# Accelerating particles to speeds infinitesimally close to the speed of light?

I'm in a freshmen level physics class now, so I don't know much, but something I heard today intrigued me. My TA was talking about how at the research facility he worked at, they were able to accelerate some certain particle to "99.99% the speed of light". I said why not 100%, and I didn't quite understand his explanation, but he said it wasn't possible. This confused me. Since the speed of light is a finite number, why can we go so close to its speed but not quite?

Edit: I read all the answers, and I think I'm sort of understanding it. Another silly question though: If we are getting this particle to 99.99% the speed of light by giving it some sort of finite acceleration, and increasing it more and more, why cant we increase it just a little more? Sorry I know this is a silly question. I totally accept the fact we cant reach 100%, but I'm just trying to break it down. If we've gotten so close by giving it larger and larger acceleration every time, why cant we just supply it with more acceleration? And how much of a difference is there between 99.99% the speed of light, and the speed of light? (I'm not really sure if "difference" is a good word to use, but hopefully you get what I'm asking).

-
That is the point. Just a little more can't do it. The difference in energy between 99.99% and 100% of the speed of light is infinite. So is the difference between 99.999999999999% and 100%. You would need infinite amount of energy to get to c from something less. –  Vagelford Dec 2 '10 at 23:41
@Vagelford So the electrons or whatnot that emit light from a flashlight won't do, because they already travel at c? –  Cees Timmerman May 7 '14 at 9:07

## 8 Answers

By special relativity, the energy needed to accelerate a particle (with mass) grow super-quadratically when the speed is close to c, and is ∞ when it is c.

$$E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - (\text{“percent of speed of light”})^2}}$$

Since you can't supply infinite energy to the particle, it is not possible to get to 100% c.

Edit: Suppose you have got an electron (m = 9.1 × 10-31 kg) to 99.99% of speed of light. This is equivalent to providing 36 MeV of kinetic energy. Now suppose you accelerate "a little more" by providing yet another 36 MeV of energy. You will find this this only boosts the electron to 99.9975% c. Say you accelerate "a lot more" by providing 36,000,000 MeV instead of 36 MeV. That will still make you reach 99.99999999999999% c instead of 100%. The energy increase explodes as you approach c, and your input will exhaust eventually no matter how large it is. The difference between 99.99% and 100% is infinite amount of energy.

-
So it's 'divergent' huh –  maq Dec 2 '10 at 21:11
"infinitely many"? I think you just mean "infinite". –  Noldorin Dec 2 '10 at 22:27
@Noldorin: Yes, thanks. –  kennytm Dec 2 '10 at 22:47
I highly recommend any programmer struggling with this write a program and try plugging in numbers. As you can see, the equations are basic algebra, so programs for them are really short and simple. I did this once; you get a real feel for the "you can't ever quite get there" nature of approaching c. –  Bob Murphy Dec 3 '10 at 0:53
@maq - Actually, it's convergent. That is, velocity converges on the speed of light as energy increases. –  WhatRoughBeast Apr 20 at 2:23

You have to understand special relativity. It's basically because newtonian mechanics breaks down at speeds close to the speed of light and F=ma is false. It's basically because your mass isn't constant, it varies based on your speed. And as you approach c, your mass has to approach infinity and thus you'll need infinite force to move accelerate from c - delta(v) to c.

This is just a basic overview, I'm sure someone will come with a much more detailed overview. But you can take a look at http://en.wikipedia.org/wiki/Special_relativity

Specifically the part about relativistic mechanics.

-
F=d(mv)/dt is what's generally applicable, it's just that everyone learns the approximation F=ma first, b/c it's close enough (for most observers) to assume that masses are constant. –  JustJeff Dec 4 '10 at 15:02
@JustJeff: actually $\mathbf{F}=\mathrm{d}\mathbf{p}/\mathrm{d}t$ is the most general form. The difference is important because in special relativity (and some other contexts), $\mathbf{p}\neq m\mathbf{v}$. –  David Z Apr 7 '11 at 18:05

There are (at least) two explanations, kinematical and dynamical.

## Dynamics

When you want to make an object accelerate you have to use energy to produce force on the object. The force is $F = ma$ (this equation is not really correct in SR but it suffices for our purposes) Now the point of SR is that the mass $m$ that the object seems to be having when it is moving with respect to you is not constant. It goes like $m = m_0 \gamma(v)$ where $m_0$ is the objects invariant mass (as seen from its own rest frame) and $\gamma(v)$ is the Lorentz factor. Now $\gamma(v) \to \infty$ as $v \to c$. So this means that the (apparent or relativistic) mass of the object becomes arbitrarily large and you would need an infinite amount of energy to get to the speed of light.

## Kinematics

From the kinematical point of view it all boils down to relativistic concept of velocity. In SR when you want to change particle's speed you have to boost it. This is described by a certain Lorentz transformation.

Now its useful to move to the dual point of view. Instead of saying that you boost the particle you can just change your reference frame in the opposite way. So instead of giving the particle speed $v$ in direction $\mathbf x$ you will look at the particle at rest from a reference frame that has speed $v$ in direction $-\mathbf x$. This transformation is also described by a Lorentz transformation.

Now every Lorentz transformation preserves the relations $v < c$, $v = c$ and $v > c$ (the middle one is actually Einstein's postulate on invariance of speed of light in every inertial frame). That means that if your velocity is less than speed of light it will be so in any reference frame. And also that if some particle was once going slower than speed of light, it will always do so.

-
I've come to not like the $m = \gamma m_0$ formulation: Yes it keeps the kinetic energy in the familiar Newtonian form and is a valid way to understand Special Relativitly, but the "relativistic mass" is not a Lorentz invariant, which makes it awkward when it comes time to make calculations. –  dmckee Dec 2 '10 at 19:29
@dmckee: I don't understand your objection. Energy $E$ is also not Lorentz invariant but that doesn't make it any less useful. Actually, $E = mc^2$ so if you don't like relativistic mass you shouldn't like energy either ;-) –  Marek Dec 2 '10 at 19:43
Marek makes a good point. Relativistic mass and energy are both very useful concepts. –  Noldorin Dec 2 '10 at 20:01
@Marek: one big problem (among other) with relativistic mass is that it makes people think that a body actually gets more massive and at some point it should collapse into a black hole. –  Igor Ivanov Dec 2 '10 at 20:29
@Marek — that's why it's better not to push people towards thinking that QM is weird. Back to mass, my experience is that using relativistic mass in explanations of relativity to a novice creates more confusion later. –  Igor Ivanov Dec 2 '10 at 21:34

It is a direct consequence of the theory of special relativity that no massive massive particle can travel at the speed of light. (And every massless particle must travel at the speed of light.)

You can consider the impossibility of accelerating a particle to precisely speed c in one of several ways, but the most obvious is:

A hypothetical massive particle traveling at speed c would have infinite mass (or mass-energy). Singularities are bad! (Or if you want, it would require an infinite force/amount of energy to accelerate the particle to c, approaching the limit.)

Side note: If you're a freshman in physics, you will most likely be studying basic special relativity quite soon. Everything should be a lot clearer after taking such a course.

-

As for why you can't go just a bit higher. It's not a problem of having the energy, the problem is transferring it the particle you want to accelerate. Those particles are accelerated using electromagnetic fields generated in superconductor devices. There is a limit on how large these fields can be made, since when the magnetic field is too big, the superconductive state is lost and all hell break loose (the temperature is not the only thermodynamic variable in superconductors, you can increase the Gibbs function by increasing the magnetic field as well). You also have other less "thermodynamically fundamental" issues, but let's forget about them.

Therefore, if you want to accelerate a bit more, you have to either make the acceleration path even longer or make the particles go in circles and pass the accelerating region many times. The first case is not doable, since the size would be longer than any laboratory we already have. The second case has limitations too. You have to keep the particles in a stable beam for a long time, the particles lose some energy while going around the circular path, and so on...

-

Special relativity does not rule out tachyons which travel faster than the speed of light and whose speed increases with decreasing energy. Also, the Alcubiere drive (and metric) allows a warp bubble to travel (expand) and supraluminal speeds (providing one ignores theoretical problems with building one:)

-
Dear @Gordon Wilson, how are you? Tachyons is precisely what relativity rules out, at least in stable worlds. According to relativity, a faster-than-light particle is physically equivalent - by a Lorentz transformation - to a particle moving backwards in time which violates causality. In quantum field theory, tachyons become excitations of a field whose potential energy has a local maximum, and when extrapolated, it's unbounded from below, indicating a lethal instability. All the best, LM –  Luboš Motl Jan 25 '11 at 8:45
@lubos: Yes, they violate causality. The particle could be also considered to be a positive energy particle moving forward in time (Feinberg reinterpretation) instead of a neg energy particle moving backwards. That is not what I meant by "does not rule out"--I meant the equations are consistent with them, not that they exist. GR equations predict singularities, and they likely don't actually exist either. –  Gordon Jan 25 '11 at 17:32
Dear Gordon, the equations are surely inconsistent with influences moving in both directions of time - that's the most serious inconsistency you can get in physics. Equations of quantum field theory show that the existence of tachyons is inconsistent with the basic stability of the vacuum. This is a very different situation from singularities which may exist, and some of them almost certainly do. Tachyons can't exist and don't exist. –  Luboš Motl Jan 25 '11 at 17:40
@lubos: Well Maxwell's equations have retarded and advanced solutions. Wheeler and Feynman also thought electrodynamic fields could be consistent with both (Wheeler-Feynman absorber theory). Singularities certainly exist as solutions to equations, but do you believe that physical singularities exist as well? It seems to me that we are about to replicate the sort of debate that occurred in the late 1800s about whether actual infinity exists in mathematics. –  Gordon Jan 25 '11 at 19:06
Just so we understand each other, I do not think that tachyons exist and accept generally what you say, except all I was saying is that they exist as solutions in special relativity. Just because solutions exist, does not guarantee physical existence. What do you mean that singularities almost certainly do? Are you using the word differently from the way it is used in GR? (ie string theory, QFT) Do you mean to say that you believe that points of 0 dimension and infinite density exist in the universe? –  Gordon Jan 25 '11 at 19:33

I'm going to try for a qualitative, equation-free version of the answer..

When you push on an object, you increase its momentum, which is the product of the object's mass and its speed. When you push on an object that is at rest, i.e., not already moving relative to you, the change in the object's momentum is realized almost entirely through a change in the speed component. This is what gives us the 'common sense' that if you push something a little bit harder, it will go a little bit faster.

But as the object's speed approaches the speed of light, the effect of applying force to the object is altered. Rather than the object's speed increasing, its mass begins to increase instead. So when the object's apparent speed is for example 99.99% of light, if you push it a little harder, while it does speed up slightly, it mainly just becomes a little bit heavier.

This changeover from effect-on-speed to effect-on-mass happens gradually (not all at once!), and there are equations in the other answers that describe it quantitatively. At everyday speed scales, the change-in-mass effect is practically immeasurable, so it seems counter-intuitive, but put particles in an accelerator, and it becomes observable fact.

-

From SR, the speed of light is always $c$ in every intertial frame. Accelerating a particle to $c$ would mean the velocity of light wasn't $c$ in the frame of the particle. The Lorentz transformations ensure you can't do this where you can show that the relationship between the acceleration of the particle measured in the lab, $a$, and the particle's frame, $a'$, is given by

$a' = \gamma^3 a$

As the velocity of the particle approaches $c$, $\gamma$ approaches infinity and $a$ approaches zero for a finite $a'$.

-

## protected by Qmechanic♦Jan 5 '13 at 0:58

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?