# Relativity Question

by Lissajoux
Tags: relativity
 P: 82 1. The problem statement, all variables and given/known data A spaceship leaves Earth at 12 noon. At that time the spaceship clock is set to be synchronised with Earth time. It travels through space in a straight line at a constant speed 0⋅60 c. A radio signal is sent to the spaceship from Earth at 2pm. A shuttle is launched from the Earth at 4pm and travels directly towards the spaceship at a constant speed of 0⋅80 c. a) In the Earth frame of reference, relative to the spaceship what are: i. the speeds of the radio message and of the shuttle. ii. the corresponding times in Earth frame for the arrival at the spaceship of the radio message and of the shuttle. b) In the spaceship frame of reference: i. at what times did the signal and the shuttle leave the Earth? ii. what are the speeds of the signal and of the shuttle? iii. what are the times for the radio message and for the shuttle to arrive at the spaceship? 2. Relevant equations Within the problem statement and solution attempt. 3. The attempt at a solution In this question can ignore the acceleration periods and all gravitational effects. Really not sure how to go about this, but this is what I have so far: a) $v=0.6c=1.8\times 10^{8}ms^{-1} , u_{x}=c=3\times 10^{8}ms^{-1}$ i. use equation: $$V=\frac{u_{x}v}{1+\frac{u_{x}v}{c^{2}}}$$ therefore: $$V_{signal}=3\times 10^{8}ms^{-1}$$ and $$V_{shuttle}=2.838\times 10^{8}ms^{-1}$$ ii. Not sure how to do this, i.e. what equations to use. Assume I need to use the results from part i. somehow. b) At the moment I'm unsure how to adapt the calculations to take into account the dilations resulting from being in the spaceship's frame of reference. Some advice here would be great to get me going with this part.
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,860 The weird relativity effects you've learned about only come into play when you go from one reference frame to another. In part (a), however, you're only talking about one reference frame, that of the Earth. In that frame, you observe that the spaceship, the signal, and the shuttle move at certain speeds. To find the relative velocities, you just subject the relevant velocities from each other, and to see when two of them meet, you do the same calculation you would have done with Newtonian physics. In contrast, in part (b), now the observer is in a new frame of reference, so you need to "translate" the quantities of the original frame to this new one, by applying the velocity-addition formula, the Lorentz transformations, etc.
 P: 82 Yes so I gathered that the first part of the question where everything is considered in the Earth's reference frame is pretty simple, as don't need to consider any relativistic effects. So I have that for a)i) the speeds to the signal and shuttle are just $c$ and $0.8c$ as given in the question brief. For part a)ii) using simultaneous equations type method I have that the signal will reach the ship after 3 seconds, and the shuttle will reach the ship after 6 seconds. All correct so far? Then getting onto part b) of the question, which is rather more tricky as need to consider relativity effects due to being in the spaceship's reference frame. Knowing that the spaceship is travelling at 0.6c can find the value of the Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}=\frac{1}{\sqrt{1-\left(0.6\right)^{2}}}=\frac{1}{\sqrt{0.64}}=1.25$$ So I need to use this value to account for the relativistic effects experienced by the spaceship. I've just considered that the ship is moving at 0.6c, do I need to consider that the signal and the shuttle are moving at c and 0.8c aswell? I assume they'll each have their own reference frame too, so just need to consider $\gamma[/tex] due to [itex]v_{spaceship}$? Know that 'moving clocks run slower'. So for b)iii) for the signal arrival time: $$t=\frac{t_{E}}{\gamma}=\frac{3}{1.25}=2.4s$$ And for the shuttle arrival time: $$t=\frac{t_{E}}{\gamma}=\frac{6}{1.25}=4.8s$$ Correct? I'm sure the calculations are, not sure if these are indeed for the arrival times or for the times they left Earth at obviously those times earlier than they were received, but they won't be due to contraction, arhh confusing. I think this is for part b)iii) but not sure how to figure out part b)i). For b)ii) if the time for the signal to reach the ship in Earth reference frame is 3s, then the distance of the ship away is calculated as $d=s\times t=c\times 3=9\times 10^{8}m[/tex]. But there's length contraction in the spaceship's reference frame, so that the distance is perceived by them as: $$d=\frac{9\times 10^{8}}{1.25}=7.2\times 10^{8}m$$ So the perceived speed: $$s=\frac{d}{t}=\frac{7.2\times 10^{8}}{2.4}=3\times 10^{8}m/s$$ Then need to do the same for the shuttle: If the time for the shuttle to reach the ship in Earth reference frame is 6s, then the distance of the ship away is calculated as [itex]d=s\times t=0.8c\times 6=1.44\times 10^{9}m[/tex]. But there's length contraction in the spaceship's reference frame, so that the distance is perceived by them as: $$d=\frac{1.44\times 10^{9}}{1.25}=1.15\times 10^{9}m$$ So the perceived speed: $$s=\frac{d}{t}=\frac{1.15\times 10^{9}}{6}=1.92\times 10^{8}m/s$$ Correct? Then for part b)i) to calculate the time the signal and shuttle were sent: Surely this is just 2.4 and 4.8 seconds, as calculated earlier. -- Hopefully that all makes sense, sorry it's a bit all over the place. I think I'm nearly there, though the method is a bit mixed up. Sure should be able to calculate everything I need in order, i.e. part i) then part ii) then part iii). Hopefully too I'm getting somewhere with all this now and not too far off some right answers! PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,860 ## Relativity Question  Quote by Lissajoux Yes so I gathered that the first part of the question where everything is considered in the Earth's reference frame is pretty simple, as don't need to consider any relativistic effects. So I have that for a)i) the speeds to the signal and shuttle are just [itex]c$ and $0.8c$ as given in the question brief. For part a)ii) using simultaneous equations type method I have that the signal will reach the ship after 3 seconds, and the shuttle will reach the ship after 6 seconds. All correct so far?
Hours, not seconds. I got a different answer for when the shuttle catches up to ship.
 Then getting onto part b) of the question, which is rather more tricky as need to consider relativity effects due to being in the spaceship's reference frame. Knowing that the spaceship is travelling at 0.6c can find the value of the Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}=\frac{1}{\sqrt{1-\left(0.6\right)^{2}}}=\frac{1}{\sqrt{0.64}}=1.25$$ So I need to use this value to account for the relativistic effects experienced by the spaceship. I've just considered that the ship is moving at 0.6c, do I need to consider that the signal and the shuttle are moving at c and 0.8c aswell? I assume they'll each have their own reference frame too, so just need to consider $\gamma[/tex] due to [itex]v_{spaceship}$?
Yes. Since you're only looking at what happens from the spaceship's point of view, that's the only gamma you have to worry about.
 Know that 'moving clocks run slower'. So for b)iii) for the signal arrival time: $$t=\frac{t_{E}}{\gamma}=\frac{3}{1.25}=2.4s$$ And for the shuttle arrival time: $$t=\frac{t_{E}}{\gamma}=\frac{6}{1.25}=4.8s$$ Correct? I'm sure the calculations are, not sure if these are indeed for the arrival times or for the times they left Earth at obviously those times earlier than they were received, but they won't be due to contraction, arhh confusing. I think this is for part b)iii) but not sure how to figure out part b)i).
These aren't correct. From the equations for a Lorentz transformation, Δt'=γ(Δt-βΔx), you can see time dilation only applies when Δx=0, but the emission and reception of the radio signal occurs at different points in space (in both frames).

Try drawing a spacetime diagram for this problem. You have enough info from part (a) to write down the spacetime coordinates in the Earth frame for the various events. You can then calculate the coordinates in the ship's frame. From those, you can figure out the answers for part (b).
 For b)ii) if the time for the signal to reach the ship in Earth reference frame is 3s, then the distance of the ship away is calculated as $d=s\times t=c\times 3=9\times 10^{8}m[/tex]. But there's length contraction in the spaceship's reference frame, so that the distance is perceived by them as: $$d=\frac{9\times 10^{8}}{1.25}=7.2\times 10^{8}m$$ So the perceived speed: $$s=\frac{d}{t}=\frac{7.2\times 10^{8}}{2.4}=3\times 10^{8}m/s$$ Then need to do the same for the shuttle: If the time for the shuttle to reach the ship in Earth reference frame is 6s, then the distance of the ship away is calculated as [itex]d=s\times t=0.8c\times 6=1.44\times 10^{9}m[/tex]. But there's length contraction in the spaceship's reference frame, so that the distance is perceived by them as: $$d=\frac{1.44\times 10^{9}}{1.25}=1.15\times 10^{9}m$$ So the perceived speed: $$s=\frac{d}{t}=\frac{1.15\times 10^{9}}{6}=1.92\times 10^{8}m/s$$ Correct? Then for part b)i) to calculate the time the signal and shuttle were sent: Surely this is just 2.4 and 4.8 seconds, as calculated earlier. -- Hopefully that all makes sense, sorry it's a bit all over the place. I think I'm nearly there, though the method is a bit mixed up. Sure should be able to calculate everything I need in order, i.e. part i) then part ii) then part iii). Hopefully too I'm getting somewhere with all this now and not too far off some right answers! P: 82  Quote by vela Hours, not seconds. I got a different answer for when the shuttle catches up to ship. This is how I figured out when they would all meet.. I plotted a quick table of how far each travels in each hour, and compared the results. I hope I can indeed do it this rather simple way, and this makes sense: Ship | Signal | Shuttle ------------------------- 1.8 | 3.0 | 2.4 3.6 | 6.0 | 4.8 5.4 | 9.0 | 7.2 7.2 | 12.0 | 9.6 9.0 | 15.0 | 12.0 10.8 | 18.0 | 14.4 12.6 | 21.0 | 16.8 So the signal reaches the ship after 3 hours and.. the shuttle reaches the ship after 3 hours. Oh, that doesn't make sense. Hmm, sure this method seemed to work earlier. Now I'm confused.  Quote by vela These aren't correct. From the equations for a Lorentz transformation, Δt'=γ(Δt-βΔx), you can see time dilation only applies when Δx=0, but the emission and reception of the radio signal occurs at different points in space (in both frames). Could you expain this a bit more? So, in the spaceship's frame of reference, the signal is perceived to leave Earth at 2pm and the shuttle at 4pm?, since the spaceship clock and the Earth clock are synchronised to the same time. Perhaps I've just completely messed that up. I'm really not sure on this bit of the question. PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,860  Quote by Lissajoux This is how I figured out when they would all meet.. I plotted a quick table of how far each travels in each hour, and compared the results. I hope I can indeed do it this rather simple way, and this makes sense: Ship | Signal | Shuttle ------------------------- 1.8 | 3.0 | 2.4 3.6 | 6.0 | 4.8 5.4 | 9.0 | 7.2 7.2 | 12.0 | 9.6 9.0 | 15.0 | 12.0 10.8 | 18.0 | 14.4 12.6 | 21.0 | 16.8 So the signal reaches the ship after 3 hours and.. the shuttle reaches the ship after 3 hours. Oh, that doesn't make sense. Hmm, sure this method seemed to work earlier. Now I'm confused. You're not accounting for the fact that the ship has a two-hour head start on the signal and a four-hour head start on the shuttle. Let t be the time the ship has been traveling. The ship's position is then given by xship(t)=0.6ct. You can also write down a function that gives the position of the signal as a function of t (for t>2): xsignal=c(t-2). Set the two equal to each other and solve for t. You can similarly solve for when the shuttle reaches the ship.  Could you expain this a bit more? Time dilation lets you compare your clock to a moving clock. What it doesn't let you do is compare the difference between two clocks at different points in space in your frame to the difference between the two clocks in the moving frame. For instance, consider two events that are simultaneous in the rest frame, so Δt=0. In a moving frame, they won't be simultaneous, so Δt'≠0. It's clear that the time dilation formula, ∆t'≠γ∆t, doesn't apply. Because relativity mixes space and time, the time difference between two events a moving observer sees depends on the spatial separation of the events. The time dilation formula tells you what happens in the special case where Δx=0.  So, in the spaceship's frame of reference, the signal is perceived to leave Earth at 2pm and the shuttle at 4pm?, since the spaceship clock and the Earth clock are synchronised to the same time. Perhaps I've just completely messed that up. I'm really not sure on this bit of the question. An observer on the ship would say the signal left Earth when a clock on Earth read 2:00 PM, as would an observer on Earth. His own clock would read 2:30 PM. From his perspective, the Earth's clock was moving, so it ran more slowly.  P: 82 So define these: [itex]x_{ship}=0.6t$ and $x_{signal}=c(t-2)$ Hence: $0.6ct=c(t-2)$ Now solve for $t$ .. except I'm having a daft moment, and I'm not sure if I've done this right! .. $$0.6ct=ct-2c\implies 2c=0.4ct\implies t=\frac{2}{0.4}=5$$ hours So the signal reaches the ship after $t=5$ hours That all correct? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Then do almost exactly the same for the shuttle and the ship: So define these: $x_{ship}=0.6t$ and $x_{shuttle}=c(t-4)$ Hence: $0.6ct=c(t-4)$ Now solve for $t$ $$0.6ct=ct-4c\implies 4c=0.4ct\implies t=\frac{4}{0.4}=10$$ hours So the shuttle reaches the ship after $t=10$ hours That all correct too? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Now the times are known for the signal and shuttle both sent from Earth to reach the ship. So that's part 3 of the question, calculated first, hmm. I'm not sure how much of the previous post is correct/incorrect, you quoted it in your reply post without any comments
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 Quote by Lissajoux So define these: $x_{ship}=0.6t$ and $x_{signal}=c(t-2)$ Hence: $0.6ct=c(t-2)$ Now solve for $t$ .. except I'm having a daft moment, and I'm not sure if I've done this right! .. $$0.6ct=ct-2c\implies 2c=0.4ct\implies t=\frac{2}{0.4}=5$$ hours So the signal reaches the ship after $t=5$ hours That all correct?
Yes. To clarify, t represents the time the ship has been traveling. The signal only takes 3 hours to go from Earth to the ship. It reaches the ship at t=5 h because it leaves Earth two hours after the ship left.
 Then do almost exactly the same for the shuttle and the ship: So define these: $x_{ship}=0.6t$ and $x_{shuttle}=c(t-4)$ Hence: $0.6ct=c(t-4)$ Now solve for $t$ $$0.6ct=ct-4c\implies 4c=0.4ct\implies t=\frac{4}{0.4}=10$$ hours So the shuttle reaches the ship after $t=10$ hours That all correct too?
No, because the shuttle does not travel at the speed of light.
 Now the times are known for the signal and shuttle both sent from Earth to reach the ship. So that's part 3 of the question, calculated first, hmm. I'm not sure how much of the previous post is correct/incorrect, you quoted it in your reply post without any comments
The rest of your post relied on a misapplication of time dilation, so it's all wrong.
P: 82
 No, because the shuttle does not travel at the speed of light.

So define these: $x_{ship}=0.6t$ and $x_{shuttle}=0.8c(t-4)$

Hence: $: 0.6ct=0.8c(t-4)$

Now solve for $t$:

$$0.6ct=0.8ct-3.2c\implies 3.2c=0.2ct\\implies t=\frac{3.2}{0.2}=16 hours$$

So the shuttle reaches the ship after $: t=16 hours$

I'm sure that's not correct, but I don't know where I'm going wrong

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Now as the value of signal arrival time correctly calculated as $t=5 years$

The speed of the signal is known to be $v_{signal}=c=3\times 10^{8}ms^{-1}$

Which equates to: $v_{signal}=c=3\times 10^{8}ms^{-1}$

Can calculate the distance of the Earth away from the ship as:

$$d=(v_{signal})(t)=(9.46\times 10^{15}m years^{-1})(5 years)=4.73\times 10^{16}m=5 light years$$

.. getting anywhere?

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Then need to do the same for the shuttle, once I know it's correct time or arrival with ship.
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 Quote by Lissajoux OK, well I'm not sure, but how about this... So define these: $x_{ship}=0.6t$ and $x_{shuttle}=0.8c(t-4)$ Hence: $: 0.6ct=0.8c(t-4)$ Now solve for $t$: $$0.6ct=0.8ct-3.2c\implies 3.2c=0.2ct\\implies t=\frac{3.2}{0.2}=16 hours$$ So the shuttle reaches the ship after $: t=16 hours$ I'm sure that's not correct, but I don't know where I'm going wrong

Your sentence "So the shuttle reaches the ship after t=16 hours" is a bit ambiguous. It can be interpreted two ways, and the more common interpretation is wrong. It sounds like you're saying the shuttle catches up to the ship after traveling for 16 hours. That's wrong. The ship has traveled for t=16 hours, but the shuttle has traveled for only t-4=16-4=12 hours.
 Now as the value of signal arrival time correctly calculated as $t=5 years$ The speed of the signal is known to be $v_{signal}=c=3\times 10^{8}ms^{-1}$ Which equates to: $v_{signal}=c=3\times 10^{8}ms^{-1}$ Can calculate the distance of the Earth away from the ship as: $$d=(v_{signal})(t)=(9.46\times 10^{15}m years^{-1})(5 years)=4.73\times 10^{16}m=5 light years$$ .. getting anywhere? - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Then need to do the same for the shuttle, once I know it's correct time or arrival with ship.
Where did you get years from? Even if you used hours instead of years, your answer is wrong for the reason I explained above regarding the interpretation of your sentence. I also explained in my previous post that while the ship has traveled for 5 hours, the signal has only traveled for 3 hours because it left the Earth 2 hours after the ship did.

So how far is the ship when the signal catches up? You can calculate it two ways. If you look at the signal, it will have propagated for three hours; in this time, it travels 3 light-hours away from Earth. If you look at the ship, it will have been traveling for five hours. Since it's moving with a speed of 0.6c, it will be at a distance of (5 hours)x(0.6 c) = 3 light-hours away from Earth. Just as you'd expect, the signal and the ship are at the same distance from Earth, which has to be the case when they meet.
 P: 82 I'll have another go at this. So according to someone in the Earth's reference frame the shuttle has travelled for $t=16hours$ but according to someone on the shuttle or someone on the ship, both in the ship's reference frame, the shuttle has only travelled for $T=t-4=16-4=12 hours$? The similar case for the signal, but person in Earth ref frame records $t=5hours[/tex] but person in the ship ref frame records [itex]T=t-2=5-2=3 hours$? Then I'm still not sure about this, something like this: Signal travels for 3 hours before it reaches the ship, which equates to a travel distance of $d=3\times0.6c=1.8 light hours$. That's how far away the Earth is from the ship when the signal arrives. Probably slightly different. Basically wasn't sure whether to use 3 or 5, and whether to use 0.6c or 0.8c, as the values in the equation. Would need to do the same thing for the signal, and maybe should get the same distance, even though the ship will have travelled further in the time between the two arrivals this should be accounted for in the calculations I assume. Hopefully this is getting there. I do appreciate the help.
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 Quote by Lissajoux I'll have another go at this. So according to someone in the Earth's reference frame the shuttle has travelled for $t=16hours$ but according to someone on the shuttle or someone on the ship, both in the ship's reference frame, the shuttle has only travelled for $T=t-4=16-4=12 hours$?
No.
 The similar case for the signal, but person in Earth ref frame records $t=5hours[/tex] but person in the ship ref frame records [itex]T=t-2=5-2=3 hours$?
No.
 Then I'm still not sure about this, something like this: Signal travels for 3 hours before it reaches the ship, which equates to a travel distance of $d=3\times0.6c=1.8 light hours$. That's how far away the Earth is from the ship when the signal arrives.
No.
 Probably slightly different. Basically wasn't sure whether to use 3 or 5, and whether to use 0.6c or 0.8c, as the values in the equation. Would need to do the same thing for the signal, and maybe should get the same distance, even though the ship will have travelled further in the time between the two arrivals this should be accounted for in the calculations I assume. Hopefully this is getting there. I do appreciate the help.
You obviously have no understanding of what the expressions for xship and xsignal represent or even what the variable t stands for. I'll admit I'm frustrated. I've already explained it a couple of times, and you seem to completely ignore what I've written. You need to spend some time thinking about it and figuring it out for yourself.
 P: 82 I've gone back to the start and read through everything again. Hopefully I've been able to get this back on track now. Part A - Earth reference frame: Speed of signal and shuttle: $v_{signal}=c$ and $v_{shuttle}=0.8c$ There's no relativistic effects to worry about here, we're in the Earth's reference frame. So the values are just those given in the question. Simple. Finding when signal and shuttle reach the ship: Still considering everything in Earth reference frame. Define $t$ as the time the ship has been travelling. Ship position given by: $x_{ship}(t)=0.6ct$ since the ship leaves the Earth at time $t$ at speed $0.6c$. Signal position given by: $x_{signal}(t)=c(t-2)$ since the signal leaves the Earth 2 hours after the ship, hence $t-2$, and travels at speed $c$. Set these expressions equal to find $t$ which gives when they meet. So: $$0.6ct=c(t-2)\implies 0.4ct=2c\implies t=\frac{2}{0.4}=5 hours$$ So the signal reaches the ship after the ship has travelled 5 hours. But the signal left 2 hours later from Earth, so it has only travelled for 3 hours. So according to the ship, the signal takes 3 hours to reach the ship from Earth. The do the same for the shuttle: Ship position given by: $x_{ship}(t)=0.6ct$ since the ship leaves the Earth at time $t$ at speed $0.6c$. This is the same as the previous case. Signal position given by: $x_{signal}(t)=0.8c(t-4)$ since the signal leaves the Earth 4 hours after the ship, hence $t-4$, and travels at speed $0.8c$. Set these expressions equal to find $t$ which gives when they meet. So: $$0.6ct=0.8c(t-4)\implies 0.2ct=3.2c\implies t=\frac{3.2}{0.2}=16 hours$$ So the shuttle reaches the ship after the ship has travelled 16 hours. But the shuttle left 4 hours later from Earth, so it has only travelled for 12 hours. So according to the ship, the shuttle takes 12 hours to reach the ship from Earth. I don't see what's wrong with any of that. Already established the method was correct for calculating ship-signal interaction time. Just need to account for that shuttle moving at $0.8c$ not $c$ within the $d=s\times t$ equation for the shuttle, and have done this. Also accounted for the departure time difference. I hope that's now all good. By the way this is the section in lecture notes I have: http://img38.imageshack.us/i/rela1.jpg So that's part A done, onto part B.. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Part B - Ship reference frame: Now need to account for the relativistic effects, due to the speed of the shuttle. The Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}=\frac{1}{\sqrt{1-\left(0.6\right)^{2}}}=\frac{1}{\sqrt{0.64}}=1.25$$ Finding when signal and shuttle left the Earth: According to Earth reference frame, the signal took $t_{signal}[E]=3 hours[/tex] to reach the ship. So using that and the value of [itex]\gamma$: $$t_{signal}[S]=\frac{t_{signal}[E]}{\gamma}=\frac{3}{1.25}=2.4hours$$ So the ship thinks the signal takes 2.4 hours to reach the ship. According to Earth reference frame, the shuttle took $t_{shuttle}[E]=12 hours[/tex] to reach the ship. So using that and the value of [itex]\gamma$: $$t_{signal}[S]=\frac{t_{shuttle}[E]}{\gamma}=\frac{12}{1.25}=9.6hours$$ So the ship thinks the shuttle takes 9.6 hours to reach the ship. So these are the times the ship thinks the signal and shuttle have been travelling, so they left the Earth the respective times earlier. Finding speeds of the signal and the shuttle: Person on Earth perceived the signal to be travelling at $v_{signal}=c$ and arrived at the shuttle in $t_{signal}[E]=3 hours$. Person on the ship perceived the signal to arrive after $t_{signal}[S]=2.4 hours$. So through a simple calculation: $$d=s\times t=c\times 3=3.24times 10^{12}m$$ So that's the distance a person on Earth perceives the ship to be away from Earth. However for the person on the ship, there is length contraction to account for: $$d[S]=\frac{d[E]}{\gamma}=\frac{3.24 \times 10^{12}m}{1.25}=2.59\times 10^{12}m$$ So the distance of Earth<->Ship appears to be less for the person on the ship. The ship perceives the signal to take 2.4hours to reach it. So again, simple calculation to get the velocity of the signal: $$v_{signal}[S]=\frac{2.59\times 10^{12}m}{2.4hours}=2.9977..\times 10^{8}m/s=0.9992c$$ Then do the same but for the shuttle, I havn't detailed that right now incase there was any issues with this method. Finding the arrival times of the signal and shuttle at the ship: The signal took $t_{signal}[S]=2.4 hours$ and the shuttle took $t_{shuttle}[S]=12 hours$, these were calculated in part i. Again, by the way, this is the section in my lecture notes: http://img98.imageshack.us/i/rela2.jpg - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Sorry for frustrating you.. I really am trying to get all this
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 Quote by Lissajoux Part B - Ship reference frame: Now need to account for the relativistic effects, due to the speed of the shuttle. The Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}=\frac{1}{\sqrt{1-\left(0.6\right)^{2}}}=\frac{1}{\sqrt{0.64}}=1.25$$ Finding when signal and shuttle left the Earth: According to Earth reference frame, the signal took $t_{signal}[E]=3 hours[/tex] to reach the ship. So using that and the value of [itex]\gamma$: $$t_{signal}[S]=\frac{t_{signal}[E]}{\gamma}=\frac{3}{1.25}=2.4hours$$ So the ship thinks the signal takes 2.4 hours to reach the ship. According to Earth reference frame, the shuttle took $t_{shuttle}[E]=12 hours[/tex] to reach the ship. So using that and the value of [itex]\gamma$: $$t_{signal}[S]=\frac{t_{shuttle}[E]}{\gamma}=\frac{12}{1.25}=9.6hours$$ So the ship thinks the shuttle takes 9.6 hours to reach the ship. So these are the times the ship thinks the signal and shuttle have been travelling, so they left the Earth the respective times earlier. Finding speeds of the signal and the shuttle: Person on Earth perceived the signal to be travelling at $v_{signal}=c$ and arrived at the shuttle in $t_{signal}[E]=3 hours$. Person on the ship perceived the signal to arrive after $t_{signal}[S]=2.4 hours$. So through a simple calculation: $$d=s\times t=c\times 3=3.24times 10^{12}m$$ So that's the distance a person on Earth perceives the ship to be away from Earth. However for the person on the ship, there is length contraction to account for: $$d[S]=\frac{d[E]}{\gamma}=\frac{3.24 \times 10^{12}m}{1.25}=2.59\times 10^{12}m$$ So the distance of Earth<->Ship appears to be less for the person on the ship. The ship perceives the signal to take 2.4hours to reach it. So again, simple calculation to get the velocity of the signal: $$v_{signal}[S]=\frac{2.59\times 10^{12}m}{2.4hours}=2.9977..\times 10^{8}m/s=0.9992c$$ Then do the same but for the shuttle, I havn't detailed that right now incase there was any issues with this method. Finding the arrival times of the signal and shuttle at the ship: The signal took $t_{signal}[S]=2.4 hours$ and the shuttle took $t_{shuttle}[S]=12 hours$, these were calculated in part i. Again, by the way, this is the section in my lecture notes: http://img98.imageshack.us/i/rela2.jpg - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Sorry for frustrating you.. I really am trying to get all this
You're just repeating the calculations you did earlier, which I pointed out earlier were wrong.
 P: 82 Unfortunately I cannot see where I am going wrong. Looking at those lecture notes, this appears to be exactly what I should be doing. Perhaps you could help with where I am going wrong. The advice really is appreciated, I don't mean for it to sound like I'm having a go if it comes across that way. I just want to know what is wrong or right and how to get things sorted out, because at the moment I don't see what to change.
 PF Patron HW Helper Sci Advisor Thanks Emeritus P: 10,860 I already told you where you're going wrong, way back in post 4, and again in posts 6 and 8.
P: 82
I've gone back to your previous posts and this is what youv'e said:

 These aren't correct. From the equations for a Lorentz transformation, Δt'=γ(Δt-βΔx), you can see time dilation only applies when Δx=0, but the emission and reception of the radio signal occurs at different points in space (in both frames).
 Time dilation lets you compare your clock to a moving clock. What it doesn't let you do is compare the difference between two clocks at different points in space in your frame to the difference between the two clocks in the moving frame. For instance, consider two events that are simultaneous in the rest frame, so Δt=0. In a moving frame, they won't be simultaneous, so Δt'≠0. It's clear that the time dilation formula, ∆t'≠γ∆t, doesn't apply. Because relativity mixes space and time, the time difference between two events a moving observer sees depends on the spatial separation of the events. The time dilation formula tells you what happens in the special case where Δx=0.
I think this is the point I went wrong from then.

So you're saying that the time dilation formula I used previously:

$$t=\frac{t_{E}}{\gamma}=\frac{6}{1.25}=4.8s$$

(that's just for an example of a previous calculation)

Or rather more generally:

$$t'=\gamma t$$

This cannot be applied to this problem?

I'll assume that's the case. Right considering what's going on again..

So the two clocks, the one of Earth and the one on the ship, are synchronised to the same time of 12 noon at the point the ship leaves the Earth. That's known. Also we're in the ship's reference frame.

The signal leaves Earth at 2pm, according to the person on Earth. The person on the ship.. also thinks the signal left Earth at 2pm?

But surely his clock won't still be reading the same time as on Earth, it'll be later on (perceives that Earth clock runs slow as Earth moving) right? So how do I calculate that he thinks it's 2pm? i.e. what equation do I use?

*click*

oh, perhaps I can use this now:

$$t=\gamma t' \implies t=(1.25)\times (2.00)=2.5\rightarrow 2:30pm$$

?

Sorry, I know you said time dilation formula cannot be applied, but how else to I calculate the time difference?

I'm sure need to account for that $\gamma$ somewhere, maybe I just wasn't using it in the right way before then. I also know I may have mixed up t and t' in the order they should be there.

So that would answer finding out what time the ship clock would read, but not really as to why they know it left at 2pm as how do I apply this backwards to deduce this? Unless I'm missing something obvious.. probably.

Obviously can do similar method with the shuttle but want to get these steps correct first, so I havn't mentioned it at the moment. Hopefully what I've said makes sense.

I am sort of understanding what you're saying now and where I'm going wrong with what I've been doing, or at least I think so. Just trying to think about the things a bit differently. It's clearly something that's taking me a while to get around.
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 Quote by Lissajoux So you're saying that the time dilation formula I used previously: $$t=\frac{t_{E}}{\gamma}=\frac{6}{1.25}=4.8s$$ (that's just for an example of a previous calculation) Or rather more generally: $$t'=\gamma t$$ This cannot be applied to this problem?
Yes and no. Yes, in the sense that you can't apply it the way you have. No, in the sense that you can use it for an appropriate situation, like the one you mention below.
 So the two clocks, the one of Earth and the one on the ship, are synchronised to the same time of 12 noon at the point the ship leaves the Earth. That's known. Also we're in the ship's reference frame. The signal leaves Earth at 2pm, according to the person on Earth. The person on the ship.. also thinks the signal left Earth at 2pm? But surely his clock won't still be reading the same time as on Earth, it'll be later on (perceives that Earth clock runs slow as Earth moving) right? So how do I calculate that he thinks it's 2pm? i.e. what equation do I use? *click* oh, perhaps I can use this now: $$t=\gamma t' \implies t=(1.25)\times (2.00)=2.5\rightarrow 2:30pm$$ ?
This calculation is correct. The formula applies because the clock on Earth doesn't move with respect to the Earth. It's always at x=0, so Δx=0.

A more general way to analyze this problem is to use the Lorentz transformations to map the coordinates of events in one frame into the other frame. In the Earth frame, the event of the signal being sent is at the coordinates (ct,x)=(2 lh, 0 lh) [lh = light-hour]. Using the Lorentz transformations, you'll find

$$ct' = \gamma(ct - \beta x) = 1.25 [2~\textrm{lh} - 0.6\cdot 0~\textrm{lh}] = 2.5~\textrm{lh}$$

$$x' = \gamma(x - \beta ct) = 1.25 [0~\textrm{lh} - 0.6\cdot 2~\textrm{lh})] = -1.5~\textrm{lh}$$

where $\beta = v/c$. What this says is from the ship observer's point of view, the signal left Earth when the ship's clock read 2:30 and the Earth was 1.5 lh away.

Earlier, you found that the signal reaches the ship when t=5 h and the ship is x=3 lh away, as observed in the Earth frame. Try calculating the coordinates of this event in the ship's frame.

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