Why can't a mass be accelerated to the speed of light if its mass does not increase with speed? [duplicate]

Question

My question is similar to this question (If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light?) except that my question 1. is not a metaphysical one, 2. is clear, 3. does not contain wrong, muddled, or dubious assumptions such as "As far as I know only the observable mass (Relativistic mass) increases but not the Proper Mass or Intrinsic Mass, right? The actual mass of the object will remain the same." The two answers to that question are inadequate: This one (https://physics.stackexchange.com/a/139821/295887) concludes thus "No body can travel faster than light, not because we can not state its speed, but because it can't. This is at least what our current state of knowledge leads to, with many experimental confirmations.", which is circular reasoning --"it can't because it cannot, is essence". The other, topvoted answer https://physics.stackexchange.com/a/139827/295887 is more interesting, and quite good, as far as it goes, but it does not really address my question which is why. It is a sort of purported proof, not an explanation, certainly not an explanation that a layman could understand. It's a bunch of formulas whose relevance is not explained.

The standard (though possibly/allegedly technically wrong/ out dated) explanation for why you can't accelerate a body to the speed of light by applying a constant force for a long time is that the mass (or "relativistic mass" or "apparent mass") of the body will increase more and more, the closer it gets to c and therefore the acceleration will be less and less. A case of diminishing returns, one could say.

The logic is that $a = F/m$. As $v$ approaches $c$, $m$ approaches infinity, and $a$ approaches zero. So when I heard repeatedly here on Physics that mass does not increase with speed, I figured that since you still can't reach $c$, there must be some other reason. I figure that you still get the diminishing returns situation, with a constant force producing less and less acceleration, but without invoking mass increase. Is that correct, and if so, what is the reason for the decrease in acceleration as c is approached?

Answer 1

The underlying cause is the geometry of spacetime. Suppose you were in a spaceship that accelerated to 0.9c relative to the Earth, at which point you dropped a marker that coasted along at that speed. Having done so, you could accelerate again to 0.9c relative to the first marker, and drop a second marker that coasts at your new speed. You can repeat those steps as often as you like, dropping marker after marker, each time accelerating again to 0.9c relative to the last marker. In each case, the amount of force you need to apply to increase your speed to 0.9c relative to the last marker you dropped remains the same. You can accelerate away from each marker as easily as you accelerated away from the first.

Suppose you did that ten times, so that ten times you increased your speed by 0.9c relative to the last marker you dropped. The geometry of spacetime is such that ten instances of increasing speed by 0.9c results in an overall speed that is still less than c relative to your starting point.

In your own frame a force of $f$ applied to your spaceship of mass $m$ always results in an acceleration a given by $f=ma$. However, in other frames moving relative to you, the increase in speed seems less and less the faster you go. It is not because your mass is increasing, but because the geometry of spacetime means that increments of speed resulting from the bursts of acceleration do not add arithmetically.

Answer 2

The quantity $E = m_i c^2 + T + U$, where $T$ is kinetic energy of translation, $ U$ is internal energy including heat in the object's rest frame, and $m_i$ is invariant mass, goes to infinity as $v \to c$.

Confusion arises from overloading the symbol $m$ and the word mass by using it for invariant mass when it's already in common use for different things also called mass. e.g. the m in gravitation and momentum is $m:=E/c^2$ whereas the m in particle physics is $m:=m_i$, and almost nobody actually uses a different symbol like $m_i$ for clarity.

People sometimes make blanket declarative statements like "mass is invariant" without noting that they are defining a different symbol from the one used by every high school teacher and most of the undergraduate instructors on the planet, which is extremely counter-productive, or more likely just indicative of their own misunderstanding.

Incidentally, $F=ma$ is an approximation using the $m:=E/c^2$ definition. If kinetic energy is a significant part of that m, you need to use $F=\frac{d}{dt}(mv)$. If you use $m:=m_i$ you need to add T and U back in:

$F=\frac{d}{dt}((m + T/c^2 + U/c^2)v)$

Or pack m and U together as a constant and use the Lorentz factor to include the kinetic energy term,

$F=(m+U)\frac{d}{dt}(\gamma v)$

Answer 3

A way to see is the impulse that gives each bullet (vector mediator of the field X) fired by a machine gun with an "unlimited reload" to a mass of 1 kg, at the beginning the flow of bullet that reaches the mass is large (particle /s.m²), the mass begins to accelerate but at the end the flow will be very low because the speed of the mass at almost the same speed as the bullets. A speed of the mass which is equal to the same speed as the balls means that the balls never reach it (it takes an infinite time to reach the mass).