Length contraction numerical problem

Click For Summary
SUMMARY

The discussion focuses on a length contraction numerical problem involving a proton with a kinetic energy of 1019 eV. The calculations utilize both non-relativistic and relativistic equations, specifically k.e. = (1/2)mv² for non-relativistic and K.E. = m₀c²(γ - 1) for relativistic scenarios. The results indicate that the speed of the proton approaches the speed of light, leading to a significant time dilation effect when crossing the galaxy, estimated at 10⁵ years in the galaxy frame and shortened in the proton frame. The final suggestion is to convert the time taken into seconds for clarity.

PREREQUISITES
  • Understanding of kinetic energy equations: k.e. = (1/2)mv² and K.E. = m₀c²(γ - 1)
  • Familiarity with the concept of relativistic length contraction: L = L₀/γ
  • Basic knowledge of special relativity and Lorentz factor (γ)
  • Ability to perform unit conversions, particularly between years and seconds
NEXT STEPS
  • Learn about the Lorentz factor (γ) and its implications in relativistic physics
  • Study the relationship between kinetic energy and relativistic momentum
  • Explore time dilation effects in special relativity
  • Investigate practical applications of length contraction in astrophysics
USEFUL FOR

Students studying physics, particularly those focusing on special relativity, as well as educators and anyone interested in the implications of relativistic effects on high-energy particles.

Pushoam
Messages
961
Reaction score
53

Homework Statement


upload_2017-7-11_10-41-45.png


Homework Equations


1) k.e. = (1/2)mv2, non- relatvistic

2)K.E. = m0 c2 (ϒ -1), relativistic

3) L=L0/ϒ , Length contraction

The Attempt at a Solution


(A) Taking 1019 eV to be the kinetic energy of the proton,
Non- relativistic calculation
(1/2)mv2 = 1019 eV = 1.6

v = √[ (3.2/1.67)* 1027 ] = 4.3 * 1013 m/s

How is kinetic energy defined relativistically?

K.E. = m0 c2 (ϒ -1)

We can't take 1019 eV as total energy of the proton as if we take so then we won't be able to calculate speed of the proton.

So, for calculating speed of the proton, I have to take 1019 eV as the kinetic energy of proton.

So,
ϒ2 = [ {1019 eV/ m0 c2}+1]2

m0 c2 = 939 MeV

Solving it gives,
1 - v2/c2 = [939 * 10-13]2

v ≈ c

Hence, w.r.t. galaxy frame the proton will take 105 years to cross the galaxy.
B)

W.r.t. the proton frame,
the length of the galaxy is shortened to 105 [939 * 10-13] light years.Hence, the time taken = 105 [939 * 10-13] yearsIs this correct?
 
Physics news on Phys.org
Pushoam said:
Hence, the time taken = 105 [939 * 10-13] years
Hi Pushoam:

Your work looks OK to me. However, I suggest converting the (B) answer to seconds.

Regards,
Buzz
 

Similar threads

  • · Replies 44 ·
2
Replies
44
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 52 ·
2
Replies
52
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K