# In E=mc2, Why C?

#### Agerhell

What else could it be? There is no other way to get a constant with units of $L^2/T^2$ from any of the other universal constants. So it has to be some multiple of $c^2$. The only other possibility would be for there to be no conversion between mass and energy.
Have you ever heard of such a thing as a dimensionless constant? You could in principle have any dimensionless number k and E=mkc2, but that would not not work with the definition of relativistic momenta as seen in the link posted in response #26. k must therefore be equal to one. Are you trying to confuse the kid?

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#### Agerhell

Einstein defined:
The velocity of light is the fastest one. His formula says: $E=mc^2$, but that theory gone down, so that formula wouldn't stay for any application, if it would then it would be wrong.
If you define $m=m_0\sqrt(1/(1-v^2/c^2))$ then the initial relation $E=mc^2$ would still be correct. Just replace the rest mass with the relativistic mass and you are OK.

#### Dale

Mentor
Have you ever heard of such a thing as a dimensionless constant? You could in principle have any dimensionless number k and E=mkc2
Sure, that is why I said "some multiple of c²". You must have the c² term simply due to the units. IMO, the reason why k=1 is interesting, but the c² seems to get the questions instead for some reason.

#### -Physician

If you define $m=m_0\sqrt(1/(1-v^2/c^2))$ then the initial relation $E=mc^2$ would still be correct. Just replace the rest mass with the relativistic mass and you are OK.
WOULD be correct but it's not, and IF we define, we can't define if we shouldn't , we can't just define it.

#### Agerhell

WOULD be correct but it's not, and IF we define, we can't define if we shouldn't , we can't just define it.
If you replace the "rest mass" with the "relativistic mass" as suggested you get an expression for the energy that would work if you want to calculate for instance the amount of energy you get when you smash a proton and an antiproton togehter at a certain velocity in an accelerator for instance.

Right?

If you use the concept of "relativistic mass" the relation still holds when there are moving masses.

Right?

#### harrylin

If you replace the "rest mass" with the "relativistic mass" as suggested you get an expression for the energy that would work if you want to calculate for instance the amount of energy you get when you smash a proton and an antiproton togehter at a certain velocity in an accelerator for instance.

Right?

If you use the concept of "relativistic mass" the relation still holds when there are moving masses.

Right?
Certainly correct - and while it may be not exactly as originally intended by Einstein, it is how my textbook applied it when I was a student. Works perfect. But it has little to do with the topic I fear...

#### DrStupid

But it has little to do with the topic I fear...
Why not? The original question was why c² and not any other proportionality factor between energy and mass. Without other information this may refer to rest energy and rest mass or to total energy and relativistic mass. This question can be answered for both cases at once:

Assuming we know that energy is linear correlated with mass (as used in Newton's definition of momentum) but we do not know the proportionality factor k:

$E = k \cdot m$

Then the change of mechanic energy is

$dE = F \cdot ds = k \cdot dm$

According to Newton's second law the force is

$F = \frac{{dp}}{{dt}} = \frac{{d\left( {m \cdot v} \right)}}{{dt}} = m \cdot \frac{{dv}}{{dt}} + v \cdot \frac{{dm}}{{dt}}$

Integration of the resulting differential equation

$\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}$

results in

$m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}$

The constant of integration m0 is the mass of the body at rest and as this equation gives rational results for k<v2 only the unknown proportionality factor k must be the square of a maximum velocity that no body can reach or exceed. In relativity there is such a velocity: the speed of light in vacuum. Therefore there is only one possibility for a linear correlation of energy and mass:

$E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{c^2}} }}$

#### harrylin

[..] The constant of integration m0 is the mass of the body at rest and as this equation gives rational results for k<v2 only the unknown proportionality factor k must be the square of a maximum velocity that no body can reach or exceed. In relativity there is such a velocity: the speed of light in vacuum. Therefore there is only one possibility for a linear correlation of energy and mass:

$E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{c^2}} }}$
Very good, I had not thought of that - that mathematical insight also has physical suggestion: it tells us how and why the inertial property of energy is determined. Thanks!

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