# Algebraic Assistance In Relativistic math

1. Feb 23, 2009

### ManyNames

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

I need to know whether the derivation is correct, thank you in advance.

2. Relevant equations

Just relativity and parallel velocities

3. The attempt at a solution

$$\frac{v^2-c^2}{c}=\sqrt{\frac{v^2}{c^2}=\frac{v}{c} -1$$

by pure algebra. Now a quick look at manipulations using logic with may not be a non-sequitor by true meaning:

$$(\frac{v}{v}-1)^2=(\frac{v^2}{c^2}) -2}$$

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),

$$\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}$$

The proof of these equations that i derived at is displayed further;

$$1- \frac{v^2}{c^2}= \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}$$

square and prove as finsalized -

$$(\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2$$

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.

$$\sqrt{E}=\sqrt{Mc^2}\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{F^2t^2}{M^2c^2}}$$

Which must come to the equation:

$$E^2=E^2\frac{v^2}{c^2}+2$$

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 $$\frac{v^2}{c^2}$$ 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.

$$\frac{v^2-c^2}{c^2}=\frac{v}{c}M(v+1)$$

Carrying on $$\frac{v}{c}m(v+1)$$ which should now deduct by simplifying;

$$\frac{v}{c}=\frac{v}{c}(v+1)$$

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:

$$\sqrt{E}= \sqrt{Mc^2}\frac{v^2}{c^2}$$

which then exhaistively leads to $$\frac{p^2}{Mc^2}$$ [0.1]

Now, this second derivtion, when combined through simple math, take the denominator of $$\frac{p^2}{Mc^2}$$ and follow the proceedure from a new perspectivication:

$$Mc^2=c(m+1) \rightarrow c(mc)$$

$$c(mc)=(mc^2+M)-2=$$

since $$p^2=\frac{mv^2}{E} +2$$

Last edited: Feb 23, 2009
2. Feb 23, 2009

### ManyNames

[3rd] equation is supposed to be:

frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}

Why isn't it showing - minus the tex. But also, eqution 4

1- \frac{v^2}{c^2}= \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}

and [5] are invalid here?

(\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2

Can a mod assist me here please?

3. Feb 23, 2009

### Redbelly98

Staff Emeritus
3rd, 4th, & 6th equations need another "}" at the end.

equation 3:

$$\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}$$

equation 4:

$$1- \frac{v^2}{c^2}= \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}$$

equation 6:

$$\sqrt{E}=\sqrt{Mc^2}\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{F^2t^2}{M^2c^2}}$$

eqn 5 shows up okay:

$$(\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2$$

p.s. try refreshing the browser screen if this post does not display properly for you.

Last edited: Feb 23, 2009
4. Feb 23, 2009

### ManyNames

God bless you.

Now, is the derivation algebraicly correct?

5. Feb 23, 2009

### ManyNames

No one/b]???

6. Feb 23, 2009

### Redbelly98

Staff Emeritus
I hadn't looked that closely earlier, beyond fixing the LaTex errors.

Equation 1 makes no sense to me. Why is there a "=" underneath the square-root symbol?

I don't understand Equation 2. Where does it come from? Is this an equation to be solved, or is this supposed to be basic algebra (in which case it is incorrect)?

I'll stop here for now.

7. Feb 24, 2009

### ManyNames

Sorry, i will fix the mistakes.