# Energy conservation in relativity

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#### zordim

• Jr. Member
• 46
##### Energy conservation in relativity
« on: 05/12/2012 11:47:48 »
What would be the GR time dilation effect in some observed point which is at equal distance r from several (i.e. cluster of 3 or 4) bodies of i.e. equal masses m, and what would be the GR time dilation effect in some observed point which is at the same distance r, but from only one body whose mass is equal to the sum of masses of bodies in the previous case?

If there is the difference, how to explain it?
If there is no difference, I would be interested in the calculation which brings to that, that there is no difference.

#### zordim

• Jr. Member
• 46
##### Re: Energy conservation in relativity
« Reply #1 on: 08/12/2012 00:13:04 »
What happened? Aren't there any scientists here who could make this simple, high-school-level calculation? Who is the GR expert here on this forum?

Ok, here is one question for both GR expert, and some expert for one, or more, of (too) many overlapped and interrelated theories of micro-(nano-pico-femto-atto-zepto-yocto)-cosmos (i.e. QM, QED, QCD, P-branes, D-branes, M-theory, string, superstring, gamma-gamma physics, Higgs-mechanism, any combination(s) of all that, etc.).
Let us imagine the following scenario (am I allowed to call it "gedanken experiment"?):  We are in a small spaceship, far away in space, in the middle of some huge intergalactic void (that is, there where the general-relativity-effects are not measurable). Our spaceship has the lab equipped to accurately measure the mass of electron, of positron, as well as the energies of gamma-photons. We have measured the masses of electron and positron. And then, we setup the collision of one very slow electron and one very slow positron (hence, there are no measurable special-relativity-effects), and we measure the energies of two gamma-photons, which are the result of that collision. If we then divide the total measured energy of gamma-photons E with the sum of masses of electron and positron m, we will get that E/m = c2. Hence, shouldn't this be the obvious proof that the equation E = mc2 is the fundamental, elementary, quantum-level equation, and not the consequence of relativity? Shouldn't this equation be the origin of relativity, and not the consequence? And, the origin-equations, that is, the cause-equations, that is, the fundamental equations, can be deduced from consequences (i.e. that is how the Planck's quantum equation was derived, that is how the Maxwell's equation for EM propagation velocity was derived). In other words, bricks are not the physical consequence of the brick-house, but opposite: the brick-house is the consequence of superpostion of bricks. Or I suffer from too much reason and common sense, which are "proved as unreliable qualities" if someone wants to deal with "real science"?

$$F = m \cdot a, F = \frac{dE}{dl}, a = \frac{dv}{dt} \Rightarrow$$   $$dE = m \cdot \frac{dv}{dt} \cdot dl = m \cdot \frac{dl}{dt} \cdot dv = m \cdot v \cdot dv$$ ;          $$dE = dm \cdot c^2$$

$$dm \cdot c^2 = m \cdot v \cdot dv \Rightarrow$$   $$\displaystyle \int_{m(v_{=0})}^{m(v)} \frac{dm}{m} = \frac{1}{c^2} \displaystyle \int_{v_{=0}}^{v} vdv$$  $$\Rightarrow$$  $$\ln \frac{m(v)}{m(v_{=0})} = \frac{1}{2} \cdot \frac{v^2}{c^2}$$  $$\Rightarrow$$  $$m=m_0 \cdot e^{\frac{1}{2} \cdot \frac{v^2}{c^2}}$$  $$\Leftrightarrow$$  $$E = E_0 \cdot e^{\frac{1}{2} \cdot \frac{v^2}{c^2}}$$

This simple derivation is physically justified in the most essential way possible, and it is also experimentally confirmed: if positron and electron have higher kinetic energy before their „annihilation“, then the photons which are the result of the positron-electron „annihilation“ have higher energies, too. Hence, each infinitesimal increase of their kinetic energy $$dE = m \cdot v \cdot dv$$, is converted into appropriate energy increase $$dE = dm \cdot c^2$$ during "annihilation" - all in accordance with the energy conservation principle.

In order to get Einstein's equation, we have to add the part which is not physically explainable:
$$dm \cdot c^2 = m \cdot v \cdot dv + dm \cdot v^2$$
$$dm \cdot (c^2 - v^2) = m \cdot v \cdot dv$$
$$\frac{dm}{m} = - \frac{1}{2} \cdot \frac{d(1 - \frac{v^2}{c^2})}{1 - \frac{v^2}{c^2}}$$
$$\displaystyle m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$
« Last Edit: 08/12/2012 00:22:17 by zordim »

#### JP

• Neilep Level Member
• 3366
##### Re: Energy conservation in relativity
« Reply #2 on: 08/12/2012 14:27:16 »
Zordim, I've moved you post to New Theories since you're definitely promoting a new theory.  Please keep such posts here in the future.

#### zordim

• Jr. Member
• 46
##### Re: Energy conservation in relativity
« Reply #3 on: 08/12/2012 17:02:46 »
No problem, JP.
I just asked a scientific question, and I did not get out of the scientific frames.
I do not mind where this is going to be discussed, but I am interested in scientific discussion, based exclusively on facts, and especially the valid scientific facts.
Perhaps, there is a mistake in my theory. But, I could not find it. The harder I tried, the more consistent it turned out to be. Some non-official comments from competent scientists were also such, but they warned me in advance that most probably I won't be allowed to present that in scientific papers. The official comments, from the prominent scientists, were completely non-scientific.

So, perhaps, some knowledgeable and honest truth seeker here will recognize it, and spread the word to similar people.
Or, perhaps, some mistake, which escaped my (and not just my) thourough checks, will be noticed by someone.
Until then, the only truly scientific theory, and the only theory which clearly and explicitly unifies the physics, is my theory. It is available, here, in the "New Theories" section. Clearly written and presented.

Regards,
Zordim

« Last Edit: 08/12/2012 22:11:30 by zordim »