Question about the x term in time dilation

[Platformance]

Do you realize that this equation is the opposite of the similar one that you presented in your first post? It's always confusing when two forms of t are used, one primed and the other unprimed. There is a better, clear and common way to present the equation and that is using the Greek letter tau, τ, to represent the Proper Time leaving t (and t') to always represent Coordinate Time. You can state it as t = γτ or its incremental version, Δt = γΔτ, which is usually more useful because if you are drawing a diagram for an observer/object/clock traveling at some speed, you can easily calculate γ and then you can calculate how far in Coordinate Time to place a dot representing some increment of Proper Time. This is what my computer program does when I specify an original IRF.

I see, I wasn't quite sure of how the time dilation equation worked when I started this thread, but I understand it more clearly now.

Good, I'm glad it was just a caculation mistake. (That's one advantage of using a computer program.) It appears that you are getting a pretty good grasp on these ideas. Since you're such a good learner, I'd like to continue on with some more ideas as time permits.

Thank you! The graphs and explanation made this concept much easier to understand, and I didn't expect to learn the relativistic doppler when I started this thread. I am still learning relativity (currently on the spacetime interval) and I am sure more ideas will appear.

 

[yuiop]

So the 0.866s represents a fraction of distance between the points for the black line and not the total distance, but the black dots are a distance 1.1547 away. Does this mean that when the planet's clock tick 1 second, the observer sees an event, but when the clock ticks 1.1547 seconds, the spaceship sees an event?

They are two different clocks. It means that both the planet observer and the spaceship observer see 1 second indicated on the planet clock and 0.866s on the spaceship clock at the same event if they are both momentarily alongside the event and each other, but their clocks read differently because they are ticking at different rates in any given reference frame. At the later event, if both observers are at the same event alongside each other, then both observers see 1.1547 seconds on the planet clock and 1 second on the spaceship clock. Note, that in order to do this we need an array of observers with synchronised clocks on the planet, so that we can conveniently have observers local to the events when they occur.

By, the way, you are getting your units mixed up. 0.866s is in units of time and you are using it to mean distance. Earlier you said x = 0.5c which is a velocity, but x is a distance. You need to be careful about units to avoid confusion.

I thought the period of time dilation changed as one frame moved away from another frame.

This is not correct. When two clocks are moving with constant sped relative to each other, their distance apart is continually changing, but the time dilation remains constant. Earlier ghwellsjr gave a worked example for a clock that started at t=0 and x=0, and ended at t=1 and x=0.5 and calculated that the time on the spaceship clock would be t' = 0.866s when t=1:

... In the mean time, this is how I would do the calculation based on the correct Lorentz Transformation equation for time:

t' = γ(t - xv/c²)
where γ = 1/√(1-v²/c²)

At v = 0.5c, γ = 1/√(1-(0.5c)²/c²) = 1/√(1-0.25) = 1/√(0.75) = 1/0.866 = 1.1547

t' = γ(t - xv/c²) = 1.1547(1 - (0.5c)(0.5c)/c²) = 1.1547(1 - 0.25) = 1.1547(0.75) = 0.866

...

Now let's see what happens if a clock in the spaceship was initially at x=1 at time t=0. This spaceship clock would initially be reading: t'₀ = γ(0s-(1ls * 0.5c)/c^2 ) = -0.57735s

This is using the convention that the origins of the frames coincided at x=0, t=0, x'=0 and t'=0.

Now after 1 second as measured in the planet ref frame (S) the spaceship clock will be at x=1.5 light seconds, so we can calculate the time indicated in the spaceship ref frame (S') as t'₁ = 1.1547(1s-(1.5ls * 0.5c)/c^2 ) = 0.28867s for this event.

The elapsed time between these two events is 1 second in frame S and Δt' = t'₁ - t'₀ = 0.28867 - (-0.57735) = 0.866 seconds in frame S'.

You can now see that the time dilation factor is still the same (0.866 seconds in S' for every second in S or a time dilation factor of 1/0.866 = 1.1547) despite the clock being further away.

Note that two calculations done above to calculate the coordinates of the events and then subtracting them to get the difference interval can be done a single calculation as:

Δt' = γ(Δt - Δx * v/c²) = 1.1547( 1 - 0.5*0.5) = 0.866s.

While the Lorentz transformation can be used to calculate transformed intervals and instantaneous coordinates, the time dilation formula can only be used with intervals. In other words the time dilation formula can tell you that one clock is running slower than another by a factor of 1.1547 but it cannot tell you that when a clock in S is reading 3PM that a clock in S' is reading 2PM, whereas the Lorentz transformation can.

 

[ghwellsjr]

They are two different clocks. It means that both the planet observer and the spaceship observer see 1 second indicated on the planet clock and 0.866s on the spaceship clock at the same event if they are both momentarily alongside the event and each other, but their clocks read differently because they are ticking at different rates in any given reference frame.

I understand what you are trying to communicate in the above sentence but since you redefined standard relativity terms without stating what you are doing, then it can only lead to confusion for someone who is trying to learn relativity.

First, you are using the word "see" differently than I have been using it in this thread. "See" means what you observe with your eyes and that's not what you mean here. Could you please define what you mean by "see"?

Second, you are using the word "event" differently than I have ever heard anyone use it in a discussion about relativity. Could you please define what you mean by "event"?

At the later event, if both observers are at the same event alongside each other, then both observers see 1.1547 seconds on the planet clock and 1 second on the spaceship clock. Note, that in order to do this we need an array of observers with synchronised clocks on the planet, so that we can conveniently have observers local to the events when they occur.

I don't see why we need this array of observers. Maybe you should explain exactly where they are located and exactly what they see and exactly when they see it.

By, the way, you are getting your units mixed up. 0.866s is in units of time and you are using it to mean distance. Earlier you said x = 0.5c which is a velocity, but x is a distance. You need to be careful about units to avoid confusion.

No, he's not getting the units mixed up. If you read carefully, you will see that he was referring to the distance on the diagram between two dots on the thick black line so that he could apply the correct ratio to establish where 0.866 seconds would be.

...

Now let's see what happens if a clock in the spaceship was initially at x=1 at time t=0. This spaceship clock would initially be reading: t'₀ = γ(0s-(1ls * 0.5c)/c^2 ) = -0.57735s

This is using the convention that the origins of the frames coincided at x=0, t=0, x'=0 and t'=0.

Usually, when we set the origins of two frames to coincide, we also synchronize the two clocks to zero at that event. I don't see where you synchronized the two clocks in your new example. Until you do that, I don't see how you can just say the spaceship clock had some reading on it at some location distant from the planet clock.

Now after 1 second as measured in the planet ref frame (S) the spaceship clock will be at x=1.5 light seconds, so we can calculate the time indicated in the spaceship ref frame (S') as t'₁ = 1.1547(1s-(1.5ls * 0.5c)/c^2 ) = 0.28867s for this event.

The elapsed time between these two events is 1 second in frame S and Δt' = t'₁ - t'₀ = 0.28867 - (-0.57735) = 0.866 seconds in frame S'.

You can now see that the time dilation factor is still the same (0.866 seconds in S' for every second in S or a time dilation factor of 1/0.866 = 1.1547) despite the clock being further away.

Note that two calculations done above to calculate the coordinates of the events and then subtracting them to get the difference interval can be done a single calculation as:

Δt' = γ(Δt - Δx * v/c²) = 1.1547( 1 - 0.5*0.5) = 0.866s.

While the Lorentz transformation can be used to calculate transformed intervals and instantaneous coordinates, the time dilation formula can only be used with intervals. In other words the time dilation formula can tell you that one clock is running slower than another by a factor of 1.1547 but it cannot tell you that when a clock in S is reading 3PM that a clock in S' is reading 2PM, whereas the Lorentz transformation can.

In the scenario described in this thread, since the clocks were synchronized to zero at the origins of the two rest frames, we can use the time dilation formula to show the relationship between the Proper Time on either clock to its Coordinate Time. I made it very clear in my descriptions that the calculations applied only to this scenario. In my last post I made the point that the incremental version of the time dilation formula is usually more useful because it can be applied in any scenario, not just the one that we have been discussing so far.

 

[yuiop]

Usually, when we set the origins of two frames to coincide, we also synchronize the two clocks to zero at that event. I don't see where you synchronized the two clocks in your new example. Until you do that, I don't see how you can just say the spaceship clock had some reading on it at some location distant from the planet clock.

I did set the two clocks at the origins of the two frames to be zero at the time the origins coincide. It is contained in the statement "This is using the convention that the origins of the frames coincided at x=0, t=0, x'=0 and t'=0.". Although I did not state it, it is usually assumed that all clocks that are at rest in a given inertial reference frame that are used to determine time coordinates in that irf, are synchronised with each other. Given that information, we can determine the time t' and location x' in S' for any event t and x in S.

Second, you are using the word "event" differently than I have ever heard anyone use it in a discussion about relativity. Could you please define what you mean by "event"?

I am not sure what you find so unusual in my use of event. Could you elaborate? As far as I am concerned an event is uniquely defined as a time coordinate and three spatial coordinates in a given inertial reference frame. In a different inertial reference frame the coordinates of the same event are different from the first irf, but but they are related by the Lorentz transformations.

I don't see why we need this array of observers. Maybe you should explain exactly where they are located and exactly what they see and exactly when they see it.

You do not "need" an array of observers, but it is convenient that if there happens to be an observer adjacent to the event that is at rest in a given irf and that has a clock synchronised with a clock at the origin of the irf, then whatever the reading on that observers clock is exactly what is predicted by the Lorentz transformations without taking light travel times into account.

First, you are using the word "see" differently than I have been using it in this thread. "See" means what you observe with your eyes and that's not what you mean here. Could you please define what you mean by "see"?

Given that I am using observers that are local to the events that are being analysed, there are no issues with light travel times so "see" and measured can be used interchangeably when we are talking about local time or spatial coordinates. Yes, there could be an issue with Doppler effects but that was not originally part of the original thread.

 

[ghwellsjr]

Usually, when we set the origins of two frames to coincide, we also synchronize the two clocks to zero at that event. I don't see where you synchronized the two clocks in your new example. Until you do that, I don't see how you can just say the spaceship clock had some reading on it at some location distant from the planet clock.

I did set the two clocks at the origins of the two frames to be zero at the time the origins coincide. It is contained in the statement "This is using the convention that the origins of the frames coincided at x=0, t=0, x'=0 and t'=0.". Although I did not state it, it is usually assumed that all clocks that are at rest in a given inertial reference frame that are used to determine time coordinates in that irf, are synchronised with each other. Given that information, we can determine the time t' and location x' in S' for any event t and x in S.

I did not know about this assumption. I didn't even know that clocks are used to determine time coordinates. I had no idea that when the OP set up this scenario in post #4 with two clocks, one on a planet and one on a spaceship passing by at 0.5c that those two clocks were being used to determine time coordinates. I thought coordinates were independent of any observers or clocks that are described by those coordinates. That's why I went into great detail in post #7 to specify all these things. I didn't know I could have just assumed them.

Can you point me to an authoritative online reference that supports your assumption?

Second, you are using the word "event" differently than I have ever heard anyone use it in a discussion about relativity. Could you please define what you mean by "event"?

I am not sure what you find so unusual in my use of event. Could you elaborate? As far as I am concerned an event is uniquely defined as a time coordinate and three spatial coordinates in a given inertial reference frame. In a different inertial reference frame the coordinates of the same event are different from the first irf, but but they are related by the Lorentz transformations.

Yes, that is a good definition of "event".

Here's what you said about the first "event":

It means that both the planet observer and the spaceship observer see 1 second indicated on the planet clock and 0.866s on the spaceship clock at the same event if they are both momentarily alongside the event and each other, but their clocks read differently because they are ticking at different rates in any given reference frame.

This "event" occurs at the coordinate time in the planet's rest frame of 1 second when the planet and the spaceship are separated by 0.5 light-seconds and yet you said they are both momentarily alongside the event and each other. It appears that you are considering the line of simultaneity at 1 second to be one "event".

Here's what you said about the second "event":

At the later event, if both observers are at the same event alongside each other, then both observers see 1.1547 seconds on the planet clock and 1 second on the spaceship clock.

This "event" occurs at the coordinate time in the planet's rest frame of 1.1547 seconds when the planet and the spaceship are separated by 0.577 light-seconds and yet you again said they are both alongside each other. Again it appears that you are considering the line of simultaneity at 1.1547 second to be one "event".

I guess I must have misunderstood what you meant but even after reading your correct definition of an event, I still have no idea what you must have meant. Maybe you could reword it.

I don't see why we need this array of observers. Maybe you should explain exactly where they are located and exactly what they see and exactly when they see it.

You do not "need" an array of observers, but it is convenient that if there happens to be an observer adjacent to the event that is at rest in a given irf and that has a clock synchronised with a clock at the origin of the irf, then whatever the reading on that observers clock is exactly what is predicted by the Lorentz transformations without taking light travel times into account.

I'm glad you changed your mind. Let me make sure I understand your new position: we don't need any observers with synchronized clocks, the Lorentz transformations are sufficient, correct?

First, you are using the word "see" differently than I have been using it in this thread. "See" means what you observe with your eyes and that's not what you mean here. Could you please define what you mean by "see"?

Given that I am using observers that are local to the events that are being analysed, there are no issues with light travel times so "see" and measured can be used interchangeably when we are talking about local time or spatial coordinates. Yes, there could be an issue with Doppler effects but that was not originally part of the original thread.

Have you change your mind back again?

And what issues are there with Doppler effects? I didn't know there were any.