Special Relativity GRE question (Where is my method flawed?)

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Homework Help Overview

The discussion revolves around a problem related to special relativity, specifically focusing on the total energy and kinetic energy of a particle. The original poster expresses confusion regarding their method of calculating total energy, which they equate to 100mc², and seeks to identify flaws in their reasoning compared to the expected answer of 10⁴(mc).

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the total energy and kinetic energy relationship but questions where their reasoning might be incorrect. Some participants clarify that the formulas used are non-relativistic approximations, valid only under certain conditions. Others explore the implications of these approximations and question the logic behind ignoring the square root in the energy equations.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing insights into the limitations of the non-relativistic approximations used. There is a focus on clarifying the assumptions behind the equations, and some participants are offering explanations regarding the Taylor series expansion related to the energy calculations.

Contextual Notes

The discussion highlights the constraints of using non-relativistic formulas in scenarios where relativistic effects are significant, particularly when the momentum is not negligible compared to the rest mass energy.

PsychonautQQ
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Homework Statement


http://grephysics.net/ans/9277/70

I'm not interested in how you get the right answer, I'm really just curious if somebody could show me where my thought process was flawed.

Total Energy = E = 100mc^2
Rest Energy = Mc^2
Total Energy = Kinetic Energy + Rest Energy
100mc^2 = p^2/2m + mc^2
99mc^2 = p^2/2m
198(mc)^2 = p^2
sqrt(198)mc = P
and yet the correct answer is
10^4(mc)

where is my method flawed?
 
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The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:
E^2 = p^2 c^2 + m^2 c^4
E = \sqrt{p^2 c^2 + m^2 c^4} = mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}
E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4}) = mc^2 + \frac{p^2}{2m}
 
phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}
E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?
 
PsychonautQQ said:
phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}
E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?
Phyzguy is just showing you where your non-relativistic approximation came from. He used the binomial expansion to approximate the square root, and retained only the first two terms in the expansion. Please notice the "approximate" symbol in his equation.
 
PsychonautQQ said:
phyzguy said:
The formula KE = p^2/(2m) and TE = p^2/(2m) + mc^2 are non-relativistic approximations, and are only valid when pc << mc^2. They come about from taking the full relativistic equation and expanding the square root, keeping only the leading term, as follows:

mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}
E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4})

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?

When I expand the square root of (1+ε) in a Taylor series and keep only the first order term, I get:

\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}

For example: \sqrt{1.01} = 1.004987 \ldots \approx 1.005

One more comment. You need to have some of these more common approximations at your fingertips in order to do physics. Some of the most important are:
\sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}
\frac{1}{1+\epsilon} \approx 1-\epsilon
sin(\epsilon) \approx \epsilon
tan(\epsilon) \approx \epsilon
cos(\epsilon) \approx 1+\frac{\epsilon^2}{2}
ln(1+\epsilon) \approx \epsilon
 
Last edited:

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