Calculating Time Dilation for Space Travelers

Click For Summary
SUMMARY

The discussion centers on calculating time dilation for space travelers moving at relativistic speeds, specifically at 0.7c and 0.9c. The elapsed time for a traveler observing a distance of 6.5 light-years (ly) is calculated using the formula Δt = 6.5 / √(1 - 0.7²), resulting in 9.11 years. A subsequent calculation for a traveler moving at 0.9c yields a time of 4.01 years, raising questions about rounding errors and the interpretation of distances versus times. The conversation clarifies that 6.5 ly is a distance measurement, not a time measurement, and emphasizes the relevance of length contraction in this context.

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and length contraction.
  • Familiarity with the Lorentz transformation equations.
  • Basic knowledge of relativistic speeds, specifically fractions of the speed of light (c).
  • Ability to perform calculations involving square roots and scientific notation.
NEXT STEPS
  • Study the Lorentz transformation equations in detail.
  • Learn about the implications of length contraction in special relativity.
  • Explore practical applications of time dilation in GPS technology.
  • Investigate the effects of relativistic speeds on time perception in different inertial frames.
USEFUL FOR

Physicists, students of relativity, and anyone interested in the implications of high-speed travel on time and space measurements.

stunner5000pt
Messages
1,447
Reaction score
5
Homework Statement
A space traveler moving at a speed of 0.70c with respect to Earth makes a trip to a distant star that is stationary relative to Earth. He measures the length of this trip to be 6.5ly. What would be the length of this same trip (in ly) as measured by a traveler moving at a speed of 0.90c with respect to Earth?
Relevant Equations
[tex] \Delta t = \gamma t_{0} [/tex]
If we took the perspective of the space traveller themselves, they are stationary and the whole universe goes past them at 0.7c. THen th elapsed time of 6.5 yr looking outside is

\Delta t =6.5 \frac{1}{\sqrt{1-0.7^2}} = 9.11 yrs

THen, when the traveller looks at the person travelling at 0.9c

the time taken should be

9.11 \times \frac{1}{\sqrt{1-0.9^2}} = 3.97 yrs

but the answer is 4.01 yr. Is this just down to rounding or is there something missing in the solution above?

thanks in advance for your help
 
Physics news on Phys.org
stunner5000pt said:
... elapsed time of 6.5 yr ...
The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.
 
Hill said:
The problem does not say anything about such time, 6.5 yr. It describes a distance, 6.5 ly.
Aha! missed that, thank you

ok so then that means the distance travelled according to the traveller is:

6.5c \times 24 \times 365 \times 60^2 = 6.14 \times 10^{16} m

when the person is in motion, how is the 6.5 ly being interpreted ? Is the noninertial observer seeing a shorter'length that the stationary length i.e. is the above number the length observed from rest?
 
No need to convert or to interpret ly. It is a unit of distance as good as any.
I don't see a "noninertial observer" in the problem. However, there is length contraction rather than time dilation here.
 
  • Like
Likes   Reactions: PeroK
stunner5000pt said:
ok so then that means the distance travelled
The distance travelled by the Earth and the destination planet.

And as has already been said, ly is a perfectly fine length unit. Preferable to meters here really as by definition c = 1 ly/y.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
18
Views
4K