Clarification on the consistency of light speed

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I apologize for the the probably gross error on my behalf, but, not as a scientist in the GR /SR field, I am a bit confused. My question is why is c constant? I am not arguing why is c constant in a specific frame, I do not understand why you couldn't, for a lack of a better term, frame stack.

So for example, if one throws a rock at 20mph in a platform going 10 mph, than the speed of the rock from a person stationary on the ground sees the rock going 30 mph. absurdly simple.

Yet for a ship traveling at or near c, and the ship had a head light pointing in the same direction of the velocity of the ship. The out side observer will see the ship, and the head light of the ship traveling at the same speed, yet the Pilot of the ship sees the light, traveling away from him at c.

What I do not understand is this. Why would the out side observer see the head light at all. Why would that light even "exist" to the out side observer. Wouldn't it make more sense, for the head light to only exist to the pilot. Since theoretically, nothing can travel faster than c, if and only if they are in the same frame, citing Einstein himself:

Tests of Einstein's two Postulates

1. The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of coordinates in uniform translatory motion.
2. Any ray of light moves in the “stationary” system of coordinates with determined velocity c, whether the ray be emitted by a stationary or by a moving body.


But, since the spaceship is already nearing the speed of light, does that not define the spaceship as a new frame, which light is free to be a constant in, but breaks down when thought about from the arbitrary observer.

If this were true, it would make lightspeed stacking possible.
 

Answers and Replies

  • #2
ehj
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"So for example, if one throws a rock at 20mph in a platform going 10 mph, than the speed of the rock from a person stationary on the ground sees the rock going 30 mph. absurdly simple."
The only reason that it's absurdly simple is because you base your "simplicity" on the world you see everyday. The only reason you find it simple, is because you're used to that's how reality works. Ever tried asking why you can add up the speeds like you do in the example? I don't think the question "why" can be answered other than "that's just how reality works".
 
  • #3
Ich
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So for example, if one throws a rock at 20mph in a platform going 10 mph, than the speed of the rock from a person stationary on the ground sees the rock going 30 mph. absurdly simple.
Simple, but wrong. Believe it or not, he sees the rock going 29.99999983 mph.
The out side observer will see the ship, and the head light of the ship traveling at the same speed, yet the Pilot of the ship sees the light, traveling away from him at c.
No, that doesn't work. The ship has to travel at less that c.
Google "velocity addition".
 
  • #4
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Thank you ICh and ehj. 30mph or 29.999999983 it doesn't matter, not because you are incorrect, but rather I do not dispute that. I understand that, and I was only using it as an example.

Even if the ship, was going .99c, then my confusion still exists. Why not treat light as a constant only for frame in use. But a non constant for multiple frames.
 
  • #5
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Special relativity is partly based on the notion that the universe is not contradictory. So two people watching the same event will agree on the physical outcome. It seems that we must impose a mximum possible relative speed between objects, and insist that light speed is not affected by the arbitrary state of motion of the emitter or receiver, in order to achieve this.

Caveat, there are probably other systems than SR that can achieve this.
 
  • #6
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I believe the question "why does c move at constant speed in all reference frames?" has no answer. We have experimented and this is what we have observed, relativity studies the consequences of c being constant in every reference frame (and thus far said consequences have not been debunked experimentally) but is not an attempt to explain why this happens.
 
  • #7
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constant speed of light??why?

Special Relativity says that two observers moving at constant speed relative to each other will observe the same speed of light.

Now this I can't understand, why? Light is having a particular speed. And two observers moving relative to each other (and to light) should observe light coming at different speeds.

Now I understand that the experiments have confirmed that light speed is constant w.r.t observer. But what is the explanation?
 
  • #8
russ_watters
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From one side, the explanation is simply that the way people think about velocity and time works in everyday situations where the velocities are low and the time calculations not precise, but as Ich said, that conventional thinking is wrong.

People have an easy time believing that people can disagree on the velocity an object is travling at. SR derives from the realization that time is part of the same system and obeys the same law, so every motion through space includes a component of motion through time.
 
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  • #9
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Russ, even if consider the objects moving at high speeds such that the relativistic effects becomes apparent(time dilation and length contraction), I think with appropriate thought experiments we can show that observers moving relative to each other will observer different light speeds.

But experimentally we have found a constant speed of light for observers moving ralative to each other (If we consider the experimental result is correct and accurate).

I am not able to resolve this conflict between experimental outcome and logic(thought experiment).

Please explain using an example why/how two obervers moving relative to each other will observe light to be moving at constant speed.
 
  • #10
Fredrik
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Special Relativity says that two observers moving at constant speed relative to each other will observe the same speed of light.

Now this I can't understand, why? Light is having a particular speed. And two observers moving relative to each other (and to light) should observe light coming at different speeds.

Now I understand that the experiments have confirmed that light speed is constant w.r.t observer. But what is the explanation?
Special relativity doesn't answer that. As far as I know, there's no theory that explains it. One way of looking at it is that if we just try to find the most general velocity addition law that's consistent with rotation and translation invariance of space, the result isn't [itex]u+v[/itex], it's

[tex]\frac{u+v}{1-Kuv}[/tex]

Where K is a constant that can't be calculated from the assumptions about the symmetries of space. Since K can't be calculated, we have to do experiments to find out what it is. You may feel that it's obvious that it must be 0, but the only reason we all feel that way at first is that K is so close to 0 that we can't see any of the effects of a non-zero K without doing experiments.

It's useful to define [itex]c=1/\sqrt K[/itex], which puts the velocity addition law in the form

[tex]\frac{u+v}{1-\frac{uv}{c^2}}[/tex]

The speed c defined this way isn't by definition the speed of light. It's just a constant with dimensions of speed (i.e. it has the same units as a speed) that describes a fundamental property of space and time. When you consider the implications of this velocity addition law, the result is special relativity.

The identification of c with the speed of light comes from Maxwell's equations, and later also from relativistic quantum mechanics, which says that all massless particles move at speed c.
 
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  • #11
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Hear to what Feynman has to say about inertia.
That's where we are in terms of relativity as far as I'm concerned. We observe something through experimentation (constancy of speed of light in certain inertial frames), assume it will be true in a broader level (constancy of speed of light in all inertial frames) and try to figure out it's consequences (relativity). Relativity is a theory based on the constancy of lightspeed on all inertial frames, it is not a theory to explain why this occurs.
 
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  • #12
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I apologize for the the probably gross error on my behalf, but, not as a scientist in the GR /SR field, I am a bit confused. My question is why is c constant? QUOTE]
My question is why is c constant?
c is constant in a stationary system S because the length of the path of a ray of light emitted by a moving system S' is the same length in the stationary system S.

For example, if the length of the path of the ray of light emitted by the moving system S' is of length L, then the time, in the moving system S', the ray of light emitted by the moving system S' takes to move along its path is L/c, and if this length L is the same in the stationary system S, then the time, in the stationary system S, the ray of light emitted by the moving system S' takes to move along its path in the moving system S' is also L/c.

But if the length L (in the moving system S') of the path of the ray of light emitted by the moving system S', is L + D in the stationary system S, then the time, in the stationary system S, the ray of light emitted by the moving system S' along its path of length L (in the moving system S') is (L + D)/(c + v), where v is the constant velocity with which the moving system S' moves along the stationary system S, and c + v the speed with which the light ray moves in the stationary system S.
 
  • #13
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arbol,

...and c + v the speed with which the light ray moves in the stationary system S.
If I understand your scenario this should be c. Using relativistic velocity addition, u+v -> c.
 
  • #14
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arbol, I just don't believe the constancy of c can be explained in terms of special relativity (which derives it's predictions based on the postulate of the constancy of c.).
 
  • #15
rbj
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Even if the ship, was going .99c, then my confusion still exists. Why not treat light as a constant only for frame in use. But a non constant for multiple frames.
because two inertial frames, even if they are moving at 0.99c relative to each other, both have equal claim to being "stationary" and that "the other guy is the one who is moving". if they have equal claim to being stationary, how are the laws of physics (including Maxwell's Equations and how they are solved to get a propagating wave equation for a perturbed electric field) different for the observers in the different frames? why should one guy get one pair of values for [itex]\epsilon_0[/itex] and [itex]\mu_0[/itex] and the other guy get a different set of values if they both have equal claim to be sitting around motionless and it's the other guy who is whizzing past at 0.99c?

that's the fundamental concept to get down for SR.
 
  • #16
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arbol,

If I understand your scenario this should be c. Using relativistic velocity addition, u+v -> c.
The original question is, "Why is c constant?"

The first time I answered this question I was careless.

The speed of light c is constant only when the length L of the path of a ray emitted by the moving system is also of length L in the stationary system.

If this length L is not length L in the stationary system, the reason the speed of light c is constant is to artificially inflate the time, in the stationary system, the ray of light emitted by the moving system takes to move along its path.

Why? I am not sure, but it may have something to do with the same reason the value of the dollar is artificially inflated; it is contrary to the result of the Michelson-Morley Experiment.
 
  • #17
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Hello arbol.

Quote:-

----The speed of light c is constant only when the length L of the path of a ray emitted by the moving system is also of length L in the stationary system.----

The speed of light is always c in a vacuum.

Length contraction and time dilation which you mention as the path length L and inflated time, are consequences of the constancy of c. Light speed in vacuum, c, does not behave to fit in with these effects. These effects are a consequence of the constancy of c.

As to the original question, i do not know why c is constant.

Matheinste.
 
  • #18
rbj
2,226
7
The speed of light is always c in a vacuum.

Length contraction and time dilation which you mention as the path length L and inflated time, are consequences of the constancy of c. Light speed in vacuum, c, does not behave to fit in with these effects. These effects are a consequence of the constancy of c.
that is, as far as this non-physicist electrical engineer can tell, the correct way to look at it.

As to the original question, i do not know why c is constant.
i think, despite the weirdness that time-dilation and length-contraction and such imply, that c is the same for all inertial observers because the alternative (c is faster for some inertial observers than it for other inertial observers) is even more weird. why should the laws of physics be different (even quantitatively different) for one inertial observer (who has an equal claim to say that he/she is not moving) from another inertial observer who is in constant-velocity motion relative to the first (who also has an equal claim to say that he/she is not moving)? why should these two observers have a different [itex]\epsilon_0[/itex] or [itex]\mu_0[/itex] in their physical law? who says that the one observer gets this one pair of values and this other observer should get a different pair (which means their laws of physics are quite the same)? that would be more weird than the consequences of SR like time-dilation, etc.
 
  • #19
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Hello rbj.

I agree with what you say that a non-constant c would give awkward, weirder consequences. But that cannot be viewed as a reason for it being constant. Nature does not care about our convenience or our views of wierdness.

Matheinste.
 
  • #20
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Weirdness we can handle, but outright contradictions and loss of causality would be chaos. Possibly the most important condition is that one event cannot affect another spatially separated event instantaneously. There is a speed limit on everything. Maybe the limit can change from one time/place to another, but it is always there.

As for 'how' the speed of light is always c, I have no idea.

M
 
  • #21
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The original question is, "Why is c constant?"

The first time I answered this question I was careless.

The speed of light c is constant only when the length L of the path of a ray emitted by the moving system is also of length L in the stationary system.

If this length L is not length L in the stationary system, the reason the speed of light c is constant is to artificially inflate the time, in the stationary system, the ray of light emitted by the moving system takes to move along its path.

Why? I am not sure, but it may have something to do with the same reason the value of the dollar is artificially inflated; it is contrary to the result of the Michelson-Morley Experiment.
Hello arbol.

Quote:-

----The speed of light c is constant only when the length L of the path of a ray emitted by the moving system is also of length L in the stationary system.----

The speed of light is always c in a vacuum.

Length contraction and time dilation which you mention as the path length L and inflated time, are consequences of the constancy of c. Light speed in vacuum, c, does not behave to fit in with these effects. These effects are a consequence of the constancy of c.

As to the original question, i do not know why c is constant.

Matheinste.
hello Matheinste.

I meant to say that the reason to inflate the time, in the stationary system, the ray of light emitted by the moving system takes to move along its path of length L may have something to do with the same reason prices are inflated, specially the price of gold and crude oil.

Yes. The speed of light is always c in a vacuum.

In Maxwell's equations, c is a constant.

By the Michelson-Morley Experiment,

2*sqrt(L^2 + D^2)/c = 2*L/c (1), but we all know that equation (1) is not possible, specially if L = c*t is the length, in the stationary system, of a rigid rod, and t is the time, in the stationary system, the ray of light emitted by the moving system takes to move along the length L.

It would be more appropriate if we say that by the Michelson-Morley Experiment,

2*sqrt(L^2 + D^2)/sqrt(c^2 + v^2) = 2*L/c.
 
  • #22
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2. Any ray of light moves in the “stationary” system of coordinates with determined velocity c, whether the ray be emitted by a stationary or by a moving body.


1. The time t', in the moving system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

L/c.

2. Let D = v*L/c.

3. The length, in the stationary system, of the path of the ray of light, emitted by the moving system, is

L + D.

4. We say that any ray of light moves in the stationary system of coordinates with determined velocity c, whether the ray be emitted by a stationary or by a moving body.

5. And we also say that the time t, in the stationary system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

(t' + v*L/c^2)/sqrt(1 - (v/c)^2)

= (t' + v*t'/c)/sqrt(1 - (v/c)^2)

= (L + D)/sqrt(c^ - v^2).

6. Yet we are not willing to say that the time t, in the stationary system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

sqrt(L^2 + D^2)/sqrt(c^ + v^2) = (L + D)/(c + v) = L/c = t' (the result of the Michelson-Morley experiment), because it would violate statement 4.

7. But we are willing to violate statement 4. with statement 5.
 
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  • #23
Doc Al
Mentor
44,880
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1. The time t', in the moving system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

L/c.
OK.
2. Let D = v*L/c.
OK.
3. The length, in the stationary system, of the path of the ray of light, emitted by the moving system, is

L + D.
Nope. The distance is (L + D)*gamma = (L + D)/sqrt(1 - (v/c)^2).
4. We say that any ray of light moves in the stationary system of coordinates with determined velocity c, whether the ray be emitted by a stationary or by a moving body.
Yep.
5. And we also say that the time t, in the stationary system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

(t' + v*L/c^2)/sqrt(1 - (v/c)^2)

= (t' + v*t'/c)/sqrt(1 - (v/c)^2)

= (L + D)/sqrt(c^ - v^2).
OK.
6. Yet we are not willing to say that the time t, in the stationary system, the ray of light, emitted by the moving system, takes to move along the length L of its path is

sqrt(L^2 + D^2)/sqrt(c^ + v^2) = (L + D)/(c + v) = L/c = t' (the result of the Michelson-Morley experiment), because it would violate statement 4.

7. But we are willing to violate statement 4. with statement 5.
Fix your error and try again.
 
  • #24
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OK.

OK.

Nope. The distance is (L + D)*gamma = (L + D)/sqrt(1 - (v/c)^2).

Yep.

OK.

Fix your error and try again.

The error is in substituting the length (L + D)/sqrt(1 - (v/c)^2) for the length L. The length (L + D)/sqrt(1 + (v/c)^2) is derived from the Theory of Relativity itself (a Theory paradoxically based and refuted by the result of the Michelson-Morley experiment).

If it is given that the length in the moving system of the path of a ray of light emitted by the moving system is the length L of a rigid rod, and that the length in the stationary system of L is rAB, then

yes, if tB - tA = rAB/(c + v) = t'A - tB = rAB/(c + v), then rAB/L is not equal to tB - tA.

But by the result of the Michelson-Morley,

tB - tA = rAB/(c + v) = t'A - tB = rAB/(c - v) = L/c.

3. According to the Special Theory of Relativity, the length L, in the stationary system is

(L + v*t')/sqrt(1 - (v/c)^2) (a function of the length L and of the time t')

= c*(L + D)/sqrt(c^2 - v^2).
 
  • #25
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The error is in substituting the length (L + D)/sqrt(1 - (v/c)^2) for the length L. The length (L + D)/sqrt(1 + (v/c)^2) is derived from the Theory of Relativity itself (a Theory paradoxically based and refuted by the result of the Michelson-Morley experiment).

If it is given that the length in the moving system of the path of a ray of light emitted by the moving system is the length L of a rigid rod, and that the length in the stationary system of L is rAB, then

yes, if tB - tA = rAB/(c + v) = t'A - tB = rAB/(c + v), then rAB/L is not equal to tB - tA.

But by the result of the Michelson-Morley,

tB - tA = rAB/(c + v) = t'A - tB = rAB/(c - v) = L/c.

3. According to the Special Theory of Relativity, the length L, in the stationary system is

(L + v*t')/sqrt(1 - (v/c)^2) (a function of the length L and of the time t')

= c*(L + D)/sqrt(c^2 - v^2).
The purpose for this reply is to correct errors in typing. The quote above ought to be written as follows:

The error is in substituting the length (L + D)/sqrt(1 - (v/c)^2) for the length L. The length (L + D)/sqrt(1 - (v/c)^2) is derived from the Theory of Relativity itself (a Theory paradoxically based and refuted by the result of the Michelson-Morley experiment).

If it is given that the length in the moving system of the path of a ray of light emitted by the moving system is the length L of a rigid rod, and that the length in the stationary system of L is rAB, then

yes, if tB - tA = rAB/(c - v) = t'A - tB = rAB/(c + v), then rAB/L is not equal to tB - tA.

But by the result of the Michelson-Morley,

tB - tA = rAB/(c + v) = t'A - tB = rAB/(c - v) = L/c.

3. According to the Special Theory of Relativity, the length L, in the stationary system is

(L + v*t')/sqrt(1 - (v/c)^2) (a function of the length L and of the time t')

= c*(L + D)/sqrt(c^2 - v^2).
 

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