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Is Einstein's Most Famous Equation Wrong?

K

Keklas Rekobah

SOC-14 1K
https://www.cnn.com/2018/05/03/opinions/alpha-cern-antihydrogen-opinion-lincoln/index.html

According to this article, "Making antimatter is extremely easy using modern technology. Giant particle accelerators, like those at Fermilab in Illinois or CERN, constantly make and study antimatter. And the process is very well understood. Energy always makes matter and antimatter in equal quantities."

So, instead of "E = m • c^2", should we instead say, "E = ((m/2 • c^2) + (-m/2 • c^2))"?

Or does it not really matter at all?
 
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According to this article, "Making antimatter is extremely easy using modern technology. Giant particle accelerators, like those at Fermilab in Illinois or CERN, constantly make and study antimatter. And the process is very well understood. Energy always makes matter and antimatter in equal quantities."

So, instead of "E = m • c2", should we instead say, "E = ((m/2 • c2) + (-m/2 • c2))"?
Click to expand...

If antimatter has negative mass, as the equation above seems to imply, then when matter and antimatter particles annihilate one another, they should wink out of existence and not produce any radiation, rather than producing annihilation quanta (gamma photons). There would be no energy for them to carry away, and conservation of energy would be violated.

Your equation implies E=0.
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No, his equation is not wrong.

It just doesn't mean what people think it means.

Here is a pbs space time vid that explains it:

https://www.youtube.com/watch?v=Xo232kyTsO0

Antimatter does not have negative mass, nor does it have negative energy.

Energy = energy (matter) + energy (antimatter)
Click to expand...
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I haven't looked at the link you posted yet, but the equation [FONT=arial,helvetica][FONT=arial,helvetica]E = ([m/2] • c2) + (-[m/2] • c2) will algebraically sum to zero, regardless.
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If I remember right, antimatter is actually the same subatomic stuff as matter it's just spinning in the opposite direction. But yeah, antimatter has mass, not negative mass and as such moves slower than light.
 
whulorigan said:
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I haven't looked at the link you posted yet, but the equation [FONT=arial,helvetica][FONT=arial,helvetica]E = ([m/2] • c2) + (-[m/2] • c2) will algebraically sum to zero, regardless.
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Click to expand...
And is therefore incorrect since energy can neither be created or destroyed.

The machines that make antimatter take a huge amount of energy and convert it into matter and antimatter, the energy doesn't sum to zero, total energy is still energy input less energy lost due to inefficiencies. Its just some of that energy is now in the matter and some in the anti-matter.
1(input energy)=1/2(energy in matter)+1/2(energy in antimatter)

1(input energy)=1/2(energy in matter)-1/2(energy in antimatter) is wrong.

edit - rereading this thread I think we agree with each other lol - I was saying Keklas' equation/[postulation is wrong, not Einstein :)
 
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David Johansen said:
If I remember right, antimatter is actually the same subatomic stuff as matter it's just spinning in the opposite direction. But yeah, antimatter has mass, not negative mass and as such moves slower than light.
Click to expand...

i was under the impression it was merely the opposite electric charge, EG a normal electron is negatively charged, the Anti matter equivalent is the positron, which has a positive charge.
 
mike wightman said:
And is therefore incorrect since energy can neither be created or destroyed.

The machines that make antimatter take a huge amount of energy and convert it into matter and antimatter, the energy doesn't sum to zero, total energy is still energy input less energy lost due to inefficiencies. Its just some of that energy is now in the matter and some in the anti-matter.
1(input energy)=1/2(energy in matter)+1/2(energy in antimatter)

1(input energy)=1/2(energy in matter)-1/2(energy in antimatter) is wrong.
Click to expand...

And yet, quantum virtual particle pairs effects have been observed in several methods (the VPP's themselves have not). Energy from effectively nowhere.
 
That's not quite what those observations imply with regards to conservation of energy laws.

When a virtual particle pair appear as if from nowhere they are actually 'borrowing' the energy from the underlying quantum fields. This has to be paid back when the particles annihilate.

Before anyone asks about Hawking radiation there is a chance of falling into the trap of thinking a virtual particle from a pair disappears into the black hole leaving behind the other particle in the pair as if from nothing. This is a lay scientist/journalist approximation which isn't quite right.

Yet another pbs video

https://www.youtube.com/watch?v=qPKj0YnKANw
 
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I used the minus sign, not to imply negative mass, but negative charge!

Now for the inevitable snark from those know-it-alls who have nothing better to do than school me on the proper methods of scientific notation.

3 ... 2 ... 1 ...
 
Keklas Rekobah said:
I used the minus sign, not to imply negative mass, but negative charge!

Now for the inevitable snark from those know-it-alls who have nothing better to do than school me on the proper methods of scientific notation.

3 ... 2 ... 1 ...
Click to expand...

Then why apply it to mass? Charge is represented by the speed of light (C) in the equation which is usually converted to electron volts = 931 MEV.
 
Keklas Rekobah said:
I used the minus sign, not to imply negative mass, but negative charge!

Now for the inevitable snark from those know-it-alls who have nothing better to do than school me on the proper methods of scientific notation.

3 ... 2 ... 1 ...
Click to expand...

If you are putting signs where they don't belong, and in general producing incorrect and confusing equations because you don't know how to correctly make the equation (including notations), then perhaps you DO need someone to correct your knowledge.

Just so we can understand what you are saying, to further the conversation and exchange of ideas, if for nothing else.
 
So, instead of "E = m • c^2", should we instead say, "E = ((m/2 • c^2) + (m/2 • -c^2))"?

[Wouldn't it have been easier to just move the '-' and show him what you meant?]
 
Nope because you can not have a negative velocity of light in a vacuum.
Either the minus sign is outside the bracket or you are making up a constant that doesn't exist.
 
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Usually this is equation is written as E = ∆m * 931MEV

That is, the change in energy released = change in mass * charge where mass is the atomic weight of the atom or particle involved
 
Enoki said:
Usually this is equation is written as E = ∆m * 931MEV

That is, the change in energy released = change in mass * charge where mass is the atomic weight of the atom or particle involved
Click to expand...

So ... how would you modify the equation to reflect that half of the energy was in the form of antimatter?
You have still left us hanging. :)
 
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