Why was the concept of aether discarded in the study of light and motion?

[Adrian07]

Am a bit confused about the speed of light being constant, does this mean that whatever speed I am going at, up to and including the speed of light, I will always measure it as going 300000000 mts/sec faster than myself?

 

[Doc Al]

Yes, you will always measure the speed of light to be c with respect to you, regardless of your motion with respect to anything else.

 

[harrylin]

Am a bit confused about the speed of light being constant, does this mean that whatever speed I am going at, up to and including the speed of light, I will always measure it as going 300000000 mts/sec faster than myself?

Welcome to PF!

Not necessarily so; and note that you can never reach the speed of light.

If you set up a standard inertial reference system* in the lab, then you will next measure the speed of light in vacuum to be nearly 3E8 m/s wrt that reference system. That is a constant: it is the same in all directions, and independent of the motion of the source. Moreover, you don't need to be at rest in your reference system, and neither has the light detector to be at rest in that system.

Many people (even teachers) confound that "constant" with a different kind of constancy, which is that the speed of light is invariant. With that is meant that if you measure light - even coming from the same source - with another standard reference system that is in uniform motion relative to the first, you will again find the same speed.

Now, if you do what you seem to suggest - take your physical system out of your lab and in your car, and measure the one-way speed of entering light rays while you are driving - then you may not find the same value. However, after you re-synchronize your on-board clocks at that velocity, then you'll have again a standard reference system if you keep approximately an inertial course. Subsequently you'll find again the standard value.

* For a simple explanation see section 1 of http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Erland]

Now, if you do what you seem to suggest - take your physical system out of your lab and in your car, and measure the one-way speed of entering light rays while you are driving - then you may not find the same value.

Are you here talking about the car as an accelerated frame of reference?

For if the car is moving with a constant velocity wrt the lab system, then, certainly, the speed of light will be measured to c inside the car.

 

[harrylin]

[..] if the car is moving with a constant velocity wrt the lab system, then, certainly, the speed of light will be measured to c inside the car.

Not necessarily so: "clock synchronization" is for each velocity different, so that it isn't a standard reference system anymore (it is not auto-correcting). Therefore I stressed the importance of synchronization at that velocity.

For others who perhaps not understand this, see: http://www.bartleby.com/173/9.html

 

[Erland]

Not necessarily so: "clock synchronization" is for each velocity different, so that it isn't a standard reference system anymore (it is not auto-correcting). Therefore I stressed the importance of synchronization at that velocity.

Do you mean that if we put two synchronized clocks at different places in a car at rest, and then accelerate the car up to a constant velocity, then, after the acceleration, the clocks are no longer synchronized, wrt an observer inside the car?

 

[Erland]

Not necessarily so: "clock synchronization" is for each velocity different, so that it isn't a standard reference system anymore (it is not auto-correcting). Therefore I stressed the importance of synchronization at that velocity.

Do you mean that if we put two synchronized clocks at different places in a car at rest, say one clock in the front seat and one in the back seat, and then accelerate the car up to a constant velocity, then, after the acceleration, the clocks are no longer synchronized, wrt an observer inside the car?

 

[harrylin]

Do you mean that if we put two synchronized clocks at different places in a car at rest, say one clock in the front seat and one in the back seat, and then accelerate the car up to a constant velocity, then, after the acceleration, the clocks are no longer synchronized, wrt an observer inside the car?

Exactly.

Edit: I doubt however that this can be detected with current technology; a car is too slow and too small.

 

[Adrian07]

So does this mean if I am emitting the light it will always leave me at c whatever speed I am doing? If I am moving at say 1/2 c and measure the light to be leaving me at c how can someone who is not moving still measure that light at c?
You mention clocks and I think I read somewhere that einstein said something about time changing how does that work?

 

[harrylin]

So does this mean if I am emitting the light it will always leave me at c whatever speed I am doing?

Yes, just like sound: the speed of the emitter does not affect the speed of propagation.
(that is called in the first paper that I linked for you the "second postulate").

If I am moving at say 1/2 c and measure the light to be leaving me at c how can someone who is not moving still measure that light at c?

I think that the question should be phrased the other way round (often a misunderstanding already exists in the question). Someone who is not moving will still measure that light at c, independent of the motion of the source; that is the second postulate, based on a well established theory of electromagnetism and radiation.
So, the question to ask is: how can you, when you are moving, also measure that light at c? And that was indeed the question around 1900. In the introduction of the paper that I linked for you this was said to be "apparently irreconcilable" with the first. How much of it did you read?

You mention clocks and I think I read somewhere that einstein said something about time changing how does that work?

Probably you mean "time dilation". However, for one-way light speed, the first (and main!) change is a man-made change, as explained in the first section of the paper to which I gave you a link. He explains clock synchronisation. Did you understand it?

And did you read my last remark about what happens when you accelerate with such a reference system to a new constant velocity (post #3)? Assuming that you use perfect clocks and that you can measure precisely enough: then without doing a new synchronisation you will not measure light leaving you at c, but at approximately* c-v for light that leaves you straight ahead. And that result is probably just what you would expect.


*"approximately": still good approximation at 1% of c, but poor at 0.5 c. I now prepared a numeric example if someone is interested

 

[Adrian07]

Have tried to read the link but it seems long winded and difficult to follow.
I assume that inertial means moving at a constant speed.
I am having problems with the fact that the speed of light is not dependant on the speed of the source, whereas normally the speed of something is dependant on its source i.e. if I fire a gun the speed of the bullet depends on the speed and direction of the gun although the speed is always the same relative to the gun. Someone standing still will measure the bullets speed as being different to that of the moving person holding the gun. With light I get the impression that both would measure it as moving at c, is this the case?
As you keep talking about synchronising clocks does time change somehow relative to the person doing the measurement so changing their perception of speed.

 

[Doc Al]

I am having problems with the fact that the speed of light is not dependant on the speed of the source, whereas normally the speed of something is dependant on its source i.e. if I fire a gun the speed of the bullet depends on the speed and direction of the gun although the speed is always the same relative to the gun. Someone standing still will measure the bullets speed as being different to that of the moving person holding the gun. With light I get the impression that both would measure it as moving at c, is this the case?

Yes, exactly.

You might find the following discussion of special relativity easier to follow: Special Relativity. That's the first lecture in a series that covers all the usual relativistic effects, such as time dilation, length contraction, and the relativity of simultaneity.

 

[Erland]

Suppose we have two inertial systems S1 and S2 and that S2 moves with velocity u wrt S1, -c<u<c. Also assume that an object O moves with a constant velocity v wrt S2, in a direction parallell to the direction of the motion between S1 och S2. Here -c<=v<=c, so O might be a photon, but it doesn't have to.

Now what is the velocity w of O wrt S1?

In classical physics, we have w= u+v. But this simple addition formula is invalid in SR, where we have the more complicated velocity addition formula:

w=(u+v)/(1+uv/c^2).

If u and v are small compared to c, then the denominator is close to 1, and then this almost reduces to the classical case. For small mundane velocities, the error in the classical formula is not detectable without very advanced instruments.

It can be shown from this SR-formula that w<=c, with equality if and only if v=c or v=-c (we assumed that -c<u<c). So the relative velocity of a light source does not matter, light speed will be measured to c in all inertial systems (yeah yeah Harrylin, the clocks must be synchronized). And a velocity greater than c will never be measured.

 

[ZikZak]

I am having problems with the fact that the speed of light is not dependant on the speed of the source, whereas normally the speed of something is dependant on its source

The speed of sound waves in the air, or water waves on the ocean don't have anything to do with the source. Generally, waves travel at a speed in the medium that depends on the characteristics of that medium, not on anything having to do with the source of the waves.

The difference is that light waves have no medium except space, and space is the same regardless of your frame of reference.

 

[harrylin]

Yes, exactly.

You might find the following discussion of special relativity easier to follow: Special Relativity. That's the first lecture in a series that covers all the usual relativistic effects, such as time dilation, length contraction, and the relativity of simultaneity.

At first sight, that's a nice introduction of the basics.

Regretfully some of it is misleading, partly because the author postpones the introduction of what it means to set up a reference system (incl. what "speed of light" here means) to after discussing the results of measurements with such a system.
For example, compare "Light travels at c relative to the observer" just before giving a one-way light speed example (as if it's a True Measurement, see also the word "Truth" in the header!), with the Wikipedia article on that same topic:
http://en.wikipedia.org/wiki/One-way_speed_of_light

 

[Doc Al]

Sorry, harrylin, but I believe you are making things even more confusing for a beginner by focusing (as you often do) on one-way speed of light issues.

 

[harrylin]

Sorry, harrylin, but I believe you are making things even more confusing for a beginner by focusing (as you often do) on one-way speed of light issues.

One-way speed is the focus of the OP and I usually try to answer the question of the OP. But I fully agree that for a complete beginner it may be better to explain two-way speed of light measurements first (Fowler's lecture mixes them up however).

Note: on top of that, that lecture spreads misinformation in the introduction concerning MMX, which only in the best case causes threads like https://www.physicsforums.com/showthread.php?t=631954.

It may be a good idea to start a wiki kind of page with links to good web lectures for beginners. Make it a topic on the forum feedback?

 

[ghwellsjr]

Sorry, harrylin, but I believe you are making things even more confusing for a beginner by focusing (as you often do) on one-way speed of light issues.

Einstein focused on the one-way speed of light issue in his 1905 paper introducing Special Relativity. His second postulate focuses on the one-way speed of light.

Beginners need to understand that when we are talking about measuring the speed of light being equal to c, we are always talking measuring the round-trip speed of light. When we are talking about the one-way speed of light, we are not talking about a measurement but rather an arbitrary assignment, an arbitrary definition, an arbitrary stipulation, an arbitrary assumption, an arbitrary postulate, an arbitrary axiom, according to Einstein.

Look at the OP's question:

Am a bit confused about the speed of light being constant, does this mean that whatever speed I am going at, up to and including the speed of light, I will always measure it as going 300000000 mts/sec faster than myself?

He's asking about measuring the one-way speed of light. We have to assume that he's asking about the speed of light in the direction that he is moving, not in the direction from where he is coming from, otherwise, he would have wondered if the light would be going slower than himself.

If we point out to him that if he measures the round-trip speed of light by putting a mirror in front of him (like Einstein discussed in his 1905 paper), he will get the same answer that he will get if he does the same measurement in the opposite direction by putting the mirror behind him. This usually surprises beginners until they realize that it's the same measurement, the only difference being the two directions of light travel happen in the opposite order.

Then they have to realize that it is impossible to know if it takes the same time for the light to traverse the distance to the mirror as it takes for the reflection to get back to the observer and this is where Einstein's arbitrary assignment of those two times being equal comes in. This is where we get the unmeasurable one-way speed of light being equal to the same value as the measured two-way speed of light--it's by assignment.

This is the foundational basis of Einstein's argument for Special Relativity, both in his 1905 paper and in his 1920 book. I don't understand why we should hide this from beginners. It's how the theory began. I support harrylin's focus and if he hadn't been here prior to now, I would have been.

 

[VegaMan]

Einstein focused on the one-way speed of light issue in his 1905 paper introducing Special Relativity. His second postulate focuses on the one-way speed of light.

Beginners need to understand that when we are talking about measuring the speed of light being equal to c, we are always talking measuring the round-trip speed of light. When we are talking about the one-way speed of light, we are not talking about a measurement but rather an arbitrary assignment, an arbitrary definition, an arbitrary stipulation, an arbitrary assumption, an arbitrary postulate, an arbitrary axiom, according to Einstein.

Look at the OP's question:


He's asking about measuring the one-way speed of light. We have to assume that he's asking about the speed of light in the direction that he is moving, not in the direction from where he is coming from, otherwise, he would have wondered if the light would be going slower than himself.

If we point out to him that if he measures the round-trip speed of light by putting a mirror in front of him (like Einstein discussed in his 1905 paper), he will get the same answer that he will get if he does the same measurement in the opposite direction by putting the mirror behind him. This usually surprises beginners until they realize that it's the same measurement, the only difference being the two directions of light travel happen in the opposite order.

Then they have to realize that it is impossible to know if it takes the same time for the light to traverse the distance to the mirror as it takes for the reflection to get back to the observer and this is where Einstein's arbitrary assignment of those two times being equal comes in. This is where we get the unmeasurable one-way speed of light being equal to the same value as the measured two-way speed of light--it's by assignment.

This is the foundational basis of Einstein's argument for Special Relativity, both in his 1905 paper and in his 1920 book. I don't understand why we should hide this from beginners. It's how the theory began. I support harrylin's focus and if he hadn't been here prior to now, I would have been.

wait so the speed of the light going to the mirror is not the same as the speed of the light being reflected from the mirror? Only that the total distance over the total time is the same? That kinda makes sense since some energy is lost due to reflecting but I always thought that the amplitude was the lost energy not the velocity

 

[Erland]

No, no. Light speed is constant and the same in all directions. What the other posters talk about is just the difficulties in defining and measuring it.

 

[ghwellsjr]

wait so the speed of the light going to the mirror is not the same as the speed of the light being reflected from the mirror?

Nobody, certainly not Einstein, said the speeds were different. The point is that we can't know. We can't measure the difference. We can't tell if they are the same or if they are different. We don't know.[/QUOTE]

Only that the total distance over the total time is the same?

With regard to what we can measure, yes, it's only the total distance divided by the total time that always comes out the same.

That kinda makes sense since some energy is lost due to reflecting but I always thought that the amplitude was the lost energy not the velocity

No, this has nothing to do with energy or amplitude, only velocity. Have you read Einstein's 1905 paper, especially the first couple of sections?

 

[harrylin]

Have tried to read the link but it seems long winded and difficult to follow. I assume that inertial means moving at a constant speed.

Only the introduction and part 1, the explanation of how to set distant clocks are essential here. And yes, "inertial" is a term that people nowadays use for "Newtonian" or "Galilean" reference systems, which are systems in uniform rectilinear motion ("frames of reference for which the equations of mechanics hold good").

Of part 1, did you understand how to "synchronise" distant clocks? By means of such synchronisation you make the one-way "speed of light" equal to the measured two-way speed of light. Please be sure to understand that.

Next you may ask how it can be that (according to special relativity) the two-way speed of light is always measured as c. This can be attributed to time dilation and length contraction. From the perspective of relativity theory, those effects take care of maintaining the relativity principle. And how exactly time dilation and length contraction operate in the two-way measurement of light speed is explained on the second page of Doc Al's reference:

http://galileoandeinstein.physics.virginia.edu/lectures/srelwhat.html

Alternatively you could also try Wikipedia:

en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity

http://en.wikipedia.org/wiki/Length_contraction

I am having problems with the fact that the speed of light is not dependant on the speed of the source, whereas normally the speed of something is dependant on its source i.e. if I fire a gun the speed of the bullet depends on the speed and direction of the gun although the speed is always the same relative to the gun.

Not always "normally": light is not material, in certain ways it's more like sound or water waves than like bullets. The successful theory on which SR is based (Maxwell's electrodynamics), models light as a kind of wave in space. An essential feature of waves is that their speed c is independent of the motion of the source (to the precision that this has been verified).

[..] With light I get the impression that both would measure it as moving at c, is this the case?

As answered before: Yes, if you first synchronised your clocks at that velocity.

As you keep talking about synchronising clocks does time change somehow relative to the person doing the measurement so changing their perception of speed.

Perception of speed relative to what?

 

[rbj]

So does this mean if I am emitting the light it will always leave me at c whatever speed I am doing? If I am moving at say 1/2 c and measure the light to be leaving me at c how can someone who is not moving still measure that light at c?

Yes, just like sound: the speed of the emitter does not affect the speed of propagation.

i have trouble with "just like sound".

it is the case that with both sound and light that the speed of the emitter does not affect the speed of propagation. what must be kept in mind is that the speed of propagation of sound is relative to the medium that sound propagates in. a medium of propagation is necessary for sound. there is no sound in a vacuum. if this medium was whizzing past you (as an observer) at some speed (like with wind), you will measure the speed of sound to be different in the direction of the movement of the medium than in other directions.

but light does not propagate in any medium. there is no aether for light to propagate in. light can propagate in nothing but empty space. there is no physical meaning in nothing nor empty space whizzing by an observer at some speed. it's just that a changing E-field is causing a changing B-field which is causing a changing E-field. etc. once emitted, that's the mechanism for the propagation of the EM radiation. it does that in a vacuum or in air. for that reason, it doesn't matter if the frame of the emitter is whizzing past you at speed c/2 or not. when you are examining this beam of light (and measuring its speed) and someone traveling along with the emitter is examining the very same beam of light, both observers are in an inertial setting, both observers have equal claim to being "at rest" and the laws of physics, including physical constants, should be the same for both observers.

that's why both observers, one riding along with the light emitter and another watching that observer and his emitter fly by, must measure the speed of light emitted to be the same. now that is not the case with sound. if one observer is moving relative to the other, then the medium in which sound propagates cannot be moving at the same velocity for both observers.

 

[harrylin]

i have trouble with "just like sound".

it is the case that with both sound and light that the speed of the emitter does not affect the speed of propagation. [..] what must be kept in mind is that the speed of propagation of sound is relative to the medium that sound propagates in. [..]

First of all, the second postulate distinguishes SR from ballistic emission theories which just couldn't be made to work correctly.
However, and that was my point but also yours, it was assumed that if a wave appears to propagate isotropically at c relative to one reference system, it cannot also appear to propagate isotropically at c relative to another reference system that is in uniform motion relative to the first. It is because of that consideration that Einstein remarked that the light postulate is "apparently irreconcilable" with the PoR. You seem to have trouble with that remark.

 

[VegaMan]

Nobody, certainly not Einstein, said the speeds were different. The point is that we can't know. We can't measure the difference. We can't tell if they are the same or if they are different. We don't know.

With regard to what we can measure, yes, it's only the total distance divided by the total time that always comes out the same.

No, this has nothing to do with energy or amplitude, only velocity. Have you read Einstein's 1905 paper, especially the first couple of sections?[/QUOTE]

I have read it (cause it was linked above) but maybe i don't quite comprehend exactly what I am reading.

Basically what I'm getting from the first 2 sections of the paper is that In order to synchronize 2 clocks the time it takes light to travel from A to B must equal the time it takes for light to be reflected from B and arrive back at A. Also, that the distance from A to B and B to A (round trip distance) divided by the time it takes light to originate from A, travel to B, reflect off of B, and travel back to A equals the speed of light. This makes sense if and only if the the distance between points A and B constantly remains the same.

My question is that if these are true then, wouldn't that technically mean that no 2 clocks in motion relative to each other could ever be synchronized?
And if the speed of light is constant, since v = d/t wouldn't that inherently mean that time and distance would end up being dependent on each other instead of independent variables? when referring to points A and B?

 

[ghwellsjr]

Basically what I'm getting from the first 2 sections of the paper is that In order to synchronize 2 clocks the time it takes light to travel from A to B must equal the time it takes for light to be reflected from B and arrive back at A. Also, that the distance from A to B and B to A (round trip distance) divided by the time it takes light to originate from A, travel to B, reflect off of B, and travel back to A equals the speed of light. This makes sense if and only if the the distance between points A and B constantly remains the same.

That is correct.

My question is that if these are true then, wouldn't that technically mean that no 2 clocks in motion relative to each other could ever be synchronized?

That is correct.

And if the speed of light is constant, since v = d/t wouldn't that inherently mean that time and distance would end up being dependent on each other instead of independent variables? when referring to points A and B?

Yes, that's the importance of Special Relativity. Time is relative, space is relative. We now talk about spacetime where space and time are not independent of each other.

 

[VegaMan]

That is correct.

That is correct.

Yes, that's the importance of Special Relativity. Time is relative, space is relative. We now talk about spacetime where space and time are not independent of each other.

So as you go faster lengths get smaller? Or does time flow faster? or both?

 

[ghwellsjr]

So as you go faster lengths get smaller? Or does time flow faster? or both?

Once you use Einstein's definition of a Reference Frame, any rigid object that is traveling is length contracted along its direction of motion and it's clocks take longer to tick which means they run slower. But if you transform everything to a different Reference Frame moving at some speed with respect to the first one, the speeds of those objects can be different, meaning their length contractions and time dilations can be different and you can always transform to a frame in which any given object is at rest in which case there will be no length contraction or time dilation.

Also, be aware that any object or ruler that is traveling with an object will have the same length contraction and time dilation so any measurement that is performed on a "moving" object with identically moving rulers and clocks will come out the same no matter which Reference Frame is used.

 

[StevieTNZ]

An interesting article here: http://phys.org/news/2012-10-physicists-special-relativity.html

 

[VegaMan]

Once you use Einstein's definition of a Reference Frame, any rigid object that is traveling is length contracted along its direction of motion and it's clocks take longer to tick which means they run slower. But if you transform everything to a different Reference Frame moving at some speed with respect to the first one, the speeds of those objects can be different, meaning their length contractions and time dilations can be different and you can always transform to a frame in which any given object is at rest in which case there will be no length contraction or time dilation.

Also, be aware that any object or ruler that is traveling with an object will have the same length contraction and time dilation so any measurement that is performed on a "moving" object with identically moving rulers and clocks will come out the same no matter which Reference Frame is used.

ok, so if I'm traveling at let's say minutely just under the speed of light. To me, from my perspective, nothing changes. Time and space go on as normal? Or is it actually possible to turn on a flashlight and watch the beam of light slowly extend outward in front of me? Wouldn't time technically stop if i were to reach the speed of light? From my perspective in a fast spaceship that reached the speed of light, would i just instantaneously skip over the speed of light "gap" or interval of time until my speed was reduced to that less than the speed of light?

man this is some freaky stuff

 

[ghwellsjr]

ok, so if I'm traveling at let's say minutely just under the speed of light. To me, from my perspective, nothing changes. Time and space go on as normal?

What you're saying is that in some particular Reference Frame, you're traveling at just under the speed of light and yes, everything is normal for you. But, of course, whatever is at rest or traveling at slow speeds in that frame will be traveling at just under the speed of light relative to you so it's not like you can't tell that you are traveling at a high speed.

Or is it actually possible to turn on a flashlight and watch the beam of light slowly extend outward in front of me?

How do you watch a beam of light? It's not like watching some kind of projectile traveling away from you for which you shine light on it and the reflected light off the surface of the projectile is what you are actually seeing, correct? Since projectiles travel at a very small fraction of the speed of light, you simply ignore the additional time it takes for the reflected light to get back to you and you approximately "see" the projectile moving away from you.

But you can't do that with light. What you have to do instead is have some portion of the light beam itself reflect off of other things placed at increasing distances away from you but now you can't ignore the additional time it takes for the reflected light to get back to you. And what will it look like? Since the light is making a round trip, when you turn on the flashlight, it will look like the beam is traveling at one-half the speed of light, do you understand that?

Now because you are traveling in the same direction that the beam is traveling, and light is defined to travel at c in the selected Reference Frame, the beam will be traveling very slowly away from you and then after it hits one of the objects out in front of you, the reflected light will travel back to you almost instantly. Remember how you can't tell if the light takes the same amount of time to go away as it does to come back? This is an example of how you can't tell. As far as you are concerned, it won't look any different than when you are at rest in the Reference Frame and the light takes the same amount of time to go away as it does to come back (by definition, not by observation).

Wouldn't time technically stop if i were to reach the speed of light?

Didn't you start off your post by saying "To me, from my perspective, nothing changes. Time and space go on as normal?" To which I agreed. So even if you had accelerated from being at rest in the Reference Frame and you spent an enormous amount of energy getting to your high rate of speed, it would be just like you were at rest and you would have to start all over again to get back to your high rate of speed. You could repeat this as often as you wish and you'd be no closer to the speed of light than before you started.

But technically, you can't reach the speed of light, so there is no meaning to your question of what would happen if you were to reach it.

From my perspective in a fast spaceship that reached the speed of light, would i just instantaneously skip over the speed of light "gap" or interval of time until my speed was reduced to that less than the speed of light?

Another meaningless question for the same reason.

man this is some freaky stuff

Not to me and hopefully not to you some day.

 

[VegaMan]

What you're saying is that in some particular Reference Frame, you're traveling at just under the speed of light and yes, everything is normal for you. But, of course, whatever is at rest or traveling at slow speeds in that frame will be traveling at just under the speed of light relative to you so it's not like you can't tell that you are traveling at a high speed.

How do you watch a beam of light? It's not like watching some kind of projectile traveling away from you for which you shine light on it and the reflected light off the surface of the projectile is what you are actually seeing, correct? Since projectiles travel at a very small fraction of the speed of light, you simply ignore the additional time it takes for the reflected light to get back to you and you approximately "see" the projectile moving away from you.

But you can't do that with light. What you have to do instead is have some portion of the light beam itself reflect off of other things placed at increasing distances away from you but now you can't ignore the additional time it takes for the reflected light to get back to you. And what will it look like? Since the light is making a round trip, when you turn on the flashlight, it will look like the beam is traveling at one-half the speed of light, do you understand that?

Now because you are traveling in the same direction that the beam is traveling, and light is defined to travel at c in the selected Reference Frame, the beam will be traveling very slowly away from you and then after it hits one of the objects out in front of you, the reflected light will travel back to you almost instantly. Remember how you can't tell if the light takes the same amount of time to go away as it does to come back? This is an example of how you can't tell. As far as you are concerned, it won't look any different than when you are at rest in the Reference Frame and the light takes the same amount of time to go away as it does to come back (by definition, not by observation).

Didn't you start off your post by saying "To me, from my perspective, nothing changes. Time and space go on as normal?" To which I agreed. So even if you had accelerated from being at rest in the Reference Frame and you spent an enormous amount of energy getting to your high rate of speed, it would be just like you were at rest and you would have to start all over again to get back to your high rate of speed. You could repeat this as often as you wish and you'd be no closer to the speed of light than before you started.

But technically, you can't reach the speed of light, so there is no meaning to your question of what would happen if you were to reach it.

Another meaningless question for the same reason.

Not to me and hopefully not to you some day.

ok got it. So hypothetically, suppose we get the technology needed to travel to Proxima Centauri. It's about 4 light years away. We go there and travel from point A (being earth) to point B (being Proxima Centauri) at half the speed of light. Going off what Einstein said, me in the spaceship would perceive time to be traveling like normal, but it wouldn't take me 8 years to get there from MY time. Back on Earth though it would be an 8 year wait (technically 12 years since the first transmission sent out once the ship reaches Proxima Centauri would take 4 years to get back to earth). This sort of doesn't make sense. The distance between here and Proxima Centauri (for the purposes of the example) doesn't change. The velocity also does not change (suppose we got a running head start and only wanted to do a flyby of point B). Yet on the ship, both the distance and the time have changed? Is this correct? Could we actually calculate how long the trip would "feel" like on the ship?

 

[ghwellsjr]

ok got it. So hypothetically, suppose we get the technology needed to travel to Proxima Centauri. It's about 4 light years away. We go there and travel from point A (being earth) to point B (being Proxima Centauri) at half the speed of light. Going off what Einstein said, me in the spaceship would perceive time to be traveling like normal, but it wouldn't take me 8 years to get there from MY time.

That is correct. It would take you 6.9282 years to get there according to your clock.

Back on Earth though it would be an 8 year wait (technically 12 years since the first transmission sent out once the ship reaches Proxima Centauri would take 4 years to get back to earth).

That is correct. In the common earth/Centauri rest frame, it takes 8 years to get there but it takes 12 years for the earthlings to see you get there.

This sort of doesn't make sense. The distance between here and Proxima Centauri (for the purposes of the example) doesn't change.

In your rest frame, the distance that Proxima Centauri is away from you is not 4 light years but rather 3.4641 light years.

The velocity also does not change (suppose we got a running head start and only wanted to do a flyby of point B). Yet on the ship, both the distance and the time have changed? Is this correct? Could we actually calculate how long the trip would "feel" like on the ship?

Yes, that is correct. Your trip will take 6.9282 years and it will feel like 6.9282 years.

 

[phyti]

harrylin #3

Now, if you do what you seem to suggest - take your physical system out of your lab and in your car, and measure the one-way speed of entering light rays while you are driving - then you may not find the same value. However, after you re-synchronize your on-board clocks at that velocity, then you'll have again a standard reference system if you keep approximately an inertial course. Subsequently you'll find again the standard value.

The 1-way, 2-way,...n-way speed of light is c.
The relative light speed v/c is the variable, but that's what can't be measured. The observer can't receive the same unidirectional signal he sends (unless he can move faster then light). If he uses a 2nd observer, a 2nd clock is needed and the synchronization problem appears. Despite being unable to measure his absolute speed (he has one, otherwise v/c would be variable and unreliable), he can still achieve a relative synchronization. This is the familiar parallelogram with the skewed axis, seen in Hermanns (aka Minkowski spacetime diagrams,...who needs verboseness). The synchronization does not alter the value of c, it makes the outbound and inbound paths equal, per Einsteins 'stipulation', i.e, it's not a rule of physics, as he clearly states, but a definition (physics by decree). Since the observer can only be coincident with the emission and detection of the reflected signal, only the round trip time is measurable. The time and location of the reflection event is speculation, thus the skewed spatial axis is bogus.
Do you think it's possible to alter distance by setting a clock?

 

[harrylin]

harrylin #3
[..] The observer can't receive the same unidirectional signal he sends (unless he can move faster then light). If he uses a 2nd observer, a 2nd clock is needed and the synchronization problem appears. [..] Do you think it's possible to alter distance by setting a clock?

Sorry but I can't follow you. A person who has set up a physical reference system uses as many clocks and detectors as he wants (the system that I referred to uses two clocks, typically as detailed in post #7). That has nothing to do with "altering distances".

 

[Adrian07]

Have been looking at the special relativity link in post 12 particlarly the diagram at the end.
Quote The observer on the spaceship will measure the blip of light to be traveling at c relative to the spaceship, the observer on the ground will measure the same blip to be traveling at c relative to the ground. That is the unavoidable consequence of the Theory of Relativity.
While the speed of light remains the same the measured speed must be different as it would take longer to travel between the sensors.
If we add to the diagram the spaceship moving at c and a blip of light fired from the back of the ship towards the front as it passes the stationary light source, from outside both blips move at the same speed, inside the blip would seem as not moving, if it was seen as moving inside the ship it would be seen as moving at a different speed to the outside one from outside the ship, so how do we keep the measured speed of light as constant?
Please keep replies in laymans terms as simple as possible please.

 

[Erland]

If we add to the diagram the spaceship moving at c.

A ship cannot move at c. It cannot, even in theory, be accelerated up to c. That is a consequence of SR. Therefore, it is not meaningful to speculate of how things will be perceived inside such a ship.

 

[ghwellsjr]

Have been looking at the special relativity link in post 12 particlarly the diagram at the end.
Quote The observer on the spaceship will measure the blip of light to be traveling at c relative to the spaceship, the observer on the ground will measure the same blip to be traveling at c relative to the ground. That is the unavoidable consequence of the Theory of Relativity.
While the speed of light remains the same the measured speed must be different as it would take longer to travel between the sensors.
If we add to the diagram the spaceship moving at c and a blip of light fired from the back of the ship towards the front as it passes the stationary light source, from outside both blips move at the same speed, inside the blip would seem as not moving, if it was seen as moving inside the ship it would be seen as moving at a different speed to the outside one from outside the ship, so how do we keep the measured speed of light as constant?
Please keep replies in laymans terms as simple as possible please.

As Erland pointed out, the spaceship cannot travel at c so your following description is not meaningful but aside from that, you can still make your point, and it's a good one, with the spaceship traveling at one half c.

The point is that the diagram is illustrating that when you want to measure the speed of light, it is always a round-trip measurement, although the author seems blissfully unaware of that because he links it to Einstein's second postulate when it is actually a consequence of the first postulate.

Here's the author's explanation of how the measurements take place:

The setup is as follows: on a flat piece of ground, we have a flashlight which emits a blip of light, like a strobe. We have two photocells, devices which click and send a message down a wire when light falls on them. The photocells are placed 10 meters apart in the path of the blip of light, they are somehow wired into a clock so that the time taken by the blip of light to travel from the first photocell to the second, in other words, the time between clicks, can be measured. From this time and the known distance between them, we can easily find the speed of the blip of light.

He doesn't say where the clock is or how the wires go but let's assume that the clock is located next to the first photocell. We will assume that the messages from the photocells are sent down the wires to the clock at the speed of light.

As soon as the blip of light hits the first photocell, it instantly registers the time on the clock since the wire is very short. Now we wait for the blip to reach the second photocell and for the message to travel down the wire back to the clock. Both observers are making the same measurement separately with their own clock, photocells and wires, although they are measuring the same blip of light. Let's assume that light actually travels at c with respect to the light source and the ground observer.

Then for the ground observer, the time it takes the light to travel between the two photocells will be equal to the time it takes for the signal to travel from the second photocell back to the clock (colocated with the first photocell). So if the ground observer is blissfully unaware (like the author) of what's going on, he will say that his measurement of the speed of light is actually one half of c. If he realizes that the message takes additional time to travel down the wire, he might be inclined to double his calculation of the measured speed of light and then he'd get the right answer.

But would the same thing happen for the spaceship observer? Well it better or Einstein's first postulate would be violated. So how can they both get the same measurement when obviously it will take a lot longer for the blip to get from the first photocell to the second one which I think is your point? Well, what if the two photocells were closer together than the spaceship observer measured them to be? And what if his clock is running slower than he thinks it is? Then his measurement can come out the same as for the ground observer. We don't have to go into the details because you're a layman but I think you can see how this would allow the spaceship observer to get the same answer as the ground observer.

Unfortunately, the diagram doesn't show this and they have a note saying they ignored this salient fact but that is the explanation.

So in conclusion, if this measurement really was of the one-way speed of light like the author claims, then both observers would get a different answer because the amount of length contraction between the photocells and the amount of time dilation of the clock that explains the same answer for the round-trip measurement would not also work for the one-way measurement.

I hope I have hit the nail on the head for you. Did I? If not, please point out where I missed your concern.

 

[Adrian07]

George many thanks. I always asumed that the speed of light was measured in one direction, if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant.This I think is not much different to measuring something as going there and back.
Do the effects of length and time dilation mean that two people could measure the same thing and get different answers or two people measure different things and get the same answer.
How can just movinf faster affect your view of the universe?

 

[harrylin]

[..] if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant. [...]

Regretfully that is wrong. In SR, light propagation is modeled like waves and not like bullets. Please make sure to understand that first; correct understanding comes step by step.

For example, you could first read up on MMX which the foregoing lecture actually discusses (and I agree with that sequence of explanation):
http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html
(warning: the section with "The only possible conclusion" is plain wrong, as mentioned in post #17 with a link to a discussion in which this is elaborated; but you can simply skip that section).

You can also find a different discussion here:
http://en.wikipedia.org/wiki/Michelson–Morley_experiment

Try if you can derive those equations, based on Newton's mechanics and Maxwell's hypothesis of "a constant speed of light": It is assumed that the speed of light in vacuum is everywhere c in all directions, independent of the speed of the light emitter (and as you see, although the required math is very basic, even Michelson made a blooper when he first did so! ).

 

[ghwellsjr]

George many thanks. I always asumed that the speed of light was measured in one direction, if it is measured as there and back then the motion of the source would make no difference to the measured speed the same way if we go back to a gun being fired, one bullet in one direction the speed of the gun makes a difference but fire one forwards and one backwards then the combined speed of the bullets moving apart becomes constant.This I think is not much different to measuring something as going there and back.

As harrylin pointed out, the ballistic model of light is incorrect but beyond that, I'm not sure what you are describing with respect to a gun firing bullets. Wouldn't you need multiple guns because if we think of the light source as being a gun, in effect, we really need a second and third gun at each of the final detectors to simulate sending the message back to the clocks, correct?

Do the effects of length and time dilation mean that two people could measure the same thing and get different answers or two people measure different things and get the same answer.

I'm not sure what you are asking here. If we go back to the linked diagram, there are two people involved, one on the ground and one on the spaceship. Do you consider them to be measuring the same thing or different things?

How can just movinf faster affect your view of the universe?

The obvious answer is that the faster you move, the faster your view of the universe appears to be moving past you. But I don't think that is what you are looking for. If not, could you please rephrase your question?

 

[Adrian07]

It just seems to get more complicated.
Going back to the spaceship and forgeting about actually measuring the speed of light. We will keep the stationary light emitter and the one at the back of the ship and the ship moving at 0.5c. We will however put the observers on treadmills synchronized at rest at say 5 km/hr. As the emitter on the ship passes the one on the ground each emits a blip of light. What will I see with regards to the 2 blips of light and the different observers. I am on the stationary treadmill.
Will the blips of light remain moving together at a constant speed?, the stationary treadmill will still be going at 5 km/hr, what will I see happen to the treadmill on the ship.
Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c.
There is also the question now about the difference between waves and photons.
Obviously waves are waves, although I don't see how they can propagate through nothing, are photons then like bullets? if so how do waves and photons with the same energy relate to each other.
Finally the MM experiment seems to assume a moving aether, would it make any difference if it was a stationary field just producing resistance so as to cause a wave to form, as waves seem to form due to the resistance of the medium or am I way out here.

 

[Drakkith]

It just seems to get more complicated.
Going back to the spaceship and forgeting about actually measuring the speed of light. We will keep the stationary light emitter and the one at the back of the ship and the ship moving at 0.5c. We will however put the observers on treadmills synchronized at rest at say 5 km/hr. As the emitter on the ship passes the one on the ground each emits a blip of light. What will I see with regards to the 2 blips of light and the different observers. I am on the stationary treadmill.

What do you mean by a "blip" of light? A single pulse or flash? A single photon? A constantly updating blip like on a radar?

Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c.

Are you saying you are trying to figure out an actual way of measuring this, or that you are attempting to understand how such a thing is possible?

There is also the question now about the difference between waves and photons.
Obviously waves are waves, although I don't see how they can propagate through nothing, are photons then like bullets? if so how do waves and photons with the same energy relate to each other.

Light is an EM wave. The field itself carries the energy and the vector of the electric and magnetic fields of the wave are what are oscillating. The wave interacts in quanta called photons. They are NOT little bullets.

 

[dubiousraves]

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave

 

[harrylin]

It just seems to get more complicated.

That is normal for physics; and it's hopeless without doing the necessary exercises.

[..]Am still trying to work out how I not moving can measure a beam of light going past me at c and someone moving in the same direction as the beam can still measure it as moving at c. [..] Finally the MM experiment seems to assume a moving aether, would it make any difference if it was a stationary field just producing resistance so as to cause a wave to form, as waves seem to form due to the resistance of the medium or am I way out here.

Instead, MMX assumes a moving interferometer in a light medium. So, if you understand waves, please present your MMX calculation, and we can take it from there.
It is like becoming familiar with multiplication in order to next understand exponents. But if you don't understand waves, then it's necessary to go back even one more step...

 

[harrylin]

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave

You only "got" 1/3 of the reasons:
- clock synchronization
- time dilation
- length contraction

See: https://www.physicsforums.com/showthread.php?t=620279
All effects together are presented in the equation of post #3 (with "relative speed", he just means "speed").

Your question is the topic of that thread, and you can still discuss more there.

 

[dubiousraves]

You only "got" 1/3 of the reasons:
- clock synchronization
- time dilation
- length contraction

See: https://www.physicsforums.com/showthread.php?t=620279
All effects together are presented in the equation of post #3 (with "relative speed", he just means "speed").

Your question is the topic of that thread, and you can still discuss more there.

Thanks. I just posted over there.

 

[Adrian07]

The spaceship example post 42 for blip read pulse if you like.

Light has particle, wave duality what is a photon then and how does it relate to the wave/particle nature. Waves and particles have energy so why do we also need photons.

All I was trying to work out at the outset was how can the speed of light be constant and independant of the motion of the emitter.
I am standing still, an emitter traveling at 0.5c emits a pulse of light as it passes me, the light moves away from me in all directions at c, relative to the emitter though the forward direction of the pulse would be 0.5c, and backwards 1.5c and sideways c. So if the emitter could measure the speed of light moving away from it it should measure the forward speed as 0.5c moving away and backwards 1.5c moving away, the speed of light remains constant but the measured speed is dependant on the motion of whoever is doing the measuring.
What I got from the answer in post 2 was that both I and the emitter would measure the speed in all directions as c.

 

[Drakkith]

The spaceship example post 42 for blip read pulse if you like.

Alright. Also, your example there is...hard to understand. You have an emitter sending out pulses of light. You have a spaceship traveling at 0.5c with an emitter at the back of it. Then you have 2 treadmills that are synchronized, but one is stationary with respect to the ship? That doesn't make any sense to me.

Light has particle, wave duality what is a photon then and how does it relate to the wave/particle nature. Waves and particles have energy so why do we also need photons.

An EM wave of a specific frequency will only interact in little chunks or packets we call photons. Each photon carries with it part of the energy of the wave, where the amount of energy is E=hf, with h being Planks constant, and f being the frequency of the EM wave. The wave can only interact in this manner.

All I was trying to work out at the outset was how can the speed of light be constant and independant of the motion of the emitter.
I am standing still, an emitter traveling at 0.5c emits a pulse of light as it passes me, the light moves away from me in all directions at c, relative to the emitter though the forward direction of the pulse would be 0.5c, and backwards 1.5c and sideways c. So if the emitter could measure the speed of light moving away from it it should measure the forward speed as 0.5c moving away and backwards 1.5c moving away, the speed of light remains constant but the measured speed is dependant on the motion of whoever is doing the measuring.
What I got from the answer in post 2 was that both I and the emitter would measure the speed in all directions as c.

Yes, both you and the emitter would measure the light to move at c. From the emitters point of view the light in all directions moves away at c. To reconcile this, you have to account for things like doppler shift, time dilation, etc. If the emitter is moving away from you then while the light still travels at c from it to you, it loses part of its energy from your frame of reference, while it retains it from the emitters frame.

Honestly your best bet to understand all this is probably to buy a book on the basics of relativity. Without getting into at least a little bit of the basic math this is a very difficult thing to understand properly. You can easily find some at any bookstore or online.

 

[harrylin]

Hi folks. Kindly indulge me in a dumb question. OK, I easily understand the constancy of c given in the case of someone "chasing" a beam of light. Say, the traveler is going .75c, and nevertheless sees the beam going c. The reason is, time contracts for the traveler proportionally, so that he always sees c as c.

BUT, what about when the the beam is heading directly at the traveler? He's going .75c straight at the beam, but still sees the beam going c. What accounts for this?

thanks,
Dave

Hi Dave,

Now that I better understand your question about "accounting for" the invariance of the speed of light, and giving into the confusion between "constant" and "invariant", I post my answer about how "the speed of light" can be invariant here.

(but how can the title in your post be just "speed of light"? )

Neither time dilation alone, nor together with just length contraction, can give the right answer on such one-way speed of light questions. I stressed that in post #3 of this thread.

The Lorentz transformations and the resulting equation in post 3 in the other thread can be thought of as being built up from a combination of effects, two of which by nature and one man-made:
- length contraction by factor γ
- time dilation by factor γ
- clock synchronization as if the measurement system is in absolute rest

BTW, that is roughly how these transformations were originally derived, with a bit of trial and error. So, here it is worked out from the point of view of "rest" system S for your co-moving system S'.

1. Only assuming time dilation.
0.75c -> γ ≈ 1.51
Result: you would measure the light to speed away from you at 0.25c*1.51≈0.38c
Not OK, contrary to your thinking here above.

2. Adding length contraction.
Result: Now you would measure the light to speed away from you at approximately 0.38c*1.51≈0.57c
Still not OK, contrary to your thinking in a more recent post.

Note that there is an error from slightly out of relative sync clocks according to S which I did not account for here; I prepared an example for post #10 with different speed, but nobody was interested.

3. You now make a "local" clock synchronization for system S', such that the one-way speed equals the (average) two-way speed.
Thus we first calculate the physical effects for light in the other direction:
1.75c*1.51≈2.65c
42.65c*1.51≈4c (= speed of light coming at you, as approximately measured by you)

Next we calculate the total time Δt' in S' over a length L' of that system:
L'/0.57c + L'/4c = 2L'/c
(I put the equal sign because the result is exact). 2L'/c is just the time that it would take if the light propagated at speed c over L' and back. The "two-way" speed according to S' is thus c.

With clock synchronization you simply make the one-way speed according to S' equal to this two-way speed. See section 1 of: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Note that the "experience" that he speaks of at the end of that section relates to the average two-way speed of light in the equation there, because the one-way speed is largely a matter of convention ("definition").

Next you will "measure" with your system S' in both directions a speed of light equal to c.