Deriving Relationships

I know you guys are good at noticing whether something has a mistake in it, so i will post this here. I've been deriving relationships, and i want to know if you can see any mistakes in it. I think its sound; i've even plugged in the correct units and derived the correct answers.

\(E=Mc^2\)

Add E to both sides

\(2E=Mc^2+E\)

Divide both sides by \(2\)

\(E=\frac{1}{2}(Mc^2+E)\)

Which is a derived equation that is similar to the Kinetic energy of a body. Manipulating through algebra gives:

\(\frac{1}{2}E-E=\frac{1}{2}Mc^2\)

And finally \(\frac{1}{2}Mc^2=\frac{1}{2}Mc^2\)

Moving on, considering \(E=\frac{1}{2}(Mc^2+E)\), plug into this equation Einsteins formula \(pc=E\frac{v}{c}\) knowing that \(p=mv=mc\), you get

\((E\frac{v}{c})^2=\frac{1}{2}(Mc^2+(E\frac{v}{c})^2)\)

Integrating the equation

\(\int (E\frac{v}{c})=\int \frac{1}{2}(Mc^2+(E\frac{v}{c})^2)\)

gives

\(E^2=\frac{1}{2}(p^2c^2+Mc^2)\)

Which is an equation similar to the relation of energy-momentum, and now using little algebra to relate the energy strictly to the products of momentum

\(E^2=\frac{1}{2}(p^2c^2+Mvc)\)

\(\sqrt{\frac{1}{2}(p^2c^2+Mvc)}=E\)

Finally, i wanted to interpret the following two equations equation

\(2cM=(\frac{E+Mc^2}{pc})Mv\)

\(\frac{E}{M}=\frac{\frac{1}{2}(Mc^2+E)}{M}\)

Then solving for \(Mv\) gives;

\(Mv=\frac{\frac{2cM}{pc}}{E+Mc^2}\)

\(Mv=(\frac{2cM}{E+Mc^2})pc\)

Collecting like terms together (symbolized by integrator)

\(\int Mv=\int (\frac{2cM}{E+Mc^2})pc\)

Gives

\(\int p=\int (\frac{2cM}{2E})E\)

For a final new equation relating energy to momentum

And finishing with the final equation i wanted to investigate,

\(\sqrt{\frac{\frac{1}{2}(Mvc+E)}{M}}=c\)

 

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Just expand out \(E = m_{0}c^{2} \gamma\) in terms of \(\frac{v}{c}\) via a Taylor expansion and you get the relativistic corrections, order by order, to the Newtonian kinetic energy equation.

For a photon (or any massless particle) you have that p = E/c but you must set M=0 to use that.

 

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Why, in your own words, is this a desirable or useful result?

 

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Oops i made a mistake in equation 13. Let me fix it, then i will ansewr you guys.

 

In SR \(p \neq Mv\) and the denominator of the second term \(E+Mc^2\) I presume you've used that \(E = Mc^2\). Note \(E + E \neq E^2\)

I notice you use \(E = Mc^2\) a lot. This equation is only valid for a particle at rest so where you've got v's in subsequent steps you should put in v = 0 to be correct. As you can see, your final equation reduces to \(E = Mc^2\) when you do this. The equation is not valid for arbitrary v.

 

Yes.

I know, where \(p^2c^2-E^2=(Mc^2)^2\) can describe a massless boson if \(M=0\).

Doh!

My fault. I have always used the integration to mark when i collect like terms to tidy the equation up.

Same reasoning. (Unless i have done it wrong.) I will check it again soon.

 

I sat for at least an hour wondering exactly that. I just put it down as a curiosity, and said if i ever needed it i could use it to plug into an equation that may require the same products of both sides.

 

Well, i would like to know, please.

I have plugged in real values to see if each side of the equations would yield the correct answers. And they did according to the algebra.

James, you are really the first to say this, so please go ahead.

 

Total quibbles.

Right, for your first analysis, i should have wrote 2, instead of 1/2. Second, yes, i forgot to put in the minus part. Third part, the equation is valid if M=0. For the fourth, i explained this was me collecting like terms, as it was how i marked where i did such a thing. (Just the way i have wrote equations for a while -- in no way does it make the equations derived, underivable).

And the last equation cannot be wrong, because if \(c^2=\frac{\frac{1}{2}(Mvc+E)}{M}\), then taking the square root of one side must eliminate the square of c.

 

(By the way, i already explained about where i wrote integrating symbols. That's why i sounded a bit secondary on the whole thing.)

 

I just altered the wording a bit too, saying that i was collecting like terms, using the integration symbol, like i have done for years. As i said, it just helps me mark out where i decide to use that process, because it really rips the equation right down.

 

I just wondered, would it have made it better if i had stabbed in \(\gamma Mc^2\), so that it could be relativistic mass we are talking about instead?

 

Oh right, i get you now; yes, you are right. It's still fine. p=mc doesn't need to be recognized. The equations needed are accounted for. I will delete the recognition of p=mc.

 

Not really. I made it quite clear i was referring to relativistic mass. I am very careful within myself which one's i was referring to a rest mass and which ones are a relativistic mass.