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What is time? (Again)


The victorious truther

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On 8/15/2020 at 5:35 PM, Markus Hanke said:

That would probably help things - it's in chapter 15 of the book. The essential train of thought is this - suppose you have an elementary 4-cube of spacetime Ω . We know that energy-momentum within that cube is conserved, so (in differential forms language):

[...]

This is the exact same as the conservation law for energy-momentum given above. So we can associate the two: [...]

So, using the concept of a moment of rotation, some elementary geometric considerations, and the "boundary of a boundary is zero" principle, the form of the Einstein equations is uniquely fixed up to a proportionality constant. MTW even obtain this constant somehow, though I don't quite follow their thoughts on this minor detail.

Anyway, that's the general idea. If you can get access to the text, it is all described in much more detail there.

Ah, I think I know what you mean with the Cartan dual. It has g tensors in the definition, doesn't it?

Don't worry, I'll take a closer look, ASAP.

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On 8/18/2020 at 3:57 PM, joigus said:

Ah, I think I know what you mean with the Cartan dual. It has g tensors in the definition, doesn't it?

Don't worry, I'll take a closer look, ASAP.

MTW never gave a formal definition for this operator, other than to say it acts only on contravariant vectors, but not on forms. I think they might just have invented it :)

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On 8/18/2020 at 4:21 PM, swansont said:

It’s the usual situation one is in. And the corrections of relativity are usually not noticeable under virtually all circumstances the average person encounters.

Since we are observing galaxies receding at high velocities, doesn't that mean that reversely, observers living on that galaxy are observing us as if we are moving at close to C velocity?

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14 minutes ago, michel123456 said:

Since we are observing galaxies receding at high velocities, doesn't that mean that reversely, observers living on that galaxy are observing us as if we are moving at close to C velocity?

Non-sequitur. We are not observing our relative motion, per se, we are observing effects of expansion. Special relativity implies situations where we don’t have to account for expansion.

Regardless, unless you are measuring the redshift, you aren’t noticing the effects, so my statement about “usually not noticeable under virtually all circumstances the average person encounters” still holds. 

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1 minute ago, swansont said:

Non-sequitur. We are not observing our relative motion, per se, we are observing effects of expansion. Special relativity implies situations where we don’t have to account for expansion.

Regardless, unless you are measuring the redshift, you aren’t noticing the effects, so my statement about “usually not noticeable under virtually all circumstances the average person encounters” still holds. 

So are you saying that we are not observing galaxies moving at close to C velocity?

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6 minutes ago, michel123456 said:

So are you saying that we are not observing galaxies moving at close to C velocity?

If you want to discuss expansion, I suggest you open a new thread rather than continuing to hijack this one 

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On 8/17/2020 at 3:00 PM, geordief said:

Is there  a problem if we just go the whole hog and accept the  interpretation that time is indeed  just the flip side of spatial distance?

Does our experience of time differ from our experience of spatial  distance in any way?

It is said that you can go "back" in space but not in time ,but is that really so?

 

Maybe we can only go forward in space as well.That is how I look at it,anyway;we never return to the sane place in space and always keep moving forward ,never back.

 

But are there ways where it can be correctly said that time is fundamentally different from space.?

(...)

To me, no time cannot be fundamentally different from space. The reason why I think so is that if they mix so well, they must have some "common stuff".

"the flip side of spatial distance" yes, some sort. One can make a parallelism between time & distance: both are always positive, both have 1 dimension. It had even come to my mind that distance & time are the same & one thing.

But:

1.Time has a negative signature. It is "the flip" of distance but what does that mean anyway??

2. As Swansont points out, Time is recognized differently from distance. The ticking of the clock remains to be explained.

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2 hours ago, swansont said:

You can be at rest with respect to space, but not time. 

It is the model that shows two objects as being at rest,isn't it?

 

In the real world they never are at rest  except as an approximation.

 

Is it possible that in extreme situations**  that that  approximation becomes less accurate and at the same time  the "un- at- restness" of time becomes closer and closer to the approximation that was occupied by  space?

 

I am just speculating (there is no hope of me  actually  learning the subject)

 

**ie extreme gravity

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2 hours ago, geordief said:

It is the model that shows two objects as being at rest,isn't it?

Normally only talking about the X, Y, Z coordinates being unchanging with respect to something to define at rest. While there is always be something increasing or decreasing in distance from you(relative motion) we also don't define being at rest in universal terms either.

We are all always 'moving' at 1second per second though in terms of time. This is still change in respect to your previous coordinates.

ie.

X, Y, Z, T

vs

X, Y, Z, T+1

 

 

 

 

 

Edited by Endy0816
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7 hours ago, geordief said:

It is the model that shows two objects as being at rest,isn't it?

One object at rest with respect to another object

Quote

In the real world they never are at rest  except as an approximation.

This is special relativity, which is a classical theory, but yes it would be an approximation.

The SR effects are negligible.

 

Quote

Is it possible that in extreme situations**  that that  approximation becomes less accurate and at the same time  the "un- at- restness" of time becomes closer and closer to the approximation that was occupied by  space?

 

I am just speculating (there is no hope of me  actually  learning the subject)

 

**ie extreme gravity

SR means no gravity. 

If you want to follow that tangent, I will give the same advice I gave Michel — start a new thread.

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39 minutes ago, michel123456 said:

You cannot, but a photon can.

Can a photon do that? In what frame of reference does that happen? 

(Answer: 'No' and 'None'; as far as I know. In relativity the photon is not a valid frame of reference)

 

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1 hour ago, Ghideon said:

Can a photon do that? In what frame of reference does that happen? 

(Answer: 'No' and 'None'; as far as I know. In relativity the photon is not a valid frame of reference)

 

In a medium  a photon is  a valid frame of reference isn't it?

Is it more correct to say that a photon is not a valid frame of reference in a vacuum (which may not exist except in the model) ?

Edited by geordief
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55 minutes ago, geordief said:

In a medium  a photon is  a valid frame of reference isn't it?

Is it more correct to say that a photon is not a valid frame of reference in a vacuum (which may not exist except in the model) ?

No, photons travel at c. "Light speed in a medium" is not "photon speed in a medium" Light propagates at c/n, but photons still travel at c. The delay is in absorption/emission involving virtual states.

A photon does not possess an inertial frame of reference.

 

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2 minutes ago, swansont said:

No, photons travel at c. "Light speed in a medium" is not "photon speed in a medium" Light propagates at c/n, but photons still travel at c. The delay is in absorption/emission involving virtual states.

A photon does not possess an inertial frame of reference.

 

Ah yes, I forgot that.

 

 

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22 hours ago, Ghideon said:

Can a photon do that? In what frame of reference does that happen? 

(Answer: 'No' and 'None'; as far as I know. In relativity the photon is not a valid frame of reference)

 

In the observer's FOR.

You are observing a particle getting away from you at near to C velocity. This particle appears time dilated. Theory tells us that if you could observe a particle going at FTL (tachyon), you would observe it going backward in time. So I am suggesting that the flip happens at C (between going normally in time & backward in time). Which means that a photon going away, as observed by you, would look as hovering still in time.

I suppose that the same counts for a photon getting close to you.

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6 hours ago, michel123456 said:

In the observer's FOR.

You are observing a particle getting away from you at near to C velocity. This particle appears time dilated. Theory tells us that if you could observe a particle going at FTL (tachyon), you would observe it going backward in time. So I am suggesting that the flip happens at C (between going normally in time & backward in time). Which means that a photon going away, as observed by you, would look as hovering still in time.

I suppose that the same counts for a photon getting close to you.

No, in the observer’s frame the photon moves at c. There is no frame of the photon, so there is no time dilation for the photon. You can only observe time dilation with respect to some other frame of reference.

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8 hours ago, michel123456 said:

In the observer's FOR.

You are observing a particle getting away from you at near to C velocity. This particle appears time dilated. Theory tells us that if you could observe a particle going at FTL (tachyon), you would observe it going backward in time. So I am suggesting that the flip happens at C (between going normally in time & backward in time). Which means that a photon going away, as observed by you, would look as hovering still in time.

I suppose that the same counts for a photon getting close to you.

Don't confuse relativistic Doppler effect with time dilation.  Relativistic Doppler effect depends on whether the object is going way or coming towards you.  Tine dilation does not.

Theory does not predict "negative" time dilation (time reversal) for FTL objects. Just look at the time dilation equation. if v exceeds c, you end up with the square root of a negative number, for which there is no answer.   For a photon, where v=c, you end up with 1/0, which is undefined.

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1 hour ago, Janus said:

  For a photon, where v=c, you end up with 1/0, which is undefined

 

When we talk about singularities which involve infinities ,it is said (I think) that the GR model has broken down.

Why don't we say the same when  you have that situation?

Or do we? Is that what saying the photon's  frame of reference is undefined or invalid can also be understood to mean?

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Division by zero is definitely undefined, and although you might think that faster than ight tachyonic motion might have some possibilities, it does create more problems than solutions. Even re-interpreting causalty breaking, FtL, time reversed motion from source to target as forward time motion from target to source has itsproblems.
The Lorentz invariant energy solution does however, lead to a negative root, and imaginary mass quantum fields.
Tachyonic filds have become important in modern theoretical physics.
The negative mass is interpreted as instability to condensation of the field,and excitations of the field ( particles ) are not tachyonic.

https://en.wikipedia.org/wiki/Tachyon_condensation

It is also used in the standard model, with associated symmetry breaks, and have read the Higgs boson field has an imaginary mass in its uncondensed phase

Edited by MigL
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Question about the (lack of) photon frame of reference in relation to the initial question. 

2 hours ago, Janus said:

For a photon, where v=c, you end up with 1/0, which is undefined.

and

4 hours ago, swansont said:

There is no frame of the photon

I clearly see how the theories tells us that massless particles are not a valid frame of reference. My question, in layman terms due to lack of knowledge of related or underlaying theories: Our models of massless particles seems to prevent us from "assigning a clock" to ride along with anything massless, but how come that it is so? Are there deeper connections between mass* and time described somewhere? Does the fact that it seems tricky to speak of "passage of time" from the point of view of something massless tell something about the nature of time, according to accepted models**?

 

*) and energy due to E=mc2
**) I'm aware of general relativity and gravitational time dilation, my question is about assigning time and frame of reference to something with zero invariant mass. Sorry if the question has been addressed already; there are some parts in the thread that I may have not fully grasped (yet).

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25 minutes ago, Ghideon said:

 My question, in layman terms due to lack of knowledge of related or underlaying theories: Our models of massless particles seems to prevent us from "assigning a clock" to ride along with anything massless, but how come that it is so? Are there deeper connections between mass* and time described somewhere? Does the fact that it seems tricky to speak of "passage of time" from the point of view of something massless tell something about the nature of time, according to accepted models**?

We lack a model to do so, and lack the ability to test any model one might propose. Our model works for massive particles.

We can only describe photons from our frame of reference.

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5 hours ago, swansont said:

No, in the observer’s frame the photon moves at c. There is no frame of the photon, so there is no time dilation for the photon. You can only observe time dilation with respect to some other frame of reference.

A light beam was send from the Earth 100 years ago (1920) into space. The first photons of this beam are today 100LY away. They are still in 1920.

Because if they hit the eye of some E.T., the E.T. will see the image of 1920.

IOW the photons are traveling in space but not in time.

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