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- What is the force on a scale in an accelerated rocket in flat spacetime in the following scenario (SR)?
[Moderator's note: Spin-off posts from previous thread have been included in this new thread. Also, the OP's re-post of the scenario for discussion has been moved to this top post for clarity.]
Physically, scales measure a force (and indirectly the energy) in their frame. Consider the same scenario in a large accelerating elevator cabin in space and describe it in an inertial frame, in which ##v_z/c## is small for a certain time (##v_x## is relativistic and constant).
##dp_z = m\gamma dv_z##
##a_z = dv_z/dt##
##F_z= dp_z/dt##
##F_z/a_z = dp_z/dv_z = m\gamma##
Source:
http://www.physikdidaktik.uni-karlsruhe.de/download/195_longitudinal_and_transverse_mass.pdf
[Moderator's note: Including later re-post of the scenario here.]
A rocket in space has at it's stern a constant proper acceleration of 9.81 m/s² (in z-direction).
On the floor at the stern of the rocket stands a scale. A toy-car with invariant mass ##m## drives with constant velocity ##v_x## over the scale. What is the force on the scale?
As reference frame for the calculation, I choose a co-moving inertial frame, which is at rest relative to the accelerating rocket at an instance of time, in which I do the calculation. I make use of the following for the inertial reference frame:
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force
So in this inertial frame - at the described instance of time - the car moves in x-direction with ##v## and accelerates together with the rocket-stern in z-direction with 9.81 m/s².
##dp_z = m\gamma dv_z##
##a_z = dv_z/dt##
##F_z= dp_z/dt##
##F_z/a_z = dp_z/dv_z = m\gamma##
=> The force on the scale is ##F_z = m\gamma \cdot 9.81 m/s^2##.
Of course, the ratio ##F_z/a_z## in this scenario should not been called "traverse mass" nor "relativistic mass", because that would only create confusion regarding the word "mass".
Is my above calculation correct and if not, why exactly?
What is a possible better way to calculate the force on the scale?
Yes.PeterDonis said:The downward proper acceleration of an object on the surface of a planet like the Earth is independent of tangential velocity
PeterDonis said:So scales don't measure relativistic mass anyway, they measure invariant mass
Physically, scales measure a force (and indirectly the energy) in their frame. Consider the same scenario in a large accelerating elevator cabin in space and describe it in an inertial frame, in which ##v_z/c## is small for a certain time (##v_x## is relativistic and constant).
##dp_z = m\gamma dv_z##
##a_z = dv_z/dt##
##F_z= dp_z/dt##
##F_z/a_z = dp_z/dv_z = m\gamma##
Source:
http://www.physikdidaktik.uni-karlsruhe.de/download/195_longitudinal_and_transverse_mass.pdf
[Moderator's note: Including later re-post of the scenario here.]
A rocket in space has at it's stern a constant proper acceleration of 9.81 m/s² (in z-direction).
On the floor at the stern of the rocket stands a scale. A toy-car with invariant mass ##m## drives with constant velocity ##v_x## over the scale. What is the force on the scale?
As reference frame for the calculation, I choose a co-moving inertial frame, which is at rest relative to the accelerating rocket at an instance of time, in which I do the calculation. I make use of the following for the inertial reference frame:
Source:Rindler said:it is always an IF, it always moves uniformly: an accelerating particle co-moves with its rest-frame for only one instant. But at that instant the rest-frame measures the particle's proper acceleration.
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Four-force_and_three-force
So in this inertial frame - at the described instance of time - the car moves in x-direction with ##v## and accelerates together with the rocket-stern in z-direction with 9.81 m/s².
##dp_z = m\gamma dv_z##
##a_z = dv_z/dt##
##F_z= dp_z/dt##
##F_z/a_z = dp_z/dv_z = m\gamma##
=> The force on the scale is ##F_z = m\gamma \cdot 9.81 m/s^2##.
Of course, the ratio ##F_z/a_z## in this scenario should not been called "traverse mass" nor "relativistic mass", because that would only create confusion regarding the word "mass".
Is my above calculation correct and if not, why exactly?
What is a possible better way to calculate the force on the scale?
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