WIKI and Time Dilation: The Possible Error in Relative Velocity

[yuiop]

I am going to stop you right here. WIKI does not admit 2 observers. There is only 1 observer and that observer is moving.

Can you explain how that 1 observer is moving under SR?

This is meaningless until you complete the phrase "There is only 1 observer and that observer is moving" relative to __________.

We are going to start worrying about you, when do not pick up on things you have been told numerous times by numerous people.

Same applies to post #147. You mention stationary or moving 6 times and never once use the phrase "relative to _____" or "according to _____".

 

[yuiop]

I am going to stop you right here. WIKI does not admit 2 observers. There is only 1 observer and that observer is moving.

Can you explain how that 1 observer is moving under SR?

In SR, when we talk of "an observer" we actually mean a whole grid of observers that are all at rest with respect to each other. These observers do not actually have to be sentient beings. At each node of the grid "an observer" could just be a camera with a proximity sensor that triggers the camera to photograph any passing object with a time stamp of when the photograph was taken. Such an array can produce a historic record of any object passing through the grid giving an indication of where it was at any given time, its velocity and length etc. just by looking retrospectively at the collection of timestamped photographs. Only one sentient being is required to analyse the collection of photographs and make sense of it. This "grid" model was described in post #52 by Jesse and included this illustration:

See all the little clocks at the intersections of the grid?

If you really don't like the grid idea, you could have a single observer with a radar device which indicates where an object is at any given time and after allowing for light travel times, can compute velocity, length etc. of a given moving object.

 

[espen180]

Agreed, your post #90 is a perfect example of reciprocal time dilation.

Except, instead of two clocks for A and B, let's have one at say -k' in the primed system.

It moves to the unprimed origin. Does this create reciprocal time dilation?

I don't understand you. Whenever two observers exist that have a nonzero velocity relative to each other, they both define their own rest frame and both measure the other's time to run slow by an equal measure. What's the problem?

 

[espen180]

SR does not admit a moving observer unless there is another I guess watching it.

As long as a frame can be defined, usually by use of some physical reference object like a clock, any entity or object, including an observer, is free to move around in that frame.

If there is just one observer and no physical objects around, that does not mean he is stationary. Why do you think that is so?

 

[Dale]

I am going to have to agree you are correct with the metric.

when the frame is taken as statonary, objects in that frame have Euclidian measurements. Those in the moving frame have Minkowski. Is that not a difference?

You are contradicting yourself. Either you correctly agree with me that all inertial frames use the Minkowski metric or you incorrectly believe (in contradiction of the first postulate) that the metric is different in some privileged frame. You cannot have it both ways.

Let's look at the physical consequences if some frame were to use the Euclidean metric: ds² = dt² + dx² + dy² + dz²

In the Euclidean frame clocks would not function and time would be measured by rulers. Light would not propagate, but would only exist at the event of emission. There would not be any identifiable future or past or causality.

Since these measurable consequences do not arise in any frame we can be certain that that all frames use the Minkowski metric and none use the Euclidean metric. Furthermore, the signature of the metric is an invariant, and the Euclidean metric has a different signature than the Minkowski metric. Therefore it is simply not logically possible for one frame to use the Euclidean metric and another to use the Minkowski.

 

[Dale]

Then you do not understand SR and you need to start more simply.
The article says,

That is to say, in a frame moving relative to the clock, the clock appears to be running more slowly.

which expresses the fact that for the moving observer the period of the clock is longer than in the frame of the clock itself.

As an observer in SR, how do you know you are moving?

You see, the Wiki authors understand that velocity is relative and so they clearly specified what the velocity of the frame is measured wrt. You would do well to learn to do the same. Anytime you use the words "stationary", "moving", "at rest", etc. you should make sure that you have specified what it is measured wrt.

I am going to stop you right here. WIKI does not admit 2 observers. There is only 1 observer and that observer is moving.

Can you explain how that 1 observer is moving under SR?

Multiple observers are not needed. Only a specification of what any velocity is measured relative to. If one observer observs that an object is moving wrt him then he automatically knows that he is moving wrt the object.

 

[Dale]

Oh, I was not saying you must take one frame in that sense. I was saying you must take a frame as stationary to calculate time dilation.

Is this false?

Yes, this is false. Let's say that frame A is not "stationary" but is moving at some unknown velocity v. And let's say that frame B is moving as some known velocity u wrt A. Then the time dilation of A as measured by B (and vice versa) is entirely goverened by u and is not a function of v at all. Therefore you can calculate the time dilation between two frames even if neither is taken as "stationary".

 

[chinglu1998]

This is amusing. An observer is only moving if another observer is watching?

How exactly would 1 observer know they are moving?

 

[JesseM]

How exactly would 1 observer know they are moving?

The word "moving" is meaningless unless it is understood to mean "moving relative to X", where X is some other object (like the clock in the wiki article) or frame (like some frame we have arbitrarily decided to label 'stationary'). Certainly an observer can know they are moving relative to some other specific object or frame, they just check if it's moving relative to themselves.

 

[chinglu1998]

But your data concerns the coordinates of events throughout spacetime, so if you plug it into LT you don't get a "space", you get another description of spacetime, where for example light from an event forms 4D cones with sides sloped at c just like in your frame.

I am in agreement.

However, the systerm take as stationary has the light sphere with the center at the center.
The LT mapped light sphere is not a sphere but an ellipsoid with is center located in the direction of travel and not at the center of the ellipsoid.

These are clearly different looking "spacetimes".


The wiki article only says the observer is moving relative to the clock, it doesn't say the observer is "moving" in your sense that we have to start from the observations of a "stationary" frame and only later calculate how things would look in a "moving" frame. Again, this rule that we must start with the data from the "stationary" or "observer" frame is your own invention, not one that's a standard notion among physicists.

No, this is not relevant. Read this from WIKI.

From the frame of reference of a moving observer traveling at the speed v.
http://en.wikipedia.org/wiki/Time_dilation

How exactly are in in a frame of reference or from a frame of reference and you are moving at v? This is an error.

So is your disagreement with the wiki article not about their equations, but just about the way the wiki started out with data in a frame different from the observer's frame? i.e. you just don't like the order in which the data and equations were presented, but have no objection to the data and equations on their own?

At first, I was in disagreement with the article suggesting the clock is stationary and a frame is moving and the clock is time dilated.

But, when I liik at the math now, the article is poorly writtne byt the math holds regardless.

The frame with the light source will always show less time than the frame without the light source regardless. Hence, from the view of the clock frame, the moving observer will show more time on the clock contrary to time dilation.

I think that is what everyone else is aying here with the math. But, the language of the article does not say this.

The usual approach is just to talk about how things look in different frames and use the LT to go between them, without any rule that we must start with data in a "stationary" frame or the frame of the "observer". When presenting a problem a textbook can start with how things look in any frame the author wants, even if it isn't the frame of the observer or is the one the author verbally refers to as "moving".

This is how I do it. But, the only data a frame has is in the stationary system or its frame.

 

[chinglu1998]

This is meaningless until you complete the phrase "There is only 1 observer and that observer is moving" relative to __________.

We are going to start worrying about you, when do not pick up on things you have been told numerous times by numerous people.

Same applies to post #147. You mention stationary or moving 6 times and never once use the phrase "relative to _____" or "according to _____".

This is not the problem. The problem is with the article.
So, apply your rules to the WIKI article.

But, I really do not have any problem understanding an observer is moving relative to the clock and the clock is moving relative to the observer.

Have you figured out yet the equation t' = tγ holds no matter which frame you take as the ? viewing frame?

 

[chinglu1998]

In SR, when we talk of "an observer" we actually mean a whole grid of observers that are all at rest with respect to each other. These observers do not actually have to be sentient beings. At each node of the grid "an observer" could just be a camera with a proximity sensor that triggers the camera to photograph any passing object with a time stamp of when the photograph was taken. Such an array can produce a historic record of any object passing through the grid giving an indication of where it was at any given time, its velocity and length etc. just by looking retrospectively at the collection of timestamped photographs. Only one sentient being is required to analyse the collection of photographs and make sense of it. This "grid" model was described in post #52 by Jesse and included this illustration:

See all the little clocks at the intersections of the grid?

If you really don't like the grid idea, you could have a single observer with a radar device which indicates where an object is at any given time and after allowing for light travel times, can compute velocity, length etc. of a given moving object.

Actually I do not see it that way.

I see a clock at any point in a Cartesian 3D space all of which are synchronized in the frame.
When LT is used, and t' is returned for a particular space-time coordinate, that gives the time on a corresponding clock in the ? moving other frame/primed frame.

 

[chinglu1998]

I don't understand you. Whenever two observers exist that have a nonzero velocity relative to each other, they both define their own rest frame and both measure the other's time to run slow by an equal measure. What's the problem?

Can observer be moving relative to one another?

 

[chinglu1998]

As long as a frame can be defined, usually by use of some physical reference object like a clock, any entity or object, including an observer, is free to move around in that frame.

If there is just one observer and no physical objects around, that does not mean he is stationary. Why do you think that is so?

1) If an observer is in a frame and moves in thatr frame, the observer is no longer in that frame.
2) If there is only one observer, how does the observer know he/she is moving?

 

[JesseM]

I am in agreement.

However, the systerm take as stationary has the light sphere with the center at the center.
The LT mapped light sphere is not a sphere but an ellipsoid with is center located in the direction of travel and not at the center of the ellipsoid.

But you don't use the LT to map events occurring only at one time, you use it to map events at all times. So, you have a light cone in your frame, then you use the LT and find a light cone in the other frame. It's true that if you pick only events on the light cone which happened at a single time in your frame and map them to the other frame, you get a set of events at different times in that frame whose positions form an ellipsoid, but it's equally true that if you pick a bunch of events on the light cone which happened at different times in your frame, their positions may form an ellipsoid.

These are clearly different looking "spacetimes".

You seem not to understand the difference between a 4D spacetime and a 3D space! the set of events on light worldlines in spacetime don't form a sphere or an ellipsoid, they form a 4D cone.

The wiki article only says the observer is moving relative to the clock, it doesn't say the observer is "moving" in your sense that we have to start from the observations of a "stationary" frame and only later calculate how things would look in a "moving" frame. Again, this rule that we must start with the data from the "stationary" or "observer" frame is your own invention, not one that's a standard notion among physicists.

No, this is not relevant. Read this from WIKI.

From the frame of reference of a moving observer traveling at the speed v.
http://en.wikipedia.org/wiki/Time_dilation

But it's clear they are talking about an observer "moving" relative to the clock frame.

How exactly are in in a frame of reference or from a frame of reference and you are moving at v? This is an error.

They never said the observer was moving at speed v in his own frame, it's clear from the context they mean the observer is moving at speed v in the clock frame.

The usual approach is just to talk about how things look in different frames and use the LT to go between them, without any rule that we must start with data in a "stationary" frame or the frame of the "observer". When presenting a problem a textbook can start with how things look in any frame the author wants, even if it isn't the frame of the observer or is the one the author verbally refers to as "moving".

This is how I do it. But, the only data a frame has is in the stationary system or its frame.

But the person reading the problem in a textbook is not a "frame", so the author can give the reader "data" from either frame, it makes no difference. There is no need to start with data only from the point of view of the "observer" in the problem, or only from the point of view of the frame that has been given the label "stationary". It's perfectly normal for a textbook, or a wiki article, to start out giving data from a frame that's different from the frame of the "observer". You may choose to adopt a different convention if you prefer, but you should understand that your convention is different from how most physicists think and not accuse them of being in "error" for using a different convention than you when presenting problems.

 

[chinglu1998]

You are contradicting yourself. Either you correctly agree with me that all inertial frames use the Minkowski metric or you incorrectly believe (in contradiction of the first postulate) that the metric is different in some privileged frame. You cannot have it both ways.

Let's look at the physical consequences if some frame were to use the Euclidean metric: ds² = dt² + dx² + dy² + dz²

In the Euclidean frame clocks would not function and time would be measured by rulers. Light would not propagate, but would only exist at the event of emission. There would not be any identifiable future or past or causality.

Since these measurable consequences do not arise in any frame we can be certain that that all frames use the Minkowski metric and none use the Euclidean metric. Furthermore, the signature of the metric is an invariant, and the Euclidean metric has a different signature than the Minkowski metric. Therefore it is simply not logically possible for one frame to use the Euclidean metric and another to use the Minkowski.

I am not trying to say the Minkowski metric does not apply to the unprimed frame.

I am trying to say when it does, it is also a Euclidian light sphere for example.

However, when I map that surface of the Euclidian light sphere to the primed frame, the metric is the same 0 but the mapped light beams are different based on direction unlike the light beams in the unprimed frame.

Therefore, for any light beam in the unprimed frame, the metric produces a constant 0 for light, however, the equation x² + y² + z² = c² t²*produces a constant c² t² for all points on the light sphere surface.

When each light beam is mapped, the metric again produces 0 but the equation x'² + y² + z² = c² t'² is such that c² t'² is not constant in all directions.

Consequently, although the metric produces 0 for any light beams in any frame translated or not, the underlying geometry is hidden because of the difference in the fact, c² t² is constant in all directions and c² t'² is not.

Therefore, the originating space is different from the mapped space.

 

[chinglu1998]

You see, the Wiki authors understand that velocity is relative and so they clearly specified what the velocity of the frame is measured wrt. You would do well to learn to do the same. Anytime you use the words "stationary", "moving", "at rest", etc. you should make sure that you have specified what it is measured wrt.

I am fine with this. But, the article is talking about time dilation.
Under time dilation, it is normal to say the moving clock is time dilation.

But, in the frame of the clock on WIKI, the "moving observer" would not be time dilated. I guess that is why that observer did not have a clock. Isn't a moving clock supposed to be time dilated?

What do you think?

 

[chinglu1998]

Yes, this is false. Let's say that frame A is not "stationary" but is moving at some unknown velocity v. And let's say that frame B is moving as some known velocity u wrt A. Then the time dilation of A as measured by B (and vice versa) is entirely goverened by u and is not a function of v at all. Therefore you can calculate the time dilation between two frames even if neither is taken as "stationary".

Good point.

 

[chinglu1998]

The word "moving" is meaningless unless it is understood to mean "moving relative to X", where X is some other object (like the clock in the wiki article) or frame (like some frame we have arbitrarily decided to label 'stationary'). Certainly an observer can know they are moving relative to some other specific object or frame, they just check if it's moving relative to themselves.

What?

An observer knows they are moving under SR?

 

[JesseM]

What?

An observer knows they are moving under SR?

Please don't misquote me, I said "moving relative to some other specific object or frame". And yes, an observer can certainly know this, for example, if you are moving at 0.8c in my frame, then I know that I must be moving at 0.8c relative to your frame.

 

[chinglu1998]

But you don't use the LT to map events occurring only at one time, you use it to map events at all times. So, you have a light cone in your frame, then you use the LT and find a light cone in the other frame. It's true that if you pick only events on the light cone which happened at a single time in your frame and map them to the other frame, you get a set of events at different times in that frame whose positions form an ellipsoid, but it's equally true that if you pick a bunch of events on the light cone which happened at different times in your frame, their positions may form an ellipsoid.

Let's take a timeout here.

Each slice of the light cone is a certain time. So, yes, you must consider all mapped light beams at a particular time in the "chosen" frame. See how you need to specifiy this?

In the stationary frame, the light sphere is a sphere. In the LT mapping for another frame it is not. You must specify which frame is stationary.

You seem not to understand the difference between a 4D spacetime and a 3D space! the set of events on light worldlines in spacetime don't form a sphere or an ellipsoid, they form a 4D cone.

OK, but, for each time t in a frame, there exists the surface of the sphere and for some reason, they map a sphere surface to a circle and then use time to make the cone.

If a light sphere is not a sphere, then just say that. But, by the relativity postulate, each frame sees a sphere and hence the geometry should present the facts. You cone does not. So, it is a simplification or a model but not reality.

But it's clear they are talking about an observer "moving" relative to the clock frame.

They never said the observer was moving at speed v in his own frame, it's clear from the context they mean the observer is moving at speed v in the clock frame.

You can justify all you want. Under SR, when you refer to a single observer, that observer is stationary. That observer does not move and thinks all other objects move. Just look at LT and that will help you udnerstand this fact.

But the person reading the problem in a textbook is not a "frame", so the author can give the reader "data" from either frame, it makes no difference. There is no need to start with data only from the point of view of the "observer" in the problem, or only from the point of view of the frame that has been given the label "stationary". It's perfectly normal for a textbook, or a wiki article, to start out giving data from a frame that's different from the frame of the "observer". You may choose to adopt a different convention if you prefer, but you should understand that your convention is different from how most physicists think and not accuse them of being in "error" for using a different convention than you when presenting problems.

I hever said the person reading the book is a frame.

 

[yuiop]

What?

An observer knows they are moving under SR?

Yes, they know their relative motion. Jesse used the important word "relative" which is important in SR which stands for Special Relativity. If you are looking at your PC screen while reading this and not the time, then one minute later is the screen is still the same distance from you, you can conclude that the screen is stationary with respect to you and you are not moving relative to the screen. If a car passing in the street is in different locations after one minute has elapsed, then it has motion relative to you AND you have motion relative to the car. SO at one the same time you are not moving relative to the screen and you are moving relative to the car. All you know is your motion relative to other objects, but not your absolute motion. You and your screen may be at rest with respect to each other and you might presume your stationary, but relative to the sun or the galaxy centre or whatever you are moving. In relativity, all motion is relative and there is no absolute sense of stationary or moving, so saying an observer or reference frame or object is stationary or moving without specifying what they are stationary or moving with respect to, is meaningless. I think Dalespam might have mentioned that somewhere already.

 

[chinglu1998]

Please don't misquote me, I said "moving relative to some other specific object or frame". And yes, an observer can certainly know this, for example, if you are moving at 0.8c in my frame, then I know that I must be moving at 0.8c relative to your frame.

The calculations of LT assume you are stationary as an observer and all other objects are moving.

What does stationary really mean?

It means you have data gathered in your frame. You then map that data with T to understand the other frame.

Otherwise, you are able to gather data, without being told, from the other frame and map to understand your data.

Since this is false, there is a distinction.

I guess that is just one of the reasons Einstein used the term stationary 62 times in his 1905 paper.

 

[chinglu1998]

Yes, they know their relative motion. Jesse used the important word "relative" which is important in SR which stands for Special Relativity. If you are looking at your PC screen while reading this and not the time, then one minute later is the screen is still the same distance from you, you can conclude that the screen is stationary with respect to you and you are not moving relative to the screen. If a car passing in the street is in different locations after one minute has elapsed, then it has motion relative to you AND you have motion relative to the car. SO at one the same time you are not moving relative to the screen and you are moving relative to the car. All you know is your motion relative to other objects, but not your absolute motion. You and your screen may be at rest with respect to each other and you might presume your stationary, but relative to the sun or the galaxy centre or whatever you are moving. In relativity, all motion is relative and there is no absolute sense of stationary or moving, so saying an observer or reference frame or object is stationary or moving without specifying what they are stationary or moving with respect to, is meaningless. I think Dalespam might have mentioned that somewhere already.

Thanks, I have this figured out.

We are not talking anything except relative motion.

 

[Vanadium 50]

The calculations of LT assume you are stationary as an observer and all other objects are moving.

You keep saying this. It is not true. The Lorentz transformations work just fine if everything is in relative motion.

 

[chinglu1998]

You keep saying this. It is not true. The Lorentz transformations work just fine if everything is in relative motion.

I assume everyting is in relative motion.

So, where do you get your data for LT? Are you going to get data from the other frame? No.

For example, assume we LT a light sphere.

So, you tell me as an observer, where do you get your data to LT?

 

[chinglu1998]

Anyway, using the WIKI article, the frame with the light source will always show less time than the other frame regardless of which frame is viewed.

What does this do to time dilation given the observer could have a clock.
that means the clock at rest with the light source will view the moving observer and clock as time expanded.

What is wrong with this?

 

[Vanadium 50]

So, where do you get your data for LT? Are you going to get data from the other frame? No.

"Get the data"? The Lorentz transformation gets you from any frame to any other frame. Have you ever used the Lorentz transformation yourself? The reason I ask is that it doesn't sound like you have much facility with it.

 

[chinglu1998]

"Get the data"? The Lorentz transformation gets you from any frame to any other frame. Have you ever used the Lorentz transformation yourself? The reason I ask is that it doesn't sound like you have much facility with it.

Yes I have. It is curious to me that you do not know what the data is.

Let me try to ask. The data for LT you need is time and a space coordinate. Where are you going to get this from?

Once you agree you can only get this from your frame, you will understand why Einstein used the term stationary 62 times in his paper.

It seems too much abstraction has crept into the theory. It is a theory that takes information from one frame and maps it to another.

That data in the stationary frame is collected according to the rules of Euclidian geometry and a constant reliable time. In reality, you collect data in your frame according to the rules of Newtonian physics.

You are not able to collect data from the other frame due to the incompatability of time.

So, you are required to gather space time coordinates in your frame and map them to other frames.

 

[JesseM]

What does this do to time dilation given the observer could have a clock.
that means the clock at rest with the light source will view the moving observer and clock as time expanded.

What do you mean "time expanded"? In the frame of the observer, the light clock will take more time to tick forward a given amount (light clock is ticking slower than observer's clock in observer's frame), and in the frame of the light clock, the observer's clock will take more time to tick forward a given amount (observer's clock is ticking slower than light clock in light clock's frame). Do you agree?