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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?

Matthew Christopher Bartsh
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    Does this answer your question? If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light? – Michael Seifert Dec 08 '21 at 19:57
  • @MichaelSeifert It doesn't answer my question. I have added a paragraph to my question explaining why. – Matthew Christopher Bartsh Dec 08 '21 at 21:15
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    My answer might be useful: https://physics.stackexchange.com/a/675896/148184 from https://physics.stackexchange.com/questions/675460/why-is-harder-to-accelerate-object-at-higher-speeds-if-speed-is-relative-term – robphy Dec 08 '21 at 21:44
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    The modern view is that "mass" is the minimum energy of an object (its rest mass), and "energy" is the current energy of the object. Energy is what has inertia; that is, the Newtonian expression $F=ma$ is better re-written as $F=\frac{E}{c^2}a$. As the velocity of an object approaches $c$, its energy increases without bound, and so more and more force is required to accelerate it. – Eric Smith Dec 09 '21 at 02:07
  • @EricSmith +1 for interest and clarity. But it seems to me that 1. mass is inertia and 2. Einstein said that mass and energy are the same thing, and even coined "massenergy" which he wanted to use to replace both "mass" and "energy". So it is not clear to me what is achieved by, as it appears to me, calling "massenergy" and "mass", "energy". What is the modern view on the idea that mass and energy are the same thing? – Matthew Christopher Bartsh Dec 10 '21 at 21:16
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    It isn't really useful to have two terms, "mass" and "energy", which both refer to the same thing. Hence modern physicists generally stick with "energy" for the thing that changes (is frame dependent), and use "mass" to mean "rest mass", i.e. the minimal possible energy of the object or system. Of course this isn't done as consistently as one might wish, and hence the confusion which leads to your question. In any case, the inertia of a body does increase without bound as its speed approaches $c$. – Eric Smith Dec 11 '21 at 02:21
  • @EricSmith +1 You've helped me a lot. But, "isn't done as consistently as one might wish" is quite an understatement. Why does physics not get rid of "rest mass" if it means the same as "mass"? Why don't physicists (on the whole) say "We don't use the concept of relativistic mass. The math is easier that way" rather than what they do say, which is "photons have no mass", "mass does not increase with velocity" and so on? By the way, can you confirm that the word "inertia" is really used in modern relativity. If so, what is the difference if any between energy and inertia? – Matthew Christopher Bartsh Dec 12 '21 at 00:04
  • This web site is full of answers that say "We don't use the concept of relativistic mass", it's pretty much the first thing anyway says in answers to SR questions involving mass. "Photons are massless" just means they have no rest mass (since they are never at rest) or equivalently there's always a frame in which light is more redshifted and hence has less energy. As for inertia, it's tricky to calculate relativistically because the effects of force on a moving object depend on whether the force is applied in the direction of motion or perpendicular to it. – Eric Smith Dec 12 '21 at 03:02
  • @EricSmith This web site is a mine of information. I was thinking of physicists generally, not here on Physics. I had never even heard about transverse mass v. longitudinal mass, until a week ago. Physicists seem to sweep that under the rug, perhaps understandably, since it is very weird and dismaying. It doesn't sound like mass or inertia of any kind, if direction plays a part in its magnitude. I find it dismaying. – Matthew Christopher Bartsh Dec 14 '21 at 08:06
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    @MatthewChristopherBartsh some hints how Einstein and Lorentz riddled about concepts of transveral and longituninal mass you can find here and here. I do not think Physicists sweep this unter the rug. They just moved on. – Kurt G. Dec 14 '21 at 10:06

3 Answers3

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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.

Gert
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Marco Ocram
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  • ... is the right anwer, IMO. +1. – Gert Dec 08 '21 at 21:29
  • My question asked about a force applied in the rest frame to a moving object. You are talking about a force applied in the frame of the moving object, perhaps by a spring. So your answer, though food for thought, doesn't really answer the question, in my honest opinion. – Matthew Christopher Bartsh Dec 08 '21 at 21:44
  • Suppose a single burst of the ship's rocket engine increases its speed by 1000km/s in its own frame. Now imagine successive bursts viewed from some frame in which the rocket is initially at rest. At first each burst will add close to 1000km/s to the ship's speed. But the geometry of spacetime means that as the total speed gets closer to c, each successive increase of 1000km/s in the ship's frame adds less and less to the overall speed viewed from the initial frame. It seems that a burst of the rocket has less and less effect... – Marco Ocram Dec 09 '21 at 08:50
  • in fact, each burst of the rocket continues to work exactly as before, and Newton's second law continues to apply from the perspective of the rocket or of anybody else moving relative to the rocket at a speed not close to c. – Marco Ocram Dec 09 '21 at 08:53
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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,

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

g s
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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).

The Tiler
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    This answer makes no sense whatsoever. And what does "(particle /s.m²)" even mean? – Gert Dec 08 '21 at 19:46
  • it is a way of seeing, and my nickname indicates it, if there is better, I am also taker , in the following link is defined the flux : j=I/A, with I=dq/dt where I took q= N : number of particles .https://en.wikipedia.org/wiki/Flux – The Tiler Dec 08 '21 at 21:13
  • ok, if we take the formula F=(m+U) d(γv)/dt , what meaning do we give to γv , I will tell you my physical vision, if an observer leaves in a direction orthogonal to the direction of propagation of a wave at the same instant when it is emitted at a point O, according to the pythagoras theorem r²+l²=R², c²t²+vt'²=ct'² , t: time to arrive at the starting point of the observer, we find t'=γt – The Tiler Dec 08 '21 at 22:00
  • we will formalize in the following way r=(R,l) , dr/dt=γ(c,v), if the speed of the observer reaches c, the light wave emitted in O never reaches the observer in uniform relativistic motion (condition to have γ in special relativity), why do we derive the speed with respect to time ? – The Tiler Dec 08 '21 at 22:01