A question about stationary reference frame

[goodabouthood]

Is this correct?

There are two clocks on Earth that are synchronized. One clock goes out on a spaceship at .8c.

Now according to the stationary or Earth FOR one year goes by for its clock but he sees that only .6 of a year has gone by on the clock of the spaceship.

Now the spaceship returns home after another year has passed on my clock.

So this means that 2 years has passed on my stationary clock and only 1.2 years has passed on the moving clock according to my FOR?

Correct?

 

[goodabouthood]

This is assuming that at the 1 year mark on my clock I see the spaceship turn around and head back to Earth.

 

[Snip3r]

anyone can get that value by plugging in the values in time dilation equation. What exactly you want to know by asking is this correct?

 

[elfmotat]

anyone can get that value by plugging in the values in time dilation equation. What exactly you want to know by asking is this correct?

He wants to know whether or not it is correct. And yes (assuming perfect conditions) it is correct.

 

[goodabouthood]

What I don't understand is where the paradox sets in.

To me it does not seem like a paradox that one clock has passed two years and the moving clock which now returns home has only passed 1.2 years.

The other thing I want to understand is what is happening in the FOR of the moving clock.

How do I use the lorentz transformation to figure this out?

 

[goodabouthood]

I think I am confusing what I mean by FOR.

In my reference frame one year has passed and in the moving observers reference frame .6 years has passed.

There can only be one clock in each reference frame, correct? So that .6 years which has passed after the 1 year passed in my own reference frame is the lorentz transformation for time? Correct?

 

[Snip3r]

How do I use the lorentz transformation to figure this out?

how did you arrive at the number 0.6 for moving clock?that is lorentz transformation

The other thing I want to understand is what is happening in the FOR of the moving clock.

This is where the paradox steps in. Relativity states that any inertial frame is equally valid. So if i am in that ship i can very well say i am stationary and you are moving at the velocity of 0.8c away from me. So when i calculate 1 year from my FOR it will be just 0.6 in yours

There can only be one clock in each reference frame, correct?

yes in your frame you have only 1 clock for you

 

[ghwellsjr]

This is very similar to the scenario you asked about in your thread called Twin paradox question except that there you had the traveling twin go out for one year according to his clock and then return the next year whereas now you have the twin go out for one year according to the Earth clock and return in another similar year. I worked out in detail on post #22 how you use the Lorentz Transform to transform events from one frame to another frame.

You asked about how you use the Lorentz Transform to figure out what happens in the FoR of the moving clock. I pointed out there that the moving clock is not stationary in a single inertial FoR and that you need two FoRs, one for the outbound portion of the trip and one for the inbound portion of the trip and I showed a way to do this without incurring any jumps in time. But I also pointed out that you should transform the entire scenario into one of these two FoRs and then do it again for the other FoR.

It is not correct that there can be only one clock in each reference frame--every clock is in every reference frame. A reference frame is a coordinate system--we use it to describe where everything is at different times. You described this very well in your first post where you talked about what happened to both clocks in the Earth FoR. Why did you then feel the need to question your good start?

 

[atyy]

What I don't understand is where the paradox sets in.

To me it does not seem like a paradox that one clock has passed two years and the moving clock which now returns home has only passed 1.2 years.

The other thing I want to understand is what is happening in the FOR of the moving clock.

How do I use the lorentz transformation to figure this out?

There is, as you say, no paradox.

The time elapsed on each clock is indpendent of the frame of reference, ie. it is the same in every frame of reference.

However, the time dilation formula only works in a Lorentz inertial frame (ie. the frame of a non-accelerating observer). An accelerating observer (such as the spaceship twin who undergoes two acelerations) has to use a different formula. A paradox only arises if one erroneously uses the time dilation formula for the accelerating observer.

You have correctly used the time dilation formula for both clocks (non-accelerating and accelerating) described from a Lorentz inertial frame.

 

[goodabouthood]

ghwellsjr,

I was under the impression that each reference frame has its own time. You are making it confusing for me.

Let me give you an example.

I am the stationary observer and I see the spaceship moving at .8c.

Now one year passes for me so that would be the clock in my reference frame correct?

Isn't it correct to say that .6 of a year elapses for the moving ship or in other words for that FOR relative to mine?

How is it possible to say that a frame of reference can have multiple clocks? That does not make sense. I say in my FOR one year has passed and in the moving ships FOR relative to me .6 of a year has passed.

It would be strange to say that the moving ship is part of my FOR and that his clock has moved .6 and mine has moved one year. I need to say in my stationary FOR one year has passed and the moving FOR has elapsed .6 of a year relative to me.

 

[goodabouthood]

There is, as you say, no paradox.

The time elapsed on each clock is indpendent of the frame of reference, ie. it is the same in every frame of reference.

However, the time dilation formula only works in a Lorentz inertial frame (ie. the frame of a non-accelerating observer). An accelerating observer (such as the spaceship twin who undergoes two acelerations) has to use a different formula. A paradox only arises if one erroneously uses the time dilation formula for the accelerating observer.

You have correctly used the time dilation formula for both clocks (non-accelerating and accelerating) described from a Lorentz inertial frame.

Alright,

So how do we flip this around?

We have to say there are two synchronized clocks on earth. One leaves Earth on a spaceship and sees 1 year on its clock and it sees .6 of a year on the Earth clock. Now he returns home and his clock says 2 years has passed but only 1.2 has passed on Earth.

Now I'm getting confused.

 

[HallsofIvy]

Alright,

So how do we flip this around?

We have to say there are two synchronized clocks on earth. One leaves Earth on a spaceship and sees 1 year on its clock and it sees .6 of a year on the Earth clock. Now he returns home and his clock says 2 years has passed but only 1.2 has passed on Earth.

Now I'm getting confused.

No- the situation is not symmertric. In this scenario, the clock was stationary on earth, then had to accelerate to .8c, then had to accelerate to turn around, then had to decellerate to rest on Earth again. There was no acceleration or deceleration for the clock that stayed on earth.

 

[Dale]

So how do we flip this around?

We have to say there are two synchronized clocks on earth. One leaves Earth on a spaceship and sees 1 year on its clock and it sees .6 of a year on the Earth clock. Now he returns home and his clock says 2 years has passed but only 1.2 has passed on Earth.

You cannot just flip it around. As has been pointed out multiple times the traveling twin is not at rest in any inertial frame during the whole journey. The traveling twin is a non-inerital observer. If you want to use the standard formulas you can only apply them in inertial frames. However, you can look at different inertial frames.

For instance, consider the inertial frame where the traveller is at rest during the first half of the journey. In that frame, the Earth travels at .8c for 5/3 2 years = 3.33 years until they are reunited. The traveling twin starts at rest for .6 years during which time he experiences no time dilation. Then, to catch up with the Earth twin he must move at .98c experiencing a time dilation factor of 41/9 for 2.73 years = .6 years. So, in that inertial frame you also wind up with 2 years on the Earth clock and 1.2 years on the traveling clock at the reunion.

 

[goodabouthood]

Well, is there really such things as inertial frames in reality?

Because even in both my examples one clock still needs to leave the Earth so it had to make an acceleration at some point.

 

[Dale]

Because even in both my examples one clock still needs to leave the Earth so it had to make an acceleration at some point.

There is nothing wrong with acceleration. You can use all of the standard formulas in an inertial frame regardless of the non-inertial acceleration of the various objects wrt that frame. The important thing is that the reference frame itself be inertial. As long as your reference frame is inertial you can have any non-inertial objects that you like.

Does that distinction between inertial frames and non-inertial objects make sense? Remember, a reference frame is basically just a coordinate system.

 

[Snip3r]

No- the situation is not symmertric. In this scenario, the clock was stationary on earth, then had to accelerate to .8c, then had to accelerate to turn around, then had to decellerate to rest on Earth again. There was no acceleration or deceleration for the clock that stayed on earth.

There is no symmetry even if you don't know the word "acceleration". Or even if the ship blames the Earth as accelerating or decelerating

 

[ghwellsjr]

ghwellsjr,

I was under the impression that each reference frame has its own time. You are making it confusing for me.

Let me give you an example.

I am the stationary observer and I see the spaceship moving at .8c.

Now one year passes for me so that would be the clock in my reference frame correct?

Isn't it correct to say that .6 of a year elapses for the moving ship or in other words for that FOR relative to mine?

How is it possible to say that a frame of reference can have multiple clocks? That does not make sense. I say in my FOR one year has passed and in the moving ships FOR relative to me .6 of a year has passed.

It would be strange to say that the moving ship is part of my FOR and that his clock has moved .6 and mine has moved one year. I need to say in my stationary FOR one year has passed and the moving FOR has elapsed .6 of a year relative to me.

Sorry to be confusing you, I will try to make it clear.

First, I recommend you study How did Einstein Define Time, especially towards the end where I explain the difference between Proper Time and Coordinate Time.

But I'll give you a synopsis here. A Frame of Reference in Special Relativity is defined as a coordinate system with the normal kind of x, y, and z components to specify a 3-dimensional distance for any point from a predefined origin and with synchronized clocks at every location which keep what is called Coordinate Time. These are normal clocks but they are not allowed to accelerate, that is, they must remain at the locations where they were synchronized to the clock at the origin. When they are synchronized, they all read zero. So we now add a fourth component to the coordinate system which is the time. The four components making up the location and time are called an "event", which doesn't have to imply that any thing actually happened there at that time, just like we can talk about the location of a point in a normal coordinate system, even if there isn't anything there. (Actually, we just imagine that these clocks exist at every location just as we imagine that there are grids of measuring rulers laid out in all directions from the origin.)

Now if we want to describe a scenario in this Frame of Reference, we specify the events for each real clock that we want to consider. Since these clocks are allowed to change locations and therefore experience acceleration and therefore experience speed and therefore experience time dilation with respect to the Coordinate Times on the clocks that are fixed to the Coordinate Locations, we have a different term to refer to the times on these real clocks which is called Proper Time. We can determine how fast any clock is ticking by knowing its speed in the Frame of Reference and applying the reciprocal of the Lorentz Factor to it. Adding up all the ticks allows us to keep track of the time after any period of Coordinate Time we want to consider.

So let's see how this works for the scenario you described in your first post. You started with two synchronized clocks on Earth. Let's assume that they are located at the origin so that the x, y, z and the time components are all zero and that the Proper Time on these two clocks is also zero. I like to use the nomenclature of [t,x,y,z] to specify events and since y and z are always zero in a simple scenario, I like to omit them and use [t,x] to specify each event. Remember that the "t" in this event is the Coordinate Time on the clock that is located at x=0, even though we have two more real clocks at this location at the start of the scenario, the two clocks you specified.

So let's refer to the "stationary" clock as clock S and the "traveling" clock as clock T. At the start of your scenario, both clocks are specified by the event [0,0] and they both read 0. Then T accelerates to a speed of 0.8c (instantly, to keep things simple) for one year. Where will it be after one year? That's easy, it will be at location x=0.8 (we are using units of time in years and distance in light years.) The corresponding event is [1,0.8] for T and for S it is [1,0]. Now the first question is what time will be displayed on these two clocks? Well T has been at a speed of 0.8c so its time will be dilated by a factor of 0.6 (you already know how to do this calculation so I won't repeat it). So keep in mind that there are two times related to clock T at the point of turn-around, one is the Coordinate Time of the event of the turn-around which is 1 year and the other is the Proper Time displayed on the clock which is 0.6 years. Meanwhile, the Proper Time on clock S is 1 year and the Coordinate Time for the corresponding event is also 1 year.

Now clock T returns and it takes another year of Coordinate time for it to get back. The event describing its reuniting with clock S is [2,0]. This event applies to both clocks but the Proper Time on clock T has advanced by another 0.6 years so it now reads 1.2 year while clock S has advanced by 1 year so that its Proper Time reads 2 years, again the same as the Coordinate Time, since it never moved.

Now I know that you know all the details with regard to the Proper Times because you described them in your first post but you may not have been aware of the concept of Coordinate Time or how they applied to events.

Then you asked about how to view things from the Frame of Reference of clock T. As I pointed out in post #8, clock T is not stationary in any Frame of Reference so you cannot use the Lorentz Transform to answer your question. It's actually stationary in three frames, the one it shares with clock S at the beginning and ending of the scenario, the one it is stationary in during the outbound portion of the trip and the one it is stationary in during the inbound portion of the trip.

So let's start by transforming three of the four events we have already specified from the first FoR into the FoR in which clock T is stationary during the first half of its trip. I'm not going to bother to transform the event for clock S corresponding to the time of the turn around for clock T because it has no bearing on anything.

Here at the three events in FoR 1:
[0,0] start of scenario for both clocks
[1,0.8] turn around event for clock T
[2,0] end of scenario for both clocks

The first event [0,0] will remain the same no matter what other FoR we transform to so I won't go through its calculation.

For the other events, the first thing we need to do to use the Lorentz Transform is calculate gamma, γ, where the speed as a fraction of c, is β = 0.8 according to:
γ = 1/√(1-β²)
γ = 1/√(1-0.8²)
γ = 1/√(1-0.64)
γ = 1/√(0.36)
γ = 1/0.6
γ = 1.667

The formulas for the Lorentz Transform are:
t' = γ(t-βx)
x' = γ(x-βt)

For the outbound portion of the trip, β=0.8, and the event of the traveling clock at the end is [1,0.8]:

t = 1
x = 0.8

t' = γ(t-βx)
t' = 1.667(1-(0.8*0.8))
t' = 1.667(1-0.64)
t' = 1.667(0.36)
t' = 0.6

x' = γ(x-βt)
x' = 1.667(0.8-(0.8*1)
x' = 1.667(0.8-0.8)
x' = 1.667(0)
x' = 0

So the turn-around event for clock T is:
[0.6,0]

We can calculate its speed by taking the difference in its starting and ending locations divided by the difference in its starting and ending times. But the starting and ending locations are both 0 so it is stationary in this FoR. So all we have to do to see how much time has advanced on clock T during the first leg of the trip is take the difference in the Coordinate Times for these two events which is 0.6-0 or 0.6 years.

Now let's jump to the last event [2,0].

t = 2
x = 0

t' = γ(t-βx)
t' = 1.667(2-(0.8*0))
t' = 1.667(2-0)
t' = 1.667(2)
t' = 3.333

x' = γ(x-βt)
x' = 1.667(0-(0.8*2)
x' = 1.667(0-1.6)
x' = 1.667(-1.6)
x' = -2.667

So the ending event for both clocks is:
[3.333,-2.667]

Now in order to calculate the speed of clock T during the inbound portion of the trip we have to do the difference thing I mentioned earlier. The coordinate location difference for clock T between the start and end of his inbound trip is -2.667-0 = -2.667 light years. The corresponding coordinate time difference for clock T is 3.333-0.6 = 2.733 years. Dividing these we get a speed for clock T of 0.9756c. Now we apply the reciprocal of the Lorentz Factor on this speed to see what time dilation clock T experiences:

1/γ = √(1-β²)
1/γ = √(1-0.9756²)
1/γ = √(1-0.9518)
1/γ = √(0.0482)
1/γ = 0.2159

Now we multiply this factor by the Coordinate Time difference of 2.733 years to get 0.6 years as the Proper Time difference for the clock T during the inbound portion of the trip (just like we got in the first FoR). Finally, we need to add the Proper Time at the end of the outbound portion of the trip to the additional time accumulated during the inbound portion of the trip to get the final Proper Time on clock T as 1.2 years.

Now let's see how much clock S has advanced in this FoR. We need to do the difference thing again but for this clock and we will do it from the beginning of the scenario to the end. The location difference is -2.667-0 or -2.667 light years. The time difference is 3.333-0 or 3.333 years. The speed is 2.667/3.333 or 0.8c so the reciprocal of the Lorentz Factor is 0.6. So the Proper Time on clock S has advanced by 0.6 times 3.333 or 2 years.

So as you can see, even in the FoR for the outbound portion of the traveling's clock trip, we still calculate the Proper Times for both clocks to be the same as they were in the original FoR. In fact, it won't matter what FoR we use, we will always get the same answers. If you want to do the transform for the inbound portion of the trip, you will see that it is similar to the outbound and ends up with the same Proper Times.

 

[goodabouthood]

Isn't coordinate time pretty much equal to what my clock says in my FOR?

Let's say there is a train in my coordinate system that is going at .5c and I want to know how much time has passed for this train after 5 years in my own coordinate system. Am I correct in saying that 5 years is the coordinate time?

Now is Proper time equal to the amount of the time that has passed for the train moving at .5c after 5 years of my coordinate time?

So the Proper Time for the train moving at .5c relative to my FOR is 4.33 years. Correct?

To figure this out I just took my Coordinate Time/The Lorentz Factor of .5c.

I still have not figured out what the Lorentz Factor exactly is. According to the wikipedia it says it's the change in coordinate time/change in proper time.

It's saying the change in proper time is equal to the square root of 1 - Bsquared. I don't really understand what that means. It is also saying that equals to the square root of c2-v2. Not really sure what that means. The speed of light - the velocity something is going = proper time?

 

[atyy]

Coordinate time is not clock time.

Clock time is "proper time", and is the same in every reference frame. It is a property of the spacetime trajectory of a clock.

Coordinate time is the "time" coordinate assigned when "space" and "time" coordinates are assigned within one reference frame to all conceivable events in spacetime.

Proper time coincides with the coordinate time of a Lorentz inertial frame for a clock that is stationary with respect to that Lorentz inertial frame.

For clocks that are moving relative to a Lorentz inertial frame, the proper time can be calculated from the coordinate time of that Lorentz inertial frame and the "Lorentz factor".

 

[ghwellsjr]

Isn't coordinate time pretty much equal to what my clock says in my FOR?

We're talking about inertial Frames of Reference which means they don't accelerate which means they don't change their speed or direction. So if you remain stationary in what you call "my FOR", then yes, as long as your clock was synchronized to the previously synchronized coordinate clocks defining your FOR, and you also never accelerate, then the Proper Time on your clock will remain synchronized to the Coordinate Time in your inertial FOR. But the "clocks" that we are talking about that keep Coordinate Time are usually "imaginary" clocks that would behave exactly like real clocks at each location that we want to consider. This is why atyy said, "Coordinate time is not clock time." We use the term "Proper Time" to refer to the time on your real clock in your FOR even if it never accelerates and so it keeps the same time as the Coordinate Time in your FOR.

Let's say there is a train in my coordinate system that is going at .5c and I want to know how much time has passed for this train after 5 years in my own coordinate system. Am I correct in saying that 5 years is the coordinate time?

Yes.

Now is Proper time equal to the amount of the time that has passed for the train moving at .5c after 5 years of my coordinate time?

Yes.

So the Proper Time for the train moving at .5c relative to my FOR is 4.33 years. Correct?

Yes.

To figure this out I just took my Coordinate Time/The Lorentz Factor of .5c.

Yes, the Lorentz Factor of .5c is 1.1547 so it looks like you did this correctly.

I still have not figured out what the Lorentz Factor exactly is. According to the wikipedia it says it's the change in coordinate time/change in proper time.

That is correct and you did everything correctly both in this example and in your first post so I don't understand why you say you have not figured it out.

It's saying the change in proper time is equal to the square root of 1 - Bsquared. I don't really understand what that means. It is also saying that equals to the square root of c2-v2. Not really sure what that means. The speed of light - the velocity something is going = proper time?

You've left some things out here. The Lorentz Factor, gamma, is given by:

γ = 1/√(1-β²)

And beta, β, is the ratio of the speed that the clock is traveling at divided by the speed of light:

β = v/c

So the other way to express gamma is:

γ = c/√(c²-v²)

But I like to use the first formula for gamma since it uses speeds as a ratio of the speed of light which is what you used also.

You said the Lorentz Factor is equal to the change in coordinate time/change in proper time. So that means the change in proper time is equal to the change in coordinate time divided by the Lorentz Factor, correct? So we could express this as:

Δτ = Δt/γ
Δτ = Δt * √(1-β²)

where Δτ (delta tau) is the change in proper time and Δt is the change in coordinate time.

So given a speed as a factor of the speed of light as in your example of β=.5, did you actually do this calculation for the Lorentz Factor?

γ = 1/√(1-β²)
γ = 1/√(1-0.5²)
γ = 1/√(1-0.25)
γ = 1/√(0.75)
γ = 1/0.866
γ = 1.1547

Now you said there was a 5-year change in the coordinate time so the change in the proper time for the train is equal to 5/1.1547 which equals 4.33 years.

Now since you got the correct answer, how did you get it if you didn't understand how to do the calculations? Where are you still confused? It looks to me like you have perfect understanding (except for the incomplete quotes from wikipedia).

Also, have you studied the calculations I gave you for doing the Lorentz Transform in my previous post? Does it all make sense? Did I answer your question about how to do the Lorentz Transform?

 

[goodabouthood]

Anybody can plug numbers into a calculation. This has nothing to do with understanding what they actually mean.

 

[goodabouthood]

Just to clarify.

I am the stationary observer. There is a moving ship going at .8c relative to me.

I experience one year on my clock. This 1 year is equal to my coordinate time and proper time, correct?

Now the Coordinate time for the moving ship would be equal to 1 year as well, correct?

The proper time would be equal to .6 years, correct? Is that what I call t'?

Would this be symmetrical? I read .6 years on his clock but he reads 1 year on his clock and .6 on mine?

 

[ghwellsjr]

Just to clarify.

I am the stationary observer. There is a moving ship going at .8c relative to me.

I experience one year on my clock. This 1 year is equal to my coordinate time and proper time, correct?

In your rest frame, yes, 1 year of coordinate time is equal to 1 year of proper time. The formula to use is not the Lorentz Transform but the one I mentioned in my previous post:

Δτ = Δt * √(1-β²)

where Δτ (delta tau) is the change in proper time and Δt is the change in coordinate time.

So because your speed in your FoR is 0, β=0 and Δτ = Δt.

Now the Coordinate time for the moving ship would be equal to 1 year as well, correct?

If you are talking about the same interval in your FoR, then yes, the coordinate time for the ship and yourself and anything else you want to consider throughout your FoR is the same 1 year.

The proper time would be equal to .6 years, correct?

Yes, since the ship has a speed of β=0.8, then applying the same formula for Proper Time:

Δτ = Δt * √(1-β²)
Δτ = 1 * √(1-0.8²)
Δτ = 1 * √(1-0.64)
Δτ = 1 * √0.36
Δτ = 1 * 0.6
Δτ = 0.6 years

Is that what I call t'?

No, t' is used to specify the time component of an event in a FoR moving with respect to another FoR. I have been assuming that you have only been referring to your FoR so we haven't applied the Lorentz Transform.

Would this be symmetrical? I read .6 years on his clock but he reads 1 year on his clock and .6 on mine?

If you're going to specify just your FoR, which I have been assuming, then during the one year of coordinate time, your clock advances by one year and his clock advances by .6 years which means the time he will read on his clock--that's what we have been calculating.

Now if you want to know how much time he reads on your clock during the same time interval that you read 1 year on your clock and .6 years on his clock, it would be .36 years. If you want to talk about a different time interval of 1 year on his clock, then, yes, he will read .6 years on your clock. For both of these situations we would switch to the ship's FoR and specify whatever time interval we want to consider, in the first instance, a proper time and a coordinate time of .6 years and we would calculate a proper time for you of .36 years using the same formula above. For a proper time and coordinate time of 1 year for the ship, the proper time for you would be .6 years.

Keep in mind that we are not using the Lorentz Transform to do these calculations where t' is applied.

 

[goodabouthood]

http://www.trell.org/div/minkowski.html

Using this interactive minkowski diagram I plugged in a relative velocity of .8c, and for event B I plugged in a (1,.8). This means in my reference frame one year has passed and event B is at location of .8light years.

It tells me that t' = .6 and x' = 0. This would mean in his reference frame he is only at .6 years and he is at his origin.

Now when I plug in 2,0 as in for the time the ship comes back to Earth I get t' = 3.33 and x' = -2.67.

So what exactly is this saying? It's saying that his time is 3.33 when he is at location -2.67. Surely I don't think this is right. I had thought he only measured 1.2 years. This must not be indicating the twin paradox correctly.

Now I plug in for the stationary frame (2,1.6) and I get t' =1.2 and x' =0.

Explain to me the differences. Why do I plug in the total distance he traveled relative to me even though he is now back at location 0 for me?

 

[ghwellsjr]

http://www.trell.org/div/minkowski.html

Using this interactive minkowski diagram I plugged in a relative velocity of .8c, and for event B I plugged in a (1,.8). This means in my reference frame one year has passed and event B is at location of .8light years.

It tells me that t' = .6 and x' = 0. This would mean in his reference frame he is only at .6 years and he is at his origin.

Now when I plug in 2,0 as in for the time the ship comes back to Earth I get t' = 3.33 and x' = -2.67.

So what exactly is this saying? It's saying that his time is 3.33 when he is at location -2.67. Surely I don't think this is right. I had thought he only measured 1.2 years. This must not be indicating the twin paradox correctly.

I explained all this in post #17. Please study that post and see if you still have questions.

Now I plug in for the stationary frame (2,1.6) and I get t' =1.2 and x' =0.

Explain to me the differences. Why do I plug in the total distance he traveled relative to me even though he is now back at location 0 for me?

What you are doing here gets the right answer of 1.2 years for the total time on the traveling clock but it's a solution to the wrong problem. This is the answer to how much time would have progressed on the traveling clock if it had continued on in a straight line for two years instead of turning around after one year and returning to its start location.

 

[darkhorror]

I think you are missing what people are having a problem with. Let's take the Earth and a ship moving away from Earth at .8c.

from the Earth's FOR the Earth is stationary and the ship is moving at .8c. also the ship's clocks are ticking slower than the earth's.

from the the ship's FOR the ship is stationary and the Earth is moving at .8c. also the Earth's clocks are ticking slower than the ships.

 

[goodabouthood]

I think you are missing what people are having a problem with. Let's take the Earth and a ship moving away from Earth at .8c.

from the Earth's FOR the Earth is stationary and the ship is moving at .8c. also the ship's clocks are ticking slower than the earth's.

from the the ship's FOR the ship is stationary and the Earth is moving at .8c. also the Earth's clocks are ticking slower than the ships.

I get this.

So doesn't that mean if I am in the FOR of the Earth and 1 year passes by for me while .6 years passes by for the ship?

What I want to know is what is going on in the FOR of the ship? Does he see his clock at one year and my clock at .6?

 

[ghwellsjr]

So doesn't that mean if I am in the FOR of the Earth and 1 year passes by for me while .6 years passes by for the ship?

In the FOR of the earth, you are at rest and the ship is traveling at 0.8c. For the first year of coordinate time in the Earth's FOR, the time on your clock advances by 1 year and the time on the ship's clock advances by 0.6 years. For the second year of coordinate time in the Earth's FOR, the ship has turned around but continues to travel at 0.8c and so it advances by another 0.6 years for a total of 1.2 years while your clock advances by another year for a total of 2 years.

What I want to know is what is going on in the FOR of the ship?

There isn't just one FOR for the ship. There is one FOR for when the ship is traveling away from the Earth and then there is another FOR for when the ship is traveling back toward the earth. And the ship is at rest in each of these FORs for only 0.6 years each so it doesn't make sense to ask about what the ship would see at one year, does it?

Does he see his clock at one year and my clock at .6?

If you want to change the problem and say that the ship continues on in the same direction for a longer period of time, then your question is legitimate and would be answered identically to your first question about the Earth's FOR. In other words, in the FOR of the ship, the ship's clock is at rest and you are traveling at 0.8c. For one year of coordinate time in the ship's FOR, the time on the ship's clock advances by 1 year and the time on your clock advances by 0.6 years.

 

[goodabouthood]

So it's only on the turn around, or when the clock enters a new FOR, that the symmetry breaks?

 

[ghwellsjr]

So it's only on the turn around, or when the clock enters a new FOR, that the symmetry breaks?

At turn around, the ship's clock exits one symmetry and enters a new symmetry, and they are the same kind of symmetry, you just have to be careful how you analyze it.

 

[goodabouthood]

Let's say I am the stationary observer and there is spaceship moving at .8c relative to me.

I see one year on my clock and I see .6 years on his clock. What time does he see on his own clock?

I want to say he sees 1 year on his clock and .6 on mine, but I'm not sure.

 

[alexg]

In either inertial frame, the spaceship twin or the Earth twin, it is valid for each to consider themselves at rest and the other as moving.

However, in order to compare the clocks, one, or the other, or both frameworks must undergo acceleration to bring them into a common frame. It is this acceleration which differentiates one inertial frame from the other.

 

[ghwellsjr]

Let's say I am the stationary observer and there is spaceship moving at .8c relative to me.

I see one year on my clock and I see .6 years on his clock. What time does he see on his own clock?

I want to say he sees 1 year on his clock and .6 on mine, but I'm not sure.

If you are changing your scenario so that the spaceship continues on in the same direction and doesn't turn around, then yes, in your FOR, when 1 year passes for you, .6 years passes for the ship and in the ship's FOR, when 1 year passes for it, .6 years passes for you.

But you should be aware that neither of you can actually see the others clock as you are asking about. When you see 1 year pass on your own clock, you will actually see 4 months (1/3 year) year pass on the ship's clock and in the same way when the ship see's 1 year pass on its own clock, it will see 4 months (1/3 year) pass on your clock. This is called the Relativistic Doppler effect and is a result of the time dilation of .6 years plus the time it takes for the image of the ship's clock to propagate across space to your clock and vice versa.

 

[goodabouthood]

so all moving bodies that are inertial are symmetrical to whatever inertial frame you choose, correct?

 

[ghwellsjr]

No, they are symmetrical to each other but you can use any inertial frame you choose to define, analyze and demonstrate what is going on.