When moving at the speed of light time stops

[Dale]

I suppose that neutrino's speed in neutrino's frame is not higher than light speed because the distance sun-earth is diminishing in neutrino's frame as well? (according to Lorentz transformation)

The neutrinos speed in the neutrinos frame is 0.

 

[Moonraker]

These are not times measured from the (nonexistent) frame of a photon, they are measured from other quite ordinary frames of reference.

No, the moment of emission is the same for the observer and for the photon.
At the absorption, the clock of the observer shows 8 minutes (Sun-Earth). 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.

 

[Moonraker]

The neutrinos speed in the neutrinos frame is 0.

This would mean that the distance is 0 as well in the neutrino's frame.

 

[Doc Al]

No, the moment of emission is the same for the observer and for the photon.
At the absorption, the clock of the observer shows 8 minutes (Sun-Earth). 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 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.

 

[Dale]

This would mean that the distance is 0 as well in the neutrino's frame.

The distance the neutrino travels is indeed 0 in the Neutrinos frame. Note that that is not the same as the distance between the Earth and the sun.

 

[Nugatory]

even if the Lorentz transformation does not apply to photons.

Be careful here... The Lorentz transforms never apply to things, be they photons moving at the speed of light in all frames, or massive particles and observers moving at less than the speed of light in all frames. They apply to the coordinate systems that we use to assign (x,t) values to events such as "the thing is at point x at time t"

 

[Moonraker]

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]

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]

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.

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]

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 traveled 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]

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 traveled 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]

... 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]

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]

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]

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.

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]

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]

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]

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."

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]

>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]

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:

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.

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.

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.

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.)

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.

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.

 

[Dale]

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)
t=\frac{1}{\sqrt{1-v^2/c^2}}\left( t' - \frac{vx'}{c^2}\right)
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:
t&#039;=t\sqrt{1-v^2/c^2}
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 \tau = t&#039;, so we can easily make that substitution when we make the simplification, giving: \tau=t\sqrt{1-v^2/c^2}, where t is still a coordinate time but now \tau is a proper time. But even so, -c<v<c.

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]

>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]

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]

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.

 

[PAllen]

There is a sense in which √(1-v^2/c^2) does apply to a photon. Time dilation is normally defined as the ratio of proper time to coordinate time for path. In this sense, the given factor is just a direct consequence of the metric for any inertial frame. Used for a photon in any inertial frame (not a photon frame), it just expressed the fact that a photon path is a null path - proper time is zero along a photon path. This much is valid. Where everything would break down is talking about the 'point of view' = frame of a photon; or distances as seen by a photon. Also, the normal meaning of proper time breaks down as gwellsjr points out - you cannot imagine a clock moving with the photon. However, as a geometric quantity, it is not only valid but required to speak of proper time=0 over a photon path.

 

[TomTelford]

>Nugatory - Thank you... will chew on this for a while.

 

[Dale]

There is a sense in which √(1-v^2/c^2) does apply to a photon. Time dilation is normally defined as the ratio of proper time to coordinate time for path. In this sense, the given factor is just a direct consequence of the metric for any inertial frame. Used for a photon in any inertial frame (not a photon frame), it just expressed the fact that a photon path is a null path - proper time is zero along a photon path. This much is valid. Where everything would break down is talking about the 'point of view' = frame of a photon; or distances as seen by a photon. Also, the normal meaning of proper time breaks down as gwellsjr points out - you cannot imagine a clock moving with the photon. However, as a geometric quantity, it is not only valid but required to speak of proper time=0 over a photon path.

I am a little bit reluctant to identify the spacetime interval along a null path with proper time.

If you have some path with a negative interval squared then that represents the proper time along the path. If you have some path with a positive interval squared then that represents the proper length of the path.

If you have a null interval then why would you identify that with proper time rather than proper length? No clock can follow that worldline, nor any ruler, so either way seems to be a stretch.

I think that I would probably just call it a null spacetime interval and not try to identify it with either proper time or proper length.

 

[PAllen]

I think that I would probably just call it a null spacetime interval and not try to identify it with either proper time or proper length.

I'll buy that. [In an inertial frame in SR:] Along a spacelike path, you can integrate √(1-c^2/v^2) dx to get proper length (v being strictly a coordinate derivative, not a physical quantity; treat as infinite when dt/dx=0). Along a timelike path you integrate √(1-v^/c^2)dt to get proper time. Along a null path, either one leads to zero interval.

 

[ghwellsjr]

There is a sense in which √(1-v^2/c^2) does apply to a photon. Time dilation is normally defined as the ratio of proper time to coordinate time for path. In this sense, the given factor is just a direct consequence of the metric for any inertial frame. Used for a photon in any inertial frame (not a photon frame), it just expressed the fact that a photon path is a null path - proper time is zero along a photon path. This much is valid. Where everything would break down is talking about the 'point of view' = frame of a photon; or distances as seen by a photon. Also, the normal meaning of proper time breaks down as gwellsjr points out - you cannot imagine a clock moving with the photon. However, as a geometric quantity, it is not only valid but required to speak of proper time=0 over a photon path.

I believe you are referring to the spacetime interval which provides another equivalent definition for Proper Time.

If two events define a time-like spacetime interval, that means an inertial clock can be present at both events and will measure out the time-like spacetime interval, because it is a time interval and it is the Proper Time accumulated on the clock. We can think of the clock as being at a fixed location in a frame and the time is also the coordinate time delta between the two events.

If two events define a space-like spacetime interval, that means an inertial ruler can be present at both events in a frame in which both events occur at the same time and the ruler will measure out the space-like spacetime interval, because it is a distance and it is the Proper Distance measured by the ruler and it is also the coordinate location delta between the two events.

If two events define a light-like spacetime interval, that means that there is no frame in which an inertial clock can be present at both events nor is there a frame in which a ruler can be simultaneously present at both events. But a photon can be present at both events and it doesn't matter what frame is used.

When you consider the coordinates of two events where one of them is changing such that a space-like spacetime interval approaches zero and then hits zero and crosses over to a time-like spacetime interval, to attach the meaning of the zero-crossing to either a clock (time-like) or a ruler (space-like) misses the point. That is why it is also called a null interval--it is not merely a zero time interval any more than it is a zero distance interval or some hybrid of the two. It is neither. It is in a class all by itself, the class that only applies to light.

EDIT: I see DaleSpam got in essentially the same points while I was composing my post.

 

[Nugatory]

Where everything would break down is talking about the 'point of view' = frame of a photon; or distances as seen by a photon..

Don't pay any attention to those people who want you to have a different reference frame for every observer. That just leads to unnecessary confusion. Any IRF can handle all the observers and all the objects

+1x2

If it were up to me, people would have to pass a test and be licensed to use the words "<something>'s reference frame"; until then they would be required to always say "a reference frame in which <something> is at rest".

Likewise, only license holders would be allowed to use the words "in a" in front of "reference frame" (not that they'd be likely to); the unlicensed would be required to say "working with coordinates calculated in" a reference frame instead.

Think of the possibilities...
WHOOP! WHOOP! WHOOP! <flashing blue lights>
"May I see your license, sir?"
"Officer, what's the problem? I just wanted to ask about the reference frame of a photon!"
"Sir, your license is restricted. You will have to restate your question, or I will forced to charge you with a license violation"
"What a stupid silly pedantic rule!"
"Sir, I am warning you that you are in violation of the law"
"OK, OK, I don't want to [STRIKE]go to jail[/STRIKE] be banned... I'll say it your way! I just want to ask about a reference frame in which a photon is at rest... Oh... wait... That does sound rather silly, doesn't it?"
"Yes, sir. That's why we do the license checks. I'll just give you a warning this time, but please [STRIKE]do drive more carefully[/STRIKE] post more precisely in the future."