If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light?

Question

I know that as the velocity increases, the mass of the object also increases so it becomes tougher and tougher to move the object which ultimately leads to a requirement of infinite energy to accelerate an object to the speed of light. But I have a doubt.

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. So if the actual mass ($m_0)$ remains the same and only the observable mass increases, why is more and more energy required? I know that even if the object exceeds the speed of light, we will not be able to say that it is moving faster than the speed of light but is it possible to make it move faster than light?

If rest mass does not change with $v$ then why is infinite energy required to accelerate an object to the speed of light?

I quote Igor Ivanov (a stackexchange user from this question Why does the (relativistic) mass of an object increase when its speed approaches that of light?) to give an insight into my question.

The mass (the true mass which physicists actually deal with when they calculate something concerning relativistic particles) does not change with velocity. The mass (the true mass!) is an intrinsic property of a body, and it does not depends on the observer's frame of reference. I strongly suggest to read this popular article by Lev Okun, where he calls the concept of relativistic mass a "pedagogical virus".

EDIT

After reading the answers I have a doubt.

SO can I say that only the overall energy of the system increases while the mass remains constant? But then if mass remains constant, then why is more and more energy gradually required? I mean there should be a reason for that requirement of infinite energy.

Answer 1

If rest mass does not change with v then why is infinite energy required to accelerate an object to the speed of light?

The momentum of a material particle, a conserved quantity, is theoretically and experimentally a non-linear function of velocity given by

$$\vec p = m \frac{\vec v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

which goes to infinity as $v \rightarrow c$. The relativistic energy, also a conserved quantity, is

$$E = c\sqrt{|p|^2 + (mc)^2}$$

When a particle is ultrarelativistic, $|p|^2 \gg (mc)^2$, this expression is approximately

$$E = |p|c $$

which has been experimentally confirmed. So, it easy to see that the particle's energy goes to infinity as $v \rightarrow c$

Answer 2

In relativity the rest mass is the mass of an object measured from a reference frame in which it is at rest. But this is not the mass involved in acceleration or inertial mass.

Inertial mass, or the opposition of the body to the change of movement (directional or in magnitude), will grow with the speed of the body:

$$m = \frac{m_o}{\sqrt{1-v^2/c^2}}$$

and this is the mass observed when the object is moving. By observed I mean the way you would measure it, which could be: applying a force $F$ to it and measure the acceleration produced $a$ and you would get $m = F/a$.

Also, it is a mistake to think:

I know that even if the object exceeds the speed of light, we will not be able to say that it is moving faster than the speed of light

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.

Answer 3

This might be a useful energy-momentum diagram (below) with coordinates $(E,p)$ [with $E$ running upward].

A particle's 4-momentum vector $\tilde P$ is drawn, as well its energy $E$ and momentum $p$ components in this frame.

The square-magnitude of this 4-momentum is given by the hyperbola $E^2 - (pc)^2 = (mc^2)^2$ , which is called the mass-shell for [invariant-] [rest-]mass $m$.

The velocity of this particle is the "slope" $$v=\displaystyle\frac{pc}{E}c.$$

While keeping $m$ fixed,

• in order to increase the velocity $v$ to approach the speed of light $c$,
  the tip of the particle's 4-momentum must travel up along the mass-shell (along the green curve),
  which requires that $E$ tends to infinity
  (and $p$ tends to infinity while constrained by $p=\frac{1}{c}\sqrt{E^2-(mc^2)^2}$)
• Note that $v \rightarrow c$, but $v$ will never reach $c$.

(My diagram was taken from my answer
https://physics.stackexchange.com/a/510241/148184
to Ultra-Relativistic and Non-Relativistic cases for energy of a particle )

Answer 4

The answer is that velocities don't add up straightforwardly in SR.

If you were in a car on clone of the Earth which was travelling at a metre per second less than c relative to Earth, you could still drive around as normal, and wouldn't notice anything odd. The amount of force required to accelerate the car to 100m/s on the clone planet would be exactly the same as the force required back on the real Earth.

The difference is that while on the clone planet the car's speed has increased to 100m/s, in the Earth's reference frame the speed of the car has hardly increased at all. So it's not harder to accelerate a speeding object in its own rest frame- it's just that the effect of that acceleration seems smaller and smaller in other frames moving faster relative to the object.

If ever you find yourself confused by SR, you might find it helpful to remember that all speed is relative and the effects of SR are symmetrical. Relative to a passing muon, you are travelling at 0.99c, but you don't experience any increase in mass as a consequence and you are not any harder to accelerate. However, if you get in your car and accelerate up the street, in the frame of the muon you hardly increase your speed at all.

Answer 5

Let's write the well known equations that are involved in this discussion: The energy-momentum relation: $$E^2=p^2c^2+m_{0}^2c^4$$ The relation between energy and mass: $$E_{0}=m_{0}c^2\quad\quad\quad \text{rest energy}$$ $$E=mc^2\quad\quad\quad \text{total energy}$$ The relation between rest mass and the so-called relativistic mass: $$m=m_{0}\gamma$$ where the gamma factor is defined as: $$\gamma\equiv \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

So if we take $E=mc^2$ and rewrite it as: $$m=\dfrac{E}{c^2}$$ we see that $m$ is actually another way of expressing the total energy of a particle (or a system). So of course you need an infinite amount of energy to accelerate it to the speed of light, because if $v\rightarrow c$ then $\gamma\rightarrow\infty$, which means that if $m_{0}\neq 0$ then $m\rightarrow\infty$.

So:

• if $m_{0}\neq 0$ and $v\rightarrow c \Rightarrow \gamma\rightarrow\infty$ thus $m\rightarrow\infty$.
• if $m_{0}=0$ and $v\rightarrow c \Rightarrow \gamma\rightarrow\infty$ but now $m_{0}\gamma$ (which is another way of expressing the total energy) gives a finite result (becasue $0\cdot\gamma$ gives something that is finite). So only massless particles like photons can travel at the speed of light (and in fact they can only travel at the speed of light).

So massive particles can travel at any speed they like except the speed of light, while massless particles can travel only at the speed of light.

Answer 6

In order to apply a force to an object to make it move faster (or slower), the force must be moving at the same velocity as the object. In an H2-O2 rocket engine, the fuel does travel at the same velocity as the rocket. When the fuel burns the pressure creates a force which drives the water molecules out the exhaust in the opposite direction to the rocket. The pressure is created by the interaction of the protons in the water with the protons in the rocket engine walls. These aren't little billiard balls that bounce off each other. They don't even get close. The interaction is essentially the Coulomb force(more complicated but you get the picture). The interaction is by Maxwell's electromagnetism.

Here comes the point: The interaction can proceed at the speed of light only. The fastest velocity at which a force can be applied is c. Every thing in our universe is made of electromagnetism.

In the rocket's frame, the exhaust is travelling at -c. An the exhaust's frame the rocket is travelling at +c. An observer in the exhaust's frame would see the rocket at the speed of light laying down a stream of water travelling at 0 m/s.

The photons doing the pushing are red shifted in the exhaust's frame to zero energy. There for to have any accelerating effect an infinite number of photons are required by the rocket at the speed of light.