Question about choice of reference frame

[Chenkel]

Hello everyone,

I might be highlighting my ignorance in this post (it might be ignorance of the particulars of simultaneity) but that's a good thing because then I might be able to figure out what I don't know.

If the Lorentz factor is 2 for a spaceship that is launched from Earth then treating Earth as the rest frame for every hour that passes for the spaceship two hours pass on Earth.

Treating the spaceship as the rest frame the spaceship measures half an hour passes for Earth when an hour passes for the spaceship.

Which calculation is more correct and why?

My intuition is that if two hours passes on Earth then one hour passes on the spaceship, but I think I'm incorrect to say that if an hour passes on the spaceship then half an hour passes on Earth.

Why does treating the spaceship as a rest frame seem to give me incorrect results for time elapsed on Earth?

I think I may not be doing the math right when treating the spaceship as the rest frame, perhaps this is why I calculate only half an hour elapsing on Earth when treating the spaceship as the rest frame?

Thanks in advance!

 

[Orodruin]

Which calculation is more correct and why?

Both are correct. They just compute different things due to the relativity of simultaneity. You keep asking variations of the same question and the answer will keep being the same.

 

[Dale]

Which calculation is more correct and why?

They are both correct.

Until you finish learning to draw spacetime diagrams, I would like to suggest that you completely discard the time dilation equation. Do not use it at all. Do not write it down, nor write nor ask anything related to it.

Instead, use only the Lorentz transformations. Write only Lorentz transform equations, and ask only Lorentz transform questions.

That does not require any new skills like drawing spacetime diagrams, it is just ordinary algebra. It will avoid all of the confusions and ambiguities and it will always include time dilation, length contraction, and the relativity of simultaneity.

 

[FactChecker]

Hello everyone,

I might be highlighting my ignorance in this post (it might be ignorance of the particulars of simultaneity)

That is the key to it all. Don't forget that all inertial reference frames will measure the speed of a light pulse as c, no matter how they are moving with respect to each other. So their clocks within their frame MUST be synchronized differently.

If the Lorentz factor is 2 for a spaceship that is launched from Earth then treating Earth as the rest frame for every hour that passes for the spaceship two hours pass on Earth.

More precisely, the rest frame Earth is observing the spaceship as it travels from the initial point to a far-distant point. So the Earth must take the time difference between its clock at the initial point versus its synchronized clock (in the Earth frame synchronization) at a far distant point.

Treating the spaceship as the rest frame the spaceship measures half an hour passes for Earth when an hour passes for the spaceship.

More precisely, the rest frame spaceship is observing the Earth as it travels from the initial point to a far-distant point. So the spaceship must take the time difference between its clock at the initial point versus its synchronized clock (in the spaceship synchronization) at a far distant point.

Which calculation is more correct and why?

They are both correct. Each one is comparing two of its frame's synchronized clocks with the other frame's traveling clock. Its two synchronized clocks are synchronized much differently from the clocks in the other frame.
So we are talking about different sets of synchronized clocks. And we know that each inertial reference frame did the synchronization differently. How else could they both measure the speed of the same light pulse as c?

 

[Chenkel]

That is the key to it all. Don't forget that all inertial reference frames will measure the speed of a light pulse as c, no matter how they are moving with respect to each other. So their clocks within their frame MUST be synchronized differently.

By light pulse are you referring to some synchnization process involving a light pulse?

Also don't we only need a clock for Earth and a clock for the spaceship?

 

[Dale]

Also don't we only need a clock for Earth and a clock for the spaceship?

No. With only those two clocks you can only compare them when they are co-located. If you want to compare them at any time other than when they are co-located then you also need a simultaneity convention. This is the key feature of a reference frame.

 

[Halc]

By light pulse are you referring to some synchnization process involving a light pulse?

Just a pulse of light, sent at some event and received (measured) at another. The pulse in this case is not being used to sync anything.

Also don't we only need a clock for Earth and a clock for the spaceship?

FactChecker is referring to an array of clocks all stationary in some frame. So in the case of the Earth frame (E), there would be the stationary clock on Earth and some clock (also stationary in E) say 2 light hours away (as measured in E) synced to that frame, perhaps by light pulses or by some other means. The ship will pass that clock at a time 1.732 hours later when the ship clock reads 0.866 (assuming both ship and Earth clock read 0 at the departure event).

Meanwhile, there could be yet another clock stationary in the ship frame (S) that is 2 light hours away in S, and is synced with the ship clock. Earth will pass that clock when the Earth clock reads 0.866 and this remote S clock reads 1.732.

 

[Chenkel]

Just a pulse of light, sent at some event and received (measured) at another. The pulse in this case is not being used to sync anything.

FactChecker is referring to an array of clocks all stationary in some frame. So in the case of the Earth frame (E), there would be the stationary clock on Earth and some clock (also stationary in E) say 2 light hours away (as measured in E) synced to that frame, perhaps by light pulses or by some other means. The ship will pass that clock at a time 1.732 hours later when the ship clock reads 0.866 (assuming both ship and Earth clock read 0 at the departure event).

Meanwhile, there could be yet another clock stationary in the ship frame (S) that is 2 light hours away in S, and is synced with the ship clock. Earth will pass that clock when the Earth clock reads 0.866 and this remote S clock reads 1.732.

I'm a little confused but I will study it, thank you.

 

[PeroK]

They are both correct.

Until you finish learning to draw spacetime diagrams, I would like to suggest that you completely discard the time dilation equation. Do not use it at all. Do not write it down, nor write nor ask anything related to it.

Instead, use only the Lorentz transformations. Write only Lorentz transform equations, and ask only Lorentz transform questions.

That does not require any new skills like drawing spacetime diagrams, it is just ordinary algebra. It will avoid all of the confusions and ambiguities and it will always include time dilation, length contraction, and the relativity of simultaneity.

As previously noted, this entails abandoning most, if not all, SR textbooks like Morin or Taylor & Wheeler.

You need to specify a text book that uses this approach. Otherwise, where does the student find source material?

 

[FactChecker]

By light pulse are you referring to some synchnization process involving a light pulse?

Not necessarily. It has been experimentally shown by a multitude of experiments, the most famous being Michelson and Morley but even more accurate ones done later, that the speed of light in a vacuum is always measured as c in any inertial reference frame. It does not matter what speed or direction the frame is moving in.

Also don't we only need a clock for Earth and a clock for the spaceship?

Only if you want to be confused. ;-)
How would you want to compare a moving clock two light-hours away (1.34 billion miles) with your one clock back at the starting point? In SR, there is already a spaceTIME coordinate system. Why not just use the time in your frame at that distant location?
To record times of anything that is moving a distance, it is simplest to record the time directly at each location. But to do that, the clocks along the path must be synchronized.

 

[Chenkel]

You guys gave me some stuff I will study, thank you.

 

[Dale]

You need to specify a text book that uses this approach. Otherwise, where does the student find source material?

There is no textbook to my knowledge that does this, but for this specific student the existing approaches are not working.

 

[PeroK]

There is no textbook to my knowledge that does this, but for this specific student the existing approaches are not working.

Chapter 11.4 of Morin is The Lorentz Transformation

Would your advice be to start afresh from there?

 

[robphy]

From a skim of the OP’s threads, I think the OP needs to first understand “events”. Without this, Lorentz Transformations are likely to be misused.

From https://www.physicsforums.com/threads/clarification-on-length-contraction.980109/post-6260356

http://aapt.scitation.org/doi/10.1119/1.17728
"Lapses in Relativistic Pedagogy" by Mermin
makes some good points

...Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it.

I think “spacetime diagrams” ( a position vs time diagram where you plot [point-]events) must be the highest priority. On them, you learn to model your various scenarios and questions about them.

From there, you can try to use the Lorentz Transformations (as in typical textbooks).

Or, as I would advocate,
is to work toward reasoning [spacetime-]geometrically and trigonometrically
by first reviewing how one
does high school 2-d Euclidean geometry
with geometry and circular-trigonometry.

(Most textbook problems in relativity are essentially asking about an unknown side or angle in a triangle in spacetime using hyperbolic-trigonometry. The $\gamma$ factor is essentially the ratio of the adjacent side to the hypotenuse in Minkowski spacetime geometry. The velocity is the slope = (opp/adj) of a worldline segment. Use this geometric reasoning to guide your newly developing spacetime-intuition...
… but you must be disciplined and patient.)

My $0.02.

 

[sdkfz]

... My intuition is that if two hours passes on Earth then one hour passes on the spaceship, but I think I'm incorrect to say that if an hour passes on the spaceship then half an hour passes on Earth. ...

I really hope I'm not muddying the waters by mentioning this at all, but I wonder if this comes from the Twins' Paradox (which I think was the first SR topic from Chenkel). After all it's the reciprocal time dilation that puts the "paradox" in it. That the returning (yes, the OP in this thread doesn't have that) spaceship crew is younger sometimes makes people think one of the frames has the "real" time dilation and the other not. e.g. (wrongly) Earth is the thing that "really" is stationary, and the spaceship is "really" the thing that moves and gets the slower time.

So just in case, I think it's worth pointing out to Chenkel that although the view from each rest frame is reciprocal while in inertial motion (i.e. without contradiction, both rest frames have 1 second per second clocks, and the others' clock is slow), the overall Twins' Paradox scenario is not reciprocal.

 

[FactChecker]

I really hope I'm not muddying the waters by mentioning this at all, but I wonder if this comes from the Twins' Paradox (which I think was the first SR topic from Chenkel).

That may be the route that brought him to realize that there was something profound to understand about the "relativity of simultaneity" and the reciprocal nature of time dilation. But, IMO, the Twin Paradox is something to worry about later, if ever. This is the really profound and basic part of SR. When Einstein described having an epiphany regarding SR, it was about this, not about the twin paradox.

 

[sdkfz]

... But, IMO, the Twin Paradox is something to worry about later, if ever. ...

Totally agree with that. But the reality of the internet and such is that people do get exposed to it, and I felt trying to see where Chenkel was coming from might help. (Edit: My intention was not to make this another Twins Paradox thread.)

 

[Sagittarius A-Star]

Which calculation is more correct ...

As @Dale mentioned, both are correct.

... and why?

Because of ...

Hello everyone,

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

After looking into it if you have two clocks P and Q that synchronize when they meet and if $\gamma$ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

 

[Chenkel]

Should I keep reading Morin's book or should I try to find a book that has more of a focus on Lorentz transformations as @Dale was pointing might be useful?

 

[Chenkel]

As @Dale mentioned, both are correct.
Because of ...

So from Earth's reference frame the Earth Clock ticking 2 hours and the spaceship clock ticking 1 hour are simultaneous.

And for the spaceship the spaceships clock ticking one hour and the Earth clock ticking half an hour are simultaneous events.

That means the Earth and the spaceship have different simultaneity between the two clocks.

 

[Hill]

if an hour passes on the spaceship then half an hour passes on Earth

Let's run some numbers using the Lorentz transformation, $t'=\gamma (t-vx/c^2)$. Let the spaceship move at $v=.6c$. $\gamma=1.25$.
As the spaceship passes the Earth, the Earth clock shows $t=0$, and the spaceship clock shows $t'=0$.
After 10 hours Earth time, at $t=10$ o'clock on Earth, the spaceship is $x=vt=6$ light-hours away from Earth. There is a clock at that location which is synchronized with the clock on Earth and is at rest relative to Earth. When the spaceship passes it, it shows the Earth time, $t=10$ o'clock, and at this exact moment it takes a picture of the passing spaceship clock. The spaceship clock shows $t'=1.25 (10-.6 \times 6)=8$ o'clock. Evidently for the Earth observer, the spaceship clock is slower than the Earth clock.
The spaceship also has clocks everywhere which are synchronized with the spaceship clock, and which move together with it, i.e., are at rest relative to the spaceship. In the spaceship frame of reference, the Earth passes one of these clocks exactly at $t'=8$ o'clock spaceship time, and at this exact moment the clock takes a picture of the passing Earth clock. The Earth clock shows $t=(8+0)/ 1.25= 6.4$ o'clock. Evidently for an observer on the spaceship, the Earth clock is slower than the spaceship clock.

 

[Chenkel]

Let's run some numbers using the Lorentz transformation, $t'=\gamma (t-vx/c^2)$. Let the spaceship move at $v=.6c$. $\gamma=1.25$.
As the spaceship passes the Earth, the Earth clock shows $t=0$, and the spaceship clock shows $t'=0$.
After 10 hours Earth time, at $t=10$ o'clock on Earth, the spaceship is $x=vt=6$ light-hours away from Earth. There is a clock at that location which is synchronized with the clock on Earth and is at rest relative to Earth. When the spaceship passes it, it shows the Earth time, $t=10$ o'clock, and at this exact moment it takes a picture of the passing spaceship clock. The spaceship clock shows $t'=1.25 (10-.6 \times 6)=8$ o'clock. Evidently for the Earth observer, the spaceship clock is slower than the Earth clock.
The spaceship also has clocks everywhere which are synchronized with the spaceship clock, and which move together with it, i.e., are at rest relative to the spaceship. In the spaceship frame of reference, the Earth passes one of these clocks exactly at $t'=8$ o'clock spaceship time, and at this exact moment the clock takes a picture of the passing Earth clock. The Earth clock shows $t=(8+0)/ 1.25= 6.4$ o'clock. Evidently for an observer on the spaceship, the Earth clock is slower than the spaceship clock.

I wish I was more knowledgeable about the full Lorentz transform to understand this on the first reading, but I will study the Lorentz transformation and what you wrote, thank you.

 

[robphy]

I wish I was more knowledgeable about the full Lorentz transform to understand this on the first reading, but I will study the Lorentz transformation and what you wrote, thank you.

Take a piece of graph paper and plot two points on it. Learn how to perform a rotation by some angle… then find the coordinates of those two transformed points.

A Lorentz transformation is a similar calculation.. but would likely be meaningless to you unless you do what I suggested in the first paragraph.

 

[PeterDonis]

Which calculation is more correct and why?

They are both correct. You have posed the same question in multiple threads now and have gotten the same answer. The answer is going to stay the same no matter how many times you ask.

Thread closed.