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20/01/2007
RengaW
Why is the speed of light the same in every frame?
This question was asked in the thread
SPEED OF LIGHT???
http://www.orkut.com/CommMsgs.aspx?cmm=
and I gave Landau's argument from The Classical Theory of Fields.
SPEED OF LIGHT???
http://www.orkut.com/CommMsgs.aspx?cmm=
and I gave Landau's argument from The Classical Theory of Fields.
30/03/2007
RengaW
Related thread:
What is the ultimate speed
http://www.orkut.com/CommMsgs.aspx?cmm=
What is the ultimate speed
http://www.orkut.com/CommMsgs.aspx?cmm=
From the first link:
showing 1-10 of 45
19/01/2007
Deepak
er....im not that good at relativity....so i dont know whether this question is a valid one or not....
"how is speed of light a constant in all frames of reference....???"
even though i read quite a bit abt the experimental proofs regarding
the speed of light being constant...i never found an article which describes why it is so...is it just sumthin dat cant be explained yet???
guys please tell ur views..........
"how is speed of light a constant in all frames of reference....???"
even though i read quite a bit abt the experimental proofs regarding
the speed of light being constant...i never found an article which describes why it is so...is it just sumthin dat cant be explained yet???
guys please tell ur views..........
20/01/2007
RengaW
Like everything else in physics, this is ultimately an experimental issue so, yes, "nature wants it that way." But here's an argument why the speed of light must be the same in every reference frame.
Let me start with a more fundamental experimental fact: it's an experimental fact that all interactions propagate with a finite speed.
(By "interaction", I mean a "means of exchanging information.")
The statement in blue above doesn't mean that all interactions propagate, necessarily, with the same speed. It could be, for example, that gravity propagated at one speed and light propagated with another. (That isn't the case, however. More on this at the end.)
It also doesn't mean that they all propagate at a constant speed. For example, the speed of sound in a gas depends on the kind of gas, as well as on its pressure and on its density. Light propagates in a optically transparent medium with a speed that depends on the medium's index of refraction.
What it does mean is that there is a maximum speed of propagation for each interaction. And since we're not assuming that all interactions have the same maximum speed, we can rank the different interactions in ascending order of their maximum speeds. Let's call the largest of those maximum speeds, C. Thus, C is the ultimate barrier, the maximum speed of propagation of any interaction.
Ok, so now imagine that Alice and Bob, at rest with respect to one another, exchange some information using the interaction that has C for its maximum speed. More specifically, imagine that Alice is the sender and Bob is the recipient. Now imagine that Charlie and Debbie are at rest inside a train moving with speed v, in the direction from Bob to Alice.
Let me start with a more fundamental experimental fact: it's an experimental fact that all interactions propagate with a finite speed.
(By "interaction", I mean a "means of exchanging information.")
The statement in blue above doesn't mean that all interactions propagate, necessarily, with the same speed. It could be, for example, that gravity propagated at one speed and light propagated with another. (That isn't the case, however. More on this at the end.)
It also doesn't mean that they all propagate at a constant speed. For example, the speed of sound in a gas depends on the kind of gas, as well as on its pressure and on its density. Light propagates in a optically transparent medium with a speed that depends on the medium's index of refraction.
What it does mean is that there is a maximum speed of propagation for each interaction. And since we're not assuming that all interactions have the same maximum speed, we can rank the different interactions in ascending order of their maximum speeds. Let's call the largest of those maximum speeds, C. Thus, C is the ultimate barrier, the maximum speed of propagation of any interaction.
Ok, so now imagine that Alice and Bob, at rest with respect to one another, exchange some information using the interaction that has C for its maximum speed. More specifically, imagine that Alice is the sender and Bob is the recipient. Now imagine that Charlie and Debbie are at rest inside a train moving with speed v, in the direction from Bob to Alice.
20/01/2007
RengaW
To them, inside the train, the signal sent from Alice to Bob would seem to travel at a speed (C+v) and, therefore, would move between Charlie and Debbie at a speed larger than C. But that contradicts the experimental fact that no signal can propagate faster than C, since C is the largest of all the maximum speeds of propagation.
Therefore, we must conclude that the signal also travels at speed C with respect to Charlie and Debbie, that is, the largest of all the maximum speeds of propagation must be the same in every reference frame. (Here I used two such frames: one, where Alice and Bob are, and another, where Charlie and Debbie are.)
Now, as it happens, it's also an experimental fact that the maximum speeds of propagation of every known interaction are all the same.And since light moves with a maximum speed of c, we conclude that C is, in fact, c. Thus, for instance, gravitational waves also propagate (in vacuum) at the speed of light in vacuum, c.
And that is the reason why the speed of light in vacuum must have the same value in all reference frames.
Of course, this is not how Einstein came up with his postulate that the speed of light is the same in all inertial frames. He reached this conclusion by analyzing the details of how electromagnetic phenomena behave in different inertial frames. In a way he was lucky, because if it was the case that different interactions had different maximum speeds, it could have been the case that light was not the fastest, in which case his postulate would have been wrong. As it happens, all interactions do have the same maximum speed and so it was ok for him to speak of the speed of light.
The entire argument I gave above also shows that relativity really has nothing to do with light, specifically. Relativity has to do with exchanging information, be it by means of light signals or anything else. That's why relativity is such a fundamental part of physics.
Therefore, we must conclude that the signal also travels at speed C with respect to Charlie and Debbie, that is, the largest of all the maximum speeds of propagation must be the same in every reference frame. (Here I used two such frames: one, where Alice and Bob are, and another, where Charlie and Debbie are.)
Now, as it happens, it's also an experimental fact that the maximum speeds of propagation of every known interaction are all the same.And since light moves with a maximum speed of c, we conclude that C is, in fact, c. Thus, for instance, gravitational waves also propagate (in vacuum) at the speed of light in vacuum, c.
And that is the reason why the speed of light in vacuum must have the same value in all reference frames.
Of course, this is not how Einstein came up with his postulate that the speed of light is the same in all inertial frames. He reached this conclusion by analyzing the details of how electromagnetic phenomena behave in different inertial frames. In a way he was lucky, because if it was the case that different interactions had different maximum speeds, it could have been the case that light was not the fastest, in which case his postulate would have been wrong. As it happens, all interactions do have the same maximum speed and so it was ok for him to speak of the speed of light.
The entire argument I gave above also shows that relativity really has nothing to do with light, specifically. Relativity has to do with exchanging information, be it by means of light signals or anything else. That's why relativity is such a fundamental part of physics.
20/01/2007
RengaW
By the way, as much as I'd love to take credit for coming up with this argument, I didn't come up with it. The essence of the argument has been described in at least one book, which is where I learned of it: Landau's The Classical Theory of Fields.
I merely added Alice, Bob, Charlie, and Debbie into the mix, to make the argument easier to understand.
I merely added Alice, Bob, Charlie, and Debbie into the mix, to make the argument easier to understand.
24/07/2008
Rakesh
Coming back to "Alice, Bob, Charlie, and Debbie "
Signals sent from Alice takes a finite time to reach Bob with a velocity C. And logically Charlie, and Debbie should percieve that signal to be C+ v. Which of course according to the theory they are not. The perceptions of the velocity of signals by Alice and Bob is C and they think that Charlie, and Debbie must be perceiving it to be 'C+v'.
But the fact is that the later two are percieving the velocity in their frame as to be same as the former two are percieving it in their frame.
This apparent paradox can ONLY happen if the perception of time is altered for Charlie, and Debbie in such a way that a velocity of C+v is percieved as being C only to them in their frame .
So we can say that this happens because diffrent perceptions of time in diffrent frames.
Signals sent from Alice takes a finite time to reach Bob with a velocity C. And logically Charlie, and Debbie should percieve that signal to be C+ v. Which of course according to the theory they are not. The perceptions of the velocity of signals by Alice and Bob is C and they think that Charlie, and Debbie must be perceiving it to be 'C+v'.
But the fact is that the later two are percieving the velocity in their frame as to be same as the former two are percieving it in their frame.
This apparent paradox can ONLY happen if the perception of time is altered for Charlie, and Debbie in such a way that a velocity of C+v is percieved as being C only to them in their frame .
So we can say that this happens because diffrent perceptions of time in diffrent frames.
25/07/2008
Utkarsh
With the solution given by RengaW, I discovered one loop hole. Let me explain.
Consider this example.
A car is moving on Highway at the Speed of 200 km/hr. Now the scene is of the night. So turns on the Headlights.
Now according to Sir Newton, the speed of light would be 3 00 003.35 km/s.
(200 km/hr = 3.35 km/s and Speed of light = c + 3.35 km/s).
But on measuring (practically) the speed of light it is found to be 3 00 000 km/s and not 3 00 003.35 km/s. Why? Here comes the explanation.
Consider a person that is standing on Highway watching this Car traveling at 200 km/hr. Now also consider that the car is to cross the person after some time. Now the driver and the stationary person on the highway both start their stopwatches at the same time to an accuracy of 1 picosecond.
When the car will cross the person, both the people stop their stopwatches at the same instant. Suppose the reading of the stopwatch in the hands of the stationary person is 10 s, the reading of the stopwatch of the driver would be say 9.9999998 s. (You did not mention this my dear RengaW). Thus the time became slow for the person in the car.
Now if the car would have been stationary, the light would have traveled 3 00 000 km/s. But the car is traveling at 200 km/hr. So the distance traveled by light in 1 second is the total of the following:
Distance traveled by light in 9.9999998 + 3.35. This would total up to exactly the speed of light as measured from the stationary frame of reference (just to say because absolutely no frame of reference is STATIONARY.)
So the speed of light remains constant!
Consider this example.
A car is moving on Highway at the Speed of 200 km/hr. Now the scene is of the night. So turns on the Headlights.
Now according to Sir Newton, the speed of light would be 3 00 003.35 km/s.
(200 km/hr = 3.35 km/s and Speed of light = c + 3.35 km/s).
But on measuring (practically) the speed of light it is found to be 3 00 000 km/s and not 3 00 003.35 km/s. Why? Here comes the explanation.
Consider a person that is standing on Highway watching this Car traveling at 200 km/hr. Now also consider that the car is to cross the person after some time. Now the driver and the stationary person on the highway both start their stopwatches at the same time to an accuracy of 1 picosecond.
When the car will cross the person, both the people stop their stopwatches at the same instant. Suppose the reading of the stopwatch in the hands of the stationary person is 10 s, the reading of the stopwatch of the driver would be say 9.9999998 s. (You did not mention this my dear RengaW). Thus the time became slow for the person in the car.
Now if the car would have been stationary, the light would have traveled 3 00 000 km/s. But the car is traveling at 200 km/hr. So the distance traveled by light in 1 second is the total of the following:
Distance traveled by light in 9.9999998 + 3.35. This would total up to exactly the speed of light as measured from the stationary frame of reference (just to say because absolutely no frame of reference is STATIONARY.)
So the speed of light remains constant!
Some replies on this page have been deleted or are under review.
The second link deals with some of the doubt clearing that occurred subsequently:
showing 1-10 of 17
29/03/2007
Gururaj
What is the ultimate speed
Hi, I always have this doubt:
What is the ultimate speed? Is it limited by the speed of Light in vaccum or is it 3*10^8 m/s?
I want to know if the utimate speed achievable is 3*10^8 m/s and light happens to have that speed or is it that the utimate speed barrier is the speed of light, whatever it might be(and in our universe, it just happens to be 3*10^8 m/s)?
What is the ultimate speed? Is it limited by the speed of Light in vaccum or is it 3*10^8 m/s?
I want to know if the utimate speed achievable is 3*10^8 m/s and light happens to have that speed or is it that the utimate speed barrier is the speed of light, whatever it might be(and in our universe, it just happens to be 3*10^8 m/s)?
29/03/2007
Amar
it is the speed of light itself
the ultimate speed in this universe is the speed at which electromagnetic radiation propagates through vacuum.
as u all must be knowing, even if we remove all the matter in a given space, and also remove the heat energy in that space by cooling it, it is impossible to eliminate the electromagnetic radiation that prevades through it.
as u all must be knowing, even if we remove all the matter in a given space, and also remove the heat energy in that space by cooling it, it is impossible to eliminate the electromagnetic radiation that prevades through it.
29/03/2007
Gururaj
What is the meaning of All pervading electromagnetic radiation? EM radiation can travel without any medium, so it can travel even if there is nothing, but that doesnt make it all pervading...
Unless there is a source of EM radiation, how do we get it?
Unless there is a source of EM radiation, how do we get it?
29/03/2007
RengaW
as u all must be knowing, even if we remove all the matter in a given space, and also remove the heat energy in that space by cooling it, it is impossible to eliminate the electromagnetic radiation that prevades through it.
The above is not the reason why the correct answer is the speed of light itself. Moreover, as you must be knowing from reading the community guidelines, Amar, the use of sms is not acceptable in Physics II.
To understand why the ultimate speed is the speed of light, refer to the following thread:
Why is the speed of light the same in every frame?
http://www.orkut.com/CommMsgs.aspx?cmm=
The above is not the reason why the correct answer is the speed of light itself. Moreover, as you must be knowing from reading the community guidelines, Amar, the use of sms is not acceptable in Physics II.
To understand why the ultimate speed is the speed of light, refer to the following thread:
Why is the speed of light the same in every frame?
http://www.orkut.com/CommMsgs.aspx?cmm=
30/03/2007
Ganesh ~
http://www.orkut.com/CommMsgs.aspx?cmm=
You explained in this forum why light travels with a constant speed with respect to all frames of reference by referring to peak speed of interactions. I've a silly doubt regarding that.
Consider the double slit experiment. If two waves of light are emitted from two slits with the same phase,(let's not enter into 'particle-wave duality" duel here), then according to relativity(and electromagnetic field theory) if light travels at the same speed of 3*10^8 m/s regardless of the speed of motion of the observer, then one wave emitted from one of the slits should travel at a speed of 3*10^8m/s with respect to the other wave(consider this wave as an observer). so in that case, one wave would've reached the screen before the other had managed to. and hence no question of interference. But the reality is the vice-versa. We would get a interference fringe in that case experimentally. So where had i gone wrong. Sorry if my question is so fundamental.
You explained in this forum why light travels with a constant speed with respect to all frames of reference by referring to peak speed of interactions. I've a silly doubt regarding that.
Consider the double slit experiment. If two waves of light are emitted from two slits with the same phase,(let's not enter into 'particle-wave duality" duel here), then according to relativity(and electromagnetic field theory) if light travels at the same speed of 3*10^8 m/s regardless of the speed of motion of the observer, then one wave emitted from one of the slits should travel at a speed of 3*10^8m/s with respect to the other wave(consider this wave as an observer). so in that case, one wave would've reached the screen before the other had managed to. and hence no question of interference. But the reality is the vice-versa. We would get a interference fringe in that case experimentally. So where had i gone wrong. Sorry if my question is so fundamental.
30/03/2007
MegaGoat
There is a mistake in your argument that the MegaGoat can perceive. The other wave cannot be the observer, because a frame of reference cannot be attached to it. Nothing travels at 'c'. So the observer can't. The speed of one ray of light with respect to another is not a physically meaningful term.
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30/03/2007
Ї∫∏ặ...JOKER-
well, as the thread is closely related to speed of light, i rather wanted to post my question here instead of starting a new thread.
hope it doesn't violate any guidelines:
why does light travel at c=2999792458 m/s and not some other value?
it may sound stupid, but i couldn't figure it out.
hope it doesn't violate any guidelines:
why does light travel at c=2999792458 m/s and not some other value?
it may sound stupid, but i couldn't figure it out.
30/03/2007
RengaW
This is where your argument is flawed:
then one wave emitted from one of the slits should travel at a speed of 3*10^8m/s with respect to the other wave(consider this wave as an observer).
EM waves (or, in the parlance of QED, photons) are not physically realizable reference frames. In other words, you cannot attach a reference frame to a photon, which is to say, you can never be at rest with respect to a photon. Thus, you cannot make a statement such as "consider the wave as an observer."
That's because a reference frame is not an abstract concept. A reference frame is actually an array of metersticks and properly synchronized clocks, positioned at every point in space.
To understand this is a little better, consider what it means to say "consider X to be an observer" where X is something (a person, a planet, a particle, whatever). What it means is that we have an array of clocks and metersticks (the reference frame) momentarily or permanently at rest with respect to the observer in question (X). For that to happen, the whole array of clocks and metersticks have to be moving momentarily or permanently with the same velocity as X itself.
But no massive object can move at the speed of light, so we conclude that one cannot have a reference frame momentarily or permanently at rest with respect to a photon. In other words, we cannot attach a reference frame to a photon - the photon is not a valid observer.
then one wave emitted from one of the slits should travel at a speed of 3*10^8m/s with respect to the other wave(consider this wave as an observer).
EM waves (or, in the parlance of QED, photons) are not physically realizable reference frames. In other words, you cannot attach a reference frame to a photon, which is to say, you can never be at rest with respect to a photon. Thus, you cannot make a statement such as "consider the wave as an observer."
That's because a reference frame is not an abstract concept. A reference frame is actually an array of metersticks and properly synchronized clocks, positioned at every point in space.
To understand this is a little better, consider what it means to say "consider X to be an observer" where X is something (a person, a planet, a particle, whatever). What it means is that we have an array of clocks and metersticks (the reference frame) momentarily or permanently at rest with respect to the observer in question (X). For that to happen, the whole array of clocks and metersticks have to be moving momentarily or permanently with the same velocity as X itself.
But no massive object can move at the speed of light, so we conclude that one cannot have a reference frame momentarily or permanently at rest with respect to a photon. In other words, we cannot attach a reference frame to a photon - the photon is not a valid observer.
30/03/2007
RengaW
why does light travel at c=2999792458 m/s and not some other value?
That's a question for which we do not have an answer yet. No current theory can predict or explain the value of fundamental constants such as c, Planck's constant h, the gravitational constant G, and others.
I have to say, though, that the actual numerical value is irrelevant. In fact, in particle physics, where particles often move at speeds close to the speed of light, it's common to use a system of units where c has the value 1, that is, c = 1.
This change of values is possible because c, being the value of a speed, has units. c is 2999792458 m/s and also 1 light-year/year.
The only meaningful way to refer to the values of fundamental constants is by means of combinations of them having no units - the so-called dimensionless constants.
For example, the fine-structure constant <http://en.wikipedia.org/wiki/Fine_stru
Thus, your question and my answer should both be reformulated in terms of fundamental dimensionless constants. For instance, "why does the fine-structure constant have the value 1/137 and not something else?" and the answer would be "we don't know; there is currently no theory that can predict its value from first principles."
That's a question for which we do not have an answer yet. No current theory can predict or explain the value of fundamental constants such as c, Planck's constant h, the gravitational constant G, and others.
I have to say, though, that the actual numerical value is irrelevant. In fact, in particle physics, where particles often move at speeds close to the speed of light, it's common to use a system of units where c has the value 1, that is, c = 1.
This change of values is possible because c, being the value of a speed, has units. c is 2999792458 m/s and also 1 light-year/year.
The only meaningful way to refer to the values of fundamental constants is by means of combinations of them having no units - the so-called dimensionless constants.
For example, the fine-structure constant <http://en.wikipedia.org/wiki/Fine_stru
Thus, your question and my answer should both be reformulated in terms of fundamental dimensionless constants. For instance, "why does the fine-structure constant have the value 1/137 and not something else?" and the answer would be "we don't know; there is currently no theory that can predict its value from first principles."
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