# How does $E=mc^2$ put an upper limit to velocity of a body?

How does $E=mc^2$ put a upper limit to velocity of a body? I have read some articles on speed of light and they just tell me that it is the maximum velocity that can be acquired by any particle. How is it so? What is violated if $v>c$ ?

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The equation $E=mc^2$ does not put an upper limit to the velocity of an object. The limit in the velocity of an object comes from the fact that the speed of light in the vacuum is the same for all the observers, this is an experimental fact. Plus, in the development of special theory of relativity one can see that $v$ must be less than $c$. And if this condition is violated then the "causality" of events is violated. This means that your mother could born after you born, in some reference system if $v>c$. – Ana Apr 5 '13 at 5:24
@Anuar that should probably be an answer :-) – David Z Apr 5 '13 at 5:29

$E_0 = m_0 c^2$ is only the equation for the "rest energy" of a particle/object.

The full equation for the kinetic energy of a moving particle is actually:

$$E-E_0 = \gamma m_0c^2 - m_0c^2,$$ where $\gamma$ is defined as $\gamma = \frac{1}{\sqrt{1 - (v/c)^2}},$

where $v$ is the relative velocity of the particle.

An "intuitive" answer to the question can be seen by noticing that the particle's energy approaches $\infty$ as its velocity approaches the speed of light. Thus, in order for the particle to move faster than the speed of light would require it to attain infinite kinetic energy, which can't happen.

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Great answer, i didn't knew that KE for relativistic speeds is $\gamma mc^2$ – ABC Apr 5 '13 at 9:58
-1: KE for relativistic speed is NOT $\gamma mc^2$! This expression gives the total energy. The kinetic energy is $\gamma mc^2 - mc^2$; the total energy minus the rest energy. – joshphysics Apr 5 '13 at 15:36
Ahh yes, I'll correct that. Thanks for the correction. – bclifford Apr 5 '13 at 16:25

To complete bclifford's answer, our current equation for energy-momentum of a particle is $E^2=p^2c^2+m^2c^4$ which is the final expression for $E=\gamma mc^2$, where $\gamma$ is the Lorentz factor obtained from his transformations.

Hence, for a particle like the photon - this equation is valid throwing $E=pc$, which says that the photon has momentum.

For particles at rest, $p=0$ which gives the rest energy $mc^2$ of the massive object.

The great necessity of this equation is that it restricts massive objects to be accelerated above $c$ as it requires infinite energy, massless particles to travel at $c$ always and also faster-than-light hypothetical particles to travel above $c$ always...

You can consider $pc$ and $mc^2$ as the opposite and adjacent sides of a right-angled triangle and the energy along the hypotenuse. No matter how fast a massive object moves, the hypotenuse is always greater than the other two sides, (i.e.) it can never quite reach $c$...

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Just to visualise the other answers, here is a plot of the kinetic energy of a body in relativistic and nonrelativistic mechanics (note the log scale on the vertical axis):

You can see that as the speed approaches the speed of light the energy required according to special relativity shoots up compared to what nonrelativistic mechanics would say. It requires an infinite amount of energy for any massive body to reach the speed of light.

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The first time I looked at that chart, it appeared that the kinetic energy is zero when $v \approx 0.05c$. Then I looked again. :) – Michael Kjörling Apr 5 '13 at 7:33

It doesn't. The equation $E = mc^2$ and the fact that no physical object can be accelerated past the speed of light are two entirely separate conclusions of special relativity.

The reason $c$ is an upper bound on the speed of an object has to do with the Lorentz transformations. These are the mathematical expressions that relate positions and times as measured by one observer to positions and times as measured by another observer. Now, suppose an object starts at rest with respect to observer A, and then accelerates until it is at rest with respect to observer B, which is moving at a speed $v$ relative to A. There has to be some Lorentz transformation you can use to convert between A's measurements and B's measurements, or equivalently, between the reference frame of the object pre-acceleration and its reference frame post-acceleration. But there is no Lorentz transformation that will take you from a reference frame in which an object is going slower than light to a reference frame where the same object is going faster than light or at the speed of light.

(Technically, that argument is a little hand-wavy, but it should get the main point across.)

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This is not really a satisfactory argument. It is possible to extend the Lorentz transformations to $v>c$ (but not $v=c$); see Recami, Riv Nuovo Cimento 9 (1986) 1. This works in $m+n$ dimensions only if $m=n$; otherwise there is a no-go theorem, V. Gorini, "Linear Kinematical Groups," Commun Math Phys 21 (1971) 150. A better argument would be that regardless of the existence or nonexistence of Lorentz-like transformations for $v>c$, no sequence of Lorentz transformations with $v<c$ produces $v>c$, so no continuous process of acceleration can boost a bradyon past $c$. – Ben Crowell Apr 5 '13 at 17:42
@BenCrowell That's exactly why I said my argument is a little hand-wavy. The question didn't seem to be looking for a rigorous proof. Although what you're saying is what I meant to write, so let me clarify my answer a little bit. – David Z Apr 5 '13 at 19:09