# The time component is $\gamma m c$, so shouldn't $E=mc$?

Basically, the book is Brian Cox's Why Does $E=mc^2$?: (And Why Should We Care?).

I just finished Chapter 5, where we derived the spacetime momentum vector (energy-momentum four vector, as he establishes the physics jargon).

Let $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

So, as we found out, the vector's spatial component is $\gamma m v$, leaving the time component as $\gamma m c$.

He went on, under the guise of making the outcome more intellgible, to saying we could happily multiply the time component by $c$ without changing it's meaning. Cool, I thought, no problem. Next, he pointed out $\gamma\approx1+\frac{1}{2}\frac{v^2}{c^2}$, so $\gamma m c^2 \approx mc^2 + \frac{1}{2}mv^2$. Et voilà, $mc^2$.

Granted, he's obviously trying to simplify things so I can reach an intuitive understanding, but from that point onwards, he uses $mc^2$ as the conversion value.

I'm confused. Could someone explain why we use $mc^2$ and not the version scaled down by a factor of $c$?

-
simple answer dimensional inconsistency!!! $mc$ only becomes $[kg^1.m^1.s^{-1}]$ where as for energy you need $[kg^1.m^2.s^{-2}]$. –  Vineet Menon Nov 15 '11 at 6:25
You scale it by $c$ so that the time becomes a distance, just like the spatial dimensions. –  queueoverflow Apr 12 '12 at 8:57

Dimensional analysis is enough to see that $mc^2$ has the same units as energy while $\gamma mc$ or $mc$ doesn't.

Concerning the components of a 4-vector, special relativity unifies the spatial and temporal components. But the 4 components of a 4-vector with "uniform units" do not necessarily enjoy the same normalization as the quantities outside relativity. Instead, you must typically multiply the time component by $c$ or $1/c$ to get the usual non-relativistic normalization of the quantity.

The position vector is $(x,y,z,ct)$. Note that all of them have the dimension of length. But the real time is $ct$, the last component, divided by $c$. Similarly, the energy-momentum vector has $m_0\gamma v_x,m_0\gamma v_y, m_0\gamma v_z,m_0\gamma c$. Again, all components have the same units but now you have to multiply the last component $m_0\gamma c$ by $c$ to get the usual normalization for the energy, $E=m_0 \gamma c^2$.

This is just a question of units. Nothing guarantees that the simplest identifications and conventions will lead to correct formulae without any extra $c$ with the units used before relativity. When we say "the time component of a vector is a particular quantity known before relativity", we mean that it contains the same information but sometimes we need to normalize it differently, add a universal factor. Before relativity, people used very unnatural units for many quantities and the temporal and spatial components didn't have the sam units even though their key information was always a part of the same 4-vector which shows that they may be "rotated into each other".

Among physicists who study relativistic phenomena (e.g. particle physicists), this is a complete non-problem and physicists often use units with $c=1$, anyway. This really means that distances are measured in light seconds and the speed of light is one light second per second, and the difference between a light second and second is suppressed. (Particle physicists typically use units in which one GeV, and not one second, and its powers are used for everything, i.e. times, distances, energies, momenta, masses, etc.: they also set $\hbar=1$ so that energy and frequency i.e. inverse time have the same units.)

-

Uh, because $mc$ is not energy? And what do you mean "time component"? Your $\gamma mc$ is momentum.

-
so what makes $mc^2$ energy? just multiplying by a constant... –  dougvk Nov 15 '11 at 6:12
Well one obvious way would be to check units. Other than that, go check out Einstein's relation: $E^2-(pc)^2 = (m_0c^2)^2$, where $m_0$ is rest-mass. –  Chris Gerig Nov 15 '11 at 6:14
I'm going to wait for a more intuitive explanation. Also $\gamma m v$ is momentum. $\gamma m c$ is time component. –  dougvk Nov 15 '11 at 6:17
Sorry I don't get what you mean be "more intuitive"... by dimensionality, $mc$ cannot be energy and $mc^2$ must be energy. Oh and now I see, your 'time component' is part of your momentum 4-vector (so it is indeed momentum!). –  Chris Gerig Nov 15 '11 at 6:28
oh okay. in the question i had it differentiated as time and spatial components. now I see they are all subsumed by the label momentum component. more intuitive just means using many more words to support and explain your equations –  dougvk Nov 15 '11 at 6:48