# Is there a maximum energy for a relativistic particle?

So I was told today that the Standard Model breaks down at really, really high energies. The lecturer mentioned particles such as electrons hypothetically having energies equivalent to that of entire stars and it got me thinking, surely the maximum theoretical energy any particle can have is limited by the speed of light. I understand I'm talking about only kinetic energy here, but I fail to see how any other form of energy is relevant at those sorts of speeds. I did wonder if massive particles at the speed of light have infinite energy (which satisfies my question if this is the case), but I don't see that sort of behaviour from Einstein's relativistic mass-energy relation.

So my question is- is there a maximum theoretical energy particles can have?

• "surely the maximum theoretical energy any particle can have is limited by the speed of light" no, not really. $\lim_{v\to 1}T(v)=+\infty$. – AccidentalFourierTransform Mar 10 '16 at 16:26
• Can you expand on that a little more? – Ben Jones Mar 10 '16 at 16:32
• Closely related: physics.stackexchange.com/q/1557 – Kyle Kanos Mar 10 '16 at 16:38
• There probably are physical limits but current theories aren't enough to find them. – Mithoron Mar 10 '16 at 20:28
• @BrockAdams no, it can't be a black hole in some reference frames. Imagine the point of view of someone moving at hyperrelativistic speed: ok, did you turn into a black hole? You're moving at that speed in his frame. – JDługosz Mar 11 '16 at 7:32

There is no maximum energy of a freely moving massive particle in special relativity.

The relativistic energy of a particle of rest mass $m$ moving in your frame at speed $v$ is given by $E=\gamma m c^2$ where $\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$. If you look closely at $\gamma$ you will see that it is not defined at $v=c$ ($c$ is the speed of light), and that $\underset{v\rightarrow c}{\lim}\gamma = \infty$. From this you conclude that the energy of a particle increases without bounds as its speed approaches that of light.

From this you see that: 1) you can always increase the energy of a massive particle by accelerating it, 2) you need more and more energy to approach $c$, so no massive particle can travel at the speed of light.

• Pedantry: It's a one-sided limit, $\lim_{v\to c^-}\gamma=+\infty$ but $\lim_{v\to c^+}\gamma=-\infty i$. – Charles Mar 10 '16 at 17:10
• @Charles In the complex plane you can define a limit of infinity to be that the magnitude of the complex number gets large without bound. And using that definition the limit is technically correct. And yours is not correct since the square root function requires a branch cut for complex numbers so whether you get a plus or a minus $i$ out of a square root of a negative number is undetermined. The number $i$ is simply a square root of $-1$ and so is its negative. And for positive numbers you can pick positive square roots, but in general there is no way to pick 1 of the 2 squares roots. – Timaeus Mar 10 '16 at 17:16

It's helpful someday to learn a correct expression for the energy of a particle. An example is $$E=\sqrt{(mc^2)^2+(\vec pc)^2}.$$

Which works for any mass (even zero) and any momentum (assuming the particle is on shell, which classical particles are). But it requires you to know the momentum. And the momentum of a massive particle at the speed of light, like the energy, is infinite.

You should expect momentum to increase without bound becasue a force isn't mass times acceleration. A force is actually (even to Newton) the time rate of change of momentum. So apply a force and the momentum changes, keep applying and the momentum keeps increasing. It's just that momentum isn't equal to $m\vec v.$ Instead $$\vec p=\vec v E/c^2,$$ or $$p=\sqrt{(E/c)^2-(mc)^2}.$$

And you might wonder why I didn't write an equation for $p$ in terms of mass and velocity. And the reason is because you can't determine the momentum from the mass and velocity alone. It simply isn't a function of mass and velocity. For a massive particle you can write $$\vec p=\frac{m\vec v}{\sqrt{1-\frac{v^2}{c^2}}},$$ but this isn't a fundamental relation because it only holds for massive particles. So fundamentally, momentum isn't determined by mass and velocity.

Momentum and energy are the actual fundamental things. Mass just tells you how they balance together through $$(mc^2)^2=E^2-(\vec pc)^2,$$ and velocity is $$\vec v=c^2\vec p/E.$$

• +1 Your first equation is the reason I was confused at first, looking at that I couldn't see how energy could be infinite until you explained with momentum – Ben Jones Mar 10 '16 at 22:49

A passive variant of Andrea Di Biagio answer is to consider a particle at the supposedly maximum energy and then to consider evaluating that energy in another reference frame that moves in the opposite direction of that particle.

## protected by Qmechanic♦Mar 10 '16 at 17:05

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