0
$\begingroup$

This question already has an answer here:

The speed of light theory predicts that as things travel faster their mass increases, so I think we if we look at a plane accelerating from mach one to mach two and measure the relativistic mass of the plane, we'll see an increase in relativistic mass regardless of how small. Is this true?

$\endgroup$

marked as duplicate by Kyle Kanos, Ben Crowell, John Rennie special-relativity Sep 25 '18 at 7:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Well, you could calculate the mass change, and then contemplate how to actually measure it while the jet is moving at Mach 1 or Mach 2... Much easier to do the experiment on electrons or ions in the lab. $\endgroup$ – Jon Custer Sep 24 '18 at 15:07
  • $\begingroup$ The mass of the plane is constantly decreasing due to the fuel being burned and exhausted. $\endgroup$ – safesphere Sep 24 '18 at 15:25
4
$\begingroup$

The relativistic mass is an obsolete and confusing concept.

The mass of an object is the rest mass.

The $\gamma$ factor in the 4-velocity and hence in the 4-momentum means that to accelerate an object you need more and more energy expenditure and that the speed limit for any object is the speed of light. In principle reachable with an infinite spending of energy.

$\endgroup$
  • $\begingroup$ If you consider "relativistic mass" to be obsolete, then why are you talking about both mass and energy? The distinction between these was made obsolete by the formula $E=mc^2$. $\endgroup$ – md2perpe Sep 26 '18 at 14:36
  • $\begingroup$ @md2perpe. The equation $E = mc^2$ is to be intended in the rest frame of a particle, where $m$ is the rest mass of the particle. In a generic inertial frame should be $E^2 = (mc^2)^2 + (pc)^2$, where $m$ is the rest mass and $p$ is the relativistic 3-momentum. No need to talk of relativistic mass. $\endgroup$ – Michele Grosso Sep 26 '18 at 16:07
  • $\begingroup$ Not everything is a particle. Consider a spring of mass $m$. You compress it adding an energy $T$. What is now the rest mass of the spring? What is the total energy? Accelerate the compressed spring to speed $v$. What is now the total energy of the spring? What is its momentum? What is its rest mass? $\endgroup$ – md2perpe Sep 26 '18 at 18:20
  • $\begingroup$ @md2perpe. In case of a system, compound of many constituents, the rest mass of the system is equal to the rest mass of the single constituents plus their kinetic energy plus the interaction energy. That is the meaning of the equation $E = m c^2$, where $E$ is the rest energy. That is also the energy that in general relativity gravitates, i.e. shapes the geometry of spacetime. $\endgroup$ – Michele Grosso Sep 27 '18 at 15:19
  • $\begingroup$ How do you convert the kinetic energy of the constituents to rest mass for the system without using $E=mc^2$ that you say is only intended in the rest frame of a particle (which the compound system is not)? $\endgroup$ – md2perpe Sep 27 '18 at 16:55
3
$\begingroup$

The real quantity that increases is the relativistic energy, which is given by $E = \gamma mc^2 $ with $\gamma$ being the Lorentz factor $\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$.

"Relativistic mass" comes from rolling the $\gamma$ into the $m$, such that you have $m' = \gamma m$ and $E=m'c^2$. Whether you do this is mostly a notational choice, since the results are the same either way, but generally it's not done because $\gamma$ shows up a lot in ways that don't look like a change in mass, so it's helpful to keep that seperate.

With that didactic note out of the way, we can estimate what the change in energy/mass for a plane moving from mach 1 to mach 2 would be. Mach 1 is about $10^{-6}\, c$, so the change in the Lorentz factor is about $10^{-12}$, which means that the change in energy for a Jumbo Jet (~100 tons) corresponds to a hundred extra micrograms; a bit less than the energy change if an extra fruit fly stowed on board. This is negligible, so classical mechanics is a good approximation at plane-speeds, as we would expect.

$\endgroup$
  • $\begingroup$ For low speeds, $\gamma \approx 1 + \frac12 \beta^2$, but I guess the factor of a half doesn't make much difference here. ;) $\endgroup$ – PM 2Ring Sep 24 '18 at 17:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.