Is time the speed of light?

Only from our frame of reference: Please see my other reply also.

 

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Its really incredible answer to me! Thank you sir !

 

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Perhaps the excitation of an electric field in 'space' (light travel time) does not actually require anything to physically move or change (other than locally) in order for light to appear to propagate. Why do you suppose that in the absence of a gravitational field, it appears to propagates in a straight line?

 

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Because spacetime is flat?

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Sure. And also because light travel time is at the same rate, in every direction, and for every inertial reference frame in the absence of a gravitational field. But that doesn't quite capture the entire reality of the situation.

If every element of a quantum field with spin zero excitations and which is entangled everywhere is somehow able to transfer rotational inertia to the process to keep it going like rotating bearings in a conveyor belt. One might also reasonably expect that the beam would spread out as it traveled, because the transfer of energy in exactly the same direction might not be perfect. So it makes sense, photons might be able to use some sort of interaction with the field in order to continue propagating in a given direction.

A previous post of yours gave me pause to think: how many crests and troughs are there, REALLY? An observer traveling along with the wave at very close to the speed of light in a vacuum sees an extreme red shift (fewer crests, troughs), and an observer traveling in the opposite direction at similar relativistic speed sees extreme blue shifts (and lots and lots of crests and troughs), but still measures a relative speed of c for the photon in either direction. In the frame that is at rest relative to the bound electrons in whatever created the wave in the first place, of course, there is a "proper" number of wavelengths per second. To interpret one direction as a "slow motion" equivalent of the opposite one doesn't quite capture the idea, since the speed of propagation is exactly the same (speed of light) in both directions.

But the red shifted end is of particular interest, because that end really can be considered to be closer to the frame of reference of the photon itself. Those interactions with the field must seem to occur at a lethargic pace, which is to say, practically at rest, relative to the FoR of the photon or whatever excitation that waveform might represent. This would render the interaction almost completely devoid of the influence of anything comparable to a magnetic field, since magnetic fields are supposedly the result of moving electric fields, not stationary ones. Unless the magnetic field is somehow already a dynamic of the spin zero excitation of the field.

 

"...that needed to be included in Einstein original papers."

"That includes blasphemy and bad language!"-Lock Stock and Two Smoking Barrels

 

A clock in higher gravitational potential ticks faster than lower gravitational potential. We had experiment about it on Earth. But had we any experiment about it on Moon ?

 

We have time dilation corrections every second of every day with GPS satellites, which are in orbit. If we did not make these corrections, within a day the location calculations would be off several meters. Within a week they would be useless.

The Shapiro experiment confirmed the gravitational time dilation due to the Sun's gravity near the orbit of Venus.

The science is settled that time dilation near gravitational fields works the way General Relativity says it does.

 

Thank you sir ! I knew that the answer will be like this. But I'm so sorry about it that I don't want the result of the experiment like this. I mean there will be a clock in higher gravitational potential of moon and another clock will be in lower gravitational potential of moon. I was trying to find out about the experiment on internet but those answer was like that you answered. But it is vary important to me and may be to us. May be it is about nonsensic that I'm talking about.

 

The speed of light weill never be reached as predicted by Einstein.

This can be proved mathematically:

X+X=2X

So then:

1=2
2=4
3=6

1=-1
2=-2
3=-3

We will never get there.

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Even if we could travel the speed of light the two would cancel each other out:

ONE+ONE=TWO

 

Waiter_2001: might be better if you avoid mathematical responses.

 

Contradictions of relativity with respect to vector math is what you are attempting to describe, and it is double tough, so I'm not surprised you can't quite capture it.

c ± v = c (v < c, bound energy only; c invariant for all inertial reference frames)

c - c = "at rest", for all inertial reference frames, also an invariant

The above two relations are basically all of the mathematics rendered by Special Relativity, in compact form. But there is more.

It is also not possible for aggregate bound energy to rotate (spin) at a velocity which exceeds c, so for any angular velocity omega,

(omega * radius) < c (not applicable to spin propagation of fundamental particles), and finally:

c (spin) - c (spin) = zero (spin) (spin invariant Higgs field only)

Both linear and spin modes of the propagation of energy at c are possible. Slowing down fundamental particles with respect to c requires the Higgs mechanism and a transfer of inertia between spin and linear modes of propagation between matter and the Higgs field.

AND this is exactly the reason, bound energy (matter) can never "get there", "there" being the speed of light. The Higgs boson itself has mass, and so it cannot propagate at c in the linear mode. The spin mode is a different matter, both for Higgs and the particles with which it interacts.

AND this is also why it is sheer folly to expect static Euclidean vector representations to add and subtract as ordinary vectors do in a relativistic space comprised of light travel time in every direction. Static Euclidean vector math respects no such limits as the speed of light in a vacuum in either linear or spin modes.

Greeks may have been clever in science and math, particularly geometry, but Pythagoras wasn't Einstein.

 

You're not talking nonsense you're asking a question that we can use the theory of relativity to answer. We'll predict the tick rate for each local proper frame you defined. The surface of the earth and the surface of the moon. Then ratio the results to find the difference.

dTau_earth = (1-2M_Earth/r_earth)^1/2 dt / dTau_moon = (1-2M_moon/r_moon)^1/2 dt

dTau_earth/dTau_moon = (1-2M_earth/r_earth)^1/2 / dTau_moon = (1-2M_Moon/r_moon)^1/2

= (1-2*0.00444 meter / 6.371E6 meter)^1/2) / (1-2*.00005455 meter) / (1,737,000 meter)^1/2

= .999999999 / 1 = .999999999 so for every tick recorded on the surface of the moon there is.999999999 tick recorded on the surface of the earth.

This is an infinitesimal difference that would actually be smaller if I carried the analysis out more than 9 decimal places.

So based on the predictions I would suspect they wouldn't have been much interested in revisiting those experiments on the surface of the moon. They we're pretty busy. But that's just a guess. What's instructive for this exercise is it's pretty much that way when you compare the proper tick rate between all local proper frames. It's a consequence of infinitesimal local spacetime curvature. Which is gravity in EInsteins theory.
BTW the mass is expressed as length in geometric units. The proper tick rate for these frames is assuming they are at rest with respect to eachother. Which is a safe approximation for planetary motion for obvious reason.

 

Quit spewing bullshit.

 

Thanks. We know that a electron will gain potential energy when we will bring it to higher gravitational potential from lower gravitational potential. But some times the electron doesn't do that to obey the law of conservation that I knew from 'exchemist' and 'danshawen' of my thread " http://www.sciforums.com/threads/th...-to-distroy-energy.154156/page-2#post-3349211" 10 years ago I thought about another kind of something in physics from Special theory of relativity. I thought that the experiment will not obey the theory on moon where the reference frame(earth) is not included to the experiment and it will be same on mars. I was trying to find out about the experiment but I found nothing about the experiment. A clock on moon surface will tick faster than a clock on earth surface. But the earth(reference frame) is included to the experiment. But why the experiment will not obey the theory that I can't talk about it before the experiment. May be it is the nonsense that I'm talking about.

 

Maybe not

 

No matter where we do that experiment it's done in the local proper frame. The tick rate and the proper length will always be very close between different local proper frames because the local spacetime curvature is 'so small'. But the frames are not identicle because 'so small' isn't equivalent from frame to frame. That's why there's a difference between the earth local proper time coordinate and the moon local proper time coordinate. For example the equation I used to calculate the the earth and moon proper time coordinate is the same for all local proper frames
dTau = (1-2M/r)^1/2 dt
What makes the predictions vary from frame to frame is the variables M and r. These are the variables that determine the value of the local spacetime curvature which is always an infinitesimal. In the metric 2M/r is the curvature component. So you can think of it like this.
For (1-2M/r) the variable component for the above equation
1 = flat spacetime
And
1-2M/r = curved spacetime

 

The observational frame of reference is the frame in which the observer is at rest. You do not have a choice in what frame you find yourself at rest.

 

Since you don't actually know what M is, nor where exactly it derives, nor WHY it causes curvature of light travel time nor time dilation, all you really seem to know is a little calculus.

We're very impressed, but all the calculus ever manipulated can't really answer the question of the OP, and that is a fact. To think otherwise doesn't really rise to the level of BS, so stop spewing calculus and pretending that it is better. It isn't, and you can't prove that it is.

 

Thank you sir to feel the possibility. Its mean that may be there was no experiment about it on moon. And I guess there is no experiment about speed of light has done on moon too where the reference frame(earth) is not included.