Basic special relativity question


by Lucy Yeats
Tags: basic, relativity, special
Lucy Yeats
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#1
Dec31-11, 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(t-βx)=1.25(1-0.36)=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(t-βx)=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!!
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ghwellsjr
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#2
Jan1-12, 03:51 AM
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Quote Quote by Lucy Yeats View Post
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?
It's important to realize that all observers, objects and clocks are in any reference frame you want to consider but when we talk about an observer's own reference frame, we always mean one in which he is at rest. However, it's totally permissible for an observer to view himself in any reference frame, even one that is moving with respect to himself or, equivalently, one in which he is moving, just don't call it HIS OWN reference frame or you'll just confuse everyone that is familiar with the normal terminology.
Quote Quote by Lucy Yeats View Post
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.
This is confusing in terms of where the observer is but I think what is important is you have a rocket and a planet that are moving away from each other with the rocket on the right and the planet on the left and the x-axis is aligned along their line of motion and increasing x is to the right and when they were together, you're calling that the origin where x=x'=0 and t=t'=0, correct?
Quote Quote by Lucy Yeats View Post
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'=?(t-x)=1.25(1-0.36)=0.8 This is correct, isn't it?
This is correct if you are starting with the rest frame of the planet and you are transforming an event into the rest frame of the rocket (you did say it was the prime frame which is the one you are transforming into). But you left out a lot of important details, so I will fill them in for anyone that doesn't see how you did this.

First off, since the rocket is moving to the right, we use a value of =0.6. From this we calculate ? as:
?=1/v(1-2)=1/v(1-0.62)=1/v(1-0.36)=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 light-hours to the right of the origin meaning it is at x=0.6. Now we can do your calculation:
t'=?(t-x)=1.25(1-(0.6*0.6))=1.25(1-0.36)=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.

Quote Quote by Lucy Yeats View Post
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'=?(t-x)=1.25 since ?x=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!!
This sounds like exactly what we just did for Question 1 but I think what you are asking is how you can start from the rest frame of the rocket and transform into the rest frame of the planet and get the same result as before, correct? (I think you may have gotten confused into thinking that the prime frame was the first frame but we don't mean prime that way, we simply mean it's the frame that has the prime marks on the variable names (t' and x'), the ones that we are going to transform into).

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.
Lucy Yeats
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#3
Jan1-12, 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.

Quote Quote by ghwellsjr View Post
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.
I didn't realise 'time dilation is frame dependent' in this way. In both cases, we were in the frame of the planet, so I was confused that I got different answers. I thought you could say objectively that time in one frame was dilated. I've clearly misunderstood this at a very basic level...

harrylin
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#4
Jan1-12, 06:01 AM
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Basic special relativity question


Quote Quote by Lucy Yeats View Post
[..] I didn't realise 'time dilation is frame dependent' in this way. In both cases, we were in the frame of the planet, so I was confused that I got different answers. I thought you could say objectively that time in one frame was dilated. I've clearly misunderstood this at a very basic level...
I think that you now understand it. Indeed, with two clocks in rectilinear, uniform motion that move relative to each other, we cannot objectively state which one has a faster rate; we can only state what their rates are according to measurements with a certain reference system ("relative to a certain frame"). However, with the twin scenario one can say objectively (according to everyone) that the traveller's clock has retarded on the stay-at-home clock; and that is thanks to the turn-around which broke the symmetry.
Lucy Yeats
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#5
Jan1-12, 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!
DaleSpam
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#6
Jan1-12, 08:32 AM
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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.
ghwellsjr
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#7
Jan1-12, 01:23 PM
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Quote Quote by Lucy Yeats View Post
I didn't realise 'time dilation is frame dependent' in this way. In both cases, we were in the frame of the planet, so I was confused that I got different answers. I thought you could say objectively that time in one frame was dilated. I've clearly misunderstood this at a very basic level...
But you did talk about the rocket frame in your first post so maybe I completely misunderstood your questions. If so, please ask again.
ghwellsjr
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#8
Jan2-12, 01:27 AM
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NOTE: This is a repost of post #2 which got all its special symbols clobbered when I edited a couple minor typos:

Quote Quote by Lucy Yeats View Post
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?
It's important to realize that all observers, objects and clocks are in any reference frame you want to consider but when we talk about an observer's own reference frame, we always mean one in which he is at rest. However, it's totally permissible for an observer to view himself in any reference frame, even one that is moving with respect to himself or, equivalently, one in which he is moving, just don't call it HIS OWN reference frame or you'll just confuse everyone that is familiar with the normal terminology.
Quote Quote by Lucy Yeats View Post
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.
This is confusing in terms of where the observer is but I think what is important is you have a rocket and a planet that are moving away from each other with the rocket on the right and the planet on the left and the x-axis is aligned along their line of motion and increasing x is to the right and when they were together, you're calling that the origin where x=x'=0 and t=t'=0, correct?
Quote Quote by Lucy Yeats View Post
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(t-βx)=1.25(1-0.36)=0.8 This is correct, isn't it?
This is correct if you are starting with the rest frame of the planet and you are transforming an event into the rest frame of the rocket (you did say it was the prime frame which is the one you are transforming into). But you left out a lot of important details, so I will fill them in for anyone that doesn't see how you did this.

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/√(1-β2)=1/√(1-0.62)=1/√(1-0.36)=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 light-hours to the right of the origin meaning it is at x=0.6. Now we can do your calculation:
t'=γ(t-βx)=1.25(1-(0.6*0.6))=1.25(1-0.36)=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.

Quote Quote by Lucy Yeats View Post
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(t-βx)=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!!
This sounds like exactly what we just did for Question 1 but I think what you are asking is how you can start from the rest frame of the rocket and transform into the rest frame of the planet and get the same result as before, correct? (I think you may have gotten confused into thinking that the prime frame was the first frame but we don't mean prime that way, we simply mean it's the frame that has the prime marks on the variable names (t' and x'), the ones that we are going to transform into).

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.
bobc2
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#9
Jan2-12, 12:55 PM
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Quote Quote by DaleSpam View Post
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.
In the space-time sketch below you can see that the twin paradox can be understood quite clearly using Lorentz transformations. Of course you must break up the trip into two different inertial frames as is done here.

And Lucy Yeates, I wouldn't be focusing so much on the acceleration as I would the difference in the paths through a 4-dimensional 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).
DaleSpam
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#10
Jan2-12, 05:56 PM
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Hi Lucy Yeates, I would second bobc2's suggestion:
Quote Quote by bobc2 View Post
And Lucy Yeates, I wouldn't be focusing so much on the acceleration as I would the difference in the paths through a 4-dimensional universe that each twin is following.
A geometric understanding of SR is very valuable for later study.
Lucy Yeats
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#11
Jan3-12, 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 light-years 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(t-βx/c)=5/3(10-6.4)=6
t=γ(βx'/c-t')=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?
Janus
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Jan3-12, 11:15 AM
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Quote Quote by Lucy Yeats View Post
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 light-years 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(t-βx/c)=5/3(10-6.4)=6
t=γ(βx'/c-t')=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?
You have to keep in mind that according to the observer in the rocket, the clock at Lookout Station 8 is already reads 6.4 years when he leaves the vicinity of the Earth.( Relativity of Simultaneity). Thus the clock at Lookout Station 8 only advances by 3.6 years during the trip, running slower than the ship clock according to the ship observer.
Lucy Yeats
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#13
Jan4-12, 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?
ghwellsjr
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Jan5-12, 09:54 PM
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Quote Quote by Lucy Yeats View Post
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 light-years from earth. The observer see the lookout station clock's date is ten years after the departure date. The rocket clock reads 6 years.
Let me first summarize what's happening here:

A rocket takes off from Earth at time zero and travels at 0.8c (β=0.8) to a Station that is 8 light-years 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 Earth-Station frame to travel 8 light-years at 0.8c. The time dilation factor is 1/γ and γ=1/√(1-β2)=1/.6 so the time dilation factor is 0.6.
Quote Quote by Lucy Yeats View Post
Applying the Lorentz transformations, I could see why this was the case: (The rocket frame is the prime frame.)
When you say the rocket frame is the prime frame, what do you mean by that?
Quote Quote by Lucy Yeats View Post
Δt'=6, Δx'=0, Δt=10, x=8c
Why do you put "Δ" in front of some of these values but not the last one?
Why do include "c" after the last one?
Where did you get the values of Δt'=6 and Δx'=0?
Quote Quote by Lucy Yeats View Post
t'=γ(t-βx/c)=5/3(10-6.4)=6
Why do you include "/c" in this formula and where did it go during the evaluation?
Quote Quote by Lucy Yeats View Post
t=γ(βx'/c-t')=5/3(-6)=-10 (I'm assuming the minus sign isn't a problem?)
It is a problem and that's why I'm asking you all these questions.
Where did you get this formula?
Quote Quote by Lucy Yeats View Post
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?
You need to understand what the Lorentz Transform is doing and how to use it before you can understand the answer to your question. I would like to help you in this process but you need to answer my questions so that I will know how to help you. Can you please answer every one of them?


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