
#1
Dec3111, 11:50 AM

P: 117

I've posted on this forum before and carefully read the answers, but I'm still confused... Also, I'm familiar with the twin paradox, and understand that acceleration is the reason the twins' frames aren't symmetrical.
Question 1: Is it okay for an observer to view HIS OWN reference frame as the moving one? For example, suppose an observer is on a planet with speed 0.6c to the right, and a rocket is at rest relative to him. The prime frame is the rocket frame. Suppose there is a time interval of one hour on the planet. I want to find the time interval in hours on the rocket's clock. t'=γ(tβx)=1.25(10.36)=0.8 This is correct, isn't it? Question 2: But it's equally valid to see the rocket as moving at 0.6c to the right and the planet as at rest. Once again, suppose there is a time interval of one hour on the planet. I want to find the time interval in hours on the rocket's clock. But this time I get t'=γ(tβx)=1.25 since Δx=0. How come I get these two different answers? Thanks in advance for answering, as I'm sure this is a very easy question for most of you I've just met special relativity so I'm very confused!! 



#2
Jan112, 03:51 AM

PF Gold
P: 4,541

First off, since the rocket is moving to the right, we use a value of ß=0.6. From this we calculate ? as: ?=1/v(1ß^{2})=1/v(10.6^{2})=1/v(10.36)=1/v0.64=1/0.8=1.25 Next we assume that the start of the hour is at the origin and the event we want to consider is one hour later when t=1. During this hour, the rocket has moved at 0.6c so it is 0.6 lighthours to the right of the origin meaning it is at x=0.6. Now we can do your calculation: t'=?(tßx)=1.25(1(0.6*0.6))=1.25(10.36)=1.25(0.64)=0.8 Now what does this mean? It means in the rest frame of the planet, when 1 hour has gone by for the planet (and all of the clocks in its rest frame), 0.8 hours has gone by for the rocket. Note that the rocket is moving in this frame but wherever it is, there is a coordinate clock that is at the same time as the planet's clock that remains at location x=0. So we can meaningfully talk about how much time has transpired on the moving rocket's clock during an interval of time for the stationary planet's clock. This also illustrates the normal time dilation factor of 1/? based on the speed of an object in a frame of reference. We use the Greek letter tau, t, to symbolize the Proper Time on a moving clock and compare it to t, the coordinate time for the reference frame with the equation t=t/?. So if we know the coordinate time interval, we can calculate the time on a moving clock, t, using this formula. Or we can go the other way around. If we know the Proper Time interval on a moving clock, we can calculate the corresponding coordinate time, t, for the reference frame. So let me rephrase your Question 2 into what I think you want: But it's equally valid to see the planet as moving at 0.6c to the left and the rocket as at rest. Once again, suppose there is a time interval of one hour on the planet. I want to find the time interval in hours on the rocket's clock.Now we have a totally different situation because the planet's clocks are not synchronized with the coordinate time of the reference frame (which is also equal to the time on the rocket's clock). So in order to determine a coordinate time corresponding to one hour on the planet, we can use the relationship between t and t expressed above and calculate t=?*t=1.25*1=1.25. Now we can quit right here because this is the same answer you got, although I'm not sure if you actually did a correct calculation because you left out all the details. So what does this mean? Well, time dilation is frame dependent, just like speed is. In the planet's rest frame, the rocket is moving, so it is the one that is experiencing time dilation, and one hour on the planet's clock is less time (0.8 hours) on the rocket's clock. But in the rocket's rest frame, the planet is moving, so it is the one that is experiencing time dilation, and one hour on the planet's clock is more time (1.25 hours) on the rocket's clock. You could also say that one hour on the rocket's clock is 0.8 hours on the planet's clock. This might show the symmetry a little better. So your final numbers are correct, but they were interchanged with the situations. 



#3
Jan112, 05:17 AM

P: 117

Thanks so much for your answer, and happy new year!
I understand the first bit of your answer, but I'm a bit confused about the final paragraphs. 



#4
Jan112, 06:01 AM

P: 3,178

Basic special relativity question 



#5
Jan112, 07:35 AM

P: 117

Ah, so the twin paradox is the 'exception to the rule' that time dilation depends on the frame. Thanks everyone!




#6
Jan112, 08:32 AM

Mentor
P: 16,489

The twin paradox scenario is not an exception to the rule. The rule is the Lorentz transform, and it is for transforming between inertial frames. If you take the twin paradox scenario and use the Lorentz transform to transform to any inertial frame you will get the same answer.
The confusion with the twin paradox scenario is simply that introductory students often do not realize that the travelling twin's reference frame is not an inertial frame so you cannot use a Lorentz transform with it. 



#7
Jan112, 01:23 PM

PF Gold
P: 4,541





#8
Jan212, 01:27 AM

PF Gold
P: 4,541

NOTE: This is a repost of post #2 which got all its special symbols clobbered when I edited a couple minor typos:
First off, since the rocket is moving to the right, we use a value of β=0.6. From this we calculate γ as: γ=1/√(1β^{2})=1/√(10.6^{2})=1/√(10.36)=1/√0.64=1/0.8=1.25 Next we assume that the start of the hour is at the origin and the event we want to consider is one hour later when t=1. During this hour, the rocket has moved at 0.6c so it is 0.6 lighthours to the right of the origin meaning it is at x=0.6. Now we can do your calculation: t'=γ(tβx)=1.25(1(0.6*0.6))=1.25(10.36)=1.25(0.64)=0.8 Now what does this mean? It means in the rest frame of the planet, when 1 hour has gone by for the planet (and all of the clocks in its rest frame), 0.8 hours has gone by for the rocket. Note that the rocket is moving in this frame but wherever it is, there is a coordinate clock that is at the same time as the planet's clock that remains at location x=0. So we can meaningfully talk about how much time has transpired on the moving rocket's clock during an interval of time for the stationary planet's clock. This also illustrates the normal time dilation factor of 1/γ based on the speed of an object in a frame of reference. We use the Greek letter tau, τ, to symbolize the Proper Time on a moving clock and compare it to t, the coordinate time for the reference frame with the equation τ=t/γ. So if we know the coordinate time interval, we can calculate the time on a moving clock, τ, using this formula. Or we can go the other way around. If we know the Proper Time interval on a moving clock, we can calculate the corresponding coordinate time, t, for the reference frame. So let me rephrase your Question 2 into what I think you want: But it's equally valid to see the planet as moving at 0.6c to the left and the rocket as at rest. Once again, suppose there is a time interval of one hour on the planet. I want to find the time interval in hours on the rocket's clock.Now we have a totally different situation because the planet's clocks are not synchronized with the coordinate time of the reference frame (which is also equal to the time on the rocket's clock). So in order to determine a coordinate time corresponding to one hour on the planet, we can use the relationship between τ and t expressed above and calculate t=γ*τ=1.25*1=1.25. Now we can quit right here because this is the same answer you got, although I'm not sure if you actually did a correct calculation because you left out all the details. So what does this mean? Well, time dilation is frame dependent, just like speed is. In the planet's rest frame, the rocket is moving, so it is the one that is experiencing time dilation, and one hour on the planet's clock is less time (0.8 hours) on the rocket's clock. But in the rocket's rest frame, the planet is moving, so it is the one that is experiencing time dilation, and one hour on the planet's clock is more time (1.25 hours) on the rocket's clock. You could also say that one hour on the rocket's clock is 0.8 hours on the planet's clock. This might show the symmetry a little better. So your final numbers are correct, but they were interchanged with the situations. 



#9
Jan212, 12:55 PM

P: 848

And Lucy Yeates, I wouldn't be focusing so much on the acceleration as I would the difference in the paths through a 4dimensional universe that each twin is following. You can directly compare the times it takes to travel each path (each twin is moving along his own world line at the speed of light). 



#10
Jan212, 05:56 PM

Mentor
P: 16,489

Hi Lucy Yeates, I would second bobc2's suggestion:




#11
Jan312, 10:55 AM

P: 117

Thanks!
I have another question I hope it's not too repetitive, but I'm still trying to grasp the basics. I've just been looking at an example in a textbook. The relative velocity between Earth and a rocket is 0.8c. On board the rocket, an observer looks out of the window and sees Lookout Station 8, 8 lightyears from earth. The observer see the lookout station clock's date is ten years after the departure date. The rocket clock reads 6 years. Applying the Lorentz transformations, I could see why this was the case: (The rocket frame is the prime frame.) Δt'=6, Δx'=0, Δt=10, x=8c t'=γ(tβx/c)=5/3(106.4)=6 t=γ(βx'/ct')=5/3(6)=10 (I'm assuming the minus sign isn't a problem?) So the point of view of each observer is that the other's moving clock is running slow. But the rocket observer sees that Δt'=6 and Δt=10, the interval for the 'moving' clock. Surely this means that they would view their own clock as the slow one? 



#12
Jan312, 11:15 AM

Emeritus
Sci Advisor
PF Gold
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#13
Jan412, 03:14 AM

P: 117

Thanks I think I'm starting to see how time dilation, length contraction, and the relativity of simultaneity are all closely related.
Would you mind showing me how to get 6.4 years in the post above? 



#14
Jan512, 09:54 PM

PF Gold
P: 4,541

A rocket takes off from Earth at time zero and travels at 0.8c (β=0.8) to a Station that is 8 lightyears away. When it gets there, the clock at the station reads 10 years and the rocket's clock reads 6 years. This all makes sense because it would take the rocket 10 years in the EarthStation frame to travel 8 lightyears at 0.8c. The time dilation factor is 1/γ and γ=1/√(1β^{2})=1/.6 so the time dilation factor is 0.6. Why do include "c" after the last one? Where did you get the values of Δt'=6 and Δx'=0? Where did you get this formula? 


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