# Special Relativity and $E = mc^2$

I read somewhere that $E=mc^2$ shows that if something was to travel faster than the speed of light then they would have infinite mass and would have used infinite energy.

How does the equation show this?

The reason I think this is because of this quote from Hawking (I may be misinterpreting it):

Because of the equivalence of energy and mass, the energy which an object has due to its motion will add to its mass. This effect is only significant to objects moving at speeds close to the speed of light. At 10 per cent of the speed of light an objects mass is only 0.5 per cent more than normal, at 90 per cent of the speed of light it would be twice its normal mass. As an object approaches the speed of light its mass rises ever more quickly, so takes more energy to speed it up further. It cannot therefore reach the speed of light because its mass would be infinite, and by the equivalence of mass and energy, it would have taken an infinite amount of energy to get there.

The reason I think he's saying that this is as a result of $E = mc^2$ is because he's talking about the equivalence of $E$ and $c$ from the equation.

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– Qmechanic Apr 18 '12 at 19:42
travel faster than the speed of light is still impossible. There is no scientific evidence. it seems speed of photons is a kind of cosmic limit – user8784 Apr 19 '12 at 20:53

I read somewhere that $E=mc^2$ shows that if something was to travel faster than the speed of light then they would have infinite mass and would have used infinite energy.

Nope, not true. For a couple of reasons, but first, let me explain what $E = mc^2$ means in modern-day physics.

The equation $E = mc^2$ itself only applies to an object that is at rest, i.e. not moving. For objects that are moving, there is a more general form of the equation,

$$E^2 - p^2 c^2 = m^2 c^4$$

($p$ is momentum), but with a little algebra you can convert this into

$$E = \gamma mc^2$$

where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$. This factor $\gamma$, sometimes called the relativistic dilation factor, is a number that depends on speed. It starts out at $\gamma = 1$ when $v = 0$, and it increases with increasing speed. As the speed $v$ gets closer and closer to $c$, $\gamma$ approaches infinity. Armed with this knowledge, some people look at the formula $E = \gamma mc^2$ and say that, clearly, if a massive object were to reach the speed of light, then $\gamma$ would be infinite and so the object's energy would be infinite. But that's not really true; the correct interpretation is that it's impossible for a massive object to travel at the speed of light. (There are other, more mathematically complicated but more convincing, ways to show this.)

To top it off, there is an outdated concept called "relativistic mass" that gets involved in this. In the early days of relativity, people would write Einstein's famous formula as $E = m_0 c^2$ for an object at rest, and $E = m_\text{rel}c^2$ for an object in motion, where $m_\text{rel} = \gamma m_0$. (The $m$ I wrote in the previous paragraphs corresponds to $m_0$ in this paragraph.) This quantity $m_\text{rel}$ was the relativistic mass, a property which increases as an object speeds up. So if you thought that an object would have infinite energy if it moved at the speed of light, then you would also think that its relativistic mass would become infinite if it moved at the speed of light.

Often people would get lazy and neglect to write the subscript "rel", which caused a lot of people to mix up the two different kinds of mass. So from that, you'd get statements like "an object moving at light speed has infinite mass" (without clarifying that the relativistic mass was the one they meant). After a while, physicists realized that the relativistic mass was really just another name for energy, since they're always proportional ($E = m_\text{rel}c^2$), so we did away with the idea of relativistic mass entirely. These days, "mass" or $m$ just means rest mass, and so $E = mc^2$ applies only to objects at rest. You have to use one of the more general formulas if you want to deal with a moving object.

Now, with that out of the way: unfortunately, the passage you've quoted from Hawking's book uses the old convention, where "mass" refers to relativistic mass. The "equivalence of energy and mass" he mentions is an equivalence of energy and relativistic mass, expressed by the equation $E = m_\text{rel}c^2$. Under this set of definitions, it is true that an object approaching the speed of light would have its (relativistic) mass approach infinity (i.e. increase without bound). Technically, it's not wrong, because Hawking is using the concept correctly, but it's out of line with the way we do things in physics these days.

With modern usage, however, I might rephrase that paragraph as follows:

Because energy contributes to an object's inertia (resistance to acceleration), adding a fixed amount of energy has less of an effect as the object moves faster. This effect is only significant to objects moving at speeds close to the speed of light. At 10 per cent of the speed of light, it takes only 0.5 per cent more energy than normal to achieve a given change in velocity, but at 90 per cent of the speed of light it would take twice as much energy to produce the same change. As an object approaches the speed of light, its inertia rises ever more quickly, so it takes more and more energy to speed it up by smaller and smaller amounts. It cannot therefore reach the speed of light because it would take an infinite amount of energy to get there.

Disclaimer: all I've said here applies to a fundamental particle or object moving in a straight line. When you start to consider particles with components which may be moving relative to each other, the idea of relativistic mass kind of makes a comeback... kind of. But that's another story.

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Thanks, I read it in Hawking's A Brief History Of Time. I think that's what he was saying. – Olly Price Apr 18 '12 at 19:44
Huh, I would expect Hawking to be reasonably accurate with these things. Though it is easy to make statements that can be misinterpreted, when talking about relativity. – David Z Apr 18 '12 at 19:56
I'll quote the book a bit later, I may be misinterpreting it. – Olly Price Apr 18 '12 at 20:01
OK, well, if you'd like to edit that quote into your question when you have a chance, I can update my answer to address it. – David Z Apr 18 '12 at 20:04
@Ron: No, but we should continue this in the chat room. – David Z Apr 20 '12 at 21:51

this equation is at rest , when moving the energy goes as $E= \frac{m_{0}c^{2}}{(1- v^{2}/c^{2})^{1/2}}$

so if $v \ge c$ and the energy must be real, then the mass should be purely imaginary.. :D

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I would blame the restriction on the speed at which objects can travel more on one of the two postulates that Einstein used to derive $E=mc^2$.

I wish Einstein's original derivation of $E=mc^2$ was taught in schools! It is such an amazing piece of work if you go through it in detail. But it's also cryptic, jumping very quickly through ideas that apparently seemed pretty "obvious" to Einstein, but which are far from that to the rest of us.

In any case, he derived it over the course of two papers. The first one defined the theory of special relativity, while the second and very short one derived his famous equation. It originally used $L$ for $E$, and Einstein never quite wrote it out the way we are used to seeing it.

His first paper began with with two simple postulates, which are:

(1) No test of mechanics or optics changes when you are moving without acceleration, and

(2) The speed of light is always constant when measured from such a moving frame.

Amazingly, that's all that is needed.[1]

Now, if you want to point fingers at where exactly the idea that you cannot travel faster than light emerges from special relativity, I'd point at the second of Einstein's postulates: Every frame sees the same speed of light.

So why is that important? Picture it this way: If light must always travel at $c$ from your perspective, what happens if you launch a rocket capable of traveling at very nearly $c$, and your rocket in turn sends out a light pulse ahead of itself?

The rocket will see that pulse as traveling at $c$. However, as the person who launched the rocket, you must see something different, since otherwise the light beam emitted by the rocket would look like it's traveling at nearly $2c$, which would violate Einstein's second postulate.

So, the light pulse in front of the spaceship must necessarily travel at $c$ from your perspective also, and that in turn means that the spaceship must always remain behind any pulse of light that can be emitted. If you draw that out on paper, you get this quirky result that from your view, objects moving closer and closer to the speed of light must nonetheless always remain behind an actual beam of light, since any other result would enable you to see light a light pulse moving faster than $c$. Objects thus wind up getting "flattened" against the barrier represented by the speed of an actual light beam.

There are other consequences of this apparent flattening, which is called the Lorentz contraction, that I won't get into here. They include slowed time and increased mass, both of which can be derived from the original simple postulates that Einstein made.

So, the bottom line: It's more accurate to blame Einstein's assumed postulate of constant light speed for limiting material objects to traveling at sub-light speed, rather than blaming $E=mc^2$. And historically, Einstein didn't even derive the $E=mc^2$ result until his second addendum paper, which he published after he had already shown the other consequences of his postulates.

[1] Actually, there's an interesting minor secret buried in Einstein's postulates: One is missing. To ensure proper scaling of the results, you must add the following third postulate: If two groups of particles diverge from each other at speed $s$ along axis $x$, the orthogonal plane defined by the remaining two orthogonal axes $y$ and $z$ must remain invariant in scale between the two groups of particles. Or a lot less formally: $y$ and $z$ don't change, even though $x$ Lorentz contracts.

That point seems so obvious that it's usually either assumed or treated as an outcome of the other two postulates. However, you can't really derive it from the other two postulates since an infinite number of profiles that meet the first two postulates are possible if you allow variable scaling of the $yz$ plane. Lorentz had noticed this, but his thoughts about it were largely forgotten after Einstein's papers. In any case, when talking about Lorentz contraction of $x$ it makes sense to be explicit about the invariance (or lack thereof) of the remaining two spatial axes.

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## protected by Qmechanic♦Apr 9 '13 at 1:22

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