# 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.

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#### CompuChip

Homework Helper
$$E^2-(pc)^2=(Mc^2)^2$$
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.

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

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