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

[ManyNames]

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

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

\rightarrow Mv(\frac{E}{M})=Mc^2(v)

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

Mv(E)(\frac{D}{v})=Mc^2.v

The Second Part

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

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

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

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

\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}=\frac{p^2}{2} \frac{E}{c^2}

which then follows

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

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

[CompuChip]

E^2-(pc)^2=(Mc^2)^2
That is true


\frac{\eta^{\mu v}p_{\mu}p_{\mu}}{2m}=\frac{p^2}{2m}
That is true.

I need a physicist to look over these derivations
No you need a mathematician. Because those are also the only equations I could find that don't have (mathematical) errors in them.

[ManyNames]

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.