Clocks, rulers... and an issue for relativity

[studiot]

 

 

 

 

So, the 4D spacetime is not full, isn't it? Like a circle is a part of a plane, the equation that you posted (s² = x² + y² + z² -c²t²) describes what kind of geometric figure?

 

 

First thanks for bringing my slip in post#16 to my attention. It is now corrected.

 

The answer to your question can be found here.

 

Wikipedia is good enough.

 

https://www.google.co.uk/search?q=ivariant+interval+in+relativity&rlz=1C1AVNG_enGB673GB673&oq=ivariant+interval+in+relativity&aqs=chrome..69i57.9579j0j8&sourceid=chrome&ie=UTF-8#q=invariant+interval+in+relativity

Edited August 16, 2016 by studiot

[michel123456]

 

 

First thanks for bringing my slip in post#16 to my attention. It is now corrected.

 

The answer to your question can be found here.

 

Wikipedia is good enough.

 

https://www.google.co.uk/search?q=ivariant+interval+in+relativity&rlz=1C1AVNG_enGB673GB673&oq=ivariant+interval+in+relativity&aqs=chrome..69i57.9579j0j8&sourceid=chrome&ie=UTF-8#q=invariant+interval+in+relativity

Wikipedia is not enough.

 

It is not mentioned that spacetime is not full.

It looks like Euclidian space with the addition of the time dimension.

But Euclidian space is full. I mean the coordinates are free, there is no constraint.

When one introduces time, the constraint appears, as you explained in your post. Which means that the 4 coordinates are not free. You posted the formula.

So, again, what is the Spacetime like? If it is not full? What kind of geometrical object?

Edited August 16, 2016 by michel123456

[studiot]

Wikipedia was only the first reference in the link I posted. But as the subject is so popular Wiki is particularly strong here.

 

You asked what is the 'shape' generated by the equation I gave (it also appears in Wiki in their usual highbrow form, they rarely go in for plain speaking)

 

It is the equation of the famous lightcone hypersurface in Minkowski space.

 

But remember that Minkowski spacetime is flat ie Euclidian ie unperturbed by gravity.

 

That is special relativity.

 

If you move to general relativity, gravity introduces further constraints that modify this.

However the lightcone contstraint is a strong one, the gravitational one quite weak in most of the universe.

 

The point is that, like the circle, we can put any number we like to any of the coordinates (x,y,z,t), the constraints simply prevent access to some of these.

So it becomes a philosophical question, as with the circle,

Is the rest of spacetime still there or for the circle is the rest of the pq plane still there?

This is interesting because it is the obverse argument to the one about the existence of a hole, which requires surrounding reality (matter) to define it.

[michel123456]

Wikipedia was only the first reference in the link I posted. But as the subject is so popular Wiki is particularly strong here.

 

You asked what is the 'shape' generated by the equation I gave (it also appears in Wiki in their usual highbrow form, they rarely go in for plain speaking)

 

It is the equation of the famous lightcone hypersurface in Minkowski space.

 

But remember that Minkowski spacetime is flat ie Euclidian ie unperturbed by gravity.

 

That is special relativity.

 

If you move to general relativity, gravity introduces further constraints that modify this.

However the lightcone contstraint is a strong one, the gravitational one quite weak in most of the universe.

 

The point is that, like the circle, we can put any number we like to any of the coordinates (x,y,z,t), the constraints simply prevent access to some of these.

So it becomes a philosophical question, as with the circle,

Is the rest of spacetime still there or for the circle is the rest of the pq plane still there?

This is interesting because it is the obverse argument to the one about the existence of a hole, which requires surrounding reality (matter) to define it.

Yes (the bold part)

And no it is not philosophical, it is physics.

Would you agree with me if I say that the parts out of the lightcone (inside and outside) are not directly observable?

Edited August 16, 2016 by michel123456

[studiot]

 

Would you agree with me if I say that the parts out of the lightcone (inside and outside) are not directly observable?

 

Isn't the lightcone the boundary of what is accessible?

[Mordred]

No thats not how lightcones work. Every observable event has its own lightcone to an observer. The different regions on the lightcone represent whether the event is spacelike,timelike or lightlike.

Yes (the bold part)

And no it is not philosophical, it is physics.

Would you agree with me if I say that the parts out of the lightcone (inside and outside) are not directly observable?

this statement though close isn't precisely accurate. The lightcones represent only what is possibly causally connected. If there is no possible causal connection you can't define a lightcone as a light signal can never reach an observer.

Edited August 16, 2016 by Mordred

[michel123456]

No thats not how lightcones work. Every observable event has its own lightcone to an observer. The different regions on the lightcone represent whether the event is spacelike,timelike or lightlike.

 

this statement though close isn't precisely accurate. The lightcones represent only what is possibly causally connected. If there is no possible causal connection you can't define a lightcone as a light signal can never reach an observer.

That is all ok to me (almost).

_The red part I do not understand. I suppose you meant that every observer at any instant has its own lightcone. Because your statement looks like the reverse and must be inaccurate: for any event they are a multitude of lightcones, a multitude of different (possible) observers.

_The bold part is what I am going to. Whatever we are observing from the universe is carried by light signals. Whatever we are observing from the universe lies upon the surface of the lightcone.

Edited August 16, 2016 by michel123456

[Mordred]

Its better to think of it as casually connected to an observer. Rather than directly observable. This becomes an important distiction when you involve expansion and contraction. Which can play havoc on lightcone worldlines.

 

Yes each observer will have his own set of possibly causally connected events. Or lightcone.

 

look at how the 45 degree line is determined as an event type boundary. This 45 degree line will change when you start adding curvature influences.

 

I'm hinting at an important distinction between SR lightcones to other types ie GR lightcones. Too often ppl set in stone the 45 degree cone and ignore influences upon that 45 degree line. Ie GR itself.

Edited August 16, 2016 by Mordred

[MigL]

Sorry String Junky, I was in a hurry this morning and didn't re-read my post.

You're right it was rarther confusing.

Meant to say if velocities match, then same frame.

If velocities don't match, so that one frame is moving with respect to the other, and one can be considered at rest, then different frames.

[Mordred]

for example the lightcone on this link is an Euclidean lightcone.

 

https://en.m.wikipedia.org/wiki/Light_cone

 

however the lightcone examples on this Cosmology page are different than the above.

 

http://www.astro.ucla.edu/~wright/cosmo_03.htm

 

cross posted with Migl

Edited August 16, 2016 by Mordred

[StringJunky]

Sorry String Junky, I was in a hurry this morning and didn't re-read my post.

You're right it was rarther confusing.

Meant to say if velocities match, then same frame.

If velocities don't match, so that one frame is moving with respect to the other, and one can be considered at rest, then different frames.

That's ok mate. I figured it out what you meant to say.

[Mordred]

Migl an important study hint to my last post. Worldline and line element in relation to lightcones.

 

https://en.m.wikipedia.org/wiki/Line_element

 

https://en.m.wikipedia.org/wiki/World_line

[michel123456]

for example the lightcone on this link is an Euclidean lightcone.

 

https://en.m.wikipedia.org/wiki/Light_cone

 

however the lightcone examples on this Cosmology page are different than the above.

 

http://www.astro.ucla.edu/~wright/cosmo_03.htm

 

cross posted with Migl

Sorry for derailing this thread but:

I know this great Ned Wright's diagram (from you 2nd link)

 

On this diagram, where are the observable galaxies?

From my understanding, the entire observable universe is upon the red pear-shaped curve. The galaxies inside are not (directly) observable, and the galaxies outside are not (directly) observable either. It corresponds to the analogy of the circle on a sheet of paper as described by Studiot previously. IOW from the result of the Big Bang we are observing only a tiny part.

[Mordred]

What event type is defined by the red line?

 

a) spacelike

b) timelike

c) lightlike

 

What event types of the above choices represent the inner cone region?

 

How does the Worldline terms and line element relate?

 

study hint if your unsure

hint How does the above relate to Pythagoras theory.?

Edited August 16, 2016 by Mordred

[michel123456]

lightlike

------------------

Do I have to split blood before getting a straightforward answer?

 

Are all the triangles on the diagram observable by the observer that stands at the summit of the cone?

Edited August 16, 2016 by michel123456

[studiot]

Thanks, Mordred for posting links to light cones, and cosmological stuff which is way beyond my knowledge. +1

 

I was going to post a discussion of light cones and once again stress that they are the boundary of a region which is fully accessible.

 

Yes the hypersurface is lightlike, Michel do you still need my contribution on this?

 

I can't help with the cosmology, except for observing that it clearly takes into account the simplistic comment I made about SR v GR and gravity.

Edited August 16, 2016 by studiot

[Mordred]

only the enclosed triangles are observable by the present day observer.

lightlike

------------------

Do I have to split blood before getting a straightforward answer?

 

Are all the triangles on the diagram observable by the observer that stands at the summit of the cone?

Edited August 16, 2016 by Mordred

[michel123456]

only the enclosed triangles are observable by the present day observer.

Oh. For example the enclosed triangles that are upon the straight vertical line, are they observable? IOW can we directly observe ourselves in our own past when looking into a telescope?

The blue ones. Are they observable directly?

 

PS: yes they are causally connected. It is our past after all. But can we observe directly ourselves in the past?

Thanks, Mordred for posting links to light cones, and cosmological stuff which is way beyond my knowledge. +1

 

I was going to post a discussion of light cones and once again stress that they are the boundary of a region which is fully accessible.

 

Yes the hypersurface is lightlike, Michel do you still need my contribution on this?

 

I can't help with the cosmology, except for observing that it clearly takes into account the simplistic comment I made about SR v GR and gravity.

Sure, please stay. You asked a very profound question about the metaphysical existence outside the boundary of the spacetime hypersurface, am I right?

Edited August 17, 2016 by michel123456

[Endy0816]

Mirror?

[Mordred]

Oh. For example the enclosed triangles that are upon the straight vertical line, are they observable? IOW can we directly observe ourselves in our own past

No you can never observe yourself in the past. Nor can you ever observe the same object in both the present and past. Its always one or the other never both. Not counting reflections

Edited August 17, 2016 by Mordred

[michel123456]

No you can never observe yourself in the past. Nor can you ever observe the same object in both the present and past. Its always one or the other never both. Not counting reflections

Thank you.

[robinpike]

Robin, try this explanation of the twins.

 

Call the stay at home twin A and the travelling twin B.

 

A never leaves Earth.

So A never arrives anywhere else.

 

This is the crux of the difference.

 

B leaves Earth

B arrives somewhere else.

B returns to Earth.

 

So B travels within the universe from one place to another, visiting places in between.

 

If we were to suggest that what B sees is equivalent to what A sees, but the other way round, then we would be just plain wrong.

If we suggest that B sees the rest of the universe recede and then return to B, it would imply that the rest of the universe must arrive somewhere else.

But B can determine when he returns, after consultation with A, that A never went anywhere.

So at least a part of the rest of the universe did not go anywhere (ie did not arrive somewhere else).

Suggesting that the universe recedes is the same as suggesting that B travels outside the universe.

 

What swansont is suggesting in post#12 is that A's journey is not through space at all.

It is only through time.

In terms of Frames, A is in frame (x,y,z,t) and B is in Frame (X,Y,Z,T)

 

A's journey is (0.0,0,0) ; (0,0,0,t₁) ; (0,0,0,t₂) ; (0,0,0,t₃) ; (0,0,0,t₄) ; (0,0,0,t₅) ; etc

 

On the other hand B sees his journey as (0,0,0,0) ; (X₁,0,0,T₁) ; (X₂,0,0,T₂) ; (X₁,0,0,T₂) ; (0,0,0,T₅)

 

Setting the X axis along B's journey to Betelgeuse, which he reckons is X₂ distant.

This makes the travel along the Y and Z axes all 0.

 

The fact the two frame origins were momentarily in coincidence at the start is the result of synchronisation, but since there is relative motion between the frames there is no reason to expect them to coincide at any other event.

 

What is happening is that they coincide in space and time the first event due to synchronisation , but the second time they coincide in space, but not in time since t₅ is not the same as T₅.

 

Here is the issue re-iterated using co-ordinates to notate the space and time of the stay at home clock and the travelling clock.

 

If we use the stay at home clock's inertial frame as the base reference in space and time, we can represent the space and time co-ordinates (x,y,z,t), say starting as these values (0,0,0,t1), the two clocks starting with the same space-time co-ordinates.

 

When the travelling clock leaves for its round trip journey, the stay at home clock's co-ordinates vary just with a steady movement through time, say as...

 

• stay at home clock: (0,0,0,t1), (0,0,0,t2), (0,0,0,t3), (0,0,0,t4), (0,0,0,t5), (0,0,0,t6), (0,0,0,t7)

 

Whereas when the travelling clock starts its journey, it accelerates into a different inertial frame of space-time, say its co-ordinates become...

 

• travelling clock: (0,0,0,t1), (X2,0,0,T2), (X3,0,0,T3)

 

Midway in its journey at X3 distance, the travelling clock accelerates into a different inertial frame of reference of space-time in order to return home, say as...

 

• travelling clock (X3,0,0,T3), (X2,0,0,T4), (X1,0,0,T5)

 

finally accelerating (de-accelerating) into the inertial frame of reference into the stay at home clock's original position in space, but NOT time

 

• travelling clock returns as (0,0,0,T6)

where the travelling clock's position in time (T6) is less than the stay at home clock's position in time (t7)

 

For simplicity I will ignore mutual reciprocal differences between inertial reference frames, and just concentrate on the final real difference, that is the less time on the travelling clock and look at explanations of how that occurred.

 

Perhaps the simplest to understand is to suggest that the rate of time slowed down while the travelling clock made its journey, but this is at odds with the premise that all inertial frames of reference are of equal standing, for this reason. Let's say as the travelling clock moved from the stay at home's inertial frame of reference, its rate of time slowed down. But then at its halfway point in its journey, when it accelerated into its return journey inertial frame of reference, its rate of time must slow down again - as there is no way to differentiate this change in inertial reference frames as any other. And finally, when it returns to the stay at home clock's inertial frame of reference, again its rate of time must slow down. But this is at odds with the final position, because although the travelling clock has lost time, its rate of time is nonetheless the same as the stay at home clock's rate of time.

 

So let's try another explanation.

 

When the travelling clock changes inertial frames of reference, its rate of time never changes, but instead its rate through space-time changes, it getting moving through space-time in fewer clock ticks than the stay at home clock. But again this has the same issue: if changing inertial frames of reference cause the travelling clock to move through space-time quicker, there is no change in inertial reference frame by which it can return to the stay at home clock's rate of movement through space-time.

 

With the premise that all inertial frames of reference are of equal standing, there seems to be no method by which the travelling clock can lose time AND return to the stay at home clock's rate of time / movement through space-time?

Edited August 22, 2016 by robinpike

[swansont]

 

Perhaps the simplest to understand is to suggest that the rate of time slowed down while the travelling clock made its journey, but this is at odds with the premise that all inertial frames of reference are of equal standing, for this reason. Let's say as the travelling clock moved from the stay at home's inertial frame of reference, its rate of time slowed down. But then at its halfway point in its journey, when it accelerated into its return journey inertial frame of reference, its rate of time must slow down again - as there is no way to differentiate this change in inertial reference frames as any other. And finally, when it returns to the stay at home clock's inertial frame of reference, again its rate of time must slow down. But this is at odds with the final position, because although the travelling clock has lost time, its rate of time is nonetheless the same as the stay at home clock's rate of time.

 

 

When the clock accelerates at the turnaround, there will be an accumulation of a difference in phase. The earthbound twin does not see the spaceship's clock running slower (edit: slow-> slower) on the return trip. The difference in phase continues to accumulate..

http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

[studiot]

I think, Robin, that this is the beginning of your difficulty.

 

 

 

Whereas when the travelling clock starts its journey, it accelerates into a different inertial frame of space-time

 

Yes the travelling clock accelerates into a different frame, whose origin (for time at least) is not the same as the origin for the stay at home clock

 

So the t0 and t1 etc for the travelling clock are not the same t0 t1 etc as for the stay at home clock.

 

That is why I used different letters for these frames.

 

Why is this?

 

Well suppose instead of being twins, B was already travelling in the travelling frame and just happened to be passing A at t0 in the stay at home frame.

 

B then makes the two way journey.

 

Can you tell what would be the difference between their clocks on B's return?

[michel123456]

You can make the scenario even funnier:

take 2 twins traveling, one to the East, the other to the West (for the ease of understanding), and a 3rd observer on Earth. Will the 2 twins have the same age when coming home?

Obviously the answer is Yes.

What will be their mutual observations while traveling?