Merged Studying Sharma's equation on Linear Field Equations

Singularitarian said:
For [latex]E=Mc^2[/latex] to be equivalant,
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You are already not making sense at this point in the statement. A single item, such as an equation can't be equivalent. What you probably mean is for "E and mc2 to be equivalent", or better yet "for E=mc2 to be true", but if you don't even understand such a simple concept, why would I read more?
 
Thing is though, i have previously stated it wrong. So what i said was actually right. ''For it to be equivalant,'' is me saying 'for it to be true.''
 
As for the wave equations in a gravitational field [latex]\phi[/latex], and those describing the statistical quantum field [latex]\psi[/latex], they came in the form identifying kinetic energy as the following.

Starting with the equation [latex]E/k^2=\hbar /2M[/latex], we apply a multiplication of the gravitational quantized force;

[latex]\hbar \frac{E}{k^2}= \frac{\hbar}{2M} \frac{GM}{c^2}[/latex]

from here, one can derive:

[latex]\hbar c^2/r (E/k^2)= \hbar/2M \phi[/latex]

Then we introduce the wave-form:

[latex]\nabla^2 \frac{\hbar c^2}{r} \frac{E}{k^2}= \frac{\hbar}{2M} \frac{\partial^2 \phi}{\partial x^j \partial x^j}= \nabla^2 \phi (\frac{-i^2 \hbar^2}{2M} \frac{p}{2M})[/latex]

This involves the kinetic relationships in a field, and we can choose according to the mathematical rules, how many dimensions in space and imaginary space we choose for the Laplacian Operator.

If we add the wave solution [latex]d^2/dx^2[/latex] to the above equation, we find:

[latex]\nabla^2 \phi ( \frac{p^2}{2M} \frac{d^2}{dx^2} )= \nabla^2 \phi ( \frac{-i \hbar^2}{2M} \frac{d^2}{dx^2} )[/latex]

If we divide the energy on the left hand side with [latex]c^2[/latex] in a modified version of above, with a change of a quantum field [latex]\psi[/latex], we have:

[latex]E/c^2 (GM/c^2) \nabla^2 \psi = M^2c^2 (-\hbar \nabla^2 \psi )[/latex]

Changing this laplacian for three dimensions, we finally conclude that

[latex]=\Delta f^3(M^2c^2 (- \hbar \psi))[/latex]
 
Singularitarian said:
In a series of four quantum mechanical steps, [latex]H^1[/latex] nuclei fuse to make a [latex]He^4[/latex] nucleus. The mass of the of the [latex]He^4[/latex] is slightly less than the mass of the [latex]H^1[/latex], and this lessness is released due to [latex]E=Mc^2[/latex].
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Why don't you write out the full equation for that reaction. Hint: it's not just H + H + H + H = 4He.
 
Singularitarian said:
Point?

Just take the post as it comes please - so you accept you are wrong?
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You don't know the full equation, do you?

OK, another tack: when we solve particle decays and transformations, there are two conservation laws which hold; a conservation law means that some quantity must be the same for the *full system* pre-reaction as for the *full system* post-reaction. What are the two laws?
 
ben m said:
You don't know the full equation, do you?

OK, another tack: when we solve particle decays and transformations, there are two conservation laws which hold; a conservation law means that some quantity must be the same for the *full system* pre-reaction as for the *full system* post-reaction. What are the two laws?
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Only two? I thought there were four.
 
You won't answer my question because you are far too proud.

Fix that out. If i am proven wrong, do i imediately set tasks to answer for the other person, because if you think that i ''special'' in any way, truth is, anyone can do that.

Grow up.
 
If it was that easy, then professional scientists must be dumb flunkers.

So, which is more likely? All those professional scientists being wrong, even the ones who build nuclear power plants, launch space ships, invent GPS devices, and study fusion chambers?

...or....

Your own understanding of physics is only approximate (and poorly so).

One way we can find out: What implications for technology would your replacement of E=mc2 have, that are empirically testable, (at least in principal)? If the answer is "I can't think of any, yet", then you better go do so!! Because, until you do, those scientists with the responsability to actually do stuff will just yawn at you.
 
Also, the quarks which make the nuclei of atoms actually have more mass than what makes the nucleus. The missing mass in this example is actually transformed into gluon energy. In this case, when the mass of quarks come together, we find the final energy to not be equivalent at all.
 
Wowbagger said:
If it was that easy, then professional scientists must be dumb flunkers.

So, which is more likely? All those professional scientists being wrong, even the ones who build nuclear power plants, launch space ships, invent GPS devices, and study fusion chambers?

...or....

Your own understanding of physics is only approximate (and poorly so).

One way we can find out: What implications for technology would your replacement of E=mc2 have, that are empirically testable, (at least in principal)? If the answer is "I can't think of any, yet", then you better go do so!! Because, until you do, those scientists with the responsability to actually do stuff will just yawn at you.
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I don't understand this. It made no sense to me.
 
Singularitarian said:
I don't understand this. It made no sense to me.
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Start with the first sentence. Do you think all the professional scientists, who study physics, are stupid? They would have to be, if they missed something this easy.

If they are not, then maybe you got something wrong.

One way we can test if you are right, is to think about how your ideas would change technology.

If E=mc2 is wrong, it will have to be replaced with something. What is your replacement, and what implications would that have for technology?


ETA: And who, exactly, is this Sharma you are referring to, (out of morbid curiosity)?
 
Singularitarian said:
Starting with the equation [latex]E/k^2=\hbar /2M[/latex]
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What is k? Wavenumber? If so, the units don't match; that is J-m^2 = J-s/kg.

[latex]\hbar \frac{E}{k^2}= \frac{\hbar}{2M} \frac{GM}{c^2}[/latex]
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You multiply the left by hbar (units J-s) and the left by GM/c2 (units meters)---what sense does that make? Do you really think hbar = GM/c^2? Anyway, all together that gets us to kg^2 m^2 s^-3 = m^3 s^-1 kg^-1.

Sing: there are no true equations in all of physics with mismatched units. None at all. Zero.
 
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