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When moving at the speed of light time stops 
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#37
Jan413, 07:44 AM

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The Lorentz transformation does not apply due to a division by 0. The time dilation formula does not share this problem. In other words: We get no information about proper speed and distances for photons, and there is even no inertial frame of photons, but we get information about the proper time of photons. There is no reason not to apply the time dilation formula (if there is please let me know), and there are many reasons in favor of application. 


#38
Jan413, 08:01 AM

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#39
Jan413, 09:09 AM

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#40
Jan413, 10:10 AM

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If I may...
It might be semantics but it could be more appropriate to say that time comes into existence when "things" move slower than the speed of light rather than saying time "freezes" at c. That time itself is related to this slowness and things that move subc. While the math may go all screwy when t'=0, the concept that light comes into existence, is absorbed/transformed and passes through the points in between on "it's" straight line all simultaneously is hugely interesting. There are implications on a photon's behaviour since it is limited to experiencing it's entire existence with no time even though we observe it to have travelled for potentially billions of years. That the conditions that allow a photon to be created and absorbed in it's frame must be correlated to those in ours in order for us to experience light(radio, gamma, etc) at all. I know I'm wandering into philosophy but a simple div/0 error should not stand in our way to understanding all of this. 


#41
Jan413, 10:39 AM

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#42
Jan413, 10:45 AM

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It doesn't Read what that simple math is saying. In fact take it a step further, and consider what the measured phenomena is. geometrically the div/0 error makes sense...and...yup...we're still discussing geometry, not trying to identify what nothing is. 


#43
Jan413, 11:17 AM

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1. Lorentz transformation depends upon the basic assumptions of relativity, one of which is that the speed of light is invariant and equal to c in all inertial frames. However, a photon may be considered as a frame but not as an inertial frame. 2. Time dilation (and also Lorentz contraction) should not be confused with Lorentz transformation. 3. Do not confuse the case v>c with v=c. v>c is ruled by a general principle, v=c is "only" ruled by the problem of division by 0. 4. Applying time dilation and Lorentz contraction to photons is not gibberish but leading to clear positive results. 


#44
Jan413, 11:21 AM

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#45
Jan413, 11:32 AM

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#46
Jan413, 11:40 AM

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Yes and yes, and many others on this topic. I disagree. We may not have a transformation function that produces a singular result but that is not necessarily a restriction on the term "meaningful definition". What I am proposing is that if the Lorentz transform cannot "define" time "meaningfully" then other methods should be developed. These terms have a natural ambiguity so ask if I am using them a certain way before telling me I'm wrong, thank you. We must allow a photon to have an "experience" as there are defined changes in its existence which can be correlated to observed changes. The whole point is to attempt to rationalize change without time. It most definitely has a "frame" of some kind even if it is difficult to define mathematically. I have read a great deal, thank you, on a great many subjects. You need to know when to make suggestive commentary and when not. Now, to reiterate: I was suggesting that instead of assuming a frame that includes time as the "normal" case and attempting to understand the "timeless" state of a photon we could attempt the exercise from the opposite direction. It could be that time itself is somehow anomalistic. If I was not clear, I apologize. 


#47
Jan413, 11:46 AM

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We got here in two steps. First, a previous poster has tried plugging the value v=c into the time dilation formula, without recognizing that the time dilation formula is derived from the relationship between the x and t coordinates in one frame to the x' and t' coordinates in another frame moving at a speed of less than c relative to the first frame. The result is seductive but altogether meaningless nonsense. The second wrong step (easy to take, because the nonsense of the first step is indeed very seductive) came when you allowed the sword, in boldface above, to lead you further astray. We all agree (I hope) that "simultaneous" means "at the same time"; but what exactly does THAT mean? Here's a slightly oversimplified definition: 


#48
Jan413, 12:18 PM

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If you don't agree with this, then please show me an example of what you mean when you say "The socalled time dilation formula compares proper time of two frames." 


#49
Jan413, 12:33 PM

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>nitsuj  Exactly. Which is what I have been pondering for quite some time. I have been looking at this time/no time dilemma from the perspective of an (infinitely) many to one relationship that occurs in data analysis from time to time. There are methods of reconciliation and I'm trying to apply them in this instance although my knowledge level in physics and related maths is not yet strong enough... getting there. It does indeed make sense at least as far as I have been able to poke at it. But there are implications of "no time"; it isn't simply a dead end.
>Nugatory  Agreed, that math seemed not to be applicable. ... the "S" word huh? Okay. I didn't know how else to express what I was thinking. However even if "simultaneous" means as much to a photon as time itself, it does permit a certain view of our frame from the photon's perspective which does involve simultaneity. Time may not be defined for the photon but location is not defined for us from it's perspective. Things are and are not at locations (or locations are or are not... very confusing). I'm not sure if this relates to the idea of d=0 which I am very skeptical of. Any readings in this direction would be appreciated. 


#50
Jan413, 12:49 PM

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#51
Jan413, 12:55 PM

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The Lorentz transform is (in part) [tex]t=\frac{1}{\sqrt{1v^2/c^2}}\left( t'  \frac{vx'}{c^2}\right)[/tex] In this equation t and t' are coordinate times in the two frames and v is the relative velocity between the primed and unprimed frames. Since both frames are inertial this v is constrained by c<v<c. When you make the simplification x'=0 then you get the equation you posted above: [tex]t'=t\sqrt{1v^2/c^2}[/tex] Here v is still the relative velocity between the primed and unprimed frames. The simplification does not change that in any way, so it is still constrained by c<v<c. It isn't a matter of whether or not the equation has a division by zero or not, it is simply that doing a little algebra does not change the valid domain of a variable. Also, t and t' are both coordinate times, not proper times, so they remain coordinate times after the simplification. However, this is a rather minor point. For a clock at x'=0 we have [itex]\tau = t'[/itex], so we can easily make that substitution when we make the simplification, giving: [itex]\tau=t\sqrt{1v^2/c^2}[/itex], where t is still a coordinate time but now [itex]\tau[/itex] is a proper time. But even so, c<v<c. 


#52
Jan413, 12:59 PM

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The only implication I intuit for "no time" in a continuum is no geometry. Note the difference between no time & time stops (edit: oppps i guess you know the difference). Perhaps it is better said there is no observable time at c, as opposed to implying it exists but merely doesn't continue "ticking" for that particular "object" compared to me. 


#53
Jan413, 01:17 PM

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This approach is more cumbersome (enough that we tend not to use it except when necessary) but enormously helpful when the more traditional view is confusing. The value of this approach is that the events, the relationship between them, and the distances between them are the same  gloriously and beautifully and simply the same  no matter whose notions of position and time you use to attach numbers to them. All of this "at the same place/not at the same place" confusion never even enters into the picture. Of course, once you have attached numbers to an event ("ten minutes past the hour, six inches above the lefthand corner of my desk") you can use the Lorentz transforms to convert your numbers/coordinates into someone else's numbers/coordinates; and you can calculate stuff like time dilation and length between you and someone else by comparing the differences between your time and space coordinates for two events and their time and space coordinates. However, this recipe cannot be applied to a "someone else" who is moving at the speed of light relative to you  there's no such thing, the Lorentz transforms generate nonsense if you pretend that there is, and if you take the nonsense seriously you'll be back to being confused. 


#54
Jan413, 01:29 PM

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