Could the Observer's Absolute Rest Explain the Constancy of Light Speed?

[rajeshmarndi]

Just a thought for the constancy of the light speed.

Since photons are massless and therefore if they do not add the velocity of the moving light source i.e doesn't follow Newtons first law of inertia. It is only the surrounding that moves wrt the observer. The observer and the point light is emitted are always at absolute rest.

And therefore the observer would always see the same speed of light irrespective of the motion i.e it is only the surrounding which moves which include the light source.

But once the light emit, it emit as if it was emitted from a stationary point in space. That is both observer and the point, when light is emitted are at rest.

So my question is, the observer never moves (I can say it is always at absolute rest wrt surrounding) and also since each emitted light is emitted from a stationary point in space. Couldn't this be one reason we always see same speed of light?

Thanks.

 

[ghwellsjr]

Just a thought for the constancy of the light speed.

Since photons are massless and therefore if they do not add the velocity of the moving light source i.e doesn't follow Newtons first law of inertia. It is only the surrounding that moves wrt the observer. The observer and the point light is emitted are always at absolute rest.

And therefore the observer would always see the same speed of light irrespective of the motion i.e it is only the surrounding which moves which include the light source.

But once the light emit, it emit as if it was emitted from a stationary point in space. That is both observer and the point, when light is emitted are at rest.

So my question is, the observer never moves (I can say it is always at absolute rest wrt surrounding) and also since each emitted light is emitted from a stationary point in space. Couldn't this be one reason we always see same speed of light?

Thanks.

We can't see the speed of light. We simply define it to be c in any and every Inertial Reference Frame.

 

[rajeshmarndi]

In the case when we say a spaceship is traveling at, say .7c and it switch on its light. Then the speed of light to a stationary observer, doesn't become more than c. Here we use the relativistic addition of velocity formula.

But is there any need to use the above formula for light.

Also in the spaceship case above , if a ball is thrown at .5c, why wouldn't it be observed by a stationary observer to travel at 1.2 c. Is it experimental verified in some way, in such scenario that the ball wouldn't be observed to exceed light speed.

 

[geordief]

My view (with very little experience ,understanding or mathematical /physical skills) is that the speed of light is a shorthand for the "speed of information".

I find it very easy to rationalize that we can know nothing faster than by the fastest known method of acquiring information.

This (fastest) method happens to be , so far as we know by probing events with beams of light.

If knowledge was ever to be transmitted faster than could have been explained by a beam of light then that would require something to travel faster than light (or via a shortcut across space-time perhaps) .

This has never been observed till now.

 

[geordief]

We can't see the speed of light. We simply define it to be c in any and every Inertial Reference Frame.

I don't understand why you say we can't "see" the speed of light.
Do you just mean visually because we can measure it and I would define that as "seeing".

Or is it that it is impossible to measure the speed of light other than from the point of origin (if that makes sense)?

 

[PeterDonis]

So my question is, the observer never moves (I can say it is always at absolute rest wrt surrounding) and also since each emitted light is emitted from a stationary point in space. Couldn't this be one reason we always see same speed of light?

No, because there is no such thing as "absolute rest".

 

[PeterDonis]

But is there any need to use the above formula for light.

Well, if you don't use the formula, you get the wrong answer, so yes. See below.

Also in the spaceship case above , if a ball is thrown at .5c, why wouldn't it be observed by a stationary observer to travel at 1.2 c. Is it experimental verified in some way, in such scenario that the ball wouldn't be observed to exceed light speed.

Not with balls and spaceships, but the validity of the Lorentz transformations, which is what underlies the velocity addition formula, has been verified in thousands of particle physics experiments.

 

[geordief]

Not with balls and spaceships, but the validity of the Lorentz transformations, which is what underlies the velocity addition formula, has been verified in thousands of particle physics experiments.

Only with particles? What about anything larger? Atoms? Can they be accelerated to relativistic speeds or do they break up or require too much energy?

 

[Bill_K]

Only with particles? What about anything larger? Atoms? Can they be accelerated to relativistic speeds or do they break up or require too much energy?

In addition to protons, the LHC collides lead nuclei at relativistic energies.

Look around on the web and you will find discussions about colliding relativistic chickens, but so far no serious proposals.

 

[rajeshmarndi]

No, because there is no such thing as "absolute rest".

Isn't any observer can consider itself as at ""absolute rest" because for it rest everything else are in motion and not itself.

 

[ZapperZ]

Isn't any observer can consider itself as at ""absolute rest" because for it rest everything else are in motion and not itself.

But that is just a play on words. You can call it a "relative rest", a "my-frame rest", a "cow rest". It doesn't change the physics.

However, if you use the word "absolute rest" to denote a special rest frame that is unique and unlike any other, then THAT has a significance in physics and which is not compatible with Relativity.

Zz.

 

[ZapperZ]

Only with particles? What about anything larger? Atoms? Can they be accelerated to relativistic speeds or do they break up or require too much energy?

Both the LHC and RHIC have accelerated and collided large nuclei as part of a nuclear physics experiment.

You cannot do this with neutral atoms. They do not respond to accelerating fields in the accelerating structures.

Zz.

 

[Nugatory]

Only with particles? What about anything larger? Atoms? Can they be accelerated to relativistic speeds or do they break up or require too much energy?

Atoms and small groups of atoms, yes, it's done routinely.

It's also done routinely for the specific case of large objects shining light in front of them. The macroscopic objects aren't moving anywhere close to the speed of light, but our measurement techniques are quite good enough to tell whether the light leaves them with speed $v+c$ or $c$, and it's always the latter. This has been done with objects as large as planets.

It's not practical to do measurements on macroscopic objects moving at relativistic speeds. To see why, you could try calculating how long of a runway you would need to accelerate a one-kilogram mass to half the speed of light, then let it cruise at that speed for a half-second or so while we measure its speed, then safely slow it back down again. You might also try calculating the amount of kinetic energy that would be released if that mass were to hit anything, even sea-level air.

 

[jtbell]

Also in the spaceship case above , if a ball is thrown at .5c, why wouldn't it be observed by a stationary observer to travel at 1.2 c.

Velocities don't "add" in the way you apparently expect:

http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/einvel2.html

For small velocities the result is very very close to (and experimentally indistinguishable from) classical velocity addition.

 

[Simplyh]

I'm back after some months off.

I think this can only be discussed after defining some of the terms. There have been lots of light speed measurements; however we can only surely measure the two-way speed of light. To be able to measure the one-way speed of light, we should be able to place synchronized clocks on each of the extremities of the light path, ie, clocks synchronized at distance.

But, because different theories define synchronization at distance in different ways, we can not place clocks at distance that are unanimously recognized as being synchronized.

Therefore we can only say that the average speed on the way forth and back is c. This comes from Lorentz explanation of MM experiment. If, on the way of Earth movement, light speed regarding the mirror is c-v and on the way back is c+v, adding both paths (L) we would get dt = (2L/c)/(1-v²/c²) = (2L/c)g² (with g = gamma = sqr(1-v²/c²). If space contracts by g and time expands by g, we get dtg = (2L/c)g²/g => dt = 2L/c. That means that the two-way speed of light is observed as c on the moving frame as much as on the Rest Frame.

Remember that LT was deducted by Lorentz, not by Einstein, in the conceptual frame of the Absolute Rest Theory, not in Special Relativity.

Einstein, accepted the mathematics of LT and gave them a different meaning at the light of the principal of relativity. Namely, being all reference frames equal, and being the one way speed of light equal to c on the so called rest frame, it should be also equal to c on each and every inertial system. That's why he considers the one way light speed as c and uses it to define synchronization of distant clocks.

But, doing so, the one-way light speed becomes tautological to the definition. A clock at distance should always be synchronized to the observer's by setting it to t0+L/c, being t0 the time of the light stroke and L the path between the two clocks. With such a definition it is no wonder that light speed is always equal to c on each and every way we may choose.

Therefore the discussion must be undertaken on a different way. Otherwise, the defenders of the Absolute Rest will always be right and the defenders of SR will always be right: just because they are both using different definitions of synchrony at distance.

Divirtam-se
Heitor

 

[Simplyh]

Sorry, not with g = gamma = sqr(1-v²/c²), but with g = gamma = 1/(sqr(1-v²/c²))