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Author Topic: Why can light traverse unequal lengths in an equal amount of time?  (Read 9950 times)

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #25 on: 29/10/2012 13:23:43 »
The condition of ‘no significant fringe’ for the experiment is true for all observers. So, the time remained equal for light to traverse both arms of unequal length judged from relative rest. True?

Nope.  From relative rest, the arms are of equal length, so the time is obviously the same.  In motion, they appear as different lengths, so the time is different.  The frequency will also change, which has to be accounted for.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #26 on: 29/10/2012 15:34:41 »
JP,

First and foremost, thank you for your time and expertise.

Secondly, as not to get bogged down by avoidable misunderstandings:

Quote
From relative rest, the arms are of equal length, so the time is obviously the same.
I take this to mean at rest with the apparatus in the relatively moving frame. Correct?

Quote
In motion, they appear as different lengths, so the time is different. The frequency will also change, which has to be accounted for.
I take this to mean judged by an observer in the frame that is at rest relative to the moving frame containing the apparatus. Correct?

Thank you,
Butch

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #27 on: 29/10/2012 16:11:58 »
There are two frames, K and K', moving with respect to each other.

The experiment is at rest if viewed from frame K'.

If an observer in frame K' observes the experiment, both arms appear to be equal, the light traveling down both arms seems to be the same frequency, and he sees no fringes.

If an observer in frame K observes the experiment, one arms will appear to be length contracted, so the time taken for the light to travel down back along the two arms will no longer be equal.  However, the frequency of the light down the two arms will no longer be equal due to the Doppler shift.  Since frequency multiplied by time determines the condition for a bright fringe, this observer will probably also see a bright fringe.  Again, I haven't worked out the math.

I'm not sure how much clearer I can be.  If you don't understand what I'm getting at here, perhaps someone else can explain better.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #28 on: 29/10/2012 16:36:19 »
Quote
I'm not sure how much clearer I can be.  If you don't understand what I'm getting at here, perhaps someone else can explain better.
You’re doing great! Can you please explain:

Quote
an observer in frame K' observes the experiment … sees no fringes.
If the apparatus did not produce fringe
Quote
an observer in frame K observes the experiment… see a bright fringe.
How could any observer see fringe from the apparatus?

Thanks,
Butch

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #29 on: 29/10/2012 17:40:51 »
I really don't know how to say it much clearer: frequency multiplied by time taken is what you measure in your Michelson interferometer.  Both time taken and frequency depend on the relative motion of the observer and experiment, so you need to take them into account.  If you don't include frequency, you're going to make wrong predictions.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #30 on: 29/10/2012 18:30:31 »
I am not trying to be nor want to appear to be argumentative.

My point is that the experiment did not produce significant fringe.

All observers see the same experiment.

No observer can see significant fringe from the experiment if the experiment didn’t produce significant fringe.

Do you see my point? Or, is there a flaw with my logic?

Thanks,
Butch

David Cooper

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #31 on: 29/10/2012 22:29:58 »
I can’t find any resources that discuss light projected in the direction of motion within a moving frame judged from relative rest. Do you or anyone else know of any? (I’m certain there aren't any)

That's because the light has to be observed by the observer at what you call relative rest in accordance with the speed limit on light in that observer's own frame. You cannot expect him to be able to see light moving at any other speed through his frame, but that's what you appear to be expecting him to do. As always, though, you can only measure the speed of light on a round trip, so what he observes isn't necessarily giving a true picture.

If an observer in frame K observes the experiment, one arms will appear to be length contracted, so the time taken for the light to travel down back along the two arms will no longer be equal.  However, the frequency of the light down the two arms will no longer be equal due to the Doppler shift.  Since frequency multiplied by time determines the condition for a bright fringe, this observer will probably also see a bright fringe.  Again, I haven't worked out the math.

No observer sees a bright fringe. The easiest way to resolve this thing is to put lots of semi-silvered mirrors into the experiment to reflect some of the light out sideways to the observer observing from another frame. When you do this, that observer will see the progress of the light through the apparatus travel in accordance with the speed limit for light of his own frame, and the frequency of the light will remain constant throughout from his point of view too, though it will be at a lower frequency than would be observed by an observer moving with the apparatus.

If instead of the semi-silvered mirrors we allow our stationary observer to stick his head into the beam as the apparatus goes past him, he will then see the frequency of the light change depending on which direction it's going in relative to the apparatus, but that has no bearing on what happens when the fringe is formed (or not formed) because at the location where that happens (or doesn't happen) we have two light beams merged together again and travelling along the same path with the same frequency.
« Last Edit: 29/10/2012 22:31:36 by David Cooper »

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #32 on: 30/10/2012 01:22:47 »
No observer sees a bright fringe
Actually, all observers see bright fringes.  The Michelson interferometer produces a fringe pattern.  The central fringe may be bright or dark depending on the relative optical path lengths (distance divided by frequency) in the two arms.

Quote
The easiest way to resolve this thing is to put lots of semi-silvered mirrors into the experiment to reflect some of the light out sideways to the observer observing from another frame. When you do this, that observer will see the progress of the light through the apparatus travel in accordance with the speed limit for light of his own frame, and the frequency of the light will remain constant throughout from his point of view too, though it will be at a lower frequency than would be observed by an observer moving with the apparatus.
The frequency shifts each time the light bounces off a moving mirror.  This is one of the ways that RADAR works to pick up the speed of targets.  If you're moving WRT the experiment, you're moving longitudinally with respect to one arm and transversely with respect to the other which seems like it should impart 2 different Doppler shifts along those two arms.  Similarly, the length contraction of those arms shouldn't be identical.

Quote
If instead of the semi-silvered mirrors we allow our stationary observer to stick his head into the beam as the apparatus goes past him, he will then see the frequency of the light change depending on which direction it's going in relative to the apparatus, but that has no bearing on what happens when the fringe is formed (or not formed) because at the location where that happens (or doesn't happen) we have two light beams merged together again and travelling along the same path with the same frequency.
I wouldn't be surprised if you added up the total path lengths for both paths and got equal answers for the moving and stationary observers (with respect to the experimental reference frame), but this argument isn't sufficient to prove it.  I completely agree that along the last path segment, both beams pick up the same phase shift, but the question is whether they pick up identical phase shifts along the prior segments for both the moving and stationary observers.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #33 on: 30/10/2012 16:24:16 »
Hypothetically, there is an indicator on the apparatus that illuminates for any condition other than ‘insignificant fringe’.

At the apparatus the indicator is never illuminated.

The indicator is never illuminated for any observer.

The condition of ‘insignificant fringe’ is true for all observers.

Is this logic realistic?

Thanks,
Butch

David Cooper

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #34 on: 30/10/2012 20:23:55 »
You certainly aren't going to get a different answer for different observers when it comes to the business of whether there's a null result or not. There will always be a null result, and at no point will anything show up that can show one frame to be a preferred frame (or closer to it) than any other.

The trouble with this whole business is that you can pick any frame and set the speed of light across the apparatus by it, thereby determining how the frequency must be changed at the mirrors and how far the light has to go through that frame to complete the trips along the arms, but all frames provide the same end result, no matter what happened along the way. By banning all discussion of simultaneous events at a distance, SR allows you to ignore the contradictions in the accounts generated by the analysis from the point of view of observers in different frames, but if you lift that ban you can actually create a version of SR which eliminates these and all the other contradictions normally found in SR simply by allowing there to be something similar to a preferred frame of reference (meaning that not all frames are equal): this still maintains the important difference between SR and LET by continuing to provide the Spacetime aspect of the model which is necessary for it to be used as a base for GR.

I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR. He is demanding that a stationary observer should be able to see light move through the moving apparatus at the speed of light relative to the frame of reference of the apparatus rather than to his own frame of reference, and he's demanding this on the basis of Einstein's claim that the speed of light is the same for all observers. At the same time, that stationary observer would obviously also have to be able to see light progressing through the moving apparatus at the speed of light relative to his own frame of reference, so contradiction is impossible to avoid. I think he's misinterpreting Einstein's claim, because although the contradictions are indeed there in SR, they aren't quite as overt as that.

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #35 on: 30/10/2012 21:34:10 »
I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.

David, the moderators have asked you many times to stop stating your opinions about SR as facts.  You have made many good contributions to the forum, so please don't make us moderate you further on this topic.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #36 on: 31/10/2012 10:57:32 »
Since the state of the hypothetical indicator is the same for all observers the frequency x time relationship of the light in the perpendicular arms of the apparatus is also the same for all observers.

Can there be any doubt that the time for light to traverse the perpendicular arms of the apparatus is the same for all observers?

Butch

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #37 on: 31/10/2012 15:14:48 »
Can there be any doubt that the time for light to traverse the perpendicular arms of the apparatus is the same for all observers?

Oh yes, the measured time for light to traverse both arms changes if you move with respect to the apparatus.  Why?  Because the light in the transverse arm has no apparent longitudinal motion when you're at rest, and your motion gives it longitudinal motion WRT you.  It travels a longer total path from your reference frame at a constant speed, so it takes longer.

I found this today.  It's basically the experiment you're setting up, with all the details of deriving how it works (and why you get a null result).  The frequency shift is important (and they account for it in that site via time dilation):
http://en.wikipedia.org/wiki/Kennedy%E2%80%93Thorndike_experiment

David Cooper

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #38 on: 31/10/2012 21:32:31 »
I think Butch is essentially trying to disprove SR on the basis of these contradictions, but they are just a side show - they demonstrate that there is some false dogma built into SR, but none of it is critical to the real functionality of SR.

David, the moderators have asked you many times to stop stating your opinions about SR as facts.  You have made many good contributions to the forum, so please don't make us moderate you further on this topic.

Sorry - I try to write everything carefully to avoid doing stating things directly as truths when I have no doubt that they are true, but I don't spot it every time because I'm hypersensitive to contradictions due to my work in A.I. where they're used to invalidate theories (or the parts of theories which generate those contradictions) - when I can see that something is wrong, I can't help seeing that it is wrong. It in an incontrovertible fact that contradictions are generated by SR in its standard form, but I don't see why this is such a big problem for you or for science as SR does not depend on the parts which generate those contradictions for any of its vital functionality. I don't see why it should cause you to switch into suppression mode whenever that comes to the surface. It will inevitably keep coming to the surface because there are millions of people out there who reject SR (and even science as a whole) on the basis of these unnecessary contradictions. In the near future, you're also going to have AGI systems making exactly the same point in posts to this forum (and to every other science forum), and they won't be programmed to make incorrect objections: they'll simply be analysing the theories and commenting on them in accordance with their rational analysis of them. Contradictions aren't good - they need to be eliminated if you are genuinely seeking truth.

[Edit was to changed "falsify" into "invalidate" - don't know why I used the wrong word there and failed to pick up on it while checking text after posting.]
« Last Edit: 01/11/2012 21:00:26 by David Cooper »

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #39 on: 31/10/2012 23:07:13 »
David, I have nothing against discussions of what constitutes good science and discussing interpretations of the mathematics of theories, but I'm feeling like a broken record on this point: let's keep such discussions to New Theories.

Future posts going down this path will be moved to New Theories without discussion.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #40 on: 01/11/2012 21:03:00 »
I looked at The Kennedy–Thorndike experiment and it became clear my original question was inadvertently misleading. The question should have been:

“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”

This is the question that I cannot find an answer for. Sorry for the confusion, JP.

Thank you,
Butch

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #41 on: 02/11/2012 01:23:17 »
I thought I had stated this clearly earlier, but reviewing my prior posts, I didn't.

“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”

It doesn't.  They take different times.  The reason you still get a null result of the experiment is explained in that link.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #42 on: 04/11/2012 02:28:20 »
Quote
It doesn't.  They take different times.  The reason you still get a null result of the experiment is explained in that link.
I understand that interferometer arms/light paths of unequal length can be set to lengths that result in the condition of no significant fringe notwithstanding their unequal lengths and light traverses those unequal lengths in different amounts of time.

Me:
Quote
“Why can light traverse the non-contracted length of a light path in the same amount of time light traverses the contracted length of that light path?”
Notice it refers to the non-contracted and contracted lengths for that (singular) light path. I am not referring to an interferometer for which the arms set to unequal lengths. I am referring to a non-contracted length and a length contracted by Lorentz–FitzGerald contraction.

To be clear, my question is in reference to the MMX only for which the arms are set to an equal length within the frame of the apparatus.

I sense somewhat understandable impatience that I think is due to a misunderstanding of the question and its purpose. The facts as I know them lead to an inconsistency in the interpretation of the null result of the MMX. I need to know whether my facts are correct or not and if not where my misunderstanding is. Additionally, I am unable to locate resources that address the properties of light in light paths contracted by Lorentz–FitzGerald contraction, the heart of my quandary. I’ve asked the Forum previously and I am asking again if anyone knows where I can find this information or credibly inform me that no such information exists. That is the purpose and, hopefully, an insight into the question.

That said, following is proof of the legitimacy and further clarification of my question with facts, as I know them. There is one caveat. I’m almost certain there is a credible reference to the following statement but I can’t locate it at the moment. Until I confirm the source the following should be presumed an assumption. “An event observed by any observer in observable by all observers.”

Definitions:
Frame K’ – the relatively moving inertial frame which contains the apparatus of the MMX
Frame K – the inertial frame, which is at relative rest

For the MMX its two arms, which are light paths, were set to an equal length. It was verified at the apparatus that light took an equal amount of time to traverse the perpendicular arms/light paths.

At the apparatus each light wave from the monochromatic source was split by a half-silvered mirror and produced a pair of light waves. One of each pair of the resultant light waves was projected onto each of the two perpendicular arms/light paths of the apparatus. Each pair of light waves was reflected directly back to the half-silvered mirror and recombined. The pair of light waves that resulted from the split at the half-silvered was the same pair of light waves that were recombined at the half-silvered mirror. For the same pair of light waves that was split from a single light wave to arrive back at the half-silvered mirror at the same time and recombine perfectly the light waves must traverse their respective arms/light paths in an equal amount of time. Since the split pairs of light waves arrived back at the half-silvered mirror at the same time at the apparatus and recombined, they arrived back at the half-silvered mirror at the same time for all observers. THE ASSUMPTION: *An event observed by any observer in observable by all observers. The event is the recombining of the same pair of light waves that resulted from the split.*

Therefore, the time for the pair of light waves to traverse the perpendicular arms/light paths was equal for all observers to include an observer in frame K that was at relative rest. But, for relative velocities >0<c for the relatively moving frame K’ which contains the apparatus, the arm/light path of the apparatus which is in the direction of motion is contracted by the Lorentz–FitzGerald contraction factor and the arm/light path perpendicular to the direction of motion is not contracted judged from frame K, which is at relative rest.

So, how can light traverse the contracted and the non-contracted arms/light paths of the apparatus in an equal amount of time judged from frame K, which is at relative rest?

Butch

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #43 on: 04/11/2012 05:26:38 »
The above distilled to its essence:

A light path can contract due to Lorentz contraction.
The light in that light path does not contract.

Can those two be reconciled?

Butch

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #44 on: 04/11/2012 13:08:57 »
I don't know how else to say this:

The light does contract, in a sense, since it's frequency shifts due to time dilation.  This is sufficient to make observers who are in K and K' agree on the results of the experiment.  To agree, the same number of wavelengths have to exist down both paths.  The paths contract, but time dilation (or if you prefer, length contraction) keeps the total number of wavelengths in each path the same.

If that doesn't make sense still, I suggest you read up a bit on how time dilation and length contraction are related, especially on that wiki link I provided previously.  Or perhaps someone else here can explain it more clearly.

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #45 on: 04/11/2012 18:26:48 »
JP,
Thank you so much for your time and patience.

The dilemma stated more exactly:

Think of this light path as the arm of the MMX that’s in the direction of motion. The proper length of this light path and the length of this light path contracted by Lorentz contraction are not equal.

Light traverses these unequal lengths in an equal amount of time.

Above, the constant speed of light and Lorentz contraction obviously conflict. AT LEAST ONE OF THE TWO IS INVALID!

Albert Einstein (1879–1955).  Relativity: The Special and General Theory.  1920.
XIV.  The Heuristic Value of the Theory of Relativity
http://www.bartleby.com/173/14.html
Quote from Einstein:
“OUR train of thought in the foregoing pages can be epitomised in the following manner. Experience has led to the conviction that, on the one hand, the principle of relativity holds true, and that on the other hand the velocity of transmission of light in vacuo has to be considered equal to a constant c. By uniting these two postulates we obtained the law of transformation for the rectangular co-ordinates x, y, z and the time t of the events which constitute the processes of nature. In this connection we did not obtain the Galilei transformation, but, differing from classical mechanics, the Lorentz transformation.
The law of transmission of light, the acceptance of which is justified by our actual knowledge, played an important part in this process of thought. Once in possession of the Lorentz transformation, however, we can combine this with the principle of relativity, and sum up the theory thus:
Every general law of nature must be so constituted that it is transformed into a law of exactly the same form when, instead of the space-time variables x, y, z, t of the original co-ordinate system K, we introduce new space-time variables x', y', z', t' of a co-ordinate system K'. In this connection the relation between the ordinary and the accented magnitudes is given by the Lorentz transformation. Or, in brief:General laws of nature are co-variant with respect to Lorentz transformations.
This is a definite mathematical condition that the theory of relativity demands of a natural law, and in virtue of this, the theory becomes a valuable heuristic aid in the search for general laws of nature. If a general law of nature were to be found which did not satisfy this condition, then at least one of the two fundamental assumptions of the theory would have been disproved.”

Someone at Cambridge should probably look into this.

The world of physics will change starting NOW.

Butchmurray
Thorntone E. Murray

JP

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #46 on: 04/11/2012 19:15:45 »
Butch, please don't use the mainstream fora to suggest that special relativity is wrong.  The New Theories forum is the proper place for such suggestions.

Thanks!

On the physics:

Regarding your claims: your assumptions are wrong, and therefore so are the conclusions you draw from them.  Time dilation and length contraction go hand in hand, and when you're trying to analyze the Michelson-Morely experiment from a moving reference frame, you need to account for them both.  It's the number of wavelengths of light that exist in each arm that matters, and that depends on both the frequency of the light and length of the arms.  If you're in a moving frame WRT the experiment, both of those change, not just the length.

Claiming that the length shifts is true.  Claiming that the length is the only thing that matters in the MM result is false.  Therefore, you're reaching incorrect conclusions.

I really can't be more clear than this.  Perhaps someone else can explain it better.  If not, I'd suggest reading up on time dilation and the Doppler shift.  It really does explain this effect fully.
« Last Edit: 04/11/2012 21:03:30 by JP »

butchmurray

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #47 on: 04/11/2012 20:55:13 »

Thank you,
Butch

sciconoclast

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Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #48 on: 04/11/2012 21:04:06 »
This might help:
Instead of stating that light appears to travel at the same speed to any observer try stating that light within any frame appears to travel at the same speed to an observer in that frame the same as light in any other frames appears to travel to observers in those frames.

The frame in most of these experiments is the Earth's gravitational field. The Sagnac experiment in which light is sent through mirrors on a turntable that represents motion independent of that of the Earth's gravitational field results in different arrival times.

The Michelson-Gale experiment, on the other hand, which uses the Earth as a turntable does not produce a phase shift.

Einstein predicted the results for both these experiments incorrectly. Later they were deemed to be in agreement with relativity. It is often said that Einstein was the worst interpreter of his own theories.

An interesting anomaly to relativity occurred when there were two separate rectangular paths with one having a larger portion parallel to the Earth"s rotation and the other a larger portion perpendicular to it. Within each course there is no phase shift. However, when the results of one are compared to the other there is a difference. This enabled the experimenters to calculate the rotational velocity of Earth which agreed with that calculated from the star field.

Getting back to the original question, the distances that appear to be the same to an observer within a frame will also appear to be traveled by light in the same amount of time to the same observer. However, to any observer in space that is not rotating with the Earth the distances will appear different and if he, or she, or it, is sent the arrival times the speed of light will appear to have varied.

Just ask the guys working with orbiting communication net works.

The Naked Scientists Forum

Re: Why can light traverse unequal lengths in an equal amount of time?
« Reply #48 on: 04/11/2012 21:04:06 »