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The famous equation for mass energy equivalence: $E=mc^2$

It cannot conserve energy or momentum without some loss in one way or another. To elaborate further if I take $1kg$ of mass and I also take electromagnetic radiation with same energy as the $1kg$ mass and ensuring its accuracy using the above equation. Then use this radiation and I use it to smash it into a "crash" mat I would get this:

$$E=mc^2$$ $$E=1.c^2$$ $$E=89,401,000,000,000,000$$ Approx.

That being the energy we then use this to calculate the momentum of the electromagnetic radiation transferred onto the "crash" mat which would be:

$$p=mv$$ $$p=mc$$ $$p=e/c^2 . c $$ $$p=e/c$$ $$p=89,401,000,000,000,000/c$$ $$p=299,000,000$$ Approx.

where as if I now take $1kg$ of matter and accelerate it to high velocity (say, 10,000 kilometers per hour ) to measure its momentum we get roughly $10,000 kg.m/sec$

That being said, momentum cannot be conserved, why is that? Next, Energy cannot be conserved.

For example if I used the same $1kg$ and I shoot it into space at a $x$ velocity and after say 1000 years I recieve it and measure mass of the object it would weigh exactly the same as 1000 years before therefore I can conclude it has same energy in its mass however if I use the mass-equivalent of electromagnetic radiation and do the same but my conclusion would be vastly different as first due to travelling in long fabric of space it would get slightly or even highly redshifted due to doppler effect. That being said the observer would get different energy reading to that of 1000 years ago. That in mind where did that energy go?

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  • $\begingroup$ "if I take 1kg of mass then I convert it into electromagnetic radiation using the above equation" It is not sufficient to say "I use an equation", you have to specify some physical technique, and then you have to justify your subsequent assumption that all the EM energy you just got is going in the same direction, and account for any momentum transfer to your apparatus that may have occurred to insure that collimation. $\endgroup$ – dmckee Apr 26 '14 at 18:45
  • $\begingroup$ Your edit did not address either issue I brought up. $\endgroup$ – dmckee Apr 26 '14 at 19:07
  • $\begingroup$ Yes, instead of a talking about the physical issue I simply used 2 objects with same equivalence one being light (electomagnetic radiation) and one being 1kg. $\endgroup$ – LogicProgrammer Apr 26 '14 at 19:38
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If you have a massive object of mass $m$ which spontaneously decays into electromagnetic radiation, then indeed the total energy of that electromagnetic radiation is $E=mc^2$. However the total momentum of that radiation would be $0$ due to momentum conservation.

From this, we can conclude that it is impossible that all that mass is converted into a single photon, because a single photon would indeed have the momentum $p=mc \ne 0$, which is forbidden by momentum conservation. However it is in principle allowed (depending on spin and other conserved quantities) that the massive object of mass $m$ decays into two photons of energy $E=mc^2/2$ each, both going into opposite direction (and thus having $\vec p_1=-\vec p_2$, thus $\vec p_1+\vec p_2=\vec 0$) and each having as absolute value of the momentum $p_i=mc/2$.

However for a body of mass 1kg (which, to be completely converted to electromagnetic radiation, would have to consist of 0.5kg matter and 0.5kg antimatter), you'd rather expect quite a lot of electromagnetic radiation going into all directions. However the total momentum of all that radiation would still be zero.

Similarly, electromagnetic radiation going into one direction cannot spontaneously convert into a massive object, because that also would violate energy or momentum conservation. It the radiation interacts with matter, it can pass a bit of its energy and momentum to that object, and then conversion into massive particles can be possible because that extra energy and momentum transfer allows to conserve both energy and momentum.

Now, as soon as we take General Relativity, and especially the expansion of the universe, into account, energy conservation indeed doesn't hold any longer on a global scale (it still holds locally, though).

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"if I now take 1kg of matter and accelerate it to high velocity (say, 10,000 kilometers per hour ) to measure its momentum we get roughly 10,000kg.m/sec"

You have confused your units. Actual calculation with $M=1 \,\mathrm{kg}$: $$\begin{align} v &=10\,000\,\mathrm{km/hr} \\ &= \frac{10\,000\,000\,\mathrm{m}}{3600\,\mathrm{s}} \\ & =2777.78\,\mathrm{m/s} \\ \\ p&=Mv \\ &= (1\,\mathrm{kg})(2777.78\,\mathrm{m/s})\\ &= 2777.78\,\mathrm{kgm/s} \,, \end{align}$$
which is not the same as $10,000\,\mathrm{kgm/s}$

"..same energy as the 1kg mass... I would get this:

$$E=mc^2$$
$$E=1 \cdot c^2$$

Remember:

The rest mass is: $$m_0 =1kg$$

& NOT
the relativistic effective mass, which is:

$$m =\gamma \cdot m_0 $$
where
$$\gamma=1/\sqrt{(1-(v/c)^2)}$$

You need to state a velocity $v$ for mass $m$

"however if I use the mass-equivalent of electromagnetic
radiation & do the same, my conclusion would be vastly
different: e.g. due to travelling in long fabric of space
it'd get slightly or even highly redshifted due to the
doppler effect.
That being said the observer would get different energy
reading to that of 1000 years ago.
That in mind - where did that energy go?"

The doppler shift spreads out the wave & so when you add up the energy along that wave it's total remains the same.
The only difference is that it takes longer for that total to end up somewhere in the universe.
Remember that the light is redshifted relative to something.
If that something were you, the event would probably have to occur outside of the galaxy because of the Milky way's gravity.

From this & the total momentum: $$\Sigma \vec p=\vec 0$$

..we reach a..

Conclusion:

Both energy AND momentum ARE CONSERVED.

Also see:
- "Energy is not conserved between different ref frames"
Conservation of energy and Doppler effect?
- "four-vector"
Is kinetic energy a relative quantity? Will it make inconsistent equations when applying it to the conservation of energy equations?
- "AJP paper"
photons in expanding space: how is energy conserved?
- "ΛCDM model"
photons in expanding space: how is energy conserved?

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  • 2
    $\begingroup$ Hi Socrates, please make use of MathJax for content like this. See meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Brandon Enright Aug 15 '14 at 21:23
  • $\begingroup$ @Socrates, Quick Question - Does cosmological red shift change the energy in the photon or is it still observer depended? $\endgroup$ – LogicProgrammer Aug 16 '14 at 1:37
  • $\begingroup$ It's observer dependent (approx). There may be a little energy lost to spacetime "foam". $\endgroup$ – Socrates Aug 16 '14 at 2:18
  • $\begingroup$ Socrates, I've done some more of the math for you, and in doing so I noticed that you are hacking the markup with html primitives. That works, but it is a lot harder than using markdown. It also results in less consistent (and often ugly) typesetting. If you intend to be a regular poster here you'll find it easier to learn the markdown dialect use by Stack Exchange sites as well as the idiomatic LaTeX math-mode used by MathJax. $\endgroup$ – dmckee Aug 16 '14 at 17:21
  • $\begingroup$ Thanks for the help. I thought using <br> was a bit iffy.. $\endgroup$ – Socrates Aug 16 '14 at 21:58

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