# A question about stationary reference frame

Tags: frame, reference, stationary
 Sci Advisor P: 7,398 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".
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P: 4,153
 Quote by goodabouthood 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.
 Quote by goodabouthood 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.
 Quote by goodabouthood 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.
 Quote by goodabouthood So the Proper Time for the train moving at .5c relative to my FOR is 4.33 years. Correct?
Yes.
 Quote by goodabouthood 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.
 Quote by goodabouthood 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.
 Quote by goodabouthood 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?
 P: 127 Anybody can plug numbers into a calculation. This has nothing to do with understanding what they actually mean.
 P: 127 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?
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P: 4,153
 Quote by 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?
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.

 Quote by goodabouthood 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.
 Quote by goodabouthood 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

 Quote by goodabouthood 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.
 Quote by goodabouthood 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.
 P: 127 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?
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P: 4,153
 Quote by 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.
I explained all this in post #17. Please study that post and see if you still have questions.
 Quote by goodabouthood 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.
 P: 137 I think you are missing what people are having a problem with. Lets 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.
P: 127
 Quote by darkhorror I think you are missing what people are having a problem with. Lets 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?
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P: 4,153
 Quote by goodabouthood 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.
 Quote by goodabouthood 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?
 Quote by goodabouthood 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.
 P: 127 So it's only on the turn around, or when the clock enters a new FOR, that the symmetry breaks?
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P: 4,153
 Quote by goodabouthood 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.
 P: 127 Let's say I am the stationary observer and there is space ship 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.
 P: 126 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.
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P: 4,153
 Quote by goodabouthood Let's say I am the stationary observer and there is space ship 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 space ship 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.
 P: 127 so all moving bodies that are inertial are symmetrical to whatever inertial frame you choose, correct?
 PF Patron P: 4,153 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.

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