Relativistic Energy Derivation math problem

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Discussion Overview

The discussion revolves around the mathematical derivation of relativistic energy, specifically focusing on the differentiation of relativistic momentum and the application of integration by substitution. Participants are examining the steps involved in the derivation as presented in a textbook, addressing specific mathematical expressions and their interpretations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions the derivation step involving the denominator term having a 3/2 index, seeking clarification on the differentiation process.
  • Another participant suggests there may be a missing square-root symbol in the expression.
  • A different participant provides an alternative expression for the derivative of momentum, indicating the differentiation of both the numerator and denominator and arriving at a similar conclusion regarding the exponent.
  • Some participants express difficulty in understanding the non-formatted mathematical expressions and question whether they should be able to perform such differentiation by hand at their academic level.
  • One participant mentions the chain rule as a relevant concept for understanding the differentiation of momentum with respect to time.
  • Another participant explains that the exponent of -3/2 arises from the derivative of the term γ(u), which is an exponent of -1/2, and relates this to the differentiation of power functions.

Areas of Agreement / Disagreement

Participants express varying levels of understanding regarding the mathematical steps involved in the derivation. There is no consensus on the clarity of the presented expressions, and some participants have differing opinions on the necessity of being able to perform such differentiation by hand.

Contextual Notes

Some participants indicate challenges with non-formatted mathematical notation, which may hinder comprehension of the derivation steps. There are also references to assumptions about the constancy of mass and the application of the chain rule, but these assumptions are not universally accepted or clarified.

michael154
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Hey,

In a derivation of relativistic energy (in Physics for Scientists and Engineers, 5th edition, Serway and Beichner) they use a method of integration by substitution:

Given that


F=\frac{dp}{dt}

and relativistic momentum is given by

p=\frac{mv}{\sqrt(1-(v^2/c^2))}

W=∫F dx=∫\frac{dp}{dt} dx

which is fine then they say

\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{\sqrt(1−(\frac{(v^2)}{c^2}))}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}


(This is about halfway through the derivation.)
its that last step that i don't understand. why does the denominator term now have a 3/2 index? and why is du/dt given like that? Any help would be appreciated.
 
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is there a missing square-root symbol?
 
michael154 said:
\frac{dp}{dt}=\frac{d}{dt}\frac{mv}{1−(\frac{(v^2)}{c^2})}=\frac{m\frac{dv}{dt}}{(1−(\frac{v^2}{c^2}))^(\frac{3}{2})}
(d/dt)(mv(1-v2)-1/2) = m(1-v2)-1/2dv/dt -1/2(-2v)(mv)(1-v2)-3/2)dv/dt
[first term is from d/dt'ing the v in the numerator, second is from d/dt'ing the denominator]

= m(1-v2 + v2)(1-v2)-3/2dv/dt = m(1-v2)-3/2dv/dt
 
Sorry but that's not very clear. I have trouble reading non formatted math like that. As a second year physics student in uni, should I be able to do that differentiation by hand?
 
michael154 said:
Sorry but that's not very clear. I have trouble reading non formatted math like that. As a second year physics student in uni, should I be able to do that differentiation by hand?

Try looking up "chain rule". Wiki has a definition, it might not be the easiest to read. Note that you want to find (d/dt) p(t) which is equal to (d/dt) p(v(t)), and that m is assumed to be constant (not a function of time).
 
Sorry I use u instead of v for velocity

michael154 said:
Sorry but that's not very clear. I have trouble reading non formatted math like that. As a second year physics student in uni, should I be able to do that differentiation by hand?

In plain English the reason you get an exponent of -3/2 is because γ(u) is an exponent of -1/2, and what's the derivative of x-1/2? It's (-1/2)x-3/2.

I actually just posted a similar question regarding this topic, but with the time component rather than space.
 
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