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ManyNames
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Homework Statement
I need to know whether the derivation is correct, thank you in advance.
Homework Equations
Just relativity and parallel velocities
The Attempt at a Solution
[tex]\frac{v^2-c^2}{c}=\sqrt{\frac{v^2}{c^2}=\frac{v}{c} -1[/tex]
by pure algebra. Now a quick look at manipulations using logic with may not be a non-sequitor by true meaning:
[tex](\frac{v}{v}-1)^2=(\frac{v^2}{c^2}) -2}[/tex]
these equations will allow me to substitute by a very important variable(s) (which will become clear in my second post, after you have analysed the basics here),
[tex]\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}[/tex]
The proof of these equations that i derived at is displayed further;
[tex]1- \frac{v^2}{c^2}= \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}[/tex]
square and prove as finsalized -
[tex](\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2[/tex]
now, we shall just improvise on proving the last equations, and then get into a little more complicated algebraic manipulation, at least, hard for me anyway.
[tex]\sqrt{E}=\sqrt{Mc^2}\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{F^2t^2}{M^2c^2}}[/tex]
Which must come to the equation:
[tex]E^2=E^2\frac{v^2}{c^2}+2[/tex]
In this second part, (not the second presentation i promised to give at the header, but rather a second installation) - i now analyze the important of [tex]\frac{v^2}{c^2}[/tex] and their compatible relationships with how to derive the classical derivation of Kinetic Energy.
Rearranging the above conclusions, we can devise math, which to some respects, can be at the choice of the observer.
[tex]\frac{v^2-c^2}{c^2}=\frac{v}{c}M(v+1)[/tex]
Carrying on [tex]\frac{v}{c}m(v+1)[/tex] which should now deduct by simplifying;
[tex]\frac{v}{c}=\frac{v}{c}(v+1)[/tex]
Albiet, as subtle as the change was. Now, we focus on the final stages of these ( i think fascinating algebraic manipulations) to the path of an equation which will proove invaluable to calculating the kinetic energy:
[tex]\sqrt{E}= \sqrt{Mc^2}\frac{v^2}{c^2}[/tex]
which then exhaistively leads to [tex]\frac{p^2}{Mc^2}[/tex] [0.1]
Now, this second derivtion, when combined through simple math, take the denominator of [tex]\frac{p^2}{Mc^2}[/tex] and follow the proceedure from a new perspectivication:
[tex]Mc^2=c(m+1) \rightarrow c(mc)[/tex]
[tex]c(mc)=(mc^2+M)-2=[/tex]
since [tex]p^2=\frac{mv^2}{E} +2[/tex]
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