• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • #1
784
11

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?
 

Answers and Replies

  • #2
phyzguy
Science Advisor
4,509
1,449
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:
[tex] E^2 = p^2 c^2 + m^2 c^4[/tex]
[tex] E = \sqrt{p^2 c^2 + m^2 c^4} = mc^2 \sqrt{1+\frac{p^2 c^2}{m^2 c^4}}[/tex]
[tex] E \approx mc^2 (1+\frac{p^2 c^2}{2m^2 c^4}) = mc^2 + \frac{p^2}{2m}[/tex]
 
  • #3
784
11
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:

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

How does this make any sense at all? ignoring the square root and adding a 2 to the denominator?
 
  • #4
20,147
4,218
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:

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

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.
 
  • #5
phyzguy
Science Advisor
4,509
1,449
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:

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

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:

[tex] \sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}[/tex]

For example: [tex] \sqrt{1.01} = 1.004987 \ldots \approx 1.005[/tex]

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:
[tex] \sqrt{1+\epsilon} \approx 1+\frac{\epsilon}{2}[/tex]
[tex] \frac{1}{1+\epsilon} \approx 1-\epsilon[/tex]
[tex]sin(\epsilon) \approx \epsilon[/tex]
[tex]tan(\epsilon) \approx \epsilon[/tex]
[tex]cos(\epsilon) \approx 1+\frac{\epsilon^2}{2}[/tex]
[tex]ln(1+\epsilon) \approx \epsilon[/tex]
 
Last edited:

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

  • Last Post
Replies
6
Views
503
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
760
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
981
  • Last Post
Replies
2
Views
983
  • Last Post
Replies
18
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
0
Views
1K
Top