- #1
JohnnyGui
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I have a question about the velocity of a mass that is being accelerated from its resting position, for deriving its energy according to ##E=mc^2##
The energy needed for accelerating a mass is: ##E = F \cdot s = m \cdot a \cdot s##. Where s is the distance over which the mass is moved.
Now, I understand that with each small increase in s, the mass would increase according to
##\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}=m##. So I should integrate
$$\int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot ds$$
Since the relativistic mass is dependent on the velocity we could write the distance ##ds## and ##a## in terms of velocity. Regarding the acceleration ##a##, we know that ##a \cdot t## will give the final velocity after time ##t##. So ##a \cdot t = v_{final}##. I’ll explain later why I’m emphasizing on this velocity being the final velocity.
The distance ##s## is, as known, calculated by ##\frac{1}{2}at^2##. Combining this with ##a \cdot t = v_{final}## would give: ##\frac{1}{2}v_{final}\cdot t##. Substituting these formulas for ##a## and ##ds## in the integration formula would give:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot \frac{v_{final}}{t} \cdot \frac{1}{2} \cdot dv_{final} \cdot t$$
Which, after simplifying gives:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot v_{final} \cdot \frac{1}{2} \cdot dv_{final}$$
I googled this and it seems to be the correct formula that needs to be integrated. However, I can’t see how this is possible.
Here’s why; I think the ##v_{final}## that you’d calculate from the acceleration and distance is different from the velocity ##v## in the Lorentz formula part of the integration. If a mass is being accelerated over a small distance ##ds##, then the mass increases based on the mean velocity in that small distance ##ds## and not based on the final velocity at the moment the mass finishes that small distance ##ds##. The ##v## in the Lorentz formula should be the velocity at half of ##ds##, so that ##v = \frac{1}{2}v_{final}##. How can the ##v## in the Lorentz term and the ##v_{final}## still be considered the same?
The energy needed for accelerating a mass is: ##E = F \cdot s = m \cdot a \cdot s##. Where s is the distance over which the mass is moved.
Now, I understand that with each small increase in s, the mass would increase according to
##\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}=m##. So I should integrate
$$\int_{0}^{d} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot a \cdot ds$$
Since the relativistic mass is dependent on the velocity we could write the distance ##ds## and ##a## in terms of velocity. Regarding the acceleration ##a##, we know that ##a \cdot t## will give the final velocity after time ##t##. So ##a \cdot t = v_{final}##. I’ll explain later why I’m emphasizing on this velocity being the final velocity.
The distance ##s## is, as known, calculated by ##\frac{1}{2}at^2##. Combining this with ##a \cdot t = v_{final}## would give: ##\frac{1}{2}v_{final}\cdot t##. Substituting these formulas for ##a## and ##ds## in the integration formula would give:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot \frac{v_{final}}{t} \cdot \frac{1}{2} \cdot dv_{final} \cdot t$$
Which, after simplifying gives:
$$\int_{0}^{v} \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \cdot v_{final} \cdot \frac{1}{2} \cdot dv_{final}$$
I googled this and it seems to be the correct formula that needs to be integrated. However, I can’t see how this is possible.
Here’s why; I think the ##v_{final}## that you’d calculate from the acceleration and distance is different from the velocity ##v## in the Lorentz formula part of the integration. If a mass is being accelerated over a small distance ##ds##, then the mass increases based on the mean velocity in that small distance ##ds## and not based on the final velocity at the moment the mass finishes that small distance ##ds##. The ##v## in the Lorentz formula should be the velocity at half of ##ds##, so that ##v = \frac{1}{2}v_{final}##. How can the ##v## in the Lorentz term and the ##v_{final}## still be considered the same?
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