you can estimate the number of particles that have escaped the rocket. Particle number is Lorentz-invariant so the particles composing the rocket will remain the same in any inertial reference frame.

If the rocket isn't propelled by fuel nor by any $\Delta m$ the number of particles composing the rocket stays the same.

The relativistic mass $m=\gamma m_0$ originating from relativistic speed is not the "real mass" of the system, thus is not associated to particle number. Instead the particle number is associated with the rest mass $m_0$ which is conserved and frame-invariant.

Lorenzo B.
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  • Let's suppose there is no fuel in the rocket and it attains this speed through a sequence (can be a large number) of gravitational slingshots, and the mass 0 we are talking about is calculated after receiving the first slingshot – ancad ancad Sep 17 '21 at 09:58
  • @ancadancad In that case there is no change in $N$. – Steeven Sep 17 '21 at 09:58
  • I agree with @Steeven, since particle number is invariant – Lorenzo B. Sep 17 '21 at 10:00
  • I totally agree with particle number being invariant, but could you explain the comment of this answer https://physics.stackexchange.com/a/666534/314115 – ancad ancad Sep 17 '21 at 10:04
  • Particle number is associated to rest mass $m_0$ (or simply mass) which is invariant as well – Lorenzo B. Sep 17 '21 at 10:09
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    @ancadancad The concept of "relativistic mass", while not "wrong" as such, is not considered useful and hasn't been for a long time. See this thread for the details. – Emilio Pisanty Sep 17 '21 at 12:37