Algebraic Assistance In Relativistic math

In summary, a participant presents their attempt at a solution to a homework problem, involving algebraic manipulations and the derivation of equations related to kinetic energy. However, there are errors in their equations and it is unclear whether the derivation is correct. They ask for assistance from a moderator.
  • #1
ManyNames
136
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:
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  • #2
[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?
 
  • #3
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:
  • #4
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?
 
  • #5
ManyNames said:
God bless you.

Now, is the derivation algebraicly correct?

No one/b]?
 
  • #6
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
Sorry, i will fix the mistakes.
 

Related to Algebraic Assistance In Relativistic math

1. What is algebraic assistance in relativistic math?

Algebraic assistance in relativistic math is the use of algebraic equations and manipulations to solve problems related to relativity, which is the branch of physics that studies the behavior of objects moving at high speeds or in strong gravitational fields.

2. Why is algebraic assistance important in relativistic math?

Algebraic assistance is important in relativistic math because it allows us to simplify complex equations and make calculations more manageable. It also helps us to better understand the underlying principles and concepts of relativity.

3. What are some common algebraic techniques used in relativistic math?

Some common algebraic techniques used in relativistic math include substitution, simplification of expressions, manipulation of equations, and solving for unknown variables.

4. How is algebraic assistance applied in real-world scenarios?

Algebraic assistance is applied in real-world scenarios such as space travel, GPS systems, and particle accelerators, where the effects of relativity play a significant role. It helps scientists make accurate predictions and design systems that take into account the principles of relativity.

5. Are there any limitations to using algebraic assistance in relativistic math?

Yes, there are some limitations to using algebraic assistance in relativistic math. In some cases, the equations may become too complex to solve using traditional algebraic methods, and numerical methods may be needed. Additionally, algebraic techniques may not be applicable in all scenarios, and a deeper understanding of the principles of relativity may be required to solve certain problems.

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