When moving at the speed of light time stops

nitsuj

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

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

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.

DaleSpam

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'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

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

+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 go to jail 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 do drive more carefully post more precisely in the future."

DaleSpam

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

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) realise 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 travelling instantaneously would be bypassing the speed of light.

ghwellsjr

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

phinds

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

yep, you got it.

PAllen

correct
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)
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.