When moving at the speed of light time stops

[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."

 

[Dale]

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".

You could get points off your license for saying two things happens simultaneously without specifying the reference frame, and similarly for specifying a length or a time. So many points and your license is revoked.

 

[pervect]

+1x2

"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?"

Laughing out loud here!

Unfortunately, it seems that "reference frames" are the source of a lot of confusion. And the confusion is so basic that it seems hard to get people unconfused about it.

 

[h1010134]

OK from what I understand, the question here is why light from the sun doesn't just magically appear on Earth after it is emmited, since it's moving at the speed of light, yes? The thing is, time dilation occurs for only the photon, and not for the observers on Earth looking at the photon. If a classic relativity example is used, this is akin to an observer experiencing normal time, while a person in a fast-moving vehicle (say a spaceship) experiences time dilation. E.g. 1 second will stilll be 1 second to the observer, whereas to the guy in the ship, 1 second would be maybe 1.5 seconds, depending on how fast he is moving. The point I'm trying to state here is that time does not change or dilate for the stationary observer

For photons, this effect is merely magnified enormously, whereby the photon would experience an infinite time dilation, and essentially time would stop for the photon. Nothing changes, however for the observer and hence everything continues as per normal. If you still don't understand, try to imagine if photons from the sun didn't move at c, maybe at 1m/s lower than the speed of light (this is absurd but bear with me). Without hitting that limit, we treat the photon as a classical object, and the problem resolves itself, yes? It would take a certain amount of time for the photon to reach Earth. Then, we add the extra second, and we (or maybe just me) realize that it would be absurd to think that with that extra m/s, the photon magically would be able to bypass this time duration and instead instantaneously appear on Earth.

As an extra, I'd just like to say that Einstein himself said that instantaneous time travel is impossible, since that would mean the thing traveling instantaneously would be bypassing the speed of light.

 

[ghwellsjr]

OK from what I understand, the question here is why light from the sun doesn't just magically appear on Earth after it is emmited, since it's moving at the speed of light, yes?

But light from the sun does appear magically on Earth. It's not like a spaceship coming from the sun at slightly less than the speed of light where we can actually watch it being launched at the sun and we can watch it traverse the space between the sun and the Earth and then watch its arrival on the Earth. If the spaceship had a clock that was visible to us on the Earth, we could see the same time on the clock that the riders see at launch and we could see the hands on the clock progress through the same time that the riders would see it progress during the trip and we would see the same time on arrival that they would see. However, we would see all this happening at a much greater rate than the riders would see it happening.

But not so with a photon. We cannot observe its emission from the sun as a separate event from its arrival. We cannot observe its progress as it travels the space between the sun and the Earth. All we can observe is the instant it actually arrives on Earth.

The thing is, time dilation occurs for only the photon, and not for the observers on Earth looking at the photon.

Time dilation does not occur for the photon and, as I just said, we cannot watch the photon's progress on its trip from the sun to the Earth. To us, it all appears instantly.

If a classic relativity example is used, this is akin to an observer experiencing normal time, while a person in a fast-moving vehicle (say a spaceship) experiences time dilation. E.g. 1 second will stilll be 1 second to the observer, whereas to the guy in the ship, 1 second would be maybe 1.5 seconds, depending on how fast he is moving. The point I'm trying to state here is that time does not change or dilate for the stationary observer

When you talk about Time Dilation, you must state what Inertial Reference Frame (IRF) you are referring to. (Didn't you read the previous posts on this subject?) Time Dilation is not something that can be observed because it is always in reference to a particular IRF and different IRF's assign different Time Dilations to each observer/clock/object. If you are assuming an IRF in which the sun and the Earth are at rest, then there is no Time Dilation for people on Earth that are also at rest in the IRF. People on the spaceship and their clocks will both be Time Dilated by the same amount so they will not be able to tell that you are using an Earth-sun based IRF. If you are using the IRF in which the spaceship is at rest, then the riders on the spaceship and their clocks will not be Time Dilated but those of us on the Earth and our clocks will be.

For photons, this effect is merely magnified enormously, whereby the photon would experience an infinite time dilation, and essentially time would stop for the photon.

No, Time Dilation does not apply for photons because time does not apply for a photon. There is nothing different for a photon when you use different IRF's. The one thing that applies to a photon in an IRF is its speed which is always c. All material objects traveling at less than c travel at different speeds in different IRF's and experience different Time Dilations. This effect is meaningless for a photon. Please read the rest of this thread.

Nothing changes, however for the observer and hence everything continues as per normal. If you still don't understand, try to imagine if photons from the sun didn't move at c, maybe at 1m/s lower than the speed of light (this is absurd but bear with me). Without hitting that limit, we treat the photon as a classical object, and the problem resolves itself, yes?

What problem?

It would take a certain amount of time for the photon to reach Earth. Then, we add the extra second, and we (or maybe just me) realize that it would be absurd to think that with that extra m/s, the photon magically would be able to bypass this time duration and instead instantaneously appear on Earth.

If something travels at less than c on its trip from the sun to the Earth, then it will take a certain amount of time for us to observe its progress. But since photons travel at c, we cannot observe their progress and their trip appears to take zero time. But this has nothing to do with Time Dilation. There is no Time Dilation that needs to be bypassed.

As an extra, I'd just like to say that Einstein himself said that instantaneous time travel is impossible, since that would mean the thing traveling instantaneously would be bypassing the speed of light.

Don't you mean it would be traveling at the speed of light and not bypassing it?

 

[phinds]

... this is akin to an observer experiencing normal time, while a person in a fast-moving vehicle (say a spaceship) experiences time dilation. E.g. 1 second will stilll be 1 second to the observer, whereas to the guy in the ship, 1 second would be maybe 1.5 seconds, depending on how fast he is moving. The point I'm trying to state here is that time does not change or dilate for the stationary observer

Just in case ghwellsjr didn't make it completely clear to you, this is incorrect. The traveler on the spaceship does not experience any time dilation. For him, his clock ticks away at one second per second, but he sees Earth as time dilated. This is becase ALL obervers are stationary in their own frame of reference. There are no stationary observers in any absolute sense.

 

[TomTelford]

Just seeking clarity on one point.

Case: Ship leaves non-gravitational location at relativistic speed. Both the observer at the origin and ship passenger see a clock in the other IRF as running slower, correct?

In SR is this solely a Doppler shift effect?

If the reverse is happening where a ship is approaching a non-massive location then do the clocks appear to be running faster?

BTW thank you gentlemen(women?) for clarifying why light cannot have an IRF... I am trying not to misuse terminology. :-)

 

[phinds]

Just seeking clarity on one point.

Case: Ship leaves non-gravitational location at relativistic speed. Both the observer at the origin and ship passenger see a clock in the other IRF as running slower, correct?

yep, you got it.

 

[PAllen]

Just seeking clarity on one point.

Case: Ship leaves non-gravitational location at relativistic speed. Both the observer at the origin and ship passenger see a clock in the other IRF as running slower, correct?

correct

In SR is this solely a Doppler shift effect?

No, even if you correct for Doppler based on known speed of relative motion and speed of light (but not using relativistic Doppler formula), you find an additional slowdown by a factor of 1/√(1-v^2/c^2)

If the reverse is happening where a ship is approaching a non-massive location then do the clocks appear to be running faster?

Yes, they visually appear faster. However, just as with separation, if you correct for Doppler in the obvious way, each computes a slowdown of the other by the same factor as above.

 

[Nugatory]

Just seeking clarity on one point.

Case: Ship leaves non-gravitational location at relativistic speed. Both the observer at the origin and ship passenger see a clock in the other IRF as running slower, correct?

In SR is this solely a Doppler shift effect?

If the reverse is happening where a ship is approaching a non-massive location then do the clocks appear to be running faster?

We can do "leaving" and "approaching" all in one thought experiment by doing the approaching first: Approach, pass, and keep on going, and now we're moving away.

Let's agree to use a strobe light that flashes periodically (and at a rate of once per second according to an observer at rest relative to it) as a standard clock, and equip all observers with these clocks.

If you mean "see" and "appear" literally - you have a telescope focused on the clock and you are counting the pulses and comparing their rate with your local clock - then yes, you will see the [STRIKE]moving clock ticking fast[/STRIKE] pulses arriving at a rate greater than once per second as you approach the other ship and then slow as you pass it and are moving away from it. This is Doppler at work. (Because the same description works if we say that we're at rest and the other ship is racing towards and past us in the other direction, for the rest of this post I'll try very hard not to refer to one of the ships as moving and the other at rest; there are just two ships moving relative to one another).

But note that you are not making a direct observation of the moving clock (and therefore the struck-out text in the previous paragraph is a subtle but dangerous misstatement). You are making direct observations of light pulses arriving at your telescope, right under your nose.

It's not possible for me to directly observe the time dilation of the other clock but there is a procedure that I can use to indirectly observe it (this procedure is how the cosmic ray muon decay observations are justified, for example). Ahead of time, long before the two ships are starting their approach, I station a very large number of observers all along the line of travel between me and the other ship. Each of these observers is at rest relative to me, and each one is carrying their own synchronized clock. Note that the clock synchronization is only possible because they're all at rest relative to me and each other; we could say that they're all part of the same inertial reference frame, although I'd rather say that they collectively define an irf. Eventually the other ship will pass by each of these observers, and as it does, that observer writes down on a slip of paper the time on his clock, his (fixed) position, and whether the light on the passing ship is on or off.

Then, much later and long after the thought experiment is done, we gather all of these slips of paper and correlate them to see exactly when and where the flashes happened; these are the positions and locations, aka coordinates, of the flash events as described in our inertial reference frame. When we do this, we will conclude that the clock on the other ship passing us was running consistently slow, both as we draw nearer and then as the two ships pass and start to move apart. That's the SR time dilation effect, and whatever is recorded on the slips of paper will be consistent with the observed arrival times of the pulses of light at our telescope.

It's worth noting that the other spaceship can set up their own chain of observers along the flight path as well. Because the other spaceship's observers will all be at rest relative to the other spaceship, they'll all be moving relative to our observers. The situation is completely symmetrical, and both sets of observers will find that the moving clock is running slow.

And finally, I can gather up all the slips of paper from one group of observers, and then use the Lorentz transformations to calculate what the other group of observers wrote down. That relationship is also symmetrical.

 

[Moonraker]

I hope you agree that the above qualifies as a good reason not to apply the time dilation formula.

Sorry, DaleSpam, but your mere derivation of the proper time formula from Lorentz transformation does not inhibit the possibility that the constraints the original formula is underlying do not rule the resulting formula (e.g. from the formula x = y*z/z derives x = y; both are not defined for z=0, but may be the second works also for z=0).

What I mentioned in the last posts I learned at school and read in books. In order to be precise, before posting I adopted the information

“For photons, their life time is an instant of 0 seconds, and in their frame space is contracted to zero.”

in its sens from a physics book. - Except the notion “frame” which I can take back. I used it because for me a “frame” was not an “inertial frame”. The last posts of forum users showed me that the word “frame” is not admitted in this forum. I hope you accept my following correction:

“The photon life time (in vacuo) is an instant of 0 seconds, the photon path is contracted to zero”.

However, I do not understand the following (please correct me if I’m wrong with these conclusions):
Photons are composing light, so light is submitted to a null space-time interval, life time of light is 0 seconds and the light path is zero.
But science found out that light takes 8 minutes from Sun to Earth, over a distance of 8 light minutes.
If time dilation is not applying to light, is there a “similar effect”, similar to the time dilation of special relativity (as we know it from the twin paradox)?

 

[Nugatory]

What I mentioned in the last posts I learned at school and read in books. In order to be precise, before posting I adopted the information

“For photons, their life time is an instant of 0 seconds, and in their frame space is contracted to zero.”

in its sens from a physics book.

Which physics book? Sadly, there are some bad ones out there... and if you cherry-pick quotations from even (especially?) the good ones you can be misled...

The last posts of forum users showed me that the word “frame” is not admitted in this forum

Nothing wrong with using it, you just have to use it correctly.

 

[Dale]

Sorry, DaleSpam, but your mere derivation of the proper time formula from Lorentz transformation does not inhibit the possibility that the constraints the original formula is underlying do not rule the resulting formula (e.g. from the formula x = y*z/z derives x = y; both are not defined for z=0, but may be the second works also for z=0).

Can you find one example in physics to support this? Specifically, any example where a general formula has some constraint, but a specific formula derived from that general formula does not have that constraint. If not, then it seems like you are just "grasping for straws".

Note also, your purely mathematical example is not a counterexample. If x = yz/z then a plot of z=0 would be a "hole" in the function. See the discussion in example C here:
http://www.freemathhelp.com/find-limit.html

What I mentioned in the last posts I learned at school and read in books. In order to be precise, before posting I adopted the information

“For photons, their life time is an instant of 0 seconds, and in their frame space is contracted to zero.”

in its sens from a physics book. - Except the notion “frame” which I can take back. I used it because for me a “frame” was not an “inertial frame”. The last posts of forum users showed me that the word “frame” is not admitted in this forum. I hope you accept my following correction:

“The photon life time (in vacuo) is an instant of 0 seconds, the photon path is contracted to zero”.

Please provide references for these quotes. They do not sound like they came from mainstream scientific sources like textbooks or peer-reviewed papers. They sound like they may have come from pop-sci books.

However, I do not understand the following (please correct me if I’m wrong with these conclusions):
Photons are composing light, so light is submitted to a null space-time interval,

All correct so far.

life time of light is 0 seconds and the light path is zero.

What is your basis for this claim?

But science found out that light takes 8 minutes from Sun to Earth, over a distance of 8 light minutes.

Yes.

If time dilation is not applying to light, is there a “similar effect”, similar to the time dilation of special relativity (as we know it from the twin paradox)?

If you don't make the claim above then there is no effect at all. That is one reason that such a claim requires some justification, which I don't see.

 

[Nugatory]

However, I do not understand the following (please correct me if I’m wrong with these conclusions):
Photons are composing light, so light is submitted to a null space-time interval, life time of light is 0 seconds and the light path is zero.
But science found out that light takes 8 minutes from Sun to Earth, over a distance of 8 light minutes.
If time dilation is not applying to light, is there a “similar effect”, similar to the time dilation of special relativity (as we know it from the twin paradox)?

You are mixing distances in space-time with distances in space and intervals of time, and that may be where the confusion is coming from.

Consider two events: Event A is a photon leaving the surface of the sun; and event B is that photon striking a photon detector on earth.

An observer at rest relative to solar system, measuring time in minutes and distances in light-minutes would say that event A is the point in space-time (x=0,t=0); and event B is the point (x=8,t=8). That is, it took eight minutes for the photon to travel the eight light-minutes between sun and earth.

An observer on a spaceship passing through the solar system at .6c relative to the solar system would say that event A is the point (x=0,t=0) and event B is the point (x'=4,t'=4). I used the Lorentz transforms to calculate this, and there is no contradiction because the two observers are moving relative to one another so they are using different coordinate systems; the little ' symbols help us not mix them up.

It's worth noting that the choice of origin is arbitrary, and the two observers don't have to pick the same one. For example, if our first observer decided to pick the origin to be the surface of the Earth three seconds before the photon was detected, then he'd say that event A was (x=-8,t=-5) and B was (x=0,t=3). The spaceship observer could still decide that he wants to call event A (x'=0,t'=0) so B is (x=4,t'=4)... Or he could choose as the origin an event long ago and far away, so that A is at (x'=10000,t'=8371) and B is (x'=10004,t'=8375)... or anything else.

Notice the following things:
1) As far as the first observer is concerned, the emission and detection events were separated by 8 minutes or time and 8 light-minutes of distance.
2) As far as the spaceship observer is concerned, the emission and detection events were separated by 4 minutes of time and 4 light minutes of distance.
3) Both observers measured the same value for the speed of light (the obvious one light-minute per minute), even though the times and distances involved were different.
4) Both observers also get the same value for the spacetime distance between the two events, defined by s^2=\Delta x^2-\Delta t^2
5) Although for both observers the space and the time distances are both non-zero, the space-time distance in #4 is zero. That's the null space-time interval you referred to above, and it does not mean that either the space or the time separations must be zero.
6) Try as you might, you will not be able to find a reference frame in which the space or time distances are zero, nor in which the speed of light will come out to be anything except oene light-minute per minute.

 

[Moonraker]

You are mixing distances in space-time with distances in space and intervals of time, and that may be where the confusion is coming from.

Yes, thank you for your explanations! ”null space-time interval” is wrong in this context, indeed. I did not adopt correctly the terms of the posts #55 and 58. Here is my correct question (modifications bold):

However, I do not understand the following (please correct me if I’m wrong with these conclusions):
Photons are composing light, so light is submitted to a null path, meaning that life time of light is 0 seconds and the light path in space is zero (see #55 and 58).
But science found out that light takes 8 minutes from Sun to Earth, over a distance of 8 light minutes.
If time dilation is not applying to light, is there a “similar effect”, similar to the time dilation of special relativity (as we know it from the twin paradox)?

 

[kweba]

If when you're moving at the speed of light time freezes, why then does it take light 8 minutes to reach the Earth from the sun?

The trick here is that in Special Relativity (which is the theory that encompasses Time Dilation), TIME is NOT universal, unlike in the previous Newtonian/Classical sense and model of space and time, where Time is independent. In SR, TIME is dependent on each object and observer in the universe.

To illustrate this, it's like saying every object in the universe has its own clock that ticks on its own rate. Some clocks tick faster, and some tick slower, than others.

So with light moving at the speed of light, TIME freezes in its own point of view. This is called "a frame of reference". Time freezes only within that light. In a sense, the light does not "age" and does not experience/feel time. (Refer to the Twin Paradox below.)

But for the rest of us (things and people especially here on Earth) who move at relatively slow everyday-speeds, time does not freeze (not even slowing down by that much for that matter), and so we experience "the passage of time".

It is us, the people, the things, and everything on Earth, who experience that 8 minutes before the light from the sun reaches us here on Earth, in our own reference frame.

Einstein himself had provided a beautiful illustration on Time Dilation with regards to reference frames: the ever famous "Twin Paradox"

To summarize the twin paradox, there are twin brothers. One brother is left at home, while the other brother is an astronaut in a space shuttle moving close or at the speed of light (to indicate a really fast velocity). When the astronaut brother returns home after some long period of time, he found that, in a way, he was still the same than his twin brother who have aged a lot more than him. So in essence, time have passed at different rates between the two twins, and that the astronaut twin remained "young".

For more info on the Twin Paradox, here's a link on wikipedia.

Note: I humbly admit I'm not an expert on this, so I would like to apologize, especially for the others, if I may have posted some wrong and misunderstood facts. Thank you.

 

[Nugatory]

However, I do not understand the following (please correct me if I’m wrong with these conclusions):
Photons are composing light, so light is submitted to a null path, meaning that time of light is 0 seconds and the light path in space is zero (see #55 and 58).

Post 55 and 58 are referring to proper time and proper distance; that's the s in s^2=\Delta x^2-\Delta t^2, not the Δx or Δt.

[edit: corrected a typo in the space-time interval formula]

 

[Moonraker]

Post 55 and 58 are referring to proper time and proper distance; that's the s in s^2=\Delta x^2+\Delta t^2, not the Δx or Δt.

Before I answer I’d like to make sure that you are really talking about s2=Δx2+Δt2 or if this is an error and you want to return to your above-mentioned formula s2=Δx2-Δt2. Which one do you mean?

 

[Nugatory]

Before I answer I’d like to make sure that you are really talking about s2=Δx2+Δt2 or if this is an error and you want to return to your above-mentioned formula s2=Δx2-Δt2. Which one do you mean?

You're right, that's supposed to be a minus sign.

 

[Dale]

Photons are composing light, so light is submitted to a null path

Yes.

meaning that life time of light is 0 seconds and the light path in space is zero (see #55 and 58).

No. This is not what null path means. Look at the forumla. A null path, \Delta s^2 = 0, clearly means that the time along the path, \Delta t, is equal to the distance along the path, \Delta x, in any inertial frame (in units where c=1). It does not imply that both are 0. Here, the time is 8 minutes and the distance is 8 light-minutes.

 

[Moonraker]

Post 55 and 58 are referring to proper time and proper distance; that's the s in s^2=\Delta x^2-\Delta t^2, not the Δx or Δt.

Answer: No, PAllen is talking about null spacetime interval, but he also describes something else. In#58 he is talking about a spacelike path and a timelike path, his two integration formulas giving 0 for proper length and proper time. So my question in #74 is justified.

 

[PAllen]

Answer: No, PAllen is talking about null spacetime interval, but he also describes something else. In#58 he is talking about a spacelike path and a timelike path, his two integration formulas giving 0 for proper length and proper time. So my question in #74 is justified.

Proper length is simply spacetime interval along a spacelike path. Proper time is simply spacetime interval along a timelike path. My integration formulas are just a convenient re-expression of the metric valid in a global inertial coordinate system in SR (I gave separate formulas only because for a spacelike path you typically want to integrate along a spatial coordinate, and for a timelike path along a time coordinate). I corrected my earlier argument that it might make sense to say proper time is zero along a light path on prodding by Dalespam and Gwellsjr. Instead, the better interpretation of the different types of spacetime paths is:

- spacelike means you can consider the path to be like a measuring tape path using some general 'simultaneity' convention. In the special case of a spacelike geodesic, you have a ruler in some inertial frame.

- timelike means it is a possible path of a clock.

- null means it can neither be a clock nor a measuring tape. It does not mean it can be both a zero length tape and a frozen clock. It means it is radiation path, or massless particle path - which cannot be treated as either a clock path or a tape measure.

 

[Moonraker]

I corrected my earlier argument...

Thank you for this important answer, but I must admit that I cannot understand very much.

Also it seems to me that it is off my subject. Space-time interval is concerning Lorentz transformation, and it is difficult or even impossible to develop statements about photons on this base. So you did not talk as you wanted about application of √(1-v^2/c^2) (proper time formula/ Lorentz contraction formula)?

For me the question is not at all meaningless because I am actually working on a model for wave-particle duality of light which shall be based on SR, and I have to find the right expression concerning the space-time path of light (proper time formula/ Lorentz contraction formula).

 

[Nugatory]

For me the question is not at all meaningless because I am actually working on a model for wave-particle duality of light which shall be based on SR, and I have to find the right expression concerning the space-time path of light (proper time formula/ Lorentz contraction formula).

Hmmm... I have to caution you that, based on the questions you're asking, you're building on a fairly shaky mathematical foundation here. It's not enough to identify the formulas, you have to understand their derivation. Without that, the most likely outcome is that you'll come up with a model that is less accurate and useful than the (astoundingly successful) relativistic model of wave-particle duality that we already have.

If you are serious about understanding SR, you might want to consider spending several serious months working your way through a reasonably math-oriented textbook. You will get a LOT of help from posters here if you find yourself stuck.

 

[Mentz114]

One can describe light as a plane-wave using the usual wave equation

ψ(t,x) = exp[i(ωt + kx)]

provided ω=k.

 

[Dale]

it is difficult or even impossible to develop statements about photons on this base.

Actually, it is fairly easy to develop statements about photons on the basis of the space-time interval. Furthermore, statements that you develop based on the spacetime interval have the advantage that they are frame-invariant.

For example: the spacetime interval of light is 0. From the formula we have:
ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2
setting ds=0 and rearranging we get
c^2 dt^2 = dx^2 + dy^2 + dz^2
which is the equation of a sphere with radius c dt. So light expands out in a sphere at speed c in all inertial frames. This is the second postulate of relativity.

 

[PAllen]

Also it seems to me that it is off my subject. Space-time interval is concerning Lorentz transformation, and it is difficult or even impossible to develop statements about photons on this base. So you did not talk as you wanted about application of √(1-v^2/c^2) (proper time formula/ Lorentz contraction formula)?

Spacetime interval is independent of Lorentz transformation. Spacetime interval is an invariant property of a spacetime path (when speaking of interval between two events in SR, implicitly you imagine a geodesic path between them). It is independent of coordinate system, reference frame, or observer. Lorentz transform is the particular form coordinate transformation between different inertial frames with standard coordinates, when there is also no translation or rotation.

The last sentence of your comment quoted above shows the confusion of separating a formula from its meaning. Length contraction and time dilation are very different things, and both are different from what I was referring to, even though they happen to have the same formula in inertial frames. Length contraction refers to disagreement between frames on measured distance between two objects (or ends of one object), each frame needing to do a series of operations to accomplish such a measurement, using their own, conflicting, simultaneity determination. Time dilation refers to the ratio of a moving clock rate to time rate in some inertial reference frame (sticking to the simple SR case). Time dilation for the same clock will differ between different inertial frames. What I was referring to was computing the proper time along a specific clock path, using some inertial frame. This will come out the same in every inertial frame (in one frame, the dilation may be greater on average, but then the coordinate time will be greater, so the integrated total will be the same; that's what it means to say proper time = time measured on some specific clock between two identifiable events in its history, is invariant. )

When I threw out the idea of trying to call the zero interval along a null path its proper time, it was correctly pointed out that I was proposing there was some type of clock that could follow along a light path - that's what proper time means. I accepted that my suggestion wasn't meaningful.