Can a physicist look at this please and tell me if i have derived this correctly?

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1. The problem statement, all variables and given/known data

I need a physicist to look over these derivations and help me see if there are any mistakes. Thank you in advance, it is much appreciated.

2. Relevant equations

- reltivistic math derivations

3. The attempt at a solution

The First Part

[tex]E^2-(pc)^2=(Mc^2)^2[/tex] where the expression [tex](Mc^2)^2[/tex] is by definition, the squared mathematical precision of an ''invariant mass'', hence, [tex]Mc^4[/tex].

[tex]\rightarrow Mv(\frac{E}{M})=Mc^2(v)[/tex]

allow [tex]v=c[/tex] then this simplifies to [tex]Mv^3=Mc^2[/tex] (Just to show that these are relativistic equivalances without the need of gamma function. This now leads me to calculate:

[tex]Mv(E)(\frac{D}{v})=Mc^2.v[/tex]

The Second Part

my equation, albiet as simple as it is, will show its importance throughout the metric work:

[1] [tex]M(1+M)=2M[/tex] if

[2] [tex]-(\frac{E}{c})^2+mv^2=p^2[/tex]

then combine by division of [1] and [2] equations, allowing the relativistic proof:

[tex]\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}=\frac{p^2}{2}\frac{E}{c^2}[/tex]

which then follows

[tex]p^2=\eta^{\mu v}p_{mu}p_{v}=-(\frac{E}{c^2})^2+p^2[/tex]

I need to go the now, i will finish this later, but sooner than later.
 
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CompuChip

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That is true

No you need a mathematician. Because those are also the only equations I could find that don't have (mathematical) errors in them.
That is true, friend.
 

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