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Relativity/Time dilation

  • Thread starter Kawakaze
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  • #1
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Homework Statement



A spaceship moves with speed vs directly towards a space station. It dispatches a package to the station with speed v with respect to itself. A second supply package is mistakenly dispatched before the spaceship’s cargo ejection system has fully recharged, and consequently recedes from the spaceship at a speed of only vr

1. How much time elapses in transit as measured by a clock travelling with the first supply package?

2. Assuming that the difference in times of dispatch, and the difference in distance travelled by the two consignments is negligible, calculate the difference in arrival times at the space station between the main course and the dessert consignments.

3. How much has each of the packages ‘aged’?)

Homework Equations



Lorrentz, time dilation.

The Attempt at a Solution



This is for an assessment, I left the numbers out. My question is which frame of reference do we use? Earlier in the question we calculated the speed of the package relative to the station. If the clock is travelling with the food package, in its inertial frame (constant velocity). Doesnt the clock run at normal speed to an observer travelling with the package?

Wouldnt observers see the clock at different times from the ship and station? The question asks for the clock from the packages frame. I would say time passes normally, the question has 7 marks available so I assume they want a calculation of some type.

Can someone please explain this.
 

Answers and Replies

  • #2
sylas
Science Advisor
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1. How much time elapses in transit as measured by a clock travelling with the first supply package?
This question tells you what reference from to use.

2. Assuming that the difference in times of dispatch, and the difference in distance travelled by the two consignments is negligible, calculate the difference in arrival times at the space station between the main course and the dessert consignments.
The only thing that makes good sense here is the time between arrivals as far as the station is concerned.

3. How much has each of the packages ‘aged’?)
I think this is asking how much each package ages during transit.

At a stretch, it could also be considered as asking how much each package has aged by the moment that the station has received them both. You can easily give both answers, with an explanation of what you are calculating.

Go ahead and crunch some numbers... what do you get?

Cheers -- sylas
 
  • #3
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Well from earlier in the question I get the speed of the first package to be 0.9091c and the second package is 0.8036c. We are also told the first package takes 5 days to arrive as measured from the space station. This is confusing me. As the 1st package is travelling at 0.9091c from the space stations frame of reference.

I ran a quick calculation with the lorrentz dilation equation. But I am still confused, wouldnt the time dilation be symmetrical? If the package took 5 days to arrive at the station, surely the package also aged 5 days, to the packages frame of reference the space station is approaching it at 0.9091c therefore preserving the symmetry?

T'=Troot(1-(v^2/c^2))

I get the time to be 0.4167T = 2.083 days
 
Last edited:
  • #4
sylas
Science Advisor
1,646
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Well from earlier in the question I get the speed of the first package to be 0.9091c and the second package is 0.8036c. We are also told the first package takes 5 days to arrive as measured from the space station. This is confusing me. As the 1st package is travelling at 0.9091c from the space stations frame of reference.
Can you tell us what numbers you have been given?

The initial post gives three variables: v, vr and vs; but no numbers.

I ran a quick calculation with the lorrentz dilation equation. But I am still confused, wouldnt the time dilation be symmetrical? If the package took 5 days to arrive at the station, surely the package also aged 5 days, to the packages frame of reference the space station is approaching it at 0.9091c therefore preserving the symmetry?
That's not a valid application of symmetry. You are considering time between two events; the package leaving the ship, and the package arriving at the station. The time, and the distance, between these two events depends on the frame in which time and distance are calculated. Since both events are "at" the package, the distance between the two events in the package frame is zero. But in the station frame, there is a distance between the events, and a time, so that distance/time = 0.9091 c.

The symmetrical case would be two events involving the station, which is not a part of the question.

T'=Troot(1-(v^2/c^2))

I get the time to be 0.4167T = 2.083 days
I cannot see how you get that number. v is 0.9091 c, is it not?

Cheers -- sylas
 
  • #5
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Yep package velocity to station is 0.909090

ship is travelling 0.80c and launches first package at 0.40c relative to its frame
station is stationary
 
  • #6
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Whoops my bad. Thanks for your patience.

Time of travel from stations frame of reference is 5 days

squareroot(1-(0.9091^2))=0.417

times by original time of 5 days = 2.086days
 
  • #7
sylas
Science Advisor
1,646
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Whoops my bad. Thanks for your patience.

Time of travel from stations frame of reference is 5 days

squareroot(1-(0.9091^2))=0.417

times by original time of 5 days = 2.086days
Actually, I was being a bit slow there myself.

I'll use units in which c = 1, for convenience. You have the velocity of the package relative to the station of 10/11.

The gamma factor is 2.4 (or 2.400397 to a few more figures)

The inverse of that is 0.4166, as you say.

So indeed, the package ages 2.083 days. Or 2.1 days, given that the original velocities were with two figures accuracy.

Proceeding... what do you have for the second package? You previously suggested that the velocity of the second package was 0.8036 c relative to the station. Is that right? It's only just faster that the ship... it must have been launched a lot more slowly!

Cheers -- sylas
 
  • #8
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The second package was launched at 0.01c (due to a mistake!), i got the relative speed to be 0.8036c

The next question is asking what is the different arrival times to the station. I assume because the first package took 5 days at 0.9091c, then its a simple matter of calculating the distance and then using

time(second package) = distance / speed (second package)

The last part of the question is 'how much less fresh' is the second package. Again I assume, we work out the time dilation experienced by the second package, and do a simple subtraction?
 
  • #9
sylas
Science Advisor
1,646
6
The second package was launched at 0.01c (due to a mistake!), i got the relative speed to be 0.8036c
Right; I got that as well, so it's all good so far.

The next question is asking what is the different arrival times to the station. I assume because the first package took 5 days at 0.9091c, then its a simple matter of calculating the distance and then using

time(second package) = distance / speed (second package)
Yes, you can get distance from time and velocity as long as all three are given in the same frame. The distance traveled is the same for both packages, in the station frame.

The last part of the question is 'how much less fresh' is the second package. Again I assume, we work out the time dilation experienced by the second package, and do a simple subtraction?
That compares the "freshness" at the time of arrival, which appears to be what is being asked.

Proceed. What do you get?
 
  • #10
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The first package arrived after 5 days. So I calculated the second package to arrive after 5.656 days, ie 0.656 days after the first.

I calculate the time elapsed to the second package (to its own frame of reference) to be 3.366 days. The time experienced by the first package was 2.083 days. Therefore the second package desserts are 1.283 days older than the main course. Non too fresh! Luckily space is cold so the ice creams wont have thawed out =)

Its mind blowing how this package could arrive 12 hours later and be 40 hours older.
 
  • #11
sylas
Science Advisor
1,646
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The first package arrived after 5 days. So I calculated the second package to arrive after 5.656 days, ie 0.656 days after the first.
Same here.

I calculate the time elapsed to the second package (to its own frame of reference) to be 3.366 days. The time experienced by the first package was 2.083 days. Therefore the second package desserts are 1.283 days older than the main course. Non too fresh! Luckily space is cold so the ice creams wont have thawed out =)
Yes... this is comparing their "freshness", or age, at their respective arrival times.

Its mind blowing how this package could arrive 12 hours later and be 40 hours older.
Indeed.

There is one minor point. Would you might stating exactly how the question describes the time of 5 days? Is it the time between launch event and arrival event in the station frame? Because if so, note that the station cannot measure that duration directly. The launch took place a long way off from the station, and it would have taken time for the light signals to get from the launch event to the station.

Cheers -- sylas
 
  • #12
144
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Okay I think Ive got this cracked. Thanks for your help! Here is the quote you requested -

If the first package travels with constant speed and takes time ΔT = 5.0 days to arrive as measured by the crew member on the space station.
 

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