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When moving at the speed of light time stops

by eyad-996
Tags: light, moving, speed, stops, time
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Moonraker
#37
Jan4-13, 07:44 AM
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Quote Quote by Doc Al View Post
You do realize that the time dilation formula is just a special case of the Lorentz transformation, don't you? So if the LT doesn't apply, neither does the time dilation formula.
My opinion as well as yours needs to be proved. The fact that one formula is a special case of another does not exclude an extension of its field of application.

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.
Doc Al
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Jan4-13, 08:01 AM
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Quote Quote by Moonraker View Post
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.
The time dilation formula and the Lorentz transformations depend upon the basic assumptions of relativity, one of which is that the speed of light is invariant and equal to c in all frames. Applied to a frame co-moving with photon, such a statement is gibberish.
ghwellsjr
#39
Jan4-13, 09:09 AM
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Quote Quote by Moonraker View Post
My opinion as well as yours needs to be proved. The fact that one formula is a special case of another does not exclude an extension of its field of application.

The Lorentz transformation does not apply due to a division by 0. The time dilation formula does not share this problem.
Yes it does. Time Dilation means a time interval is getting larger for a moving object in a given reference frame. You have to divide by zero to find how long any interval is for a photon.
Quote Quote by Moonraker View Post
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.
In Special Relativity, Einstein defines time as that which a clock measures. A clock cannot be made out of just photons, it requires massive particles. Massive particles cannot travel at the speed of light. Therefore a clock cannot travel at the speed of light and there is no definition for time at the speed of light. It's a meaningless concept.
TomTelford
#40
Jan4-13, 10:10 AM
P: 13
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 sub-c.

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.
ghwellsjr
#41
Jan4-13, 10:39 AM
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Quote Quote by TomTelford View Post
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 sub-c.

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.
It's not just an issue of dividing by zero. Did you read my previous post or the rest of this thread? There is no meaningful definition for time applied to a photon. A photon has no experience of any kind. A photon has no frame. You need to read before you write.
nitsuj
#42
Jan4-13, 10:45 AM
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Quote Quote by TomTelford View Post

... a simple div/0 error should not stand in our way to understanding all of this.

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.
Moonraker
#43
Jan4-13, 11:17 AM
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Quote Quote by Doc Al View Post
The time dilation formula and the Lorentz transformations depend upon the basic assumptions of relativity, one of which is that the speed of light is invariant and equal to c in all frames. Applied to a frame co-moving with photon, such a statement is gibberish.
That is not precise:

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.
WannabeNewton
#44
Jan4-13, 11:21 AM
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Quote Quote by Moonraker View Post
2. Time dilation (and also Lorentz contraction) should not be confused with Lorentz transformation.
4. Applying time dilation and Lorentz contraction to photons is not gibberish but leading to clear positive results.
Time dilations and length contractions are special cases of a general lorentz transformation. I'm not sure how to respond to #4 because I don't see what your argument is in support of the (meaningless) concept of applying time dilation or length contraction "to photons". You just keep repeating that statement over and over without any physical justification.
Moonraker
#45
Jan4-13, 11:32 AM
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Quote Quote by ghwellsjr View Post
Yes it does. Time Dilation means a time interval is getting larger for a moving object in a given reference frame. You have to divide by zero to find how long any interval is for a photon.
The so-called time dilation formula compares proper time of two frames. The photon has a proper time, the observer has a proper time, they are comparable, and no time is divided by 0 (see above-mentioned formula). Sure, please do not climb onto the photon for measuring the time of the observer! This will not work (division by 0). The photon is not an inertial frame.

Quote Quote by ghwellsjr View Post
It's a meaningless concept.
For photons, their life time is an instant of 0 seconds, and in their frame space is contracted to zero. This is fitting harmoniously with the rest of the special relativity.
TomTelford
#46
Jan4-13, 11:40 AM
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Quote Quote by ghwellsjr View Post
It's not just an issue of dividing by zero. Did you read my previous post or the rest of this thread? There is no meaningful definition for time applied to a photon. A photon has no experience of any kind. A photon has no frame. You need to read before you write.
If not, then elaborate, but that seemed to be the point upon which this thread was piling up on.

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.
Nugatory
#47
Jan4-13, 11:46 AM
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Quote Quote by TomTelford View Post
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 "its" straight line all simultaneously is hugely interesting.
.....
I know I'm wandering into philosophy but a simple div/0 error should not stand in our way to understanding all of this.
When the math goes all screwy, that's the math trying to tell us that we're misusing it; and the way to advance our understanding is to stop torturing the math in an effort to force a confession out of it.

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 s-word, 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:
If two events have the same Minkowski t coordinate in some reference frame, then we say that they are simultaneous in that frame.

(footnote 1: I specified "Minkowski" because I don't want to go anywhere near the rathole of GR and simultaneity conventions and coordinate time. Please, please, please don't send this thread down that rathole? Please?

footnote 2: This definition works just fine for classical Newtonian physics as well; I'm not making up some weird non-intuitive definition of simulataneity here)
But note that according to this more precise definition of "at the same time time", the emission, passage, and absorption of a light signal is NEVER simultaneous.
ghwellsjr
#48
Jan4-13, 12:18 PM
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Quote Quote by Moonraker View Post
Quote Quote by ghwellsjr View Post
Yes it does. Time Dilation means a time interval is getting larger for a moving object in a given reference frame. You have to divide by zero to find how long any interval is for a photon.
The so-called time dilation formula compares proper time of two frames. The photon has a proper time, the observer has a proper time, they are comparable, and no time is divided by 0 (see above-mentioned formula). Sure, please do not climb onto the photon for measuring the time of the observer! This will not work (division by 0). The photon is not an inertial frame.
Frames don't have Proper Time, they have Coordinate Time. The Time Dilation formula compares the Proper Time of a material object to the Coordinate Time of an Inertial Reference Frame in which the object is moving at some speed. If you transform the coordinates of events from one IRF to another one moving with respect to the first one, you can get a different Time Dilation for the same object. Events associated with a photon will always transform between frames such that the photon continues to have a speed of c in all IRF's. Material objects can have different speeds in different IRF's and therefore different Time Dilations. The concept of Time Dilation is meaningless for photons. I can point you to numerous examples of how this works in other threads.

If you don't agree with this, then please show me an example of what you mean when you say "The so-called time dilation formula compares proper time of two frames."
Quote Quote by Moonraker View Post
Quote Quote by ghwellsjr View Post
It's a meaningless concept.
For photons, their life time is an instant of 0 seconds, and in their frame space is contracted to zero. This is fitting harmoniously with the rest of the special relativity.
The propagation speed of light (or photons) is fundamental (it's the second postulate) to Special Relativity. But, as I said before, a precise definition of time and space are also fundamental and your ideas of time and space applied to photons are not defined and do not fit harmoniously with the rest of SR nor is there any need for them. It's not like there is a hole in SR that needs to be filled in. The track you are going down will lead you astray to understanding SR. Furthermore, it is speculation that is not permitted on this forum and if you continue, you will likely get banned.
TomTelford
#49
Jan4-13, 12:33 PM
P: 13
>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.
ghwellsjr
#50
Jan4-13, 12:49 PM
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Quote Quote by TomTelford View Post
Quote Quote by ghwellsjr View Post
It's not just an issue of dividing by zero. Did you read my previous post or the rest of this thread? There is no meaningful definition for time applied to a photon. A photon has no experience of any kind. A photon has no frame. You need to read before you write.
If not, then elaborate, but that seemed to be the point upon which this thread was piling up on.
I did elaborate in my previous post. Here, read this:
Quote Quote by ghwellsjr View Post
In Special Relativity, Einstein defines time as that which a clock measures. A clock cannot be made out of just photons, it requires massive particles. Massive particles cannot travel at the speed of light. Therefore a clock cannot travel at the speed of light and there is no definition for time at the speed of light. It's a meaningless concept.
Quote Quote by TomTelford View Post
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.
The Lorentz transform does not define time at all. It was defined by Einstein in a two-step process. First, as I said before, it's what a clock measures at a particular location. Then to define time at a remote location a second clock is placed and the two clocks are synchronized using the definition of light propagating at c. This process, along with rigid (material) rulers define the concept of an Inertial Reference Frame. None of this can apply to a photon. And there is no need for it to. You are creating a problem where none exists. Furthermore, your proposal is against the rules that you agreed to when you signed on to this forum and I don't want to be part of this kind of activity. This forum is to learn relativity, not to add to it with your own personal concepts. I'm warning you to stop or you will likely get banned.
Quote Quote by TomTelford View Post
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 is not possible to observe a traveling photon or the propagation of any light, let alone, observe any changes. Where did you get the idea that there are "defined changes in its existence"? What do you mean by "The whole point is to attempt to rationalize change without time"? I doubt that you will get up to ten posts before you get banned.
Quote Quote by TomTelford View Post
It most definitely has a "frame" of some kind even if it is difficult to define mathematically.
Unless you can do it, how can you say that it can definitely be done? (I see a ban coming on.)
Quote Quote by TomTelford View Post
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.
I'm making a strong warning: stop this nonsense or you will get banned.
Quote Quote by TomTelford View Post
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.
Instead of apologizing, I suggest you delete all your posts quickly before you get banned so that you can continue to learn what relativity is all about. If you do it quickly enough, I will delete mine and hopefully others will too. Maybe you can still survive.
DaleSpam
#51
Jan4-13, 12:55 PM
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Quote Quote by Moonraker View Post
The “clock” of the photon shows 0 seconds, according to the above-mentioned time dilation formula:
T‘ = T * sqrt (1-v2/c2),
even if the Lorentz transformation does not apply to photons.
You have been given some good advice, which you seem reluctant to take, so let me try as well.

The Lorentz transform is (in part)
[tex]t=\frac{1}{\sqrt{1-v^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{1-v^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{1-v^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.

Quote Quote by Moonraker View Post
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.
I hope you agree that the above qualifies as a good reason not to apply the time dilation formula.
nitsuj
#52
Jan4-13, 12:59 PM
P: 1,103
Quote Quote by TomTelford View Post
>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.
In some respect it is, continue to imagine the photon doesn't experience time. It also doesn't experience length. To that end it is a dead end for geometry from the perspective of the photon.

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.
Nugatory
#53
Jan4-13, 01:17 PM
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Quote Quote by TomTelford View Post
Things are and are not at locations (or locations are or are not... very confusing).
Teach yourself to think in terms of points in spacetime (aka "events"), not times and locations (aka "coordinates"). It's natural to use locations and times to say something like "at ten minutes past the hour, on my lab table - the light signal hit the detector", but I could keep them out of the description by saying "the point where the path of the light signal through spacetime intersected the path of the detector through spacetime" instead.

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 left-hand 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.
nitsuj
#54
Jan4-13, 01:29 PM
P: 1,103
Quote Quote by Nugatory View Post
Teach yourself to think in terms of points in spacetime (aka "events"), not times and locations (aka "coordinates").
I think that's a perfect way to interpret these "distances". It's well said.


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