Length contraction and direction of tavel

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Length contraction occurs only in the direction of motion, affecting both the front and back of an object as perceived by a moving observer. The speed of light remains invariant due to the Lorentz transformations, which account for time dilation and length contraction. When considering two beams of light, one approaching and one receding, the distances they travel differ based on the observer's motion, yet both are measured to travel at the speed of light. The confusion arises from trying to apply time dilation and length contraction independently, rather than using the comprehensive Lorentz transformations. Understanding these concepts is crucial for reconciling how light speed remains constant despite the relativistic effects of motion.
  • #91
The reason for the differential aging of the twins is due to the difference in length of the space-time paths taken by the two twins. That is all the reason there is. It is your destination if you are attempting to understand the twin paradox scenario. Dismissing it as "far away from your destination" is just daft - why ask the question if you aren't going to listen to the answer?

There are basically two sets of transforms between inertial frames that are consistent with the notion that physics is the same in all of them: these are the Galilean transforms and the Lorentz transforms. The Lorentz transforms are consistent with observation while the Galilean ones are not. So the Lorentz transforms are the right ones (but we were fooled for centuries because we didn't have sensitive enough experiments to spot the errors in the Galilean transforms).

In a Euclidean space, the distance between points is given by Pythagoras' Theorem, and the answer does not dependent on the choice of coordinates. I might say that two points are ##\Delta x=L## meters apart in the x direction and ##\Delta y=W## meters apart in the y direction. You might say that they ae ##\Delta x'=W## meters apart in the x' direction and ##\Delta y'=-L## meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other. However, we will both agree that the distance between the points is ##\sqrt{L^2+W^2}## meters.

The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - but it does not follow Euclidean geometry. In space-time, the equivalent of Pythagoras' theorem is ##c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}##. This quantity does not depend on the choice of coordinates - you and I might disagree on ##\Delta x## (that would be length contraction) or ##\Delta t## (that would be time dilation), but we will always come up with the same ##\Delta\tau## for any given path.

It's easy to see that, in your rest frame, ##\Delta\tau=\Delta t##, because in your rest frame you are not moving so for you ##\Delta x=\Delta y=\Delta z=0##. Someone at rest in one frame is moving in another, though, so in general ##\Delta\tau## is the time experienced by someone moving at constant speed from point A to point B, separated in space by (##\Delta x,\Delta y,\Delta z##), in time ##\Delta t##.

It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the ##\Delta\tau##s are different - ##\sqrt{10^2-0^2}=10## years for me, ##\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8## years for you.

I think that's a complete explanation of the twin paradox from top to bottom. Certainly you should learn the maths and be able to derive the Lorentz transforms if you intend to study SR. It will show you that what I have written above is self-consistent. But it will not give you any further insight into the twin paradox - this is the whole of the "why" there is differential aging.
 
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  • #92
Ibix said:
The reason for the differential aging of the twins is due to the difference in length of the space-time paths taken by the two twins. That is all the reason there is. It is your destination if you are attempting to understand the twin paradox scenario. Dismissing it as "far away from your destination" is just daft - why ask the question if you aren't going to listen to the answer?

There are basically two sets of transforms between inertial frames that are consistent with the notion that physics is the same in all of them: these are the Galilean transforms and the Lorentz transforms. The Lorentz transforms are consistent with observation while the Galilean ones are not. So the Lorentz transforms are the right ones (but we were fooled for centuries because we didn't have sensitive enough experiments to spot the errors in the Galilean transforms).

In a Euclidean space, the distance between points is given by Pythagoras' Theorem, and the answer does not dependent on the choice of coordinates. I might say that two points are ##\Delta x=L## meters apart in the x direction and ##\Delta y=W## meters apart in the y direction. You might say that they ae ##\Delta x'=W## meters apart in the x' direction and ##\Delta y'=-L## meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other. However, we will both agree that the distance between the points is ##\sqrt{L^2+W^2}## meters.

The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - but it does not follow Euclidean geometry. In space-time, the equivalent of Pythagoras' theorem is ##c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}##. This quantity does not depend on the choice of coordinates - you and I might disagree on ##\Delta x## (that would be length contraction) or ##\Delta t## (that would be time dilation), but we will always come up with the same ##\Delta\tau## for any given path.

It's easy to see that, in your rest frame, ##\Delta\tau=\Delta t##, because in your rest frame you are not moving so for you ##\Delta x=\Delta y=\Delta z=0##. Someone at rest in one frame is moving in another, though, so in general ##\Delta\tau## is the time experienced by someone moving at constant speed from point A to point B, separated in space by (##\Delta x,\Delta y,\Delta z##), in time ##\Delta t##.

It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the ##\Delta\tau##s are different - ##\sqrt{10^2-0^2}=10## years for me, ##\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8## years for you.

I think that's a complete explanation of the twin paradox from top to bottom. Certainly you should learn the maths and be able to derive the Lorentz transforms if you intend to study SR. It will show you that what I have written above is self-consistent. But it will not give you any further insight into the twin paradox - this is the whole of the "why" there is differential aging.
Thank you Ibix for your answer.
It's not that I don't want to listen to explanations. It's just that sometimes I can't fully understand.
It seems that understanding SR is very difficult, not just like in discovery channel where the host says "..., so the other twin ages slower..."
And since joining this forum, I've learned (or read) about
Worldine, simultaneity events, and now space time path.
It takes time to study all those things.
Yesterday I just understood how Hendrik Lorentz formulated this formula ##\gamma = \frac{1}{\sqrt{1-v^2}}## trough Janus' train simulation.
Perhaps you should know that I have little (if not at all) background in Physics and Math.
And thanks for this answer, too. I'll contemplate it.
 
  • #93
Ibix said:
You might say that they ae ##\Delta x'=W## meters apart in the x' direction and ##\Delta y'=-L## meters apart in the x direction. We are using coordinates rotated by 90° with respect to each other.
Did you mean in the y' direction?
 
  • #94
Stephanus said:
Did you mean in the y' direction?
Yes. Typo - sorry.
 
  • #95
Stephanus said:
Sorry, I didn't mean to rush everybody with my (endless) questions :smile:.
It's just a rhetoric statement.
Actually I want to know about twins paradox and why there's twins paradox but the universe HAS NO frame of reference.
But to get there, I've been asking about [..]
Hi Stephanus, I notice a glitch in where you want to get with all your questions!

Probably you mean that the universe has no "absolute" frame of reference. I don't know who told you that, but we can only say that WE have no "absolute" frame of reference; we are ignorant if the universe has one. There is no twins paradox if the universe does have an absolute frame of reference. There is also no twin paradox if the universe is what is called a "block universe", and many people like that model more. And there may be other explanations as well. After long discussions and even debates about such models of SR on this forum, discussions in which people argue in favour or against them have been banned*. But the old discussions give enough information, you can search the forum for "block universe". :smile:

* See item 11 in this forum's FAQ: https://www.physicsforums.com/threads/relativity-faq-list.807523/
 
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  • #96
harrylin said:
Hi Stephanus, I notice a glitch in where you want to get with all your questions!

Probably you mean that the universe has no "absolute" frame of reference. I don't know who told you that, but we can only say that WE have no "absolute" frame of reference; we are ignorant if the universe has one. There is no twins paradox if the universe does have an absolute frame of reference. There is also no twin paradox if the universe is what is called a "block universe", and many people like that model more. And there may be other explanations as well. After long discussions and even debates about such models of SR on this forum, discussions in which people argue in favour or against them have been banned*. But the old discussions give enough information, you can search the forum for "block universe". :smile:

* See item 11 in this forum's FAQ: https://www.physicsforums.com/threads/relativity-faq-list.807523/
I mean this.
If we sit at the back seat of the plane which flies at 250 m/s and we call a flight attendance in front of us. Our sound will travel at 330m/s from OUR frame in the plane.
But an observer on Earth if somehow he/she can see the sound wave travels, will see that our sound wave is traveling at 580 m/s, right?
Or imagine a sport car running at 200 m/s. It has a device that in a certain time produces an ultrasonic sound (not that its engine is silence). The ultrasonic sound will travel at 330 m/s no matter how fast the car moves. And observer on the podium if somehow can see the sound wave, he/she will see the sound wave travels at 330 m/s. But the driver will see that the sound wave travels at 130 m/s from his point of view/frame. But if the driver is communicating with it's paddock crew, assuming he/she's driving a Ferrari Testarosa, not Ferrari F1, his/her sound will travel at 530 m/s from the ground point of view, but from his/her frame it's 330 m/s.
Light is wave, sound is wave.
Light is affected by doppler effect, so is sound.
But...
If it's about light then...
But an observer on Earth if somehow he/she can see the sound wave travels, will see that our sound wave is traveling at 580 330 m/s, right?

But the driver will see that the sound wave travels at 130 330 m/s from his point of view/frame.

But if the driver is communicating with it's paddock crew, assuming he/she's driving a Ferrari Testarosa, not Ferrari F1, his/her sound will travel at 530 330 m/s from the ground point of view, but from his/her frame it's 330 m/s.
It's always 330m/s, 330m/s, 330m/s ever.
Argghh, these "light" things will surely make me go crazy.
I have to contemplate it slowly, really slow.
It's not that your and every other mentor explanations are blur or unclear, I think this thing have to be self understood. Just like riding a bike. There's no trainer in the world that can teach you how to ride a bike. You have to try it yourself, the trainer can only teach you WHAT you have to do.
So far I know how Hendrik Lorentz formulated ##\gamma \text { is } \frac{1}{\sqrt{1-v^2}}##, and why ##\gamma \text { has to be } \frac{1}{\sqrt{1-v^2}}##.
The rest I should have absorb slowly.
Thanks.
 
  • #97
Stephanus said:
If it's about light then...
It's not about light vs. sound, but fast vs. slow. The speed of anything that travels at relativistic speed in one frame, will transform in a significantly non- Galilean way to other frames.
 
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  • #98
A.T. said:
It's not about light vs. sound, but fast vs. slow. The speed of anything that travels at relativistic speed in one frame, will transform in a significantly non- Galilean way to other frames.
Really? Is it not about light only? Because the speed of sound varies, but not light.
 
  • #99
Stephanus said:
Is it not about light only?
All large speeds don't transform to other frames in the Galilean way you assume sound speed does. Even for sound that is just an approximation.
 
  • #100
A.T. said:
Speeds slightly below c also don't transform to other frames the way sound speed does.
Are you trying to say that in plane,
when we call the attendants from the back seat,
our sound wave travels at
##\frac{s + p}{1+(s * p)/c^2}##?
s = speed of sound
p = speed of plane
c = maximum speed in the universe, I refrain myself to use the world "light"
And incidently, light travel at the top speed of the universe?
 
  • #101
Stephanus said:
Are you trying to say that in plane,
when we call the attendants from the back seat,
our sound wave travels at
s+p1+(sp)/c2\frac{s + p}{1+(s * p)/c^2}?
That is how the speed transforms to other frames.
 
  • #102
Stephanus said:
And incidently, light travel at the top speed of the universe?
Not necessarily. Light speed is just a universal speed "barrier": those which have "positive" invariant mass always travel at subluminal speeds, whereas those with "negative" invariant mass always travel at superluminal speeds (tachyons). It's just that tachyons have never been observed, so it is generally assumed that nothing can travel faster than light. A tachyon will violate causality, and it's mass does not appear to have any physical meaning.
 
  • #103
Stephanus said:
Are you trying to say that in plane,
when we call the attendants from the back seat,
our sound wave travels at
s+p1+(s∗p)/c2\frac{s + p}{1+(s * p)/c^2}?
No. The air in the plane is also moving along with the plane. Remember what I said earlier? The sound of a wave is always relative to the medium (except for EM waves); the speed of sound in the plane measured by a passenger onboard will have the familiar value of around 330 m/s (at standard room temperature of course).
 
  • #104
And if a rocket travels at 0.6c for example and it shine a laser at the front,
so the speed of the laser plus the speed of the rocket is...
##\frac{r+l}{1+(r*l)/c^2}##
r is the speed of the rocket
l is the speed of the laser, which is equal to the speed of light
c is the speed of light.
So we can rewrite the formula into...
##\frac{r+c}{1+(r*c)/c^2} = \frac{r+c}{1+r/c} = \frac{r+c}{c/c+r/c} = \frac{r+c}{(r+c)/c} = c*\frac{r+c}{r+c} = c##
Is that so?
 
  • #105
PWiz said:
No. The air in the plane is also moving along with the plane. Remember what I said earlier? The sound of a wave is always relative to the medium (except for EM waves); the speed of sound in the plane measured by a passenger onboard will have the familiar value of around 330 m/s (at standard room temperature of course).
I mean from the ground observer, the speed of sound is this?
##\frac{s+p}{1+(s*p)/c^2}##
S: the speed of sound
P: the speed of plane
C: the speed of light
 
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  • #106
Stephanus said:
And if a rocket travels at 0.6c for example and it shine a laser at the front,
so the speed of the laser plus the speed of the rocket is...
##\frac{r+l}{1+(r*l)/c^2}##
r is the speed of the rocket
l is the speed of the laser, which is equal to the speed of light
c is the speed of light.
So we can rewrite the formula into...
##\frac{r+c}{1+(r*c)/c^2} = \frac{r+c}{1+r/c} = \frac{r+c}{c/c+r/c} = \frac{r+c}{(r+c)/c} = c*\frac{r+c}{r+c} = c##
Is that so?
Yes, but again it should be no surprise that this works - the velocity addition formula is derived from the Lorentz transformation, which in turn is based on the premise that ##c## is constant for all inertial observers. It's essentially a "reverse" argument.
Stephanus said:
I mean from the ground observer, the speed of sound is this?
##\frac{s+p}{1+(s*p)/c^2}##
S: the speed of sound
P: the speed of plane
C: the speed of light
Yes, that's correct.
 
  • #107
@Stephanus I can see that you have some confusion regarding multiple concepts in SR. I've tried to compile a checklist that you can use to understand stuff better. Try to understand the concepts in the order in which I'm describing them, and you will probably get some clarity on the subject.

Since you know about the velocity composition law, I recommend that you look up the relativistic momentum derivation. If you have the necessary background in calculus and a little bit of conics (check if your calculus knowledge is sufficient by using this link: http://tutorial.math.lamar.edu/Classes/CalcII/CalcII.aspx ), then you should be able to understand the derivation for ##E=mc^2## here: http://www.emc2-explained.info/Emc2/Deriving.htm#.VXq9OMu3TqB .

After this, try undestanding what a "manifold" really is and why a concept of invariant "distance" between two points on a manifold is useful. Since spacetime is a manifold on which each point is described as an "event", you should be able to prove for yourself that ##ds^2 - dt^2## works as an expression which computes the same spacetime interval (spacetime version of invariant "distance") between 2 points, regardless of the inertial frame from which it is viewed (the expression is invariant under the Lorentz transformation; additionally, ##ds^2 ## is the spatial distance between two events in the expression above, and it can take different forms depending on the type of coordinates you're using and has the familiar form of ##dx^2 + dy^2 + dz^2 ## when Euclidean coordinates are used).

Are you familiar with spacetime diagrams? If no, then learn them now. If yes, then draw the hyperbola represented by the spacetime interval equation for the case of one dimensional motion (1D because its easy). The twin paradox is most easily understood graphically, and the same applies to time dilation and length contraction. Search up older threads here at PF or do some Googling on it, and you'll marvel at the usefulness of spacetime diagrams. Twin paradox confusion down!

Now you're all set for the 4 vector formalization. Understand that the Lorentz transformation can be used for calculating the components of any 4 vector in another reference frame. There are plenty of notes on four-velocity, four-acceleration, four-momentum and four-force out there on Google. Drink them up - you'll bag in relativistic dynamics right there.

A simple application of four-momentum vectors for light will give you the relativistic version of the Doppler effect. Two down!

...whoops, I forgot what else was confusing you :-p At any rate, I recommend that you do this first - I'm sure 99% of your problems will dissolve along the way! Of course, other users are welcome to suggest changes to this little chronology of concepts. I by no means proclaim that this list is comprehensive or exhaustive - the invaluable inputs of other users can help tailor this list, and you can finally say "I nailed SR!"
Remember that search engines are your friends.

Note: if explanations of key concepts online prove to be too difficult for you, buy an undergraduate level book on Relativity. I recommend Schutz's.
 
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  • #108
Thank you very much PWiz.
This the answer that I want! :smile:
You've given me the direction and what I have to learn.
 
  • #109
Ibix said:
The Lorentz transforms imply that space and time are one four-dimensional whole called space-time - [..]the equivalent of Pythagoras' theorem is ##c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}##. [..]
It's then easy to see that if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) then to (10,0,0,0) (i.e., take five years to get three light years away in the x direction, then turn round and come back) that the ##\Delta\tau##s are different - ##\sqrt{10^2-0^2}=10## years for me, ##\sqrt{5^2-3^2}+\sqrt{5^2-(-3)^2}=8## years for you.
Thanks Ibix for your explanations!.
It's clear (at least I can understand some :smile:) for me.
Okay...
This equation: ##c\Delta\tau = \sqrt{c\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2}##
So...
##\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}## is phytagorean hypotenuse in 3D.
##c\Delta t## looks like spatial dimension. Speed (of light) multiplied by time.

Let me see if I understand this. Come on, I'm not Albert who could devise a very complicated equation and KNEW that it was (is) TRUE!
if I move from an event at (t,x,y,z)=(0,0,0,0) to (t,x,y,z)=(10,0,0,0) (i.e., stay put for ten years) while you travel from (0,0,0,0) to (5,3,0,0) [..]]
It implies that distance units here are in time light speed takes. Not in miles, yard, km, feet.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
How much speed should he take to reach 3 ly wrt me and come and in 8 years me?
What is his clock after he comes back?

Okay...
##8 = \sqrt{t^2-3^2} + \sqrt{t^2-3^2}##
##8 = 2 * \sqrt{t^2-9}##
##t = 5##

A. Is that HOW I should solve the question/problem?
B. What does it means? 8 years for me is 10 years (5+5) for him?
C. His speed is ... to travel 3 ly (3 ly in my frame) he takes 5 years (5 years in my frame), so it is 0.6c (wrt me)? or
D. His speed is ... to travel 3 ly (3 ly in his frame) he takes 5 years (5 years in his frame), so it is 0.6c (wrt me)?

or
-------------------------------------------------
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
-------------------------------------------------
is the wrong question.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (4,3,0,0,0) comeback...
T should be locked to 4
And the question is...
What is his speed?
What is his time?
then...
His speed is 0.75c
His time...
##2 * \sqrt{4^2 - 3^2} = 5.291503##
Is that true?
 
  • #110
Stephanus said:
It implies that distance units here are in time light speed takes. Not in miles, yard, km, feet.
Yes - if I measure distance in light years and time in years, then the speed of light is 1ly/y (also light seconds and seconds, light months and months etc etc). It's a "smart" choice of unit that makes calculations easier because multiplying by c is trivial.

Stephanus said:
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
Well, you're gaining dimensions as you go, but presuming you mean (8,0,0,0) and (t,3,0,0) then ok...

Stephanus said:
How much speed should he take to reach 3 ly wrt me and come and in 8 years me?
What is his clock after he comes back?

Okay...
##8 = \sqrt{t^2-3^2} + \sqrt{t^2-3^2}##
##8 = 2 * \sqrt{t^2-9}##
##t = 5##

A. Is that HOW I should solve the question/problem?
No. ##\Delta\tau## is the time on the traveller's clock; the quantity you have called t (##\Delta t## would have been better) is the coordinate time in your rest frame when the traveller arrives at 3ly. So what you have calculated is that, if the traveller is to experience 8 years total, then you would have to experience 5 years on each leg (a 10 year trip, from your perspective).
Stephanus said:
B. What does it means? 8 years for me is 10 years (5+5) for him?
No - 8 years for him is 10 for you.
Stephanus said:
C. His speed is ... to travel 3 ly (3 ly in my frame) he takes 5 years (5 years in my frame), so it is 0.6c (wrt me)? or
This is correct - but not what you said above. He takes five years on each leg, according to you, so his speed is 0.6c.
Stephanus said:
D. His speed is ... to travel 3 ly (3 ly in his frame) he takes 5 years (5 years in his frame), so it is 0.6c (wrt me)?
No. In his rest frame, his speed is zero. His destination comes to him at 0.6c, and takes 4 years (half of the 8 year total trip time), which will let you calculate the length-contracted distance it travels.
Stephanus said:
or
-------------------------------------------------
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (t,3,0,0,0) comeback...
-------------------------------------------------
is the wrong question.

For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
and someone dashes off 3 lights years away from me and (4,3,0,0,0) comeback...
T should be locked to 4
And the question is...
What is his speed?
What is his time?
then...
His speed is 0.75c
His time...
##2 * \sqrt{4^2 - 3^2} = 5.291503##
Is that true?
This seems to be correct.
 
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  • #111
Thank you, thank you Ibix for your answer.
And thanks for your previous formula :smile: ##c\Delta\tau = \sqrt{c\Delta t ^2 - \Delta x^2 - \Delta y^2 - \Delta z ^2}##
Ibix said:
Stephanus said:
For instance, I stay put for 8 years (0,0,0,0) to (8,0,0,0,0)
Well, you're gaining dimensions as you go, but presuming you mean (8,0,0,0) and (t,3,0,0) then ok...
Wow, did I type four zeros?

Ibix said:
No. ##\Delta\tau## is the time on the traveller's clock;
Ahhh, thank you, thank you... If only I had pondered two more minutes before I started calculating!

Ibix said:
No. In his rest frame, his speed is zero. His destination comes to him at 0.6c[...]
PeroK said:
But, A is always stationary in its own reference frame. A cannot travel at 0.8c in its own reference frame. You need to rethink your questions based on that.
Need confirmation here. Must all the calculation in SR be based on the obeserver REST FRAME? Must the observer assume or see that EVERYTHING is moving, and he STAYS?
Because I haven't got confirmation, yet.
PeroK said:
There's no way to measure absolute motion. No observer can say "I'm travel absolutely at 100m/s or 0.8c or whatever". All an observer can say is that they are moving with respect to something else
Should we change to this?
"All an observer can say is that they everything else are moving with respect to something else the observer"

Ibix said:
(4,3,0,0)
This seems to be correct.
Of course! I do understand now!

Thanks for your help so far. I know these questions are not important. I'm just a computer programmer, not a physicist, teacher much less scientist. But these twins paradox and universe absolute frame of reference are always haunting me at night. Before I go to sleep, it's always, why, why, why :smile:
Perhaps, if an amateur like me can understand this theory, those mentors, professors would know how to teach their students.
What path should they take to make their students understand.
Math are important of course. NGT( Neil Degrasse Tyson) once (only once?) said that "Math is the language of the universe". And so far with all these questionings and answers, I haven't met differential integral, yet. Perhaps the integral nightmare will pop up somewhere later :smile:
 
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  • #112
Stephanus said:
[..]
Need confirmation here. Must all the calculation in SR be based on the obeserver REST FRAME? Must the observer assume or see that EVERYTHING is moving, and he STAYS?
Because I haven't got confirmation, yet.Should we change to this?
"All an observer can say is that they everything else are moving with respect to something else the observer"
No, the observer can choose to be in a "moving" frame. You can choose any inertial frame you like and pretend that to be "rest frame". For example GPS calculations are the easiest by choosing the "Earth Centered Inertial" frame. Therefore, when you look at your GPS coordinates when you are sitting in your garden chair, your GPS receiver assumes that you are in a "moving frame". It gives your position and velocity (zero) relative to your moving frame.
 
  • #113
@harrylin
Yep, Perok has confirmed me.
Thanks!
PeroK said:
And, in fact, when doing classical problems, choosing the best reference frame often makes the problems easier to solve. For example, if you have two objects moving towards each other, you could study this from the point of view (rest frame) of either object; or, the reference frame where their centre of mass is at rest - which is often very useful. Or, of course, from your reference frame as a "stationary" observer.
 

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