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grav
2009-May-30, 06:03 PM
I am having trouble wrapping my head around something all of a sudden. We all know the twin paradox for a barn / polevaulter experiment, but what about light itself? Let's say we have two lights set up at a large distance outside of either end of a ten light-second long ship and an observer outside the ship that is always stationary to the lights while the ship is also stationary for the time being. Observers are placed at every light-second within the ship with synchronized clocks, so eleven in all, nine within the ship and one at each end. Now the light at one end of the ship is flashed. The ship observers record the times that they receive the flashes, one per second, and a short while later reproduce the flashes to the outside observer, one per second, and that is the rate the outside observer receives them. Then the same thing is done from the other end.

Okay, now let's say the ship has a relative speed of .6 c to the stationary observer. All of the clocks within the ship are again synchronized to each other. The outside observer remains stationary with the lights. The experiment is repeated. According to the ship observers, light from either direction still passes each observer at c, passing one observer per second, and so a short while later, the ship observers flash their own light to the outside observer once per second in both cases also, as far as I can tell. The ship is time dilated to the outside observer by a factor of .8, so the flashes are received by the outside observer once per 1.25 second each in both cases, so after the initial pulse, all pulses will be received in 12.5 seconds according to the outside observer in either case.

Now let's look at what the outside observer measures. The light coming from behind the ship is travelling through the ship at c, while the ship moves forward at .6 c, and the ship is contracted by .8, so that ray should take t = (8 light-seconds) / (c - .6 c) = 20 seconds to travel through, and that is how the pulses from the ship observer should also be received, since they match the rate the pulses pass each observer. From in front of the ship, the light should take (8 light-seconds) / (c + .6 c) = 5 seconds to travel through the ship and that is mimicked by the ship observers, so the pulses from the ship observer should be received in that amount of time to the outside observer. So neither case matches what should be observed, although the sum of the times is the same. What's going on?

grav
2009-May-30, 06:44 PM
I had the ship observers resending their own pulses separately in order to avoid time of flight effects from each of them, but I think we all know by now it is not necessary to take time of flight effects into consideration anyway, just the synchronized time of the frame in general, so let's simplify that scenario a bit. Let's just say that each of the ship observers directly reflects some of the ray coming from either direction to the outside observer. According to their own frame, then, one pulse will be reflected each second from either direction. That means the outside observer should receive one pulse every 1.25 seconds from the light coming from either direction as well. But the outside observer doesn't, but measures a pulse reflected every 2 seconds for the light coming from the back, and one pulse reflected every .5 seconds for the light coming from the front. What's up?

grant hutchison
2009-May-30, 09:35 PM
Hint: simultaneity.

Grant Hutchison

grav
2009-May-30, 10:32 PM
Hint: simultaneity.

Grant HutchisonThat works for the pole vaulter travelling at less than the speed of light through the ship and finding simultaneity differences for when events occur at one end of the ship as compared to the other when the ship observers consider them simultaneous, but in this case, the two rays of light don't have to occur simultaneously to the ship observers or outside observer, so they don't have to agree upon when the events took place as it concerns this scenario, as far as I can see, only the rate of pulses observed during each event. All that matters is the rate that the outside observer receives the pulses, and if the ship observers measure the same rate of time for the light travelling either way, then they will send the pulses to the outside observer at the same rate each time and the outside observer should receive them at the same rate each time as well, although time dilated by the same amount for each case. The outside observer, however, disagrees that the time that the light takes to pass through the ship is the same in either direction, however, so should not be receiving the pulses at the same rate each time, either as they are sent by the ship observers or directly reflected by them, which should be the same difference as to how the outside observer receives them, and I'm not sure otherwise how any simultaneity effects would figure into the scenario.

grant hutchison
2009-May-30, 10:47 PM
This one is pretty obvious if you draw a space-time diagram. (That's why I keep telling you how helpful they are!)

You only really need a pair of clocks, at the front and back of the ship, and an external observer. The light makes the front-to-back journey faster than the back-to-front journey, by the external observer's measurements. But the ship's clocks measure both journeys as having the same duration.
What is it about the ship's clocks that causes this difference?

Grant Hutchison

grav
2009-May-30, 11:15 PM
This one is pretty obvious if you draw a space-time diagram. (That's why I keep telling you how helpful they are!)

You only really need a pair of clocks, at the front and back of the ship, and an external observer. The light makes the front-to-back journey faster than the back-to-front journey, by the external observer's measurements. But the ship's clocks measure both journeys as having the same duration.
What is it about the ship's clocks that causes this difference?

Grant HutchisonWell, according to the external observer, there is a time lag between the clocks on the ship. So maybe I shouldn't have included clocks. Let me ask it this way, so maybe I can pinpoint where the problem lies.

According to the ship observers, the light takes the same time to travel one way as the other, right?

So the ship observers will signal to the external observer at the same rate both times, correct?

So the external observer will receive the signals at the same rate each time, only time dilated?

The rate that the ship observers signal the external observer is exactly the same as the rate the light is reflected at each point along the ship according to the ship observers?

The reflected light that leaves the ship leaves at the same rate as the signals the ship observers send, so are identical to the signals themselves?

So the reflected light that the external observer sees arrives at the same rate in either case, only time dilated, same as the signals?

But the rate at which the external observer sees the light reflected should be different from each direction from his point of view?

grant hutchison
2009-May-30, 11:22 PM
According to the ship observers, the light takes the same time to travel one way as the other, right? Yes.


So the ship observers will signal to the external observer at the same rate both times, correct?According to their own measurements, yes.


So the external observer will receive the signals at the same rate each time, only time dilated?Not at the same rate.


The rate that the ship observers signal the external observer is exactly the same as the rate the light is reflected at each point along the ship according to the ship observers?Yes.


The reflected light that leaves the ship leaves at the same rate as the signals the ship observers send, so are identical to the signals themselves?Yes.


So the reflected light that the external observer sees arrives at the same rate in either case, only time dilated?Not at the same rate.


But the rate at which the external observer sees the light reflected should be different from each direction from his point of view?And is.

Grant Hutchison

grav
2009-May-30, 11:29 PM
So the external observer will receive the signals at the same rate each time, only time dilated?
Not at the same rate.Why not? I'm thinking of a central ship observer here, that measures the light as passing each ship observer in each direction at the same rate, so sends the flash signals to the external observer at the same rate each time also.

grant hutchison
2009-May-30, 11:54 PM
Why not? I'm thinking of a central ship observer here, that measures the light as passing each ship observer in each direction at the same rate, so sends the flash signals to the external observer at the same rate each time also.Oops. Such a central observer would need instantaneous communication with his various ship observers. Better to leave those ship observers generating their own signals, which are received by an external observer so distant that light-path differences are negligible.

But lets think about the central observer on the ship, who later collates the information from the various other observers and determines what rate of signalling they produced. He has eleven data points; eleven times, from eleven separate clocks. How does he derive a rate from that information, unless the clocks have been previously synchronized?

Grant Hutchison

grav
2009-May-31, 12:09 AM
Yes. My OP was originally going to include a central computer that can subtract for time of flight effects and just emit the signals at the same rate as the ship observers receive them. The ship observers can later verify that the signals were sent correctly according to the rate that the light was received by each of them, according to the general overall record that the observers took for the frame. So the ship's computer sends signals at the same rate for either way the light travels, right? Wouldn't the ship observers say that the flash signals and the reflected light leave the ship at the same rate, then?

grant hutchison
2009-May-31, 12:20 AM
So the ship's computer sends signals at the same rate for either way the light travels, right?Right.

Wouldn't the ship observers say that the flash signals and the reflected light leave the ship at the same rate, then?Since the flash signals and the reflected signals are locally synchronized, they would leave the ship at the same rate.

But the only way to work out the rate at which signals are generated by eleven different pieces of apparatus at eleven different places is to somehow time-stamp the signal events at each location, so that when the data are brought together, the rate can be calculated. The time-stamp needs some process of synchronization. The synchronization requires a simultaneity convention. The external observer and the ship observers disagree about simultaneity. So the external observer receives the signals at different rates (front-to-back versus back-to-front), but observes that the ship observers, when comparing their signal time-stamps, derive identical rates (front-to-back versus back-to-front). Why? Because (according to the external observer) there's a systematic error in how the time-stamp process was set up, with a variation along the length of the ship.

Grant Hutchison

grav
2009-May-31, 12:31 AM
Right.
Since the flash signals and the reflected signals are locally synchronized, they would leave the ship at the same rate.

But the only way to work out the rate at which signals are generated by eleven different pieces of apparatus at eleven different places is to somehow time-stamp the signal events at each location, so that when the data are brought together, the rate can be calculated. The time-stamp needs some process of synchronization. The synchronization requires a simultaneity convention. The external observer and the ship observers disagree about simultaneity. So the external observer receives the signals at different rates (front-to-back versus back-to-front), but observes that the ship observers, when comparing their signal time-stamps, derive identical rates (front-to-back versus back-to-front). Why? Because (according to the external observer) there's a systematic error in how the time-stamp process was set up, with a variation along the length of the ship.

Grant HutchisonRight, okay. So now let's bring back the clocks to establish that simultaneity convention. If consecutive observers on the ship receive the light every second either way, then the computer will be sending out signals of one flash per second according to their frame, right? And the rate at which the reflected light leaves the ship will be the same as that for the signals according to the ship observers as well, so that they measure one reflection per second leaving the ship also, correct? So if both sets of signals are sent from the ship at the same rate, the external observer will receive both sets at the same rate, although time dilated, is that so? And if the reflected light leaves the ship at the same rate as the signals, then the external observer will receive both of them at the rate also, correct?

grant hutchison
2009-May-31, 12:53 AM
Right, okay. So now let's bring back the clocks to establish that simultaneity convention. If consecutive observers on the ship receive the light every second either way, then the computer will be sending out signals of one flash per second according to their frame, right?Right.

And the rate at which the reflected light leaves the ship will be the same as that for the signals according to the ship observers as well, so that they measure one reflection per second leaving the ship also, correct?Correct.

So if both sets of signals are sent from the ship at the same rate, the external observer will receive both sets at the same rate, although time dilated, is that so?No. Not if (as you now seem to want) one set is sent out by local reflections progressively along the length of the ship, and the other is sent out from a single source.

And if the reflected light leaves the ship at the same rate as the signals, then the external observer will receive both of them at the rate also, correct?No. See above.

If the external observer receives signals from a ripple of sources along the length of the ship, he'll get different rates when the light signal is propagating front-to-back and back-to-front. But if a ship-board observer collates the signals and sends out a delayed set of pulses from one location on the ship, then the external observer will receive signals at the same rate, whether the original propagation was front-to-back or back-to-front: he'll just see a time-dilated rate, which is the mean of the two different rates he would have received in the original scenario.

Grant Hutchison

grav
2009-May-31, 01:17 AM
Oh, wait. Simultaneity doesn't matter, but time of flight does in this case, ironically, and direction. It depends upon the position of the external observer as to which way the light is reflected and at what rate the reflected light leaves the ship, even according to the ship observers themselves. For instance, if the external observer is directly behind the ship, then the light travelling from front to back of the ship will be reflected in the same direction the light itself is travelling, and so all of the reflected pulses will leave the ship together, regardless of the timing the ship observers measure between each reflection. The light travelling from back to front, reflected toward the back also, will leave the ship at half the rate that the ship observers send their signals. So time of flight effects do matter in such a case, and I will have to rethink this. Thanks for your help so far, Grant. :)

grant hutchison
2009-May-31, 01:32 AM
Oh, wait. Simultaneity doesn't matter, but time of flight does in this case ...Simultaneity does matter. Time of flight is a red herring. Get the simultaneity sorted out and you'll understand the solution to your paradox. Time of flight will just mess up the picture.

Have you drawn that space-time diagram yet? :)

Grant Hutchison

grav
2009-May-31, 04:46 AM
Well, that scenario only succeeded in confusing me more, so let me try another approach. This one's a little closer to home for what I'm really trying to find anyway.

Let's say we have a ship observer and an external observer. Light beams are sent from either direction along the length of the ship from a far away source. The frequency of each beam is the same, which is low, and photons are sent one by one through the ship from each direction between long intervals. Now as each photon travels through the ship from the front, the ship observer steadily pours water into bowl A during the time the photon is actually within the ship until the photon leaves the other end of the ship. For photons coming in from the back, the ship observer steadily pours water into bowl B during the time those photons travel through.

When the ship is stationary to the external observer with the same frequency of photons coming from each beam, the water is poured into both bowls at the same rate. Now let's give the ship some relative speed to the external observer. According to the external observer, the rate at which photons from the front beam now enter the ship while the ship also moves forward is w = c / f = ct + vt, then t = c / f / (c + v), so f' = 1 / t = f (c + v) / c = f (1 + v/c). So the rate of photons entering becomes greater by a factor of c + v. Now the ship observer pours the water into the bowl during the time the photon takes to pass through, so according to the external observer, the light is entering the ship at a relative speed of c + v, so the time to pass through the ship is t = Z d / (c + v), where d is the original length of the ship and Z is the length contraction. So if photons are entering the front of the ship at a rate that is c + v faster, than c + v more water would be poured, except that the time they take to pass through is Z / (c + v) smaller according to the external observer, so the amount of water that should be poured for the beam of light coming from that direction is Z smaller than when stationary. Likewise, the rate of photons entering the back of the ship according to the external observer is c - v smaller and the time the photons take to travel through is Z / (c - v) larger, so the amount of water that should be poured for that direction is Z smaller also, and each bowl should be filled at the same rate.

According to the ship observer, however, the photons take the same amount of time to pass through from either direction, so the water is poured for the same amount of time per photon. The rate of photons entering from the front per second is now sqrt[(c + v) / (c - v)] greater than it was when the ship was stationary, so the water gets poured into bowl A at that much a greater rate than before. From the back, the rate of photons entering per second is now smaller by a factor of sqrt[(c - v) / (c + v)], so bowl B gets filled that much slower, and the bowls will never be the same level. The bowls are side by side, so they coincide in the same spot, and the ship observer and the external observer can compare them easily. So why would the external observer say the bowls are always filled to the same level while the ship observer says they never are?

grant hutchison
2009-May-31, 11:42 AM
So why would the external observer say the bowls are always filled to the same level while the ship observer says they never are?I think you meant that to be the other way around. :)
It's simultaneity again. How does the ship observer know when a photon enters and when it leaves? She must get timed signals from both ends of the ship. (Maybe better make it a light-pulse containing many photons: something that can be sampled without destroying it.)
The external observer, understanding the simultaneity problem, will predict that the ship observer fills the bowls equally full. The external observer, who measures different transit times front-to-back and back-to-front, would fill his own bowls differently.

Grant Hutchison

grav
2009-May-31, 04:38 PM
I think you meant that to be the other way around. :)I don't think so. The ship observer sees more photons coming from the front than the back due to Relativistic Doppler (although that includes flight of light but the result is real), blueshifted in front and redshifted from the back by a factor of sqrt[(c+v) / (c-v)] / sqrt[(c-v) / (c+v)] = (c+v) / (c-v), but they travel through at the same speed from each direction over the same distance, so in the same time according to the ship observer, so more photons from the front means more water in bowl A than bowl B by a factor of (c+v) / (c-v). The external observer sees more photons coming from the front than the back proportional to the relative speeds of c+v to c-v, so also by a factor between front and back of (c+v) / (c-v) as the ship observer does, but sees the photons travel faster across the ship from front to back than back to front by the same factor, c+v from the front for a time of 1 / (c+v) and c-v from the back for a time of 1 / (c-v) over the same distance for each, so the same amount of water should be poured either way. The time the photons take to pass through the ship according to the ship observer is the same either way, but different to the external observer, so the time it takes for photons to pass trhough the ship from front to back as compared to back to front according to each observer determines the rate the bowls are filled.


It's simultaneity again.Most likely, but I can't seem to catch on to where it should be applied for some reason.


How does the ship observer know when a photon enters and when it leaves? She must get timed signals from both ends of the ship. (Maybe better make it a light-pulse containing many photons: something that can be sampled without destroying it.)Yes, okay. The way I'm doing it, we also have the extra factor of the ship observer controlling the bowls, so they would only be filled according to how the ship observer says they should anyway, so let's try something else that is even closer to what I'm trying to find. Let's say the photons provide a steady force that acts on the ship as they pass through it, in the same direction that they are travelling. In that case, the ship observer says that the force coming from the front will be greater than that from the back, so the ship should be slowing down. The external observer, however, says that equal force is applied in either direction and the ship should continue inertially.


The external observer, understanding the simultaneity problem, will predict that the ship observer fills the bowls equally full. The external observer, who measures different transit times front-to-back and back-to-front, would fill his own bowls differently.Hopefully I'm not confusing you at this point, which it looks like I might be, but actually the results of both of those statements would seem to be true the way they are stated as this scenario is applied so far, except perhaps for the "understanding the simultaneity" part :), but yet they can't be. The bowls coincide in the same place, so we can't have one observer seeing them filled to different levels while the other sees them filled the same, and with our new scenario where force acts, we can't have one observer seeing the ship slow down while the other sees it continue inertially.

grav
2009-May-31, 08:36 PM
(Maybe better make it a light-pulse containing many photons: something that can be sampled without destroying it.)In that case, the external observer would measure the same wavelength between photons coming from the light in either direction, and so the number of photons along each beam within the length of the ship will be the same, so there will be the same number of photons within the ship at any given time from the light travelling in each direction, and so the force will be the same, and the ship stays inertial. According to the ship observer, however, the beam coming from the front is blueshifted and the one from the back is redshifted, so the front beam has a larger frequency than the beam coming from the back, so a smaller wavelength between photons, and there will always be more photons per length of the ship for the beam coming from the front than from the back, so more photons within the ship at all times travelling from the front than from the back, and thereby the ship should slow down.

grav
2009-May-31, 10:13 PM
Well, I have applied simultaneity, but now I have an additional dilemma! Let's see if we can resolve it. Let's say the ship is 10 light-secs long according to the ship observer and that it is travelling directly away from the external observer at .6 c so that the external observer now measures the length of the ship as 8 light-secs long, with a length contraction of L = .8. The wavelengths of the light from the beams is also 8 light-secs long according to the external observer, so when one photon is incident with the front of the ship from a beam in either direction, another photon will be incident with the back.

Now, if we place clocks at the front and the back of the ship that are synchronized according to the ship's frame, the time lag between the front and the back of the ship is tl = L d v / (c^2 - v^2) = 7.5 seconds according to the external observer, so the front clock reads T=0 and the back is T=+7.5 seconds. So according to the ship observer, for the beam coming from the front, one photon will be incident with the front of the ship at T=0 and the previous one incident with the back at T=+7.5 seconds, so at T=0, the first photon will be incident with the front and the previous one wherever it was 7.5 seconds before it reaches the back, so at 2.5 light-seconds from the front of the ship, and the wavelength between photons coming from the frontal direction is then w_f = 2.5 light-secs. For light coming from the back, a photon will be incident with the front at T=0 and the next one incident with the back at T=+7.5 seconds, so at T=0, one photon is incident with the front and the next whatever distance it would be from the back of the ship 7.5 seconds earlier, which is 7.5 light-seconds away from the back of the ship, so is 17.5 light-seconds away from the previous photon, and the wavelength for the beam coming from the back is 17.5 light-seconds.

Now, the original wavelength according to the ship observer when the ship was stationary to the external observer and the source of the two beams is 8 light-seconds, same as the external observer measures it. Once the ship is moving, it is travels at v = .6 c toward the source of the front beam, so according to Relativistic Doppler, the new wavelength the ship observer measures should be w_f' = sqrt[(c - v) / (c + v)] w = 4 light-seconds. For the beam coming from the back, it should be w_b' = sqrt[(c + v) / (c - v)] w = 16 light-secs. But that is not what we got using the effects of simultaneity, which should be the same. So where did I go wrong?

grav
2009-May-31, 10:27 PM
Oh, wait. I'll leave that last post on here to show that I am working with simultaneity :), but the time lag should have included the time dilation between frames as well, since the time lag I found is as the external observer would measure, so we must convert that to what the ship's clocks would read in the ship's frame, and so becomes tl = Z L d v / (c^2 - v^2) = 6 seconds, not 7.5, giving a wavelength for the front beam of w-f' = 10 - 6 = 4 light-seconds and that for the back beam of w-b' = 10 + 6 = 16 light-seconds, just as Relativistic Doppler shows. Dilemma solved for that, so back to the main deal.

grav
2009-May-31, 10:50 PM
Okay, yes. Upon trying to devise a method for containing and measuring the photons, I can see how simultaneity effects would come into play, at least for this particular method. Let's say there are holes at either end of the ship that the beams pass through. Periodly the ship will close both the holes at both ends simultaneously with detectors in order to measure the number of photons currently travelling through the ship at the moment of closing the holes. The detectors will measure more coming from the front than the back. Now to the external observer, a simultaneity shift occurs so that the external observer sees the hole close at the back of the ship first while the hole at the front stays open and more photons continue to enter from that direction until finally that hole is closed also. So both observers agree that more photons have entered from the front of the ship and will provide more force toward the back when contained this way, and the ship slows down.

Now, the idea here, though, is that both holes are closed simultaneously to the ship observer and not simultaneous to the external observer, pretty simple. But would something like that still apply if no such simultaneous act takes place and the photons just steadily produce force on the ship as they pass through? It seems likely, I guess, but why?

macaw
2009-Jun-02, 05:23 AM
I am having trouble wrapping my head around something all of a sudden. We all know the twin paradox for a barn / polevaulter experiment, but what about light itself? Let's say we have two lights set up at a large distance outside of either end of a ten light-second long ship and an observer outside the ship that is always stationary to the lights while the ship is also stationary for the time being. Observers are placed at every light-second within the ship with synchronized clocks, so eleven in all, nine within the ship and one at each end. Now the light at one end of the ship is flashed. The ship observers record the times that they receive the flashes, one per second, and a short while later reproduce the flashes to the outside observer, one per second, and that is the rate the outside observer receives them. Then the same thing is done from the other end.

Okay, now let's say the ship has a relative speed of .6 c to the stationary observer. All of the clocks within the ship are again synchronized to each other. The outside observer remains stationary with the lights. The experiment is repeated. According to the ship observers, light from either direction still passes each observer at c, passing one observer per second, and so a short while later, the ship observers flash their own light to the outside observer once per second in both cases also, as far as I can tell. The ship is time dilated to the outside observer by a factor of .8, so the flashes are received by the outside observer once per 1.25 second each in both cases, so after the initial pulse, all pulses will be received in 12.5 seconds according to the outside observer in either case.

Now let's look at what the outside observer measures. The light coming from behind the ship is travelling through the ship at c, while the ship moves forward at .6 c, and the ship is contracted by .8, so that ray should take t = (8 light-seconds) / (c - .6 c) = 20 seconds to travel through, and that is how the pulses from the ship observer should also be received, since they match the rate the pulses pass each observer. From in front of the ship, the light should take (8 light-seconds) / (c + .6 c) = 5 seconds to travel through the ship and that is mimicked by the ship observers, so the pulses from the ship observer should be received in that amount of time to the outside observer. So neither case matches what should be observed, although the sum of the times is the same. What's going on?



Grant really explained this to you, it is called "relativity of simultaneity". From the perspective of the observers in the ship, the flashes arrive at the two ends of the ship simultaneously.
From the perspective of the outside observer, they do not.
This is simple relativity, you can't continue constructing your ATM if you don't learn it first.

grav
2009-Jun-02, 01:48 PM
Grant really explained this to you, it is called "relativity of simultaneity". From the perspective of the observers in the ship, the flashes arrive at the two ends of the ship simultaneously.
From the perspective of the outside observer, they do not.
This is simple relativity, you can't continue constructing your ATM if you don't learn it first.What's strange is that it's not so much that the photons arrive at the two ends simultaneously in one frame and not in the other as it is that there seems to be more photons present within the ship one way than the other at any given time according to the ship observer and always the same either way according to the external observer, although it looks like it would be for the same reason simultaneity provides, yes.

grav
2009-Jun-02, 02:02 PM
I guess it would be similar to two trains entering a tunnel from either end at the same time according to the frame of the tunnel. When a tunnel observer sees the trains enter from each direction, they reach the ends of the tunnel simultaneously, so according to to the tunnel observer, when each train just reaches each entrance to the tunnel, the tunnel is still empty. But according to a train observer, when the front of the train reaches an entrance to the tunnel, the other end of the tunnel is future-forward, so the other train has already travelled into the tunnel some distance, and so the tunnel is not empty.

macaw
2009-Jun-02, 03:42 PM
there seems to be more photons present within the ship one way than the other at any given time according to the ship observer and always the same either way according to the external observer, although it looks like it would be for the same reason simultaneity provides, yes.

No, it isn't. You are making stuff up without any regard to actual physics.

mugaliens
2009-Jun-02, 05:10 PM
Observers are placed at every light-second within the ship with synchronized clocks...

All of the clocks within the ship are again synchronized to each other...

Clock synchronization is core to your points/questions, yet without detailing how the clocks are "synchronized," this has little meaning.

For example, if the clocks are sychronized at one end of the ship, the very act of spreading the clocks throughout the ship changes their time relative to an infinately distant observer perpendicular to the length of the ship (therefore effectively equidistance from all clocks).

grav
2009-Jun-02, 11:12 PM
Clock synchronization is core to your points/questions, yet without detailing how the clocks are "synchronized," this has little meaning.

For example, if the clocks are sychronized at one end of the ship, the very act of spreading the clocks throughout the ship changes their time relative to an infinately distant observer perpendicular to the length of the ship (therefore effectively equidistance from all clocks).Clock synchronization an ideal situation, but getting them as close as possible is a simple matter. For instance, we can synchronize them all in one place, and then move them very, very slowly to their destinations throughout the ship.

grav
2009-Jun-02, 11:39 PM
there seems to be more photons present within the ship one way than the other at any given time according to the ship observer and always the same either way according to the external observer, although it looks like it would be for the same reason simultaneity provides, yes.No, it isn't. You are making stuff up without any regard to actual physics.Sure it is. I remember now trying to do something similar with the M-M experiment after the last twin paradox/simultaneity thread, and finding that while an observer would find the number of photons along an arm in each direction the same, an observer moving relative to the arm would measure a different number of photons each way. I wish I had remembered that earlier. It might have saved some work in this thread.

Below is a diagram I drew up for illustration. It's probably about as close as I will get to a space-time diagram. :o Anyway, the row on top demonstates how the observer stationary with the arm will measure the distancing between photons, the same in both directions, with the same number of photons each way for a steady beam. The bottom row shows how an observer travelling at .6 c away from the arm (down the page) will measure it. To the moving observer, the apparatus is length contracted by L = .8 for an arm with a length of 5 light-seconds to the stationary observer, so 4 light-seconds to the moving observer, and with the same time dilation of Z = .8, so the time lag along the arm will be tl = Z L d v / (c^2 - v^2) = 3 seconds.

The frequency of the beams is one photon per second to the stationary observer, so the wavelength is one light-second. To the moving observer, that translates to w' = w sqrt[(c-v) / (c+v)] = .5 light-seconds for the beam travelling away and w" = w sqrt[(c+v) / (c-v)] = 2 light-seconds for the beam travelling toward the moving observer. We can see that as the beams travel away and back, each photon still coincides with each of the others at the same times read on clocks placed along the arms, which are synchronized in the stationary frame. For instance, photon A coincides with photon F in the center of the arm at T=7.5 in both rows, photon B coincides with photon G in the center at T=8.5 in both rows, photon A reaches the end at T=10, photon C reaches the top at T=7, etc., even though the number of photons is different for the length of the arm each way while they are the same for the stationary observer.

macaw
2009-Jun-03, 01:59 PM
Sure it is. I remember now trying to do something similar with the M-M experiment after the last twin paradox/simultaneity thread, and finding that while an observer would find the number of photons along an arm in each direction the same, an observer moving relative to the arm would measure a different number of photons each way. I wish I had remembered that earlier. It might have saved some work in this thread.

Below is a diagram I drew up for illustration. It's probably about as close as I will get to a space-time diagram. :o Anyway, the row on top demonstates how the observer stationary with the arm will measure the distancing between photons, the same in both directions, with the same number of photons each way for a steady beam. The bottom row shows how an observer travelling at .6 c away from the arm (down the page) will measure it. To the moving observer, the apparatus is length contracted by L = .8 for an arm with a length of 5 light-seconds to the stationary observer, so 4 light-seconds to the moving observer, and with the same time dilation of Z = .8, so the time lag along the arm will be tl = Z L d v / (c^2 - v^2) = 3 seconds.

The frequency of the beams is one photon per second to the stationary observer, so the wavelength is one light-second. To the moving observer, that translates to w' = w sqrt[(c-v) / (c+v)] = .5 light-seconds for the beam travelling away and w" = w sqrt[(c+v) / (c-v)] = 2 light-seconds for the beam travelling toward the moving observer. We can see that as the beams travel away and back, each photon still coincides with each of the others at the same times read on clocks placed along the arms, which are synchronized in the stationary frame. For instance, photon A coincides with photon F in the center of the arm at T=7.5 in both rows, photon B coincides with photon G in the center at T=8.5 in both rows, photon A reaches the end at T=10, photon C reaches the top at T=7, etc., even though the number of photons is different for the length of the arm each way while they are the same for the stationary observer.



You got your explanation wrong again. In the MMX experiment there is NO Doppler shift detected at the point of return. The reason is simple. The photon emitted from the center of the apparatus arrives redshifted at the mirror at the end of the "horizontal" arm:

w' = w * sqrt[(c-v) / (c+v)] (1)

When it bounces back off the mirror, it comes back blueshifted to the observer:

w" = w' sqrt[(c+v) / (c-v)] (2)

Combining (1) and (2) you get w"=w. No Doppler shift. You really need to take a class, making up things is not the way to go. Especially when what you make up turns up incorrect every time.

grant hutchison
2009-Jun-03, 03:03 PM
I think you meant that to be the other way around. :)I don't think so.So, you're timing more than one photon, and accumulating water with each successive photon? It's not exactly clear from your original description. :)
Simultaneity is still the solution to your problem.

Grant Hutchison

grant hutchison
2009-Jun-03, 03:24 PM
... there seems to be more photons present within the ship one way than the other at any given time according to the ship observer and always the same either way according to the external observer, although it looks like it would be for the same reason simultaneity provides, yes.A spacetime diagram makes it obvious why, if an external observer measures an equal density of photons travelling in each direction, an observer aboard a moving ship will find the ship contains more rearward-moving photons than forward-moving photons, at any given instant of shipboard time.
Think about how "an instant of shipboard time" is specified under SR, and how that differs from an instant of the external observer's time. (Simultaneity again.)

Grant Hutchison

grav
2009-Jun-03, 10:39 PM
A spacetime diagram makes it obvious why, if an external observer measures an equal density of photons travelling in each direction, an observer aboard a moving ship will find the ship contains more rearward-moving photons than forward-moving photons, at any given instant of shipboard time.
Think about how "an instant of shipboard time" is specified under SR, and how that differs from an instant of the external observer's time. (Simultaneity again.)

Grant HutchisonI drew "a diagram" for post #29. Have you seen it? :) Not quite a space-time diagram, but it'll do. I can see what is taking place, so the ship observer will fill each bowl differently and the external observer the same, as you said earlier. As far as a force that is applied to the ship itself, where it would slow down according to the ship observer and remain inertial to the external observer, since it can't be both, I can only conclude that it must be physically impossible to design a force that will act upon the ship according to only these characteristics.

grav
2009-Jun-03, 11:36 PM
You got your explanation wrong again. In the MMX experiment there is NO Doppler shift detected at the point of return. The reason is simple. The photon emitted from the center of the apparatus arrives redshifted at the mirror at the end of the "horizontal" arm:

w' = w * sqrt[(c-v) / (c+v)] (1)

When it bounces back off the mirror, it comes back blueshifted to the observer:

w" = w' sqrt[(c+v) / (c-v)] (2)

Combining (1) and (2) you get w"=w. No Doppler shift. You really need to take a class, making up things is not the way to go. Especially when what you make up turns up incorrect every time.Okay well, first of all, I did not make any claims in that post about whether or not the Doppler shift cancels out at the point of return. That is your claim, so who's making things up? Furthermore, your claim is incorrect. Doppler along a single arm has nothing to do with how frequencies cancel out in the MM experiment, and the photons would be perceived as travelling along the single arm in Relativity just as I have indicated. In the MM experiment, however, the beam is split into two perpendicular arms in the frame of the observer moving with the apparatus and brought back together. Then the lengths of the arms are purposely adjusted so that the frequencies have destructive interference and cancel out. But the null result is found when the apparatus is turned in various directions and the resulting frequency remains the same.

I'm sorry, but you cannot continue to claim that anything that is not based upon relativistic Doppler as you perceive it is incorrect and/or made up. There is more to it than that and since you apparently have nothing else to contribute to the discussions, being that I have not seen any physics from you whatsoever, contrary or otherwise, or any logical disagreements other than vague boasts, you should really pay more attention to what is being discussed in some of these threads if you want to learn more about Relativity.

grant hutchison
2009-Jun-03, 11:46 PM
I drew "a diagram" for post #29. Have you seen it? Well, I saw it. :)

I was thinking more of something like this (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/photons.jpg).
We have a spacetime diagram from the point of view of your external observer. Two photons (red) are crossing from left to right, one behind the other. Two equally spaced photons (blue) are moving in the opposite direction. The worldlines of the front and back of the ship are marked by the grey dashed lines: the ship is moving left to right at some appreciable fraction of lightspeed.
The external observer measures the length of the ship along the horizontal black bar: at the instant marked, we have the red photons just entering and leaving the ends of the ship, and the blue photons doing likewise while moving in the opposite direction. Any photons travelling between these "marker" photons are contained entirely within the ship, at this moment of the external observer's time.
But the ship observer measures the ship's length along a simultaneity line that is tilted in this diagram: the green bar. So we see that, from the shipboard observer's point of view, the ship easily accommodates both blue photons, but fails to accommodate either of the red photons. From the reference frame of the ship, both blue photons are within the ship simultaneously; but one red photon has already departed, and the other is yet to arrive.

Simultaneity is the problem. :)

Grant Hutchison

grav
2009-Jun-04, 12:12 AM
Well, I saw it. :)

I was thinking more of something like this (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/photons.jpg).Ah, yes. I can visualize the diagram this time with your explanation, although it will still take some practice getting used to. My mind keeps wanting to quantify it, and I tell that can be done but is unnecessary when expressed in this manner for a simple comparison of the simultaneity effects between frames, and much simpler and quicker and more gradual than drawing it out separately for each frame with each tick of the clock and then comparing them as I did. Thank you again, Grant. :)

grant hutchison
2009-Jun-04, 12:20 AM
My mind keeps wanting to quantify it ...Quantify away. There's a hyperbolic scale for length and time along lines that are inclined to the principal axes, so that takes a little getting used to. But for problems in which observers disagree about events, especially when moving objects pass through or around each other, a quick spacetime diagram lets you see where the interesting bit is, and target your calculations accordingly.

Grant Hutchison

grav
2009-Jun-04, 03:25 AM
Quantify away.Okay, here is my first attempt to quantify the space-time diagram with what I had in post #29, seen in the image below. The unit of time is one second and the unit of space is one light-second, so the paths of the photons will angle at 45 degrees in the diagram for any frame within a single dimension for the line of travel, in both directions, drawn as red and blue and spaced at one light-second each according to the moving observer. The arm is 4 light-seconds long to a moving observer, drawn in brown, while travelling at a relative speed of .6 c to the arm, so the path of the arm angles three units of space for every five units of time according to the moving observer, drawn in purple.

Now the dilemma. I have drawn an angle in black that intercepts exactly two wavelengths from one end of the arm to the other according to an observer moving with it. The problem is that it also intercepts twelve wavelengths coming the other way instead of eight as it should be. Also, if I use the dimension of time the same as another dimension of space, then the ship is way too long as measured by the observer moving with the arm, and shows a time lag between one end of the arm and the other of five seconds to the moving observer, not three. I have also drawn another angle in gray, however, which would intercept two wavelengths in one direction and eight in the other. There would also be a three second time lag along the arm, which is correct for an observer travelling away from the arm for the way it is angled, although since the time is in the frame of the moving observer, I'm not even sure that's right either, and the arm is still too long. What do I do about all of that?

grant hutchison
2009-Jun-04, 12:35 PM
What do I do about all of that?You need to remember the hyperbolic scaling. None of your initial measurements apply to the moving observer, only the stationary observer who "owns" the orthogonal spacetime axes.
Your angle in grey is a true simultaneity line for your spaceship, since it slopes three time units for five space, while the worldlines of the spaceship slope five time units for three space (that is, the ship is moving at 3/5c. It does, however, need to be extended to join the worldlines of the two ends of the spaceship if it is to be the length of the ship measured by a person aboard the ship. It then needs to be measured in its own units, not those of the orthogonal axes.

I have no idea what the physical significance of your black sloping line would be.

Grant Hutchison

grav
2009-Jun-04, 11:34 PM
You need to remember the hyperbolic scaling. None of your initial measurements apply to the moving observer, only the stationary observer who "owns" the orthogonal spacetime axes.
Your angle in grey is a true simultaneity line for your spaceship, since it slopes three time units for five space, while the worldlines of the spaceship slope five time units for three space (that is, the ship is moving at 3/5c. It does, however, need to be extended to join the worldlines of the two ends of the spaceship if it is to be the length of the ship measured by a person aboard the ship. It then needs to be measured in its own units, not those of the orthogonal axes.

I have no idea what the physical significance of your black sloping line would be.

Grant HutchisonI see what I did wrong now. Apparently I mixed up the scenario for the ship where the external observer sees the same wavelength from each direction with that for the arm where the moving observer sees the wavelengths Doppler shifted. That is where I messed up with the black line, since I was trying to find where two wavelengths would cross at the line in the diagram for the arm according to the moving observer in the last scenario. So let's figure it's the ship again, then, and the external observer sees the same wavelength in both directions, which fits the graph as presented. Okay, so the ship observer will see the wavelengths Doppler shifted in this case, and since the external observer measures the length of the ship as four light-seconds, the ship observer measures it as five. So if the external observer measures the wavelengths as 1 light-second each way, the ship observer measures them as two light-seconds for the light coming from the back and half a light-second for the light coming from the front of the ship.

Now, the ship observer measures the length of the ship as five light-seconds, so two and a half wavelengths will be seen within the ship for the light from back to front, and ten wavelengths for the light from front to back. If we extend the gray line to the line for the front of the ship in the diagram, it looks like that is what we will receive for each of those. So that works out now. And this part I don't quite understand yet, and I suppose it has something to do with the hyperbolic scaling that you mentioned, but I notice that if we then contract that gray line back by the same amount of length contraction of .8 that is applied, the gray line will then stop where it currently stops in the diagram, which gives a length of the ship as seen by the ship observer of five light-seconds (along the space direction only), over a time lag as seen by the external observer of three seconds (along the time direction) between the front and back of the ship according to the shortened gray line, which are also both correct.

grant hutchison
2009-Jun-05, 02:16 AM
And this part I don't quite understand yet, and I suppose it has something to do with the hyperbolic scaling that you mentioned, but I notice that if we then contract that gray line back by the same amount of length contraction of .8 that is applied, the gray line will then stop where it currently stops in the diagram, which gives a length of the ship as seen by the ship observer of five light-seconds (along the space direction only), over a time lag as seen by the external observer of three seconds (along the time direction) between the front and back of the ship according to the shortened gray line, which are also both correct.Remember the shipboard observers measure the length of the ship along the grey line (which is their simultaneity line), not parallel to the external observer's simultaneity line.
The hyperbolic scaling means that distances along the simultaneity line scale by γ√(1+β). You've divided by γ, leaving the scale factor at √(1+β). Now, that's just the ratio of hypotenuse to base for the right triangle which supports the simultaneity line. So the base of your triangle now matches the proper length of your spaceship, as you observe. It falls out of the properties of right triangles and rectangular hyperbolae, and I doubt if it says anything particularly deep about spacetime. I don't think it's particularly to be recommended as way of deriving proper lengths. :)

Grant Hutchison

grav
2009-Jun-05, 03:23 AM
Remember the shipboard observers measure the length of the ship along the grey line (which is their simultaneity line), not parallel to the external observer's simultaneity line.
The hyperbolic scaling means that distances along the simultaneity line scale by γ√(1+β). You've divided by γ, leaving the scale factor at √(1+β). Now, that's just the ratio of hypotenuse to base for the right triangle which supports the simultaneity line. So the base of your triangle now matches the proper length of your spaceship, as you observe. It falls out of the properties of right triangles and rectangular hyperbolae, and I doubt if it says anything particularly deep about spacetime. I don't think it's particularly to be recommended as way of deriving proper lengths. :)

Grant HutchisonAh, but now that you've shown me that it is indeed the proper length of the ship, I can continue to use it that way. A simple way of doing so, then, would be just to draw a vertical line down to the base from where the full length of the gray line intercepts the front of the ship, find the distance along the space dimension and multiply that by .8 to get the proper length, and then draw vertically back up from there to where it again intercepts the gray line to find the time lag across the length of the ship according to what the external observer sees for the clocks on the ship. Of course, I can also find those simply by using the formulas for them, but anywho... it's still a cool way of laying things out. :) Thanks.

grav
2009-Jun-05, 03:35 AM
Remember the shipboard observers measure the length of the ship along the grey line (which is their simultaneity line), not parallel to the external observer's simultaneity line.Actually, wouldn't the ship observers still measure their proper length only along the space dimension? I mean, if the gray line is the simultaneity line of the ship, then the space dimension would give the proper length of the ship and the time dimension would give the time lag observed along that length according to the external observer's frame, wouldn't they?

grant hutchison
2009-Jun-05, 12:12 PM
A simple way of doing so, then, would be just to draw a vertical line down to the base from where the full length of the gray line intercepts the front of the ship, find the distance along the space dimension and multiply that by .8 to get the proper length, and then draw vertically back up from there to where it again intercepts the gray line to find the time lag across the length of the ship according to what the external observer sees for the clocks on the ship.Simple?


Actually, wouldn't the ship observers still measure their proper length only along the space dimension?That's what they do, by following their own simultaneity line. And they measure time along their worldline. These lines mark "pure space" and "pure time" in their moving coordinate system.

Grant Hutchison

grav
2009-Jun-05, 12:54 PM
Simple?Well, I figure it's at least much simpler than trying to measure the length along the diagonal simultaneity line, scaling it at y sqrt(1 + B^2), and then running that back across the diagonal line again, which would probably require at least a ruler placed across the diagonal line or something unless one also worked it out for across the base after scaling that way and ran it up from that point. ;)


That's what they do, by following their own simultaneity line. And they measure time along their worldline. These lines mark "pure space" and "pure time" in their moving coordinate system.But if all of the clocks are synchronized in the ship's frame, scaling at y sqrt(1 + B^2) won't tell us anything about the time in the ship's frame, will it? Although I suppose combining the time and length scales, it would give the proper length of the ship and then equally the time it takes the light to travel that length, but that can also be done across the same length across the base in that case. Scaling at y gives us the proper length of the ship and the time lag in the external observer's frame, so I figure that's the better better way to go since it adds more information. Scaling anything isn't a perfect system, I guess, but as long as it works, I figure the more information that can be gained the better, and scaling at y is a two in one shot. :) I guess sometimes it just takes a fresh perspective to notice something different, so maybe now you'll have something new to get used to also, but useful? :p

grant hutchison
2009-Jun-05, 01:23 PM
But if all of the clocks are synchronized in the ship's frame, scaling at y sqrt(1 + B^2) won't tell us anything about the time in the ship's frame, will it?We plot distance along the simultaneity line. The locus of constant proper distance sweeps out a rectangular hyperbola as we move from one simultaneity line to the next (that is, as we move across the axes appropriate for observers in different states of motion). So, yes, the scaling is essential.


Although I suppose combining the time and length scales, it would give the proper length of the ship and then equally the time it takes the light to travel that length, but that can also be done across the same length across the base in that case. Scaling at y gives us the proper length of the ship and the time lag in the external observer's frame, so I figure that's the better better way to go since it adds more information.While throwing away a lot of more important information. You've found a quick and dirty way of projecting out a single bit of information, in a way that obscures the hyperbolic geometry of invariants under Minkowski spacetime.

The point of Minkowski diagrams is to understand the relationships, not to be able to read off numbers. There are, as you say, easier ways to find the numbers. I'd submit that your problem with SR at present is in understanding the relationships, not in generating numbers.

Grant Hutchison

grav
2009-Jun-05, 01:35 PM
We plot distance along the simultaneity line. The locus of constant proper distance sweeps out a rectangular hyperbola as we move from one simultaneity line to the next (that is, as we move across the axes appropriate for observers in different states of motion). So, yes, the scaling is essential.

While throwing away a lot of more important information. You've found a quick and dirty way of projecting out a single bit of information, in a way that obscures the hyperbolic geometry of invariants under Minkowski spacetime.

Grant HutchisonOh, well we haven't gotten into all of that yet, so nevermind. ;) Anyway, put it this way. Additional information can be gained by scaling in a different way as well as the hyperbolic scaling. :) Although I just realized that I'm claiming that, but I haven't actually proven it yet. It just looks that way in the diagram for that particular example, so I will still need to do that also.

grant hutchison
2009-Jun-05, 04:06 PM
Although I just realized that I'm claiming that, but I haven't actually proven it yet. It just looks that way in the diagram for that particular example, so I will still need to do that also.Notice that my discussion of your result with reference to scaling hyperbolae and right triangles involved only β and things derived from β. It's going to be a general result.

Grant Hutchison

grav
2009-Jun-05, 10:32 PM
Notice that my discussion of your result with reference to scaling hyperbolae and right triangles involved only β and things derived from β. It's going to be a general result.

Grant HutchisonI can't tell. Are you saying the y scaling does work out that way, with the proper length of the ship across the base and the time lag across the ship according to the external observer's frame along the time dimension? I hope so, because I'm not even sure where to start applying calculations to the diagram yet to find out.

Sorry about earlier when I thought that y might replace y sqrt(1 + B^2) as a better way to scale. :( I got quite a bit ahead of myself there when I don't even know what the hyperbolic scaling could be used for except what few little things I can see in the diagram for inertial observers. :silenced: If the y thing works, that should suit my purposes, though, for whatever simple quick sketching I might do, I think. Trying to sketch a bunch of twisting and turning would probably only give me a headache. Besides, my sketching skills aren't all that great. :)

grant hutchison
2009-Jun-05, 11:15 PM
I can't tell. Are you saying the y scaling does work out that way, with the proper length of the ship across the base and the time lag across the ship according to the external observer's frame along the time dimension? I hope so, because I'm not even sure where to start applying calculations to the diagram yet to find out.Yes, it's bound to work out that way.
The actual hyperbolic scaling is a function of velocity alone, as is the slope of the simultaneity line. Your Pythagorean trick works by applying the scaling factors in an unconventional way, using the slope of the simultaneity line: it's another way of doing the same maths. Either way, it all reduces to the same function of velocity.


Sorry about earlier when I thought that y might replace y sqrt(1 + B^2) as a better way to scale. :( I got quite a bit ahead of myself there when I don't even know what the hyperbolic scaling could be used for except what few little things I can see in the diagram for inertial observers.Thing is, the spacetime diagram is supposed to work by forming a fixed array of spacetime events. We then coordinatize that in various ways so that we can see how different observers would measure the same events. So we measure proper lengths along appropriately scaled simultaneity lines that intersect the events of interest, because that's where the observers are and what they do. The hyperbolic scaling also reminds us that we're not dealing with Euclidean space, but with Minkowski spacetime, where the interval, x - ct, is the important invariant.
Your construction using a right triangle steps outside the rules of the diagram, because it places a measurement in an impossible place, using means that are impossible for the spacetime that is being depicted. Whereas drawing calibration marks on the simultaneity line itself effectively hands the observer a metre-rule and lets him get on with it.

Grant Hutchison

macaw
2009-Jun-15, 03:27 PM
Sorry about earlier when I thought that y might replace y sqrt(1 + B^2) as a better way to scale.

Unfortunately for you physics doesn't work this way, you can't make up whatever formulas you want :lol:

macaw
2009-Jun-15, 03:31 PM
Okay well, first of all, I did not make any claims in that post about whether or not the Doppler shift cancels out at the point of return.

Seems that you don't understand your own claims (http://www.bautforum.com/1500368-post29.html):

"The frequency of the beams is one photon per second to the stationary observer, so the wavelength is one light-second. To the moving observer, that translates to w' = w sqrt[(c-v) / (c+v)] = .5 light-seconds for the beam travelling away and w" = w sqrt[(c+v) / (c-v)] = 2 light-seconds for the beam travelling toward the moving observer.We can see that as the beams travel away and back, each photon still coincides with each of the others at the same times read on clocks placed along the arms, which are synchronized in the stationary frame. For instance, photon A coincides with photon F in the center of the arm at T=7.5 in both rows, photon B coincides with photon G in the center at T=8.5 in both rows, photon A reaches the end at T=10, photon C reaches the top at T=7, etc., even though the number of photons is different for the length of the arm each way while they are the same for the stationary observer."

macaw
2009-Jun-15, 03:48 PM
Doppler along a single arm has nothing to do with how frequencies cancel out in the MM experiment,

The frequencies do not "cancel out in the MM experiment". Something else does. Do you even know what that "something else" is?


Then the lengths of the arms are purposely adjusted so that the frequencies have destructive interference and cancel out.

Umm, no. You do not understand MMX. You might want to study it a little better. BTW: frequency has nothing to do with "destructive interference". What creates destructive interference?



But the null result is found when the apparatus is turned in various directions and the resulting frequency remains the same.

Why is that?


I'm sorry, but you cannot continue to claim that anything that is not based upon relativistic Doppler as you perceive it is incorrect and/or made up.

...only in the case of your posts (http://www.bautforum.com/1509385-post51.html) :lol:

grav
2009-Jun-16, 12:00 AM
Unfortunately for you physics doesn't work this way, you can't make up whatever formulas you want :lol:Yes, I'm aware of that, adsar, but thank you for pointing it out. :) I am learning, as you should be doing also. That equation happens to work out for my needs, although apparently not as intended.

grav
2009-Jun-16, 12:03 AM
Seems that you don't understand your own claims (http://www.bautforum.com/1500368-post29.html):

"The frequency of the beams is one photon per second to the stationary observer, so the wavelength is one light-second. To the moving observer, that translates to w' = w sqrt[(c-v) / (c+v)] = .5 light-seconds for the beam travelling away and w" = w sqrt[(c+v) / (c-v)] = 2 light-seconds for the beam travelling toward the moving observer.We can see that as the beams travel away and back, each photon still coincides with each of the others at the same times read on clocks placed along the arms, which are synchronized in the stationary frame. For instance, photon A coincides with photon F in the center of the arm at T=7.5 in both rows, photon B coincides with photon G in the center at T=8.5 in both rows, photon A reaches the end at T=10, photon C reaches the top at T=7, etc., even though the number of photons is different for the length of the arm each way while they are the same for the stationary observer."I see nothing there that claims anything about the Doppler shift cancelling out, only that the photons are observed to coincide in the same way from both frames of reference.

grav
2009-Jun-16, 12:08 AM
The frequencies do not "cancel out in the MM experiment". Something else does. Do you even know what that "something else" is?Why don't you enlighten me?

macaw
2009-Jun-16, 12:32 AM
Yes, I'm aware of that, adsar, but thank you for pointing it out. :) I am learning...


You aren't learning since you are repeating the same error of making up formulas with no basis whatsoever.

macaw
2009-Jun-16, 12:35 AM
Why don't you enlighten me?

You need to learn on your own. You claim to know physics but your posts prove the opposite. Frequency has nothing to do with destructive interference and nothing to do with the outcome of MMX, you need to consult your high school manuals and you'll find out what other physical quantity does. I think it is taught in 9-th grade, under wave theory. :lol:

grav
2009-Jun-16, 02:06 AM
Okay, getting back to the topic, I looked back through the OP and decided to try to work that out. Now, if each of the ship observers flashed a signal at the same time they see the light pass each detector they coincide with, the signals wouldn't just leave the ship immediately, since they still have to pass some distance through the ship. Likewise, if we had a central computer detect the light from each detector, it would still take time to pass through the wires from the detectors to the computer. So I decided to mount another detector at a distance x across the front of the ship which each of the other detectors flashes a signal at (or mirrors the light), so that all signals leave the ship from that same point, and we only have to determine the frequency of the flashes that are received there and see if they match what the external observers would measure as compared to what the ship observers measure.

In order to make the frequency of the light that is received at the main detector mounted at x receive one pulse per second according to the ship observers, the distance of each detector along the length of the ship must be found with the distance from the front of the ship to each detector y, plus the diagonal length to the main detector sqrt(x^2 + y^2), and they must do so such that if the first detector is set at the front of the ship also and so the light crosses the distance x in one second, so x is one light-second across, then the next pulse is receives in two seconds and so on, so y + sqrt(x^2 + y^2) = n x, x^2 + y^2 = (n x - y)^2, x^2 + y^2 = n^2 x^2 - 2 y n x + y^2, 2 y n x = (n^2 - 1) x^2, y = (n^2 - 1) x / (2 n), and with v = .6 c, the external observer will see that length contracted to y' = .8 (n^2 - 1) x / (2 n).

According to the external observer, the light travels the length y' in a time of y' / (c + v) and diagonally to the main detector in a time of (y' + v t)^2 + x^2 = (c t)^2, (c^2 - v^2) t^2 - 2 y' v t - (y'^2 + x^2) = 0, so solving quadratically, we get t = [2 y' v +/- sqrt(4 y'^2 v^2 + 4 (c^2 - v^2) (y'^2 + x^2))] / (2 (c^2 - v^2)), t = [y' v + sqrt(x^2 (c^2 - v^2) + y'^2 c^2)] / (c^2 - v^2), and the total time for the path of the light becomes t_y = y' / (c + v) + [y' v + sqrt(x^2 (c^2 - v^2) + y'^2 c^2)] / (c^2 - v^2) = [y' (c - v) + y' v + sqrt(x^2 (c^2 - v^2) + y'2 c^2)] / (c^2 - v^2) = [y' c + sqrt(x^2 (c^2 - v^2) + y'^2 c^2)] / (c^2 - v^2).

The ship observers receive the pulses once each second so the external observer should receive them once every 1.25 seconds. So referring to the formulas found, for n=1, y' = a = 0, and t_a = [0 + sqrt(.64 + 0)] / (.64) = 1.25 seconds. For n = 2, y' = b = .6 light-seconds, and t_b = [.6 + sqrt(.64 + .36)] / (.64) = 2.5 seconds. For n = 3, y' = c = 1.066... light-seconds, and t_c = [1.066... + sqrt(.64 + 1.1377...)] / (.64) = 3.75 seconds. For n = 4, y' = d = 1.5 light-seconds, and t_d = [1.5 + sqrt(.64 + 2.25)] / (.64) = 5 seconds, etc. In this way, the distance across the ship to a main detector is also included and simultaneity does the trick. Thanks again for the help Grant. :)

macaw
2009-Jun-16, 03:17 AM
According to the external observer, the light travels the length y' in a time of y' / (c + v)

No, it doesn't. You got the term c+v wrong. If you want to know why you will really,really need to learn physics. :lol:

grav
2009-Jun-16, 03:32 AM
No, it doesn't. You got the term c+v wrong. If you want to know why you will really,really need to learn physics. :lol:There is nothing wrong with it, adsar. From the perspective of the external observer, the light travels from the front of the ship to the detector at c while the ship moves away from the observer at v, so if the external observer measures the distance from the front of the ship to the detector as d, then c t = d - v t, c t + v t = d, and t = d / (c + v). This is the time the external observer measures for the light to reach the detector from the front of the ship.

macaw
2009-Jun-16, 04:06 AM
There is nothing wrong with it, adsar. From the perspective of the external observer, the light travels from the front of the ship to the detector at c while the ship moves away from the observer at v, so if the external observer measures the distance from the front of the ship to the detector as d, then c t = d - v t, c t + v t = d, and t = d / (c + v). This is the time the external observer measures for the light to reach the detector from the front of the ship.

You are mixing two frames, a classical error made by people who don't understand relativity. Look at your y' . :lol:
Like I said, it is never too late to go back to school and study, posting stuff on the internet and waiting for others to correct you is not the best way of learning.

grav
2009-Jun-16, 12:11 PM
In this way, the distance across the ship to a main detector is also included and simultaneity does the trick.Oh wait. The first scenario doesn't necessarily use simultaneity effects, since it doesn't require determining a time lag across the ship and only considers a single set of coinciding photons and so forth, so it is only time of flight of light after all, in terms of the distance travelled to different parts of the ship once we have positioned the main detector, with Lorentz contraction figured in.

macaw
2009-Jun-16, 01:37 PM
Oh wait. The first scenario doesn't necessarily use simultaneity effects, since it doesn't require determining a time lag across the ship and only considers a single set of coinciding photons and so forth, so it is only time of flight of light after all, in terms of the distance travelled to different parts of the ship once we have positioned the main detector, with Lorentz contraction figured in.

Why would you be applying Lorentz contraction to the length of the rocket only and not to the terms ct and vt? :lol:

grav
2009-Jun-16, 10:37 PM
Why would you be applying Lorentz contraction to the length of the rocket only and not to the terms ct and vt? :lol:That's a good question. Well, t is the variable here, which is what we are attempting to find according to the external observer's clock, so when we find that, it will already be according to what the external observer measures. c is the speed of light as the external observer measures it, as it is in all frames, and v is the speed of the ship according to what the external observer measures as well, so everything is already expressed in terms of what the external observer would measure except for the length of the ship itself, in which case I originally used the length that the ship observers would measure, so that has to be Lorentz contracted to the length the external observer would measure for that also. If I had just used the length as the external observer would measure in the first place, then it wouldn't have needed to be contracted.

macaw
2009-Jun-16, 11:58 PM
That's a good question. Well, t is the variable here, which is what we are attempting to find according to the external observer's clock, so when we find that, it will already be according to what the external observer measures. c is the speed of light as the external observer measures it, as it is in all frames, and v is the speed of the ship according to what the external observer measures as well, so everything is already expressed in terms of what the external observer would measure


NOT. ct and vt represent lengths, just like y. Think some more.



except for the length of the ship itself,

...which should be y, not y' as you are using. I told you that you are mixing frames, a classical error made by people that do not understand relativity.



If I had just used the length as the external observer would measure in the first place, then it wouldn't have needed to be contracted.

You really need to study relativity. Making up "stuff" is not the way to go.

grav
2009-Jun-17, 12:19 AM
NOT. ct and vt represent lengths, just like y. Think some more.For a minute there, I thought you might actually be asking a sincere question and was interested in learning the answer. Oh well, at least I gave you the benefit of the doubt. Anyway, just in case there's still a slim chance you are, ct is the distance the external observer sees the light travel toward him while vt is the distance the ship travels away in the same time t. So we subtract the distance the ship moves the detector forward vt from the original distance from the front of the ship to the detector d during the same time that the light travels in the opposite direction and reaches the detector over a distance ct. In other words, the original distance the external observer measures between the front of the ship where the light enters and the detector is d, the detector moves forward vt over a time t, so the distance the light travels is what part of the original distance that is left ct after the detector moves forward over the same time t, thereby ct = d - vt. y' is d in the example I gave, the distance from the front of the ship to each detector (or reflector) as the external observer would measure it.

macaw
2009-Jun-17, 12:37 AM
For a minute there, I thought you might actually be asking a sincere question and was interested in learning the answer. Oh well, at least I gave you the benefit of the doubt. Anyway, just in case there's still a slim chance you are, ct is the distance the external observer sees the light travel toward him while vt is the distance the ship travels away in the same time t. So we subtract the distance the ship moves the detector forward vt from the original distance from the front of the ship to the detector d during the same time that the light travels in the opposite direction and reaches the detector over a distance ct. In other words, the original distance the external observer measures between the front of the ship where the light enters and the detector is d, the detector moves forward vt over a time t, so the distance the light travels is what part of the original distance that is left ct after the detector moves forward over the same time t, whereas ct = d - vt.

All this does not clear your misunderstandings about when to apply length contraction.
Why do you apply length contraction to the distance y ?
Why do you use the y' variable and not the y variable?
Why do you jump frames? (hint: because you don't know relativity :lol:)

grav
2009-Jun-17, 12:50 AM
Why do you apply length contraction to the distance y ? y is the length along the ship according to the ship observers where detectors must be placed in order for the main detector to receive a signal once every second, and then that would be the frequency of the photons as they are received by the main detector according to the ship observer's frame.


Why do you use the y' variable and not the y variable?The external observer measures the same distances from the front of the ship to each of the detectors as the ship observer's do, but just Lorentz contracted by sqrt[1 - (v/c)^2].


Why do you jump frames?Since we are attempting to find the frequency the external observer receives as compared to what the ship observer's receive, we must switch to his frame.

macaw
2009-Jun-17, 04:12 AM
Since we are attempting to find the frequency the external observer receives as compared to what the ship observer's receive, we must switch to his frame.

But you are mixing calculations from BOTH frames. Correct application of relativity requires that you stick to ONE frame ONLY. Like I said so many times, you need to learn.

grav
2009-Jun-17, 12:10 PM
But you are mixing calculations from BOTH frames. Correct application of relativity requires that you stick to ONE frame ONLY.That's how we compare what an observer in one frame observes to what one in another frame observes. It's perfectly legitimate. For length contraction, we just take the length one observer that is stationary to the length measures, apply the Lorentz contraction to the part of the length that lies along the line of motion according to another observer using the relative speed along the same line between the frames, and that will be the length as observed by the other frame. That is what the Lorentz contraction does. It is a translation of observed length between one frame and another.

macaw
2009-Jun-17, 01:52 PM
That's how we compare what an observer in one frame observes to what one in another frame observes.

...only for people that do not know relativity. For people that actually know relativity this is just an elementary error.




It's perfectly legitimate.

No, it's not. You are mixing variables from two frames.


This is hopeless, you will not be able to get a valid proof of your theory if you refuse to learn some physics.
Let's try a different angle, you also have an error in elementary geometry: explain the y+sqrt(x^2+y^2) in your post 59. You say :
"the distance of each detector along the length of the ship must be found with the distance from the front of the ship to each detector y, plus the diagonal length to the main detector sqrt(x^2 + y^2),"

This cannot be, since y and sqrt(x^2+y^2) are lengths of vectors that have different directions, so you can't add them up like scalars.

grav
2009-Jun-17, 11:53 PM
No, it's not. You are mixing variables from two frames.This is what the Lorentz contraction is used for. Let's say you have an observer in a ship that measures the length of the ship to be ten meters long. You also have another observer outside the ship that sees the ship moving directly away at v = .6 c . What would the external observer say the length of the ship is?



Let's try a different angle, you also have an error in elementary geometry: explain the y+sqrt(x^2+y^2) in your post 59.This is the total distance the light travels from the front of the ship to the main detector according to the ship observer. The light travels the distance y along the length of the ship to where the light is reflected, and then sqrt(x^2 + y^2) from that point to the main detector, which is located at the front of the ship at a distance of x across the width of the ship.

macaw
2009-Jun-18, 01:35 AM
This is what the Lorentz contraction is used for. Let's say you have an observer in a ship that measures the length of the ship to be ten meters long. You also have another observer outside the ship that sees the ship moving directly away at v = .6 c . What would the external observer say the length of the ship is?

You do not understand what mixing variables from two separate frames is. You do not understand relativity and I can't spend my time teaching you, you need to take a class.



This is the total distance the light travels from the front of the ship to the main detector according to the ship observer. The light travels the distance y along the length of the ship to where the light is reflected, and then sqrt(x^2 + y^2) from that point to the main detector, which is located at the front of the ship at a distance of x across the width of the ship.

It can't be. The two vectors point in different directions, so you can't add their lengths. This is elementary geometry, draw a diagram and you'll figure out your error.

grav
2009-Jun-18, 01:58 AM
You do not understand what mixing variables from two separate frames is. That is just accidently transferring the same thing that is measured in one frame to another when it does not apply in the same way. I have not done that.


You do not understand relativity and I can't spend my time teaching you, you need to take a class.Try using Lorentz contraction.


It can't be. The two vectors point in different directions, so you can't add their lengths. This is elementary geometry, draw a diagram and you'll figure out your error.It is finding the distance around the perimeter for two sides of a right triangle. That's all. The full perimeter would be x + y + sqrt(x^2 + y^2) with two perpendicular sides x and y.

macaw
2009-Jun-18, 02:51 AM
That is just accidently transferring the same thing that is measured in one frame to another when it does not apply in the same way. I have not done that.
No, it isn't.
You don't know relativity, so you would not be aware of the errors you are making.



It is finding the distance around the perimeter for two sides of a right triangle.

This is exactly my point, light travels in a straight line, not along a perimeter.