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The speed of light is also the speed of force

Quote from: DavidThe speed of light is also the speed of forceWith aether, once a moving laser would emit a light pulse, the laser would progressively leave the place in aether where the pulse had been emitted, and it is at that place that an observer at rest would see the laser beam if it had been pointing at him at emission. This is what happens when we see a plane: we hear it where it was, not where we see it, and in the same token, we don't really see it where it is because it takes time for light to reach us, so if some laser light was sent from the plane to somewhere else than where we are, it would simply miss us. It is so for sound because sound waves go straight line in air after having been emitted, so light waves should behave the same in aether if it exists. But this is not what the inertial frame principle means: a moving inertial frame can be considered not to be moving, in such a way that light can be considered to travel sideways through space in the light clock mind experiment, so it manifestly doesn't fit with aether. With aether, a laser beam aimed at 90 degree between our two moving cars would miss us if we would travel fast enough or if we were far enough from one another. I understand what you mean when you say that the speed of light would affect the forces though, but to me, with no doppler effect and no aberration to account for inside the same moving frame, whatever the speed of that frame, the light waves exchanged between two observers at rest inside the frame would always appear to carry the same energy and would always appear to come from the actual position of the source. Of course, the intensity of a wave lessens with distance, so the forces should lessen a bit for bodies moving through aether, but that effect has a lot less impact on energy than frequency, and frequency would not be affected. In my theory on motion here, the frequency of the steps also stays the same while they get longer, thus while the speed increases, but light also loses some intensity while it travels from one particle to the other, and that loss cannot be accounted for by the steps, so I figured it could be attributed to gravitation because it would necessarily escape from the standing wave between the two particles. Here are those steps caused by the limited speed of the information between two bonded particles.

Quote from: DavidThe speed of light is also the speed of forceWith aether, once a moving laser would emit a light pulse, the laser would progressively leave the place in aether where the pulse had been emitted, and it is at that place that an observer at rest would see the laser beam if it had been pointing at him at emission.

This is what happens when we see a plane: we hear it where it was, not where we see it,

and in the same token, we don't really see it where it is because it takes time for light to reach us, so if some laser light was sent from the plane to somewhere else than where we are, it would simply miss us.

It is so for sound because sound waves go straight line in air after having been emitted, so light waves should behave the same in aether if it exists.

But this is not what the inertial frame principle means: a moving inertial frame can be considered not to be moving, in such a way that light can be considered to travel sideways through space in the light clock mind experiment, so it manifestly doesn't fit with aether. With aether, a laser beam aimed at 90 degree between our two moving cars would miss us if we would travel fast enough or if we were far enough from one another.

I understand what you mean when you say that the speed of light would affect the forces though, but to me, with no doppler effect and no aberration to account for inside the same moving frame, whatever the speed of that frame, the light waves exchanged between two observers at rest inside the frame would always appear to carry the same energy and would always appear to come from the actual position of the source.

Of course, the intensity of a wave lessens with distance, so the forces should lessen a bit for bodies moving through aether, but that effect has a lot less impact on energy than frequency, and frequency would not be affected.

In my theory on motion here[/url], the frequency of the steps also stays the same while they get longer, thus while the speed increases, but light also loses some intensity while it travels from one particle to the other, and that loss cannot be accounted for by the steps, so I figured it could be attributed to gravitation because it would necessarily escape from the standing wave between the two particles. Here are those steps caused by the limited speed of the information between two bonded particles.

If you are moving in the same direction as the plane and at the same speed, you would (if you could somehow eliminate the racket from air rushing past your ears) hear the sound of the plane as coming from where you see the plane rather than from a long way behind it, even though the sound is coming to you from a long way behind where the plane now is.

The total length of 7.5 cell lengths for light to travel its cycle. This leaves us with 8.66 cell lengths for the side ways mirror. So we haven't reached the peak of the triangle at 7.5 cell lengths of time.

Quote from: GoC on 25/05/2017 22:30:55The total length of 7.5 cell lengths for light to travel its cycle. This leaves us with 8.66 cell lengths for the side ways mirror. So we haven't reached the peak of the triangle at 7.5 cell lengths of time.If you 7.5 cell lengths you mean just under 75cm, then you're getting there, but where do you get your 8.66 cell length figure from for the perpendicular light clock? The distance travelled by the train during the time taken for light to go from one mirror to the other on the perpendicular light clock is only 17.32cm (because tan(60)=opp/10cm, so opp=10tan(60)cm).

First of all I understand your physical contraction of mass and how that corrects for equilateral light paths on a 90 degree path to motion. That is what I was taught. I do not agree with what I was taught. All you are doing is force fitting the paths of light to be the same in both directions. This fudge factor is just that a fudge factor.

Lets go back to the MMX. The most accurate way to measure distance is with light. So you measure the direction of travel and you get a mile. You measure the other distance perpendicular to the direction of travel and you measure a mile. When you do the test you get a null result. This is no different than measuring the same speed of light in any frame. Of course you get a null result. Your set up was meaningless to begin with. None of your tests can pass the test of significance. So your physical contraction is based on an insignificant measurement.

Now lets look at sound again as relative to the Aether. If we have the air molecules perfectly still except for the electrons moving to transfer the sound we get a ratio of sound to distance in all directions.

We have a stationary observer on the ground. When the plane is physically overhead the sound just starts while the plane continues to go forward. We can use the same geometry with light. When the sound reaches you the visual position of the plane is forward of the position you hear.

Now if you have two planes in parallel going the speed of sound and the sound being created in back of the pilots neither pilot could hear the sound of the other pilots engine. He might hear his own engine because he carries his own air inside the plane.

Its unlikely that you can carry the Aether inside of a spaceship. So two pilots can go fast enough in space not to be able to view the other right next to you.

If you looked forward in the spaceship the front of you would appear to go from 7.46 meters to 0.53 meters. Not because the distance changed physically but because light only took that distance to reach you. Your view would be magnified also because of the inverse square law would change your viewing distance.

Quote from: Le Repteux on 25/05/2017 20:59:15Quote from: DavidThe speed of light is also the speed of forceWith aether, once a moving laser would emit a light pulse, the laser would progressively leave the place in aether where the pulse had been emitted, and it is at that place that an observer at rest would see the laser beam if it had been pointing at him at emission.Correct, and a moving observer who is for a moment at the same location as the stationary observer and who is keeping pace with the laser would see the same laser light as coming from where the laser now is rather than where the the light actually came from.

QuoteThis is what happens when we see a plane: we hear it where it was, not where we see it,It's worth remembering that if the light from the plane traveled at the speed of sound, you would see the plane further back too.

Quoteand in the same token, we don't really see it where it is because it takes time for light to reach us, so if some laser light was sent from the plane to somewhere else than where we are, it would simply miss us.But to reach us, the laser has to be pointed slightly behind us (although planes aren't really far enough away for that to show up).

QuoteIt is so for sound because sound waves go straight line in air after having been emitted, so light waves should behave the same in aether if it exists.If you are moving in the same direction as the plane and at the same speed, you would (if you could somehow eliminate the racket from air rushing past your ears) hear the sound of the plane as coming from where you see the plane rather than from a long way behind it, even though the sound is coming to you from a long way behind where the plane now is.

If a powered "bullet" (one that can maintain speed and direction without being slowed by drag) was sent out from the plane at the speed of sound perpendicular to the direction of travel of the plane while you are moving along parallel to the plane's path and level with it, you can hold a two-layer target for the bullet to pass through (perhaps with the layers several yards apart) - the bullet will make two holes through the target, and when you look through those holes, you will see the plane. It's difficult to set out a really good parallel scenario though as the space fabric doesn't replicate the drag aspect.

QuoteBut this is not what the inertial frame principle means: a moving inertial frame can be considered not to be moving, in such a way that light can be considered to travel sideways through space in the light clock mind experiment, so it manifestly doesn't fit with aether. With aether, a laser beam aimed at 90 degree between our two moving cars would miss us if we would travel fast enough or if we were far enough from one another.No - it would not miss. If the cars are on parallel paths and level with each other, they could be a whole lightyear apart and the laser light sent out from a perpendicular (to the direction of travel of the cars) laser would still hit the other car, as would a bullet sent out from a gun set up with the same alignment as the laser, though the bullet would follow a much more deviated angle than the light and would hit the other car the best part of a million years later. If you think the light has to follow the perpendicular just because the laser is pointing along the perpendicular, you are breaking the laws of nature by changing the momentum of the system.

QuoteI understand what you mean when you say that the speed of light would affect the forces though, but to me, with no doppler effect and no aberration to account for inside the same moving frame, whatever the speed of that frame, the light waves exchanged between two observers at rest inside the frame would always appear to carry the same energy and would always appear to come from the actual position of the source.So why would you want to tinker with the alignment of the laser in my diagrams if you understand that?

QuoteOf course, the intensity of a wave lessens with distance, so the forces should lessen a bit for bodies moving through aether, but that effect has a lot less impact on energy than frequency, and frequency would not be affected. Just as light is focused forwards more strongly and cancels out the difference, so must force be if the square room round a bulb is to retain the right shape for the walls to remain evenly lit. The MMX suggests that matter length-contracts in a manner consistent with forces and light behaving the same way in that regard.

but the light intensity should adjust automatically for the speed of the system through means of the headlights effect, so there shouldn't be any apparent change to account for.

Quote from: GoC on 26/05/2017 16:07:11QuoteIts unlikely that you can carry the Aether inside of a spaceship. So two pilots can go fast enough in space not to be able to view the other right next to you.That never happens - they always see each as other side by side.QuoteIf you looked forward in the spaceship the front of you would appear to go from 7.46 meters to 0.53 meters. Not because the distance changed physically but because light only took that distance to reach you. Your view would be magnified also because of the inverse square law would change your viewing distance.That is nonsense - we would be able to detect that kind of visual change easily in the lab even at the relatively low speed the Earth goes round the sun, but no such visual warping occurs. Your problem is that you have come up with a wonderful theory of your own and you're determined to try to make nature conform to it rather than allowing nature to dictate the form of your theory. You are going against what nature does.If an image (light) only goes 0.53 back relative to 7.46 in distance looking behind you, somehow you believe the one way distance for light is the same as the two way distance for a light image. Now that would be magic.

QuoteIts unlikely that you can carry the Aether inside of a spaceship. So two pilots can go fast enough in space not to be able to view the other right next to you.That never happens - they always see each as other side by side.QuoteIf you looked forward in the spaceship the front of you would appear to go from 7.46 meters to 0.53 meters. Not because the distance changed physically but because light only took that distance to reach you. Your view would be magnified also because of the inverse square law would change your viewing distance.That is nonsense - we would be able to detect that kind of visual change easily in the lab even at the relatively low speed the Earth goes round the sun, but no such visual warping occurs. Your problem is that you have come up with a wonderful theory of your own and you're determined to try to make nature conform to it rather than allowing nature to dictate the form of your theory. You are going against what nature does.

Incidentally, the same phenomenon would occur between the two planes: because of aberration, an observer on one of the planes would see the other plane where it actually is, so if he would aim its laser directly at it, he would miss it.

Your bullet would travel sideways through air because it would add the motion of the plane to its own motion, but sound doesn't, so the bullet wouldn't have to suffer aberration, and sound would.

thenakedscientists.com/forum/index.php?topic=53171.msg467891#msg467891

Here is the link to the animation: ...

It's astounding to observe the small divergences people have on certain ideas. I hope you don't mind if I insist though, because I still see contradictions when I try to apply the reference frame principle to light.

...and the laser in your MM simulation sends its light perpendicularly to its motion, not towards the future position of the mirror. If we replace the laser by a point source, the light emitted perpendicularly would not reach the mirror, and the one emitted at its future position would: two directions for light, two different paths in aether, but not in the space defined by the postulates of SR.

Simply because I think light cannot travel this way, but I don't expect you to change your mind about that. In fact, I think our ideas are like bodies resisting to acceleration: ideas cannot change without resisting automatically to the change, thus subconsciously, no matter how intelligent we are. We will thus probably go on discussing until we get tired, or until one of us suffers an intuition, which I think are due to a random process similar to mutations happening only in human minds. Good luck to us if I am right! :0)

With aether, whether the room is contracted or not, it seems to me that the light would travel less distance to reach the approaching wall than the one that is getting away from it, and if it is so, the approaching wall should absorb a more intense light than the other. Would it?

The headlight effect is due to aberration, and if the source is not moving with regard to the room, precisely because of aberration, that source would always appear to be at its actual position for any observer in the room, so I don't see how any of them could observe that effect. In this citation from wiki, relativistic beaming is called doppler beaming, and there is no doppler effect to observe in your moving room example.

Wiki says that aberration causes the photons to be emitted in the direction of motion, but they say that to conform to the relativistic definition of aberration, which comes from the idea that we cannot differentiate if it is the source or the observer that is moving. With aether we can, and we thus can see that only the photons that would have been sent towards the future position of the observer would hit him. With aether, there is no aberration either at the observer if he is at rest in aether while the source is moving, whereas there is with SR.

If an image (light) only goes 0.53 back relative to 7.46 in distance looking behind you, somehow you believe the one way distance for light is the same as the two way distance for a light image. Now that would be magic.

Quote from: GoC on 26/05/2017 22:29:11If an image (light) only goes 0.53 back relative to 7.46 in distance looking behind you, somehow you believe the one way distance for light is the same as the two way distance for a light image. Now that would be magic.If you want to project an image, you'll find that forward movement at 0.866c of the projector turns a forward-pointing lens into a stronger telephoto which means that even though the image has to go nearly seven and a half times as far to hit the screen it will display the image the same size as it would with the system at rest. Turn the projector round to point at the rear screen instead and the lens serves as a wide angle lens, again automatically correcting the size of the image. You can test that by working out where the lens is when the light catches it and drawing straight lines from the corners of the original image in the projector through the centre of the lens and on to where the screen will be when the light catches up with that, at which point you'll find the image size to be unchanged (even if you forget to deal with length contraction) - all speeds of travel of the system lead to the same size of image on the screen. Everything conspires to hide the real speed of travel of the system from all observers, and that's what makes the maths of relativity so fascinating.

Quote from: Le Repteux on 26/05/2017 19:44:55Incidentally, the same phenomenon would occur between the two planes: because of aberration, an observer on one of the planes would see the other plane where it actually is, so if he would aim its laser directly at it, he would miss it.No amount of repeating that can make it true. You are changing the momentum of the system.

You are also making it impossible for a perpendicular laser to function if it's moving along at high speed because the light wouldn't be allowed to travel through it without hitting the side.

Quote from: Le RepteuxYour bullet would travel sideways through air because it would add the motion of the plane to its own motion, but sound doesn't, so the bullet wouldn't have to suffer aberration, and sound would.The sound front would be angled, but it would still hit the hearer perpendicular in the frame in which the planes are at rest, so it is just like the bullet in that regard.

QuoteHere is the link to the animation: ...I gave up waiting for the adverts to disappear and for the actual gif to show any sign of existing.

QuoteWith aether, whether the room is contracted or not, it seems to me that the light would travel less distance to reach the approaching wall than the one that is getting away from it, and if it is so, the approaching wall should absorb a more intense light than the other. Would it?No, because the light send out backwards is weaker, having a longer wavelength, but when the wall hits it it runs into it fast enough to boost the perceived power back up to what it would be if the room was at rest. Likewise, the light going forward has a very high frequency, but when it hits the wall that's moving away from it, its power is not felt to be higher at all.

The headlights effect is caused by conservation of momentum and it is not visible to people in the moving room with the light bulb. If light behaved the way you want it to, the rear wall would become brighter and the leading wall dimmer, so the people in the room could measure the room's speed of travel through space with a lightmeter. The Doppler effect aspect of this is also hidden from them, but an observer at rest who sees some of the light escaping through holes in the wall will see considerable shifts in its frequency.

Any mechanism that throws a photon out eastwards from a stationary source would, if the source was moving north at any speed, automatically send that photon out some way to the north of eastwards such that it keeps pace northwards with the source and maintains the momentum of the system.

We are discussing 3d shapes not a projected image.

If the physical shape was one cell length in the front and the same physical object in the back one cell length there would be a different view of length with the observer in the middle when at relative rest. The physical object in back would appear to be 7.461 cell lengths long because the front of the image reaches you and the rear of the image takes 7.461 cell lengths to reach you with the constant speed of a ship at 0.866 c. Now the front image would reach you and the back of the physical object takes 0.53 cell lengths so it appears compressed length wise. Physically they are the same size at relative rest. These are your numbers we agreed on for travel distances for light. Light coming towards you from the back would reflect the image for 7.461 cell lengths for depth of view. From the front only 0.53 cell lengths are reflected.

Quote from: Thebox on 25/05/2017 09:19:06Ok, before I start , you do realise that if the train travels down a slope it makes it easier to measure the light?No, I don't realise that, and your numbers don't appear to realise it either.Quote from: Thebox on 25/05/2017 09:31:16Now I have given the answer to the angle I provided without knowing the speed of the carriage, however I am sure with it being maths, I can get the results to fit a speed of the carriage.How can you possibly calculate the angle without involving the speed of the carriage in the calculation?QuoteHowever before we move on from the last assignment , you still avoided my question, I am still waiting for the proof of the physical contraction you claim exists. I have not observed anything of the wow factor thus far, this is pretty basic stuff.I've answered that many times before, but you clearly weren't ready to take it in. Remember the five points that I made in post #134:-(A) A moving clock records fewer ticks than a stationary one in a given length of time, so it is not recording time, but is merely counting cycles. That is what all clocks do.(B) A light clock aligned perpendicular to its direction of movement therefore records fewer ticks than a stationary clock.(C) An uncontracted light clock aligned with its direction of travel (rather than perpendicular to it) will record fewer ticks than a light clock co-moving with it which is aligned perpendicular to their direction of travel.(D) A correctly length-contracted light clock aligned with its direction of travel will record the same number of ticks as a light clock co-moving with it which is aligned perpendicular to their direction of travel.(E) The null result of the MMX shows that the real universe length-contracts things in their direction of travel.Once you have generated the right numbers for assignment 2, you'll be able to compare the two moving light clocks and see that the perpendicular one ticks more often than the other. The MMX shows that this is not how the universe works because real light clocks which move along together always tick at the same rate as each other regardless of their alignment. We also know that the slowing of real clocks matches up to the perpendicular clock's predicted behaviour and not to the predicted behaviour of uncontracted light clocks aligned with their direction of travel.Quotep.s the answer to assignment 2 is a variate relative to the dimensions of the carriage.Your perpendicular light clock should be the same length as the one used in assignment 1 so that you can easily compare the numbers you get from the two assignments. Your job here is to work out how far the train moves at 0.5c while the light travels from mirror A to the far mirror (B') and back to the first one again (A"), and you need to know how to calculate the angle that the light moves away from the perpendicular. I told you how the angle can be calculated and I was hoping that you'd let me know whether you understand that method or not. If you want to find your own method for doing it, that's fine, but for it to be valid it will need to generate the same numbers. We will be stuck here until you understand the method I've spelt out to you, so I'll go through it again with a new diagram.We want to calculate the angle x. We have a right angled triangle, but the only side of that triangle with a known length is the vertical line B'A' which has a length of d [= 299,792,458m]. However, we do know the ratio of the lengths of AA' to AB' because we know the ratio of the speed of the train to the speed of light, and that ratio is 0.5:1.The trig rule soh tells us that sine(x) = AA'/AB'. We don't yet know the lengths AA' or AB', but we do know their ratio, so we can simply use the numbers from that ratio: sine(x) = 0.5/1. Because we're dividing a number by 1, the result will be unchanged, so we now have sine(x) = 0.5. All we need to do now is use a calculator to find x by typing in a sequence such as 0.5 inv sin or shift sin 0.5 (depending on the order the calculator wants you to press the keys in.Having done that and got the number 30 degrees from the calculator, we can now set about working out the actual lengths of the lines AA' and AB'. The only real length we know for the triangle is the vertical line A'B', but that's enough to do the rest now as we also have an angle to work with. All we need to do is apply more trig: tan(x) = AA'/A'B', and cosine(x) = A'B'/AB'. We can rearrange those a little before we pick up the calculator (and note that I'm going to use an asterisk for the multiplication symbol to avoid confusion with the angle x): AA' = tan(x) * A'B', and AB' = A'B'/cos(x). Now the calculator is used: AA' = 0.57735, and AB' = 1.1547. Note that I have used 1 as the length of A'B' because my length unit is d, and d = 299,792,458m. To fit with all your previous calculations, you should use 299792458 instead of 1 so that your answers come out in metres.If you double 0.57735, you'll get 1.1547, so these answers fit with the 0.5:1 ratio that we expect them to. The lower mirror has moved from A to A" by the time the light returns to it from the top mirror, so the train has moved 1.1547d and the light has moved 2.3094d. To convert those to metres, again you'd need to multiply them by 299792458. One tick of this moving clock will take 2.3094t, and as my time unit is a second long, that's 2.3094s.We now have three tick rates. The stationary clock ticks once every 2 seconds. The moving clock from assignment 1 ticks once for every 2 2/3 seconds of the stationary clock. The moving clock from assignment 2 ticks once for every 2.3094 seconds of the stationary clock.Now, lets see how many pages of posts it takes you to get up to speed with that. On the up side though, it looks as if you might reach the finish line before GoC, which no one reading this thread at the start would ever have predicted.

Ok, before I start , you do realise that if the train travels down a slope it makes it easier to measure the light?

Now I have given the answer to the angle I provided without knowing the speed of the carriage, however I am sure with it being maths, I can get the results to fit a speed of the carriage.

However before we move on from the last assignment , you still avoided my question, I am still waiting for the proof of the physical contraction you claim exists. I have not observed anything of the wow factor thus far, this is pretty basic stuff.

p.s the answer to assignment 2 is a variate relative to the dimensions of the carriage.

Lets look at it another way. We have a ship in space 10 meters long with a reflective mirror in the front and back to reflect on both sides. This ship is going 0.866 c. There is a bulb in space stationary to the ship. The ship approaches the bulb. Light reaches the front mirror and bounces back to the observer at relative rest with the bulb. The ship moves forward ~4.6 meters and reflects light off of the back mirror. The returned length of light with just one photon would measure the ship as ~5.3 meters long.

Now the ship moving away from the bulb at rest compared to 0.866c. The light reaches the back mirror and co-moves forward with the ship. Light gains 0.133 for every length of the ship. So the light reaches the front mirror in 74.6 meters compared to the observer at rest. So the observer at rest would measure the ship moving away as ~74.6 meters long.

Now what would be the perpendicular view by the observer at rest?

For that we can use the Lorentz contraction and get 0.5 relative to rest. So the view is the hypotenuse between the two legs of a created right triangle for how light actually moves through space shortens the view by Half.

We can never view true perpendicular only the angle from behind our position if we are the one in motion or forward of our position if we are at rest. Its the same view from either position.