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In E=mc2, Why C?

by Mileman10
Tags: emc2
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Agerhell
#55
Feb12-12, 01:32 PM
P: 152
Quote Quote by -Physician View Post
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
#56
Feb12-12, 01:39 PM
P: 3,187
Quote Quote by Agerhell View Post
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
#57
Feb12-12, 03:50 PM
P: 476
Quote Quote by harrylin View Post
But it has little to do with the topic I fear...
Why not? The original question was why cē 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:

[itex]E = k \cdot m[/itex]

Then the change of mechanic energy is

[itex]dE = F \cdot ds = k \cdot dm[/itex]

According to Newton's second law the force is

[itex]F = \frac{{dp}}{{dt}} = \frac{{d\left( {m \cdot v} \right)}}{{dt}} = m \cdot \frac{{dv}}{{dt}} + v \cdot \frac{{dm}}{{dt}}[/itex]

Integration of the resulting differential equation

[itex]\frac{{dm}}{m} = \frac{{v \cdot dv}}{{k - v^2 }}[/itex]

results in

[itex]m = \frac{{m_0 }}{{\sqrt {1 - \frac{{v^2 }}{k}} }}[/itex]

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:

[itex]E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{c^2}} }}[/itex]
harrylin
#58
Feb12-12, 04:23 PM
P: 3,187
Quote Quote by DrStupid View Post
[..] 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:

[itex]E = m \cdot c^2 = \frac{{m_0 \cdot c^2 }}{{\sqrt {1 - \frac{{v^2 }}{c^2}} }}[/itex]
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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