High School What reference frame is used for the velocity v in the Lorentz transformation?

[Huangdongcheng]

TL;DR

There is no such thing as an absolutely stationary reference frame, and thus the Lorentz transformation makes no sense.

When a spacecraft leaves the earth, its velocity is in the opposite direction of the earth's revolution around the sun. When the speed of the spacecraft relative to the earth is exactly equal to the linear speed of the earth's revolution, the spacecraft is stationary relative to the sun, and the earth revolves around the sun. Of course, spacecraft need to control power to avoid flying to the sun. People on the ground think that the spaceship is moving, and the clock on the spaceship is slower than the clock on the earth. But when the sun is the reference, the spaceship is stationary, and the earth is moving, and the clock on the earth is slower than the clock on the spaceship. So, when a spaceship returns to the earth after the earth has made a circle around the sun, which clock is slower, the clock on the spaceship or the clock on the ground? Why do we have to use the earth as a reference?

 

[Ibix]

You are considering a problem where gravity is a significant factor, which means you need to consider GR not SR. In this particular case you can get away with just SR, but be aware that it's not always possible.

In fact, this scenario is just a messier version of the Twin Paradox, and as with that scenario the problem you have is naively applying formulae that apply in global inertial frames to the case of the Earth, which is not at rest in any inertial frame. You can use the naive calculations when you work in the Sun's rest frame; in that case the Earth is moving and its clocks tick more slowly than those on the spaceship. Thus we predict that its clocks will have ticked less when they meet up again, which is the correct solution. The Earth is the "travelling twin" in this scenario.

The Earth is, in fact, a Langevin observer so you can't just say "principle of relativity, so the spaceship's clocks tick slowly according to the Earth". If you do so you are forgetting the relativity of simultaneity, and failing to account for the way the Earth's notion of simultaneity differs from an inertial observer's notion. If you want to consider the Earth as at rest, you must first formally define a coordinate system where the Earth is at rest and which covers all of its orbit in order to be able to define "now" in the phrase "what rate is the spacecraft clock ticking at now". Once you've done that you can answer that question, and then perform an integral over one orbit to get the elapsed time on the spacecraft clock - which you will find is the same as the simple calculation in the Sun rest frame.

If you want to perform the calculations I'd suggest using radar coordinates. They have a clean physical motivation and work well even in this accelerating case. The maths will be fairly messy but should be in reach if you want to try.

 

[Nugatory]

The Lorentz transformations are used to transform between the coordinates used with one inertial frame and the coordinates used with another inertial frame. However, none of the three frames you are considering here ( the frame in which the hovering spacecraft is at rest, the frame in which the orbiting earth is at rest, the frame in which the sun is at rest while the earth orbits around it) are inertial so we need a lot more math than just the Lorentz transformations to properly analyze the situation.

But before you take on that problem, you should take on the much more basic question in your thread title: What reference frame is the velocity v in the Lorentz transformation measured?

If I am at rest and you are moving relative to me at speed $v$, I can use the Lorentz transformation to transform between the coordinates of the frame in which I am at rest and the coordinates of the frame in which you are at rest. So $v$ is always relative to a point that we consider to be at rest in some frame, and the Lorentz transformations will convert from coordinates in that frame to coordinates in a frame in which that object is not at rest.

 

[Huangdongcheng]

If there is a clock on the spaceship and a clock on the ground, you only need to give a definite answer as to whether the clock on the ground is slower or the clock on the spaceship is slower. Is it necessary to say so much?

 

[Nugatory]

Is it necessary to say so much?

Yes, because you have posed a complicated problem, likely much more complicated than you realize. It might be helpful if you could tell us what the question behind the question is, that is, what would you conclude from the answer if we told you which clock "runs faster"?

Also note that "runs faster" assumes that there is a meaningful way of comparing the tick rates of the two clocks. That is NOT the same thing as saying that one clock measures less time than the other between the two comparisons.

 

[Ibix]

If there is a clock on the spaceship and a clock on the ground, you only need to give a definite answer as to whether the clock on the ground is slower or the clock on the spaceship is slower. Is it necessary to say so much?

There is a short answer to which one will show more elapsed time between meetings - I gave it in my previous post. But "which clock ticks slower" is a much more complicated question that requires a more complete understanding of relativity. Unless you just mean "slower on average between meetings", in which case the answer is as above.

 

[Herman Trivilino]

If there is a clock on the spaceship and a clock on the ground, you only need to give a definite answer as to whether the clock on the ground is slower or the clock on the spaceship is slower. Is it necessary to say so much?

A definitive answer was given in the first reply. Adding an explanation is what makes it a discussion. Is it necessary for you to complain about it?

 

[Huangdongcheng]

Yes, because you have posed a complicated problem, likely much more complicated than you realize. It might be helpful if you could tell us what the question behind the question is, that is, what would you conclude from the answer if we told you which clock "runs faster"?

Also note that "runs faster" assumes that there is a meaningful way of comparing the tick rates of the two clocks. That is NOT the same thing as saying that one clock measures less time than the other between the two comparisons.

A definitive answer was given in the first reply. Adding an explanation is what makes it a discussion. Is it necessary for you to complain about it?

According to the theory of relativity or the Lorentz transformation formula, time passes more slowly for a moving object than for a stationary one, and the faster the object moves, the slower time passes. So, is it wrong for me to ask whether the clock on the spaceship runs slower or the clock on Earth does? If it is impossible to determine which object's time passes more slowly, it proves that there is a problem with the theory of relativity.

 

[weirdoguy]

it proves that there is a problem with the theory of relativity

No, it proves you didn't understand both theory of relativity AND answers you've been given. Read them again.

 

[Ibix]

According to the theory of relativity or the Lorentz transformation formula, time passes more slowly for a moving object than for a stationary one, and the faster the object moves, the slower time passes.

That is a simplification of a special case. Your scenario is not that special case.

So, is it wrong for me to ask whether the clock on the spaceship runs slower or the clock on Earth does? I

You can ask the question, but as we've already said the answer to "which clock shows more elapsed time when the spacecraft returns to Earth" is straightforward. But "which clock is running slow" is a more complicated question.

If it is impossible to determine which object's time passes more slowly, it proves that there is a problem with the theory of relativity.

It isn't impossible. The answer to the simpler question is as I already stated in post #2, and the answer to the more complex version depends on details of your coordinate system (or at least your synchronisation convention) that you haven't specified.

 

[Ibix]

Do also note, @Huangdongcheng, that you are simply proposing a messier version of the twin paradox scenario. Your use of circular motion instead of a linear out-and-back trip changes nothing except to make the maths harder.

The twin paradox was first proposed and resolved by Einstein in his 1905 paper where he first set out what became known as the special theory of relativity. It isn't a problem for relativity. The resolution has been known literally as long as the theory itself.

 

[Nugatory]

According to the theory of relativity or the Lorentz transformation formula, time passes more slowly for a moving object than for a stationary one, and the faster the object moves, the slower time passes.

That is a very common misunderstanding, one that unfortunately has been repeated over and over by well-meaning people trying to explain relativity when they don’t completely understand it themselves.

So, is it wrong for me to ask whether the clock on the spaceship runs slower or the clock on Earth does? If it is impossible to determine which object's time passes more slowly, it proves that there is a problem with the theory of relativity.

If we consider A to be at rest and B to be moving, we will correctly conclude that B’s clock is running slower than A’s clock; but if we consider B to be at rest and A to be moving we will just as correctly conclude that A’s clock is one that is running slowly.
The apparent paradox here is the result of two common errors: being imprecise about exactly what it means to compare the rate of two clocks that are at not colocated; and failing to consider the relativity of simultaneity (if you are not completely comfortable with that concept, go google for it right now and do not proceed until you understand it thoroughly - it is essential to making sense of relativity).

We have a number of older threads showing how this apparent paradox is resolved. You will find some of them if you search this forum for “time dilation simultaneity” and if I have time I’ll link a few of them in a moment.

Edit: https://www.physicsforums.com/threa...on-question-w-spaceships.883303/#post-5552719

https://www.physicsforums.com/threads/time-the-special-theory-of-relativity.846242/#post-5307390

 

[Huangdongcheng]

That is a very common misunderstanding, one that unfortunately has been repeated over and over by well-meaning people trying to explain relativity when they don’t completely understand it themselves.If we consider A to be at rest and B to be moving, we will correctly conclude that B’s clock is running slower than A’s clock; but if we consider B to be at rest and A to be moving we will just as correctly conclude that A’s clock is one that is running slowly.
The apparent paradox here is the result of two common errors: being imprecise about exactly what it means to compare the rate of two clocks that are at not colocated; and failing to consider the relativity of simultaneity (if you are not completely comfortable with that concept, go google for it right now and do not proceed until you understand it thoroughly - it is essential to making sense of relativity).

We have a number of older threads showing how this apparent paradox is resolved. You will find some of them if you search this forum for “time dilation simultaneity” and if I have time I’ll link a few of them in a moment.

Edit：https://www.physicsforums.com/threa...on-question-w-spaceships.883303/#post-5552719

https://www.physicsforums.com/threads/time-the-special-theory-of-relativity.846242/#post-5307390

Before a spacecraft leaves Earth, can we compare the speeds of two clocks? After the Earth completes one orbit around the Sun and the spacecraft returns to Earth, can we compare the speeds of the two clocks? Which clock would be slower, according to your calculations? Furthermore, if there are two objects moving far from the solar system, how can we determine which is moving faster and which is moving slower? Isn't the magnitude of the speed independent of the chosen reference frame? Would the results be the same if we chose a different reference frame?

 

[Huangdongcheng]

Does the original version of the twin paradox only allow for linear motion? Doesn't the formula break down when an object moves on a curved path? No matter how complex a formula is, if it doesn't fit real life, it's useless.

 

[Herman Trivilino]

According to the theory of relativity or the Lorentz transformation formula, time passes more slowly for a moving object than for a stationary one, and the faster the object moves, the slower time passes.

That's an explanation often seen, but it can be misleading. To begin to understand this you must realize that to reach that conclusion the two clocks must share the same location, and that can happen at most only one time in scenarios where the Lorentz transforms apply.

Why is that? Because the Lorentz transformations apply to two reference frames that are moving relative to each other at a steady speed in a straight line.

If you are moving west and I am moving east we can cross paths at most only once. That's our only chance to directly compare our clocks. Since we'll never meet again what can we do to determine if my clock is ticking faster or slower compared to yours?

I can set up another clock at some distance behind me and synchronize my two clocks. Then, when you pass by my other clock it can be compared to yours. When I do that, I conclude that your clock is running slow compared to my clocks. But if you do the same you will conclude that my clock is running slow compared to your clocks.

This is a very confusing thing! Einstein said that it bothered him so much that one night as he was lying in bed thinking about it, and the solution came to him, it caused him to sit upright in his bed.

So, is it wrong for me to ask whether the clock on the spaceship runs slower or the clock on Earth does?

It's not wrong, it's just that you've simply been clever enough, as many others have before you, to arrange a scenario where there are two meetings. But all such scenarios involve motion that's not in a straight line at a steady speed, so the Lorentz transforms don't apply. This situation is so famous it has a name, the Twin Paradox.

If it is impossible to determine which object's time passes more slowly, it proves that there is a problem with the theory of relativity.

Nope. It just means you're missing the realization that Einstein had when he sat upright in his bed.

You see, in the scenario I described above you will conclude that I made an error synchronizing my two clocks, and that is what led me to conclude that your clock is running slow. Likewise, I will conclude that you made an error synchronizing your clocks and that is what led you to conclude that my clock is running slow.

But in fact no errors were made by either of us. The Lorentz transforms will show this. It's the notion that simultaneity is relative. Things (like the readings on your two clocks) that are simultaneous in your rest frame are not in mine. And vice-versa.

In my opinion you have to work through the math to understand all this. Words alone won't do it. But the good news is you're not on your own. Lots of textbook authors have written worked examples. You just have to read and understand them.

 

[robphy]

In my opinion you have to work through the math to understand all this. Words alone won't do it.

Indeed… Draw a spacetime diagram.

 

[Dale]

is it wrong for me to ask whether the clock on the spaceship runs slower or the clock on Earth does?

Yes, it is wrong to ask that. The rate a clock ticks depends on the reference frame. So asking “which clock runs slower?” is wrong to ask because it is incomplete.

A complete question would be “which clock runs slower in the sun’s frame?”. Without specifying the frame, the question is wrong.

If it is impossible to determine which object's time passes more slowly, it proves that there is a problem with the theory of relativity.

Once you ask a complete question then a valid answer is possible. The impossibility of an answer to an incomplete question is hardly a problem with any theory.

 

[Nugatory]

Does the original version of the twin paradox only allow for linear motion?

No. the original form, published in the same 1905 paper in which Einstein introduced special relativity, involved circular motion (and in fact looks like a more clearly stated version of your original post). Since then most introductory presentations have used linear motion, because that special case illustrates the physics just as well but needs only high school algebra instead of multivariable and tensor calculus.

Doesn't the formula break down when an object moves on a curved path?

No, the formula works just fine no matter how complicated and twisty the path of the object is.

 

[pervect]

Because to OP is talking about "reference frames", they are most likely doing Newtonian physics and not actually doing GR. If they were doing GR, they would be talking about what metric they were using rather than about "reference frames". But I'll try to answer the question both from a Newtonian and a GR perspective. I expect that the full GR treatment will be too brief to be useful, but I hope to at least convey something about the "flavor" of the treatment.

From a Newtonian perspective, the reference frame one uses is essentially heliocentric. The heliocentric idea occured historically before Newton - with Copernicus, Kepler, and Gallileo being important authors. Rather than get the details wrong, I'll mention that Newton came later.

A better degree of approxmation would anchor the reference frame used for Newtonian calculations to the barycenter of the solar system rather than the sun, but I won't go into this refinement.

The basic equations of motion used in the Newtonian approach would be F=ma, with the gravitational force F being equal to GmM/r^2.

The basic equations used for the GR approach would be the geodesic equations. See https://en.wikipedia.org/wiki/Solving_the_geodesic_equations. Geodesics are still an approximation in GR, but they are a good one - actual planets would depart from the geodesic for various reasons, including the emission of gravitational waves, and numerically larger effects such as the transfer of momentum to the Earth from solar radiation.

The basic geodesic equations would be
$$\frac{d^2 x^a}{d \tau^2} + \Gamma^a{}_{bc} \frac{dx^b}{d\tau} \frac{dx^c}{d\tau} = 0 \quad \Gamma^a{}_{bc} = \frac{1}{2} \partial g^{ad} \left( \frac{\partial g_{cd} }{\partial x^b} + \frac{\partial g_{bd} }{\partial x^c} - \frac{\partial g_{bc} }{\partial x^d} \right)$$

F=ma is not used, thats the Newtonian approach - the GR approach is fundamentally different and based on different concepts.

These equations are probably not familiar to the OP and would require some study to disentangle enough to actually use. Among other issues one would need famiiartiy with the convention of summation over repeated indicies. The point of the equations though is that they are differential equations (just like the ones from the Newtonian equations) derived from the metric, $g_{ab}$. So, specifying the metric gives the solutions for the orbits of test masses.

While there are several metrics one could use, the Schwarzschild metric is the usual choice. There are some alternatives one might use, but getting into the details is beyond the scope of what I want to write here.

There are a number of refinements that can be made to the GR analysis that I've outlined here - it has made some approximations, but this post is already long enough. The point is that it should not be at all surprising that we use a sun-centered coordinate system to work the problem in GR just as we did with Newton. The Sun, being the most massive object in the solar system, tends to dominate the physics. Again, refinements of this heliocentric model are possible in GR, but it would take an extended discussion to clarify this, I'll just drop the words "PPN approximations" and leave it at that.

I'll add one more thing. Rather than attempt to learn enough about GR to compute the orbits, it would be simplest to just assume that it gives a circular orbit. Then one can concentrate on the original question, by computing and integrating the proper time $\tau$ for the orbital motion, the "hovering" motion, and even answering the question of what trajectory maximizes the age of the stay-at-home twin. Perhaps I should have started here, but I didn't - and it would be another thread to do this justice. Of course, some famliarity with the concept of what proper time is is needed - perhaps another thread might be needed, to explain what proper time is and why it answers the question, before we would go about computing it :(.

 

[PeterDonis]

it would be another thread to do this justice

Actually, no, it doesn't. The computation is very simple. If we have a "hovering" observer at radius $r$ in Schwarzschild spacetime and a second observer making a circular orbit at the same radius $r$ (so that the two meet once per orbit), then their respective time dilation factors are $\sqrt{1 - 2M / r}$ for the hovering observer and $\sqrt{1 - 3M / r}$ for the observer in the circular orbit. Since the latter is smaller than the former, the circular orbit observer measures less elapsed time than the hovering observer does between two successive meetings.

 

[Nugatory]

The computation is very simple.

It can be even simpler (and may have been what @Ibix was doing way back in #2 of this thread): Remove the earth and its gravitational field from the problem. Likewise remove the sun and its gravitational field.
Now we have a flat spacetime, no general relativistic complications, can use Minkowski coordinates and the Lorentz transformations that OP was asking about in their first post. We have two clocks, one undergoing continuous circular motion under the influence of some force, and the other at rest at some point on that circle. They will meet once every rotation of the moving clock and the question is which clock counts more seconds between consecutive meetings.

I’m thinking that OP would be well-served by understanding this case and how it is just the twin paradox restated before moving out of flat spacetime.

 

[Huangdongcheng]

Because to OP is talking about "reference frames", they are most likely doing Newtonian physics and not actually doing GR. If they were doing GR, they would be talking about what metric they were using rather than about "reference frames". But I'll try to answer the question both from a Newtonian and a GR perspective. I expect that the full GR treatment will be too brief to be useful, but I hope to at least convey something about the "flavor" of the treatment.

From a Newtonian perspective, the reference frame one uses is essentially heliocentric. The heliocentric idea occured historically before Newton - with Copernicus, Kepler, and Gallileo being important authors. Rather than get the details wrong, I'll mention that Newton came later.

A better degree of approxmation would anchor the reference frame used for Newtonian calculations to the barycenter of the solar system rather than the sun, but I won't go into this refinement.

The basic equations of motion used in the Newtonian approach would be F=ma, with the gravitational force F being equal to GmM/r^2.

The basic equations used for the GR approach would be the geodesic equations. See https://en.wikipedia.org/wiki/Solving_the_geodesic_equations. Geodesics are still an approximation in GR, but they are a good one - actual planets would depart from the geodesic for various reasons, including the emission of gravitational waves, and numerically larger effects such as the transfer of momentum to the Earth from solar radiation.

The basic geodesic equations would be
$$\frac{d^2 x^a}{d \tau^2} + \Gamma^a{}_{bc} \frac{dx^b}{d\tau} \frac{dx^c}{d\tau} = 0 \quad \Gamma^a{}_{bc} = \frac{1}{2} \partial g^{ad} \left( \frac{\partial g_{cd} }{\partial x^b} + \frac{\partial g_{bd} }{\partial x^c} - \frac{\partial g_{bc} }{\partial x^d} \right)$$

F=ma is not used, thats the Newtonian approach - the GR approach is fundamentally different and based on different concepts.

These equations are probably not familiar to the OP and would require some study to disentangle enough to actually use. Among other issues one would need famiiartiy with the convention of summation over repeated indicies. The point of the equations though is that they are differential equations (just like the ones from the Newtonian equations) derived from the metric, $g_{ab}$. So, specifying the metric gives the solutions for the orbits of test masses.

While there are several metrics one could use, the Schwarzschild metric is the usual choice. There are some alternatives one might use, but getting into the details is beyond the scope of what I want to write here.

There are a number of refinements that can be made to the GR analysis that I've outlined here - it has made some approximations, but this post is already long enough. The point is that it should not be at all surprising that we use a sun-centered coordinate system to work the problem in GR just as we did with Newton. The Sun, being the most massive object in the solar system, tends to dominate the physics. Again, refinements of this heliocentric model are possible in GR, but it would take an extended discussion to clarify this, I'll just drop the words "PPN approximations" and leave it at that.

I'll add one more thing. Rather than attempt to learn enough about GR to compute the orbits, it would be simplest to just assume that it gives a circular orbit. Then one can concentrate on the original question, by computing and integrating the proper time $\tau$ for the orbital motion, the "hovering" motion, and even answering the question of what trajectory maximizes the age of the stay-at-home twin. Perhaps I should have started here, but I didn't - and it would be another thread to do this justice. Of course, some famliarity with the concept of what proper time is is needed - perhaps another thread might be needed, to explain what proper time is and why it answers the question, before we would go about computing it :(.

Why choose the Sun as the reference frame? The solar system is also in motion within the Milky Way. Is it reasonable to choose the Sun as the reference frame? What if the spacecraft is moving far away from the solar system? Then what should be chosen as the reference object to measure the speed of the spacecraft? My point is that there is no absolutely stationary reference frame. We can choose any reference frame to measure the speed of an object. The Lorentz transformation then becomes meaningless.

 

[PeterDonis]

We can choose any reference frame to measure the speed of an object.

That's correct.

The Lorentz transformation then becomes meaningless.

That's wrong.

The $v$ in the Lorentz transformation is the relative velocity between the two frames.

 

[Ibix]

We can choose any reference frame to measure the speed of an object.

True. The advantage of the Sun rest frame is that it is the only one in which the speed of both ship and Earth is constant, so it is the only one in which you can apply the time dilation formula without having to do integrals. If you are happy with integration you can use any inertial frame, with the $v$ being the instantaneous velocity of the Earth or ship in that frame.

The Lorentz transformation then becomes meaningless.

First, a meta observation. Special Relativity is over a century old and the mathematics underlies literally all of modern physics. We cannot explain how your phone (or anything else that uses electromagnetism) works without it, nor could we have expected nuclear reactors or weapons to work - yet they do. There's an FAQ linked in a sticky thread in this forum on the experimental testing that has been done. So which do you think is more likely - the core mathematics of an extremely well tested theory that you (possibly unknowingly) rely on every day is meaningless, or you are misunderstanding it?

Second, the point you are misunderstanding is that the Lorentz transforms relate the coordinates of events as measured in one inertial frame to the coordinates in another. The $v$ is the velocity of the second frame with respect to the first. That is, it is explicitly a relative velocity. The problem you have is that one of your objects is not at rest in any inertial frame, so you can only do naive calculations based on the maths of inertial frames if you pick a frame where its speed is constant. You can do non-naive calculations in any frame, but as we've said you're going to need integrals and quite possibly have to take careful account of the relativity of simultaneity. The usual constant speed linear out-and-back version of the twin paradox is a gentler introduction to this, because it requires developing understanding of the relativity of simultaneity but does not require any integrals.

 

[Ibix]

It can be even simpler (and may have been what @Ibix was doing way back in #2 of this thread): Remove the earth and its gravitational field from the problem. Likewise remove the sun and its gravitational field.

I did start re-writing the problem as a clock on a string whirled about your head, but actually if you idealise the Earth as a test particle in a circular orbit the maths of SR works exactly anyway. At least, it does as long as we don't get into stuff like detailed observations between the spacecraft and the Earth during the flight, where GR effects like Shapiro delay and gravitational lensing would kick in, not to mention the gravitational effect of the Earth itself, which is why I was careful to add an introductory paragraph noting that this is really a GR problem.

I agree it may be necessary to simplify to a clock on a string if we ever get past the OP's "Lorentz transforms are meaningless" error. The maths of rotating reference frames is quite messy enough without having to deal with numerical analysis of light paths in gravitational fields.

 

[Huangdongcheng]

True. The advantage of the Sun rest frame is that it is the only one in which the speed of both ship and Earth is constant, so it is the only one in which you can apply the time dilation formula without having to do integrals. If you are happy with integration you can use any inertial frame, with the $v$ being the instantaneous velocity of the Earth or ship in that frame.

First, a meta observation. Special Relativity is over a century old and the mathematics underlies literally all of modern physics. We cannot explain how your phone (or anything else that uses electromagnetism) works without it, nor could we have expected nuclear reactors or weapons to work - yet they do. There's an FAQ linked in a sticky thread in this forum on the experimental testing that has been done. So which do you think is more likely - the core mathematics of an extremely well tested theory that you (possibly unknowingly) rely on every day is meaningless, or you are misunderstanding it?

Second, the point you are misunderstanding is that the Lorentz transforms relate the coordinates of events as measured in one inertial frame to the coordinates in another. The $v$ is the velocity of the second frame with respect to the first. That is, it is explicitly a relative velocity. The problem you have is that one of your objects is not at rest in any inertial frame, so you can only do naive calculations based on the maths of inertial frames if you pick a frame where its speed is constant. You can do non-naive calculations in any frame, but as we've said you're going to need integrals and quite possibly have to take careful account of the relativity of simultaneity. The usual constant speed linear out-and-back version of the twin paradox is a gentler introduction to this, because it requires developing understanding of the relativity of simultaneity but does not require any integrals.

Perhaps I really don't understand relativity, but that doesn't mean we can't question and doubt authoritative theories. Aristotle's ideas persisted for over two thousand years, and eventually they were rejected.

 

[PeroK]

Perhaps I really don't understand relativity, but that doesn't mean we can't question and doubt authoritative theories. Aristotle's ideas persisted for over two thousand years, and eventually they were rejected.

There are different levels of "doubt". Suppose you started learning pure mathematics and immediately didn't understand mathematical logic. You have a choice.

1) Presume all mathematics is wrong or meaningless and that mathematicians are fools or conspirators. This is easy!

2) Presume that you, as a new student, will have to work hard to understand these ideas.

Ancient science, for all its merits, was build on shaky foundations. It progressed very slowly, with centuries passing between significant developments.

Modern science (mathematics, physics, chemistry, biology, geology, astronomy, medicine, engineering) is based on an enormous volume of empirical evidence. There is still a lot to discover, but the idea that it's all fundamentally wrong, is not likely to be a productive approach.

 

[Ibix]

Perhaps I really don't understand relativity, but that doesn't mean we can't question and doubt authoritative theories.

Of course. But you aren't doing anything except repeating "this doesn't make sense to me personally". You need to do some maths to actually get anywhere with understanding the theory or showing a problem with it, and you aren't doing that. Try calculating the elapsed time for Earth and the spaceship, for example, and we can help show you where you've gone wrong. I repeat my advice that this scenario doesn't add much except mathematical complexity to the linear version of the twin paradox, and I'd recommend working that one first.

Special Relativity is ridiculously simple when you get right down to it, and it is internally consistent. You might show by an actual experiment that the universe does not follow it (and indeed it doesn't - General Relativity is needed when gravity is a factor) but thought experiments will only show you that you do or don't understand the theory.

I notice I forgot to paste in the URL to the Experimental Basis of SR FAQ in my last post so the link won't work - I'll correct that now.

 

[weirdoguy]

authoritative theories

There are no authoritative theories in physics. We do not "belive" in any theory because someone said we should. We "belive" in them because they are backed up by thousands/milions of experiments. Physics is empirical science.

Aristotle's ideas

Which are not scientific theories by modern standards. Any other examples? No.

 

[Huangdongcheng]

Of course. But you aren't doing anything except repeating "this doesn't make sense to me personally". You need to do some maths to actually get anywhere with understanding the theory or showing a problem with it, and you aren't doing that. Try calculating the elapsed time for Earth and the spaceship, for example, and we can help show you where you've gone wrong. I repeat my advice that this scenario doesn't add much except mathematical complexity to the linear version of the twin paradox, and I'd recommend working that one first.

Special Relativity is ridiculously simple when you get right down to it, and it is internally consistent. You might show by an actual experiment that the universe does not follow it (and indeed it doesn't - General Relativity is needed when gravity is a factor) but thought experiments will only show you that you do or don't understand the theory.

I notice I forgot to paste in the URL to the Experimental Basis of SR FAQ in my last post so the link won't work - I'll correct that now.

You solve this problem using mathematics: If a pair of twins were 20 years old at 8:00 on January 1, 2015. Then the elder brother went on a space journey and returned to Earth, and his clock showed 8:00 on January 1, 2020. So did he witness the events that happened on January 1, 2020 on Earth or the events that occurred on Earth in 2025? Could he see his younger brother on January 1, 2025?