# Time Explained

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From the first post.

A second isn’t some slice of spacetime, it’s just nine billion motions of a caesium atom.

I think I disagree here. A second is a slice of spacetime, although I prefer to call spacetime nothing at all. The tick and tock of any measure constitute markers for the non-event. It is the non-event that is sensed. In other words we have tick (an event) then no event at all, and then tock (another event). Time lies between the events. Events stand as the device if you will, for measure of nothing at all.

Granted we could not sense time in the absence of the markers, but that only implies an infinity of time, with no device for measure.

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So which one do you actually mean? Is C constant, or not?

You always measure it to be the same whatever reference frame you're in, so in this respect it's constant. But it's actually different in different reference frames. It's just that you don't notice it. So it's not really constant.

Imagine you blast off away from me and earth at .99c. I know from √(1-V2/C2) that your time experience is dilated by a factor of seven. I get you to measure the speed of light, in a tranverse direction to avoid any length-contraction complications. You find it's still the same old 300,000km/s. But your second lasts seven of my seconds. So your c is a seventh of mine.

It's similar if you're in a high-gravity situation and I'm not. Your c is less than mine, but you don't notice it. And at all points between us c has intermediate values, hence the local gradient.

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Pi: from what I've read, IMHO Einstein ended up thinking that spacetime was really space.

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Pi: from what I've read, IMHO Einstein ended up thinking that spacetime was really space.

I'll try to be a bit more specific. It is time that I refer to as being nothing at all. Space would be a combination of nothing, and geometric markers, and matter is essentailly the same as space. The difference between matter and space is that matter holds the foci of these geometric representations, while space is the extension of those geometries.

In this sense the universe is the definition of nothing. This is to say that nothing exist by way of the markers on a geometric scale. The markers can be considered points, wherein nothing (the non-event) between these points has meaning on your time scale (biological clock). We sense for instance the time of a horse race by comparison to this clock. The beginning of the horse race represents a marker, and the end of the race is another marker. What is between those markers is compared to the likewise non-event between your biological markers, by which a sense of time is presented.

Thats partly my understanding of time. It comes by way of the understanding that the universe is nothing more than conceptual geometric representations of non-existence, by which existence is the logical fallout by way of no other possibilty.

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You always measure it to be the same whatever reference frame you're in, so in this respect it's constant. But it's actually different in different reference frames. It's just that you don't notice it. So it's not really constant.

Imagine you blast off away from me and earth at .99c. I know from ?(1-V2/C2) that your time experience is dilated by a factor of seven. I get you to measure the speed of light, in a tranverse direction to avoid any length-contraction complications. You find it's still the same old 300,000km/s. But your second lasts seven of my seconds. So your c is a seventh of mine.

It's similar if you're in a high-gravity situation and I'm not. Your c is less than mine, but you don't notice it. And at all points between us c has intermediate values, hence the local gradient.

Wow, this is just spectacularly wrong.

You can't get away from length contraction issues in your measurement. Because of the motion, if you trace the path of the light that is transverse to the motion, it will have a component in the direction of motion, in the external viewer's frame. One object that would fit this description is the familiar light clock; the time is dilated, but the path is longer by the same factor.

c is constant and the same in all inertial frames

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Wow' date=' this is just spectacularly wrong.

You can't get away from length contraction issues in your measurement. Because of the motion, if you trace the path of the light that is transverse to the motion, it will have a component in the direction of motion, in the external viewer's frame. One object that would fit this description is the familiar light clock; the time is dilated, but the path is longer by the same factor.

c is constant and the same in all inertial frames[/quote']

If you're an external observer looking at a subject moving past you at .99c you see the lightclock like this /\/\/\/\. However if you're the subject you see it like this this |. You don't consider yourself to be moving. You measure c to be the same old 300,000,000 m/s. The transverse metre is unchanged, and the seconds are time-dilated sevenfold. Now if meters are the same but seconds are different, and c is metres per second, where does that leave c?

If you consider time dilation to be real, you must consider the different c to be real too. Yes, you measure c to be the same in whatever inertial reference frame you're in. But the only way it's the same across these different reference frames is if you're measuring it in terms of motion through a preferred reference frame, an absolute space. Space instead of spacetime. That's the TIME EXPLAINED position.

But let's park that for now, because we were talking about gravity as a tension gradient and a variable c. Edtharan, have you done any reading yet?

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If you're an external observer looking at a subject moving past you at .99c you see the lightclock like this /\/\/\/\. However if you're the subject you see it like this this |. You don't consider yourself to be moving. You measure c to be the same old 300,000,000 m/s. The transverse metre is unchanged, and the seconds are time-dilated sevenfold. Now if meters are the same but seconds are different, and c is metres per second, where does that leave c?

If you consider time dilation to be real, you must consider the different c to be real too. Yes, you measure c to be the same in whatever inertial reference frame you're in. But the only way it's the same across these different reference frames is if you're measuring it in terms of motion through a preferred reference frame, an absolute space. Space instead of spacetime. That's the TIME EXPLAINED position.

But let's park that for now, because we were talking about gravity as a tension gradient and a variable c. Edtharan, have you done any reading yet?

In your own frame, you do not see a dilated time. As you say, you don't consider yourself to be moving. Your own clock is unchanged. If variable c is the basis of your thesis, then your thesis is wrong.

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You always measure it to be the same whatever reference frame you're in, so in this respect it's constant. But it's actually different in different reference frames. It's just that you don't notice it. So it's not really constant.

But the external observer would measure a different value of C for you.

And as C is used in E=MC^2, then if a person in a gravity well (or travelling fast) annihilated some matter and anti matter, then the person doing the annihilating will measure one value for the energy released and the distant observer will measure a different value.

If this was a bomb, then the observers should see a different sized crater (not to mention the different trajectories of all the ejecta - so one observer might see a rock crush their house where as the other observer would not see it crush the house).

So, here again, we have two different, and mutually exclusive results occurring if we accept a variable value for C.

If you consider time dilation to be real, you must consider the different c to be real too.

No we don't. If we accept everything else to be relative, then there is no problem. If we accept Time to be a physical dimension as is Space and that, using geometry in the 4 Dimensions, then we do not need to have a variable value of C. In fact to get any sensible results, we need to have a constant value for C.

Yes, you measure c to be the same in whatever inertial reference frame you're in. But the only way it's the same across these different reference frames is if you're measuring it in terms of motion through a preferred reference frame, an absolute space.

Can you mathematically prove that? The maths and experiments in relativity indicate that this is not the case. The mathematics that underpin relativity do not come to this conclusion, so could you show us how it could?

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Let's try it another way:

Swanson, I give you a metre-wide light-measuring device that you hold transversely while a black box takes you for a trip. Edtharan, you stay in the lab with a similar device.

When Swanson gets back, his device shows that light went back and forth across his transverse metre a trillion times. As far as Swanson is concerned it travelled a trillion metres. Edtharan's device says the light went back and forth seven trillion times. As far as he was concerned it travelled seven million metres.

Swanson and Edtharan are both here now. But Swanson's light travelled a trillion metres instead of seven trillion metres. So Swanson's light travelled slower than Edtharan's. Swanson didn't notice it during the trip, because the devices are akin to light clocks. His whole body was similarly affected. He experienced less time because his light was running slower.

Like I said, you can never measure your local c to be anything other than 300,000km/s. You only see the difference when you compare frames in experiments like this.

Swanson's trip might have been a ride at .99c. Or it might have been to a high-gravity area. There's not much difference between the two, but in the latter situation there's less room for dispute about how far the light "really" travelled in some absolute reference frame.

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But the external observer would measure a different value of C for you.

Yep. Let's set Twin's Paradox symmetry to one side for a minute. If we had some kind of magic instant viewscreen that showed us the inside of Swanson's black box, we'd see his light going seven times slower than yours.

And as C is used in E=MC^2, then if a person in a gravity well (or travelling fast) annihilated some matter and anti matter, then the person doing the annihilating will measure one value for the energy released and the distant observer will measure a different value.

Yep. Look up electron volts. An electron at rest will annihilate with a positron releasing gamma rays of 511keV apiece. A relativistic electron annihiliating with a relativistic positron will release gamma rays of more than 511keV as far as the stationary observer is concerned. Don't get stuck on rest mass here - I take the view that something that has energy has mass. For example, a photon has mass of hf/c2.

If this was a bomb, then the observers should see a different sized crater (not to mention the different trajectories of all the ejecta - so one observer might see a rock crush their house where as the other observer would not see it crush the house).

No they wouldn't. You're going off on one here. If a rock hit a stationery observer it would cut his head a bit, but if it hit a relativistic observer it would take his head clean off.

So, here again, we have two different, and mutually exclusive results occurring if we accept a variable value for C. No we don't. If we accept everything else to be relative, then there is no problem. If we accept Time to be a physical dimension as is Space and that, using geometry in the 4 Dimensions, then we do not need to have a variable value of C. In fact to get any sensible results, we need to have a constant value for C.

I can only try to explain to the best of my ability. Look, here's the situation. The image below presents an illusion. The illusion is that square A is a different colour to square B. The reality is that square A is the same clour as square B. You're stuck thinking that the illusion is the reality. You can learn how to see that the squares are the same colour. And you can learn to see the simple truth of what I'm saying.

See http://www.echalk.co.uk/amusements/OpticalIllusions/illusions.htm and use the swatch to prove to yourself that A and B are the same colour.

Can you mathematically prove that? The maths and experiments in relativity indicate that this is not the case. The mathematics that underpin relativity do not come to this conclusion, so could you show us how it could?
Maybe. But it's a lot of hard latex, and it would be easier for me to find something somebody else has already done. But you haven't read any of the papers I've linked to, so maybe not.
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So, despite your claims to the contrary, you do not agree with special relativity.

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I do so agree with Special Relativity. But I also see what it's telling us. I see why the postulates hold, what they mean, and how it evolves. To the best of my knowledge I think the same as Einstein did. Not in 1905, but in 1949.

In his response to Godel's paper in the Schilpp volume, Einstein acknowledged that "the problem here disturbed me at the time of the building up of the general theory of relativity." This problem he described as follows: "Is what remains of temporal connection between world-points in the theory of relativity an asymmetrical relation (like time, intuitively understood, and unlike space), or would one be just as much justified to assert A is before B as to assert that A is after B? The issue could also be put this way: is relativistic space-time in essence a space or a time."

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Yep. Look up electron volts. An electron at rest will annihilate with a positron releasing gamma rays of 511keV apiece. A relativistic electron annihiliating with a relativistic positron will release gamma rays of more than 511keV as far as the stationary observer is concerned. Don't get stuck on rest mass here - I take the view that something that has energy has mass. For example, a photon has mass of hf/c2.

I was not talking about the Electron an Positron being relativistic, but the observers being relativistic. How can the way I am moving (or if I am near a gravitating object) change the energy output of an explosion in a different location?

If you were on Earth doing this annihilation experiment and I was in the International Space Station. How can my position in a lower gravity effect your experiment?

No they wouldn't. You're going off on one here. If a rock hit a stationery observer it would cut his head a bit, but if it hit a relativistic observer it would take his head clean off.

If the energy from an explosion were to create a crater, then lots of ejecta, not just 1 rock, would be created and the energy contained in that explosion would be spread around all that ejecta. This means that a rock from an explosion would not necessarily be travelling at relativistic speeds.

If, instead of just looking at the results of a hit by some of this ejecta, but instead looked at the trajectory. A 1 tone rock given 1,000,00 newtons of force would have a different trajectory to the same rock given 500,000 newtons of force.

If we accepts a variable C, then that rock would be given one or the other amounts of energy dependant on your frame of reference (stationary, moving, accelerating, in a gravity well, etc). This means that two different observers in two different frame of reference will see that rock travel a different distance. If they then went to the location of that rock, they should be at the same location, but because they saw the same rock travel in 2 different trajectories, they will not agree on where that rock landed and so end up at two different location (the rock would therefore need to exist in two different locations simultaneously.

As such a result is ridiculous in the real universe, we can conclude that any theory that has this kind of result must be wrong. As having a variable value of C leads to this conclusion (that the rock must exist in two different locations simultaneously), we can determine that C can not be variable.

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I do so agree with Special Relativity. But I also see what it's telling us. I see why the postulates hold, what they mean, and how it evolves. To the best of my knowledge I think the same as Einstein did. Not in 1905, but in 1949.

In his response to Godel's paper in the Schilpp volume, Einstein acknowledged that "the problem here disturbed me at the time of the building up of the general theory of relativity." This problem he described as follows: "Is what remains of temporal connection between world-points in the theory of relativity an asymmetrical relation (like time, intuitively understood, and unlike space), or would one be just as much justified to assert A is before B as to assert that A is after B? The issue could also be put this way: is relativistic space-time in essence a space or a time."

If you have a variable c and a preferred reference frame, you most decidedly do not agree with SR. Einstein's philosophy has no bearing on the science, and attempts to appeal to authority have no bearing on the validity of it, either.

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Edtharan: I don't feel I've got your full attention regarding the material I've posted or linked to. And yours is just about the only attention my essays have received here. There have been other comments, but many indicate no attention to the subject or my responses. So I've decided to spend my time more productively elsewhere. Apologies.

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It was you who introduced the preferred reference frame, swansont, here:

http://www.scienceforums.net/forum/showpost.php?p=319827&postcount=55

That was before you changed the subject and delivered another snipe. Look, you won't read this essay or engage sincerely in the subject, so I'll have to go elsewhere. Perhaps one day you'll have cause to reflect upon your stance here.

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It was you who introduced the preferred reference frame, swansont, here:

http://www.scienceforums.net/forum/showpost.php?p=319827&postcount=55

That was before you changed the subject and delivered another snipe. Look, you won't read this essay or engage sincerely in the subject, so I'll have to go elsewhere. Perhaps one day you'll have cause to reflect upon your stance here.

Sorry, I see no preferred reference frame there. I mention two frames, neither of which is preferred.

As far as I can tell, what you consider a "snipe" is someone telling you that you are wrong on matters of science, and "sincere" engagement is discussing the metaphysics, in which I don't have much interest. I have no problem with my stance here.

Try the boards at physorg. They may be more to your liking.

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Edtharan: I don't feel I've got your full attention regarding the material I've posted or linked to. And yours is just about the only attention my essays have received here. There have been other comments, but many indicate no attention to the subject or my responses. So I've decided to spend my time more productively elsewhere. Apologies.

Admittedly I didn't understand a lot of the maths that was used so my understanding of what you linked to is lacking.

However I have been using your explanations of it and how they are supposed to relate to your essay as the basis for my discussions.

As I have said before, I am willing to enter these kinds of discussion because I am willing to change my position. However, not everyone is like this. Also some people will not enter into a discussion if they think the ideas are wrong, they might make a few posts and then when their issues are not addressed well they will leave.

Also, this is the 3rd (or is that 4th) "Time Explained" essay that you have posted. Some people will have been put off after the first one (the first one received much more attention than the later ones). It would have been better if you had just stuck to the one thread and posted a new revision of the essay in that (or just put in the title a revision number).

It is these that probably have contributed to the lack of interest in you essays rather than anything else.

Before you go, In would like one question about your essay cleared up. And that is of my last inquiry.

If C is variable, then two observed in different frames of reference will disagree over the energy (due to E=MC^2) contained in the matter of an object (the object does not have to be at relativistic speeds or anything).

Specifically in this scenario: An observe (A) in orbit around Earth is in a lower gravity gradient than one on Earth (B). According to your claims, Observer A will have a higher value for C than Observer B.

Because Observer B is in a higher gravitational field than Observer A, Observer B will have a lower value for C.

This means that a matter/antimatter explosion that occurs near Observer B on Earth will produce different energy values.

As observer A sees "Time" moving slower for Observer B and concludes that C is lower relative to himself, in the equation E=MC^2, C will have a different observed value to what Observer B would see (as you said Observer B will see C at the same 300,000km/s locally as Observer A would measure locally). However Observer A is not measuring the value of C locally, but as it is at Observer B relative to himself and so gets a different value of C than he would if he measured it Locally.

This means that the Energy that is observed should be different for what Observer A sees and what Observer B sees.

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There have been other comments, but many indicate no attention to the subject or my responses.

Then maybe you can read my first post, in your first thread, in a new light now.

Also, this is the 3rd (or is that 4th) "Time Explained" essay that you have posted.

Farsight have only started two "Time Explained" threads.

If you count the other "Xxxxx Explained" threads then this is the fifth.

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Farsight have only started two "Time Explained" threads.

If you count the other "Xxxxx Explained" threads then this is the fifth.

Ahh, that was what I meant, thanks.

But still, the number of XXXX Explained threads can be a negative influence.

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• 4 weeks later...
If C is variable, then two observed in different frames of reference will disagree over the energy (due to E=MC^2) contained in the matter of an object (the object does not have to be at relativistic speeds or anything).

Not if they're smart they won't. Think about that cannonball travelling at 1000m/s. It has considerable kinetic energy. When you change frames to travel alongside it, the kinetic energy is now zero. But you know the cannonball didn't change at all, so you know that the energy content didn't really change.

Specifically in this scenario: An observer (A) in orbit around Earth is in a lower gravity gradient than one on Earth (B). According to your claims, Observer A will have a higher value for C than Observer B. Because Observer B is in a higher gravitational field than Observer A, Observer B will have a lower value for C.

Yep, no problem.

This means that a matter/antimatter explosion that occurs near Observer B on Earth will produce different energy values.

No. The amount of energy contained within the matter and antimatter masses at that location is what it is. It's like the cannonball.

As observer A sees "Time" moving slower for Observer B and concludes that c is lower relative to himself, in the equation E=MC^2, C will have a different observed value to what Observer B would see (as you said Observer B will see C at the same 300,000km/s locally as Observer A would measure locally. However Observer A is not measuring the value of c locally, but as it is at Observer B relative to himself and so gets a different value of c than he would if he measured it locally.

OK, he concludes that c is lower on the surface.

This means that the energy that is observed should be different for what Observer A sees and what Observer B sees.

You're getting confused between observed and concluded here. If observer A is smart enough to calculate a lower value of c, he will also be smart enough to calculate that the annihilation energy released on the surface is less than the annihilation energy that would have been released in orbit. He will calculate that the difference equates to the kinetic energy of the masses falling from position A to position B. He will realise that the kinetic energy doesn't come from nowhere. It doesn't come from the "gravitational field" or from the earth. It comes, via the reducing c, out of the falling masses themselves.

Imagine being on earth and kicking the cannonball into orbit. You gave it kinetic energy, and this disappears into "potential energy" within the cannonball, commensurate with the higher value of c up there. The same sort of thing happens in special relativity when there's no gravity: the canonball's relativistic mass increases when you kick it to some higher velocity with respect to yourself.

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Not if they're smart they won't. Think about that cannonball travelling at 1000m/s. It has considerable kinetic energy. When you change frames to travel alongside it, the kinetic energy is now zero. But you know the cannonball didn't change at all, so you know that the energy content didn't really change.

Yes, relative to you the cannon ball is at rest. This however doesn't really cover my question. I never asked if the cannon ball's inertia is relative to the frame of reference.

I was not questioning the moving frame of reference, but the gravitational frame of reference.

You're getting confused between observed and concluded here. If observer A is smart enough to calculate a lower value of c, he will also be smart enough to calculate that the annihilation energy released on the surface is less than the annihilation energy that would have been released in orbit. He will calculate that the difference equates to the kinetic energy of the masses falling from position A to position B. He will realise that the kinetic energy doesn't come from nowhere. It doesn't come from the "gravitational field" or from the earth. It comes, via the reducing c, out of the falling masses themselves.

You are mixing frames of reference here.

You are saying that the frame of reference where the explosion is on Earth is the same as it is in Space. This is not so. It is experiencing a different gravitational field and is therefore in a different frame of reference.

because of this, you forgot that the fact that the Earth observer, now is a different frame of reference to the explosion will disagree and give mutually exclusive results to the observer in space. You haven't answered the question at all. all you have done is shift the frame of reference and ignored/not factored in the observer on Earth.

Actually, from experiments done, the annihilation energy is the same, not different, so this rules out that C is variable. In different gravitational potentials, the annihilation energy is the same. In annihilations involving moving frames of reference, the excess energy is accounted for due to the energy put into the system (by particle accelerators). There is no unaccounted for energy that indicates a variable value of C.

Current particle accelerators exist at different altitudes, and therefore at different gravitational potentials. They do not need to adjust the values of their results to account for the different gravitational frame of reference. If C was variable, they would have to.

No. The amount of energy contained within the matter and antimatter masses at that location is what it is. It's like the cannonball.

But if I change my frame of reference to the event, then according to your proposition, I should see a different energy value than one who doesn't change their frame of reference.

If the energy of the annihalation is differnet and the energy was used to power something (say a car), then the remaining observer will see the car able to move X number of metres because of it, and I would see it move Y metres.

now, we get together and measure the distance of the car on the surface with a single ruler. How can I measure Y, and you measure X, on the same ruler at the same time.

This is the major flaw of your argument. It requires that two mutually exclusive results must both occur.

Lets try a different one:

Imagine you have a Cat in a box. If the energy of the explosion is X joules, then the cat will be killed by a bottle of poison gas, if Y joules are measured then the Cat will be allowed to live.

Now, the person next to the box uses His detector and measures Y (lucky for that cat), but the person in orbit, measures X and therefore sees the cat killed.

But hang on, how can that cat be both killed and saved at the same time?

No, because of the necessity in your proposition that C be variable, you end up with inconsistent results. The observer in different frame of reference can not agree on the results and therefore will see different Universe evolve from then on.

Imagine being on earth and kicking the cannonball into orbit. You gave it kinetic energy, and this disappears into "potential energy" within the cannonball, commensurate with the higher value of c up there.

No it doesn't. There has been no energy loss. All the energy can be accounted for without a variable value of C. Why then do you insist that all these experiments are not calculating the correct energy?

You seem to be equating potential energy as to being equivalent to a change in C. But you can have other sorts of potential energy. What about the potential energy in a chemical reaction, is this due to a change in C, or the potential energy in a spring. Potential energy is a kind of stored energy, a tension if you will. How does a change in C deliver a store of energy?

The other thing is that the difference in energy of a Matter/Antimatter annihilation will be different to the change in energy of a cannonball launched into orbit. This means that for the same frame of reference you can get 2 (or more) different values for the change in C. So even if your proposition was correct, the same observer in the same frame of reference will get different results of their calculated local value of C, depending on what they are actually observing.

So it even fails for self consistency.

So not only does it give results that are internally inconsistent, it does not give results that match to reality. On these ground your proposition, as it stands and you presented, can not possibly be true in this reality.

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You are mixing frames of reference here. You are saying that the frame of reference where the explosion is on Earth is the same as it is in Space.
No I'm not. That why I stressed the difference between conclude and observe.

This is not so. It is experiencing a different gravitational field and is therefore in a different frame of reference. Because of this, you forgot that the fact that the Earth observer, now is a different frame of reference to the explosion will disagree and give mutually exclusive results to the observer in space. You haven't answered the question at all. all you have done is shift the frame of reference and ignored/not factored in the observer on Earth.
I've forgotten nothing. You might be on earth and I might be in orbit, and because I need to keep adjusting my watch I conclude that your c is slower than mine. But if I try to measure your c I actually measure my c.

Actually, from experiments done, the annihilation energy is the same, not different, so this rules out that C is variable. In different gravitational potentials, the annihilation energy is the same. In annihilations involving moving frames of reference, the excess energy is accounted for due to the energy put into the system (by particle accelerators). There is no unaccounted for energy that indicates a variable value of C.
This is more of the same. You will not measure a different energy.

Current particle accelerators exist at different altitudes, and therefore at different gravitational potentials. They do not need to adjust the values of their results to account for the different gravitational frame of reference. If C was variable, they would have to.
You always measure c to be the same old 300,000km/s regardless of your reference frame. You never measure 290,000km/s. But you can conclude that there is a difference in absolute terms. Time dilation and the transverse ruler tells you this.

But if I change my frame of reference to the event, then according to your proposition, I should see a different energy value than one who doesn't change their frame of reference...
No, you won't see a difference. You conclude there's a difference.

This is the major flaw of your argument. It requires that two mutually exclusive results must both occur.
No Ed. You've gone astray from the off here. Read what I said again.

You seem to be equating potential energy as to being equivalent to a change in C. But you can have other sorts of potential energy. What about the potential energy in a chemical reaction, is this due to a change in C, or the potential energy in a spring. Potential energy is a kind of stored energy, a tension if you will. How does a change in C deliver a store of energy?..
Sure thing. I thought you'd assume the gravitational. The change in c is caused by a local tension gradient orthogonal to mass/energy stress. See ENERGY EXPLAINED and MASS EXPLAINED.

The other thing is that the difference in energy of a Matter/Antimatter annihilation will be different to the change in energy of a cannonball launched into orbit. This means that for the same frame of reference you can get 2 (or more) different values for the change in C. So even if your proposition was correct, the same observer in the same frame of reference will get different results of their calculated local value of C, depending on what they are actually observing.
Everybody measures c to be the same. If they end up with some mismatch in their calculations they've done something wrong.

So it even fails for self consistency. So not only does it give results that are internally inconsistent, it does not give results that match to reality. On these ground your proposition, as it stands and you presented, can not possibly be true in this reality.
Sorry Ed, you've reached that conclusion because you've gone off at a total tangent. Try it again. Try to ask smaller questions instead of blatting out a whole proof that falls down because of a mistaken initial premise.
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No I'm not. That why I stressed the difference between conclude and observe.

But if your conclusions don't match up with observations, then you conclusions must be wrong.

The conclusions are an attempt to predict the outcome of a system, that is what you would observe is the system was allowed to run. If the system is allowed to run and you conclusions don't match up with what the system actually does, then your theory does not describe reality.

Your conclusionsmust match up with Observation. If they don't then the conclusions are wrong.

I've forgotten nothing. You might be on earth and I might be in orbit, and because I need to keep adjusting my watch I conclude that your c is slower than mine. But if I try to measure your c I actually measure my c.

If we only considder this situation, the explanation of that "Time is running slower on Earth" is just an equally valid explanation of this situation. You offer no reason why that could be wrong and yours must be right. When I press the point on this, you just seem to come back with "But you can conclude that it is different".

But, that is not a reason that you are right. You offer now explanation other than the "according to my theory it must be like this". You are using circular reasoning here: If your theory is right, you can conclude that C is variable. If C is variable, then your theory is right.

What that circular reasoning doesn't take into account is: If your theory is wrong, then there is another explanation, that is very well tested that give consistent results that match observations.

So, one one hand we have a tested theory that matches observations, and on the other hand we have an untested theory that gives inconsistent results that so far have not show to match with reality.

Which one would you choose as the most accurate explanation of reality?

This is more of the same. You will not measure a different energy.

Yes you will, if C is variable then:

E=100g * 300,000km/s^2

gives a different answer for E than:

E=100g * 299,999km/s^2

Do the maths. E will be different if the value of C changes.

Therefore, if you have a variable C, then the energy will be different.

You always measure c to be the same old 300,000km/s regardless of your reference frame. You never measure 290,000km/s. But you can conclude that there is a difference in absolute terms. Time dilation and the transverse ruler tells you this.

But you said in an earlier post that you would observe different values of C in different frame of reference. therefore my conclusions from one frame of reference will not match the conclusions from another frame of reference. So, as an ability to predict what would happen to a system, a variable value of C fails. It leads to two different people concluding two different and mutually exclusive results from the same observed event.

For instance:

A person on the Moon and a person on Earth both observe a particle accelerator experiment where they smash an Proton and an anti proton together.

The Moon Observer is in a different frame of reference and concludes that C is 299,999km/s on Earth where the experiment is taking place.

The Earth Observer, being in the same frame of reference as the experiment concludes that the value of C is 300,000km/s

The Moon observer then uses the value of C to conclude that the energy release in the experiment would be:

2 * 1.672 × 10^27 kg * 299,999km/s^2

Where as the Earth Observer would conclude:

2 * 1.672 × 10^27 kg * 300,000km/s^2

This therefore give different and mutually exclusive results. Both the Moon Observer and The Earth Observer can not both be correct in their conclusions.

So, which answer is the correct one?

Remember, this is just the conclusions, noit the observations of the experiment.

What this all means is that if we have a variable value for c, then E=MC^2 will give inconsistent conclusions between people in different frames of reference.

This means that if we have a variable value of C. The equation E=MC^2 must therefore be wrong.

As E=MC^2 has been tested quite well and it has not show to be wrong, then you theory must be wrong.

Only one or the other can be right. And as E-MC^2 has been tested and has not been found to be wrong, then it must therefore be the other proposition (that C is variable) that must be wrong.

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But if your conclusions don't match up with observations, then you conclusions must be wrong.

Are you being mendacious here? Take a look at the opening paragraphs of Time Explained again. We observe colour, heat, sound, et cetera, but we do not conclude that these things are fundamental properties of the world. Look again at the transverse metre rule, where you measure time t for light to travel distance s and I measure the same value. You calculate c=s/t=300,000km/s as do I, but we know that your t is different to my t by a time dilation factor. We see a similar time dilation factor on the surface of the planet. So what do we conclude Edtharan? Here's a clue: c=s/t and t ain't what it was.

This means that if we have a variable value of C the equation E=MC2 must therefore be wrong.
No it doesn't. Where did that come from? Did you even read my previous post? I always measure c to be 300,000km/s, and so do you. Come on now, pay attention, when an object falls to earth where does the kinetic energy come from? Here's a clue: E=MC2 and M didn't change.

The Moon observer then uses the value of C to conclude that the energy release in the experiment would be:

2 * 1.672 × 10^27 kg * 299,999km/s^2

Where as the Earth Observer would conclude:

2 * 1.672 × 10^27 kg * 300,000km/s^2

This therefore give different and mutually exclusive results. Both the Moon Observer and The Earth Observer can not both be correct in their conclusions.

Groan. This is utterly facile. You've taken a conclusion and turned it into a measurement, then back into a conveniently "mutually exclusive" conclusion. Why do you do this? Why duck and dive with all this dishonesty that can't explain anything and doesn't want to see anything explained? Come on, show me I'm wrong. Explain to me what gives if c=s/t and the t is dilated. Does s contract even though it was a transverse metre rule? Come on Ed, think. Yours is the circular argument, and as ever it's going nowhere.

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