Algebraic Assistance In Relativistic math

Click For Summary

Homework Help Overview

The discussion revolves around algebraic derivations related to relativistic physics, specifically focusing on equations involving velocities and energy. The original poster seeks validation of their derivations and manipulations involving relativistic equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive relationships between velocity, force, and energy using algebraic manipulations. Some participants question the validity of specific equations and their formatting, while others express confusion about the origins and correctness of certain expressions.

Discussion Status

The discussion is ongoing, with participants providing feedback on the clarity and correctness of the equations presented. Some guidance has been offered regarding formatting issues, but there is no explicit consensus on the correctness of the derivations themselves.

Contextual Notes

Participants note issues with LaTeX formatting that may hinder the clarity of the mathematical expressions. There is an emphasis on ensuring that all equations are properly formatted and complete for accurate analysis.

ManyNames
Messages
136
Reaction score
0

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]
 
Last edited:
Physics news on Phys.org
[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, equation 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?
 
3rd, 4th, & 6th equations need another "}" at the end.

equation 3:

[tex]\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}[/tex]

equation 4:

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

equation 6:

[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]

eqn 5 shows up okay:

[tex](\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2[/tex]

p.s. try refreshing the browser screen if this post does not display properly for you.
 
Last edited:
Redbelly98 said:
3rd, 4th, & 6th equations need another "}" at the end.

equation 3:

[tex]\frac{v^2}{c^2}=1+ \frac{\frac{F^2t^2}{M^2c^2}}{1+ \frac{f^2t^2}{M^2c^2}}[/tex]

equation 4:

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

equation 6:

[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]

eqn 5 shows up okay:

[tex](\frac{v^2}{c^2} - 1)=\frac{v^4}{c^^4}-2[/tex]

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

God bless you.

Now, is the derivation algebraicly correct?
 
ManyNames said:
God bless you.

Now, is the derivation algebraicly correct?

No one/b]?
 
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.
 
Sorry, i will fix the mistakes.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
Replies
2
Views
2K
Replies
8
Views
2K
Replies
13
Views
2K
Replies
10
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
46
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K