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?

 

[JesseM]

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

No they don't, using the equations only requires us to assume that you (the person reading the problem, in a textbook or a wiki article) know the coordinates of some event in one frame, then the LT gives you the corresponding coordinates in the other frame. There is no assumption that you are an actual physical observer in the scenario the problem is describing, who has determined the coordinates using rulers and clocks at rest relative to yourself.

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

Maybe instead of inventing a fantasy version of Einstein who coincidentally thought just the same way you do (but for some reason never explained this thinking explicitly), you should just pay attention to what he actually said and take him at his word that he was using "stationary" purely as a way of verbally distinguishing one frame from others.

 

[Dale]

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.

Your conclusions are completely incorrect. Just look at the math:

In the unprimed frame the metric is: ds² = -c²dt² + dx² + dy² + dz²

For ds² = 0 we obtain the equation of a sphere of radius c dt: c²dt² = dx² + dy² + dz²

In the primed frame the metric is: ds² = -c²dt'² + dx'² + dy'² + dz'²

For ds² = 0 we obtain the equation of a sphere of radius c dt': c²dt'² = dx'² + dy'² + dz'²

There is no difference between the frames in any of this. The Minkowski metric produces what you call a light sphere and what everyone else calls a light cone in all reference frames.

 

[Dale]

Isn't a moving clock supposed to be time dilated?

What do you think?

Moving wrt what? Time dilated in which frame? Whenever you are talking about a relative quantity you must specify what the quantity is relative to. How many times do we need to repeat this same point before it sinks in? The idea that velocity is a relative quantity has been around since Galileo in 1632. You are almost 4 centuries out of date when you write about velocity without specifying a reference and you are more than 1 century out of date when you do the same for time dilation.

And yes, a clock is time dilated in any reference frame in which it is moving. Note how easy it is to specify wrt what the relative quantities are measured.

 

[Doc Al]

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?

I think you're getting sidetracked by the labels "moving" and "rest", which here refer to the observed speed of the light clock itself. Better to just call them the "unprimed" and "primed" frames. Δt is the time for one cycle of the 'light clock' in a frame in which the light clock is at rest. Δt' is the time for that same cycle as measured from a frame in which that 'light clock' is moving. So, the infamous 'time dilation' maxim that "moving clocks run slow" is exactly what is described by the given equation Δt' = γΔt. All is well.

(As seen in the primed frame, the light clock is a moving clock.)

 

[espen180]

Can observer be moving relative to one another?

Imagine a univarse in which they could not, and imagine what a garbage theory SR would be is it did not allow this. Any physical or nonphysical object can play the role of observer in SR. You and I are both observers of the universe. Does that mean we can't move wrt each other?

A more direct answer: Nothing in the maths or philosophy of Einsteinian mechanics (or Newtonian, or Galilean) prevent observers to move wrt each other.

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?

1) This is nonsense. Look what this logic leads to. Let's sa you and I are on one side of a football field. You start running toward one of the goals. Due this this action on your part, you suddenly ceased to exist in my frame. Notice the absurdity?

On a serious note, I guess the underlying claim on your part is that an observer cannot make measurements in a frame wrt which he/she is moving. Yes, this is in some sense true. On the other hand, the observer can use Lorentz transformations to translate the measurements he/she does in his/her frame to any other frame.

2) How does the observer know he/she is not moving (wrt what)?

 

[chinglu1998]

Imagine a univarse in which they could not, and imagine what a garbage theory SR would be is it did not allow this. Any physical or nonphysical object can play the role of observer in SR. You and I are both observers of the universe. Does that mean we can't move wrt each other?

A more direct answer: Nothing in the maths or philosophy of Einsteinian mechanics (or Newtonian, or Galilean) prevent observers to move wrt each other.



1) This is nonsense. Look what this logic leads to. Let's sa you and I are on one side of a football field. You start running toward one of the goals. Due this this action on your part, you suddenly ceased to exist in my frame. Notice the absurdity?

On a serious note, I guess the underlying claim on your part is that an observer cannot make measurements in a frame wrt which he/she is moving. Yes, this is in some sense true. On the other hand, the observer can use Lorentz transformations to translate the measurements he/she does in his/her frame to any other frame.

2) How does the observer know he/she is not moving (wrt what)?

1) You are the one that claimed if you move in a frame, then you are still in that frame. that is false, you are in another frame.

2) An observer does not know they are moving. They view themselves as at rest and everything else is moving. So, I do not know what you mean.

 

[chinglu1998]

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?

How about I ask you. What is your math in the view of the clock frame for the the times of each frame.

Please make sure you use light aberration in tyhe "observer frame" because this is a function of which frame has the light source and the clock frame has the light source. From the view of the clock frame, what do you get?

 

[chinglu1998]

No they don't, using the equations only requires us to assume that you (the person reading the problem, in a textbook or a wiki article) know the coordinates of some event in one frame, then the LT gives you the corresponding coordinates in the other frame. There is no assumption that you are an actual physical observer in the scenario the problem is describing, who has determined the coordinates using rulers and clocks at rest relative to yourself.

I am fine with your typing above. But, now take the next step as I have.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/

In my view, this is cruciel in understanding SR. When you take a frame as stationary, your measurements are Euclidian and your physics is Newtonian. The other frame when mapped with LT is not.

So, this is not a simple verbal distinction, but a logical one.

 

[chinglu1998]

Your conclusions are completely incorrect. Just look at the math:

In the unprimed frame the metric is: ds² = -c²dt² + dx² + dy² + dz²

For ds² = 0 we obtain the equation of a sphere of radius c dt: c²dt² = dx² + dy² + dz²

In the primed frame the metric is: ds² = -c²dt'² + dx'² + dy'² + dz'²

For ds² = 0 we obtain the equation of a sphere of radius c dt': c²dt'² = dx'² + dy'² + dz'²

There is no difference between the frames in any of this. The Minkowski metric produces what you call a light sphere and what everyone else calls a light cone in all reference frames.

Your math is correct but is missing something.

Where in your math have you shown the mapped light sphere by LT is a light sphere?

The only think you can show is each mapped light beam measures c.

So, you have to look inside the problem to find the distinction.

Hence, for all light beams in the stationary frame, each measures ct in all directions for some t.

This same set of light beams mapped by LT do not measure a constant distance from the light emission point in the moving frame.

These are mathematical facts.

 

[JesseM]

How about I ask you. What is your math in the view of the clock frame for the the times of each frame.

What does "times of each frame" mean? You can only compare the times between a specific pair of events, comparing "time" without specifying events is totally meaningless. What events do you want to look at? If we look at two events on the worldline of an object at rest in the unprimed frame, then the time between them in each frame is given by t' = t γ. If we pick two events on the worldline of an object at rest in the primed frame, the time between them in each frame is given by t = t' γ. If that doesn't answer your question you'll have to be more specific about what you're asking. But please don't avoid my own question, which was asking what you meant by "time expanded" in this comment:

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.

Please make sure you use light aberration in tyhe "observer frame"

Obviously light that is sent by a light source moving relative to the observer will be aberrated in the observer frame, but beyond this I have no idea what you mean by "use" light aberration. Again, you'll have to be more specific.

From the view of the clock frame, what do you get?

"Get" for what specific quantity? Again, are you asking about the time between some events in the clock frame?

 

[chinglu1998]

Moving wrt what? Time dilated in which frame? Whenever you are talking about a relative quantity you must specify what the quantity is relative to. How many times do we need to repeat this same point before it sinks in? The idea that velocity is a relative quantity has been around since Galileo in 1632. You are almost 4 centuries out of date when you write about velocity without specifying a reference and you are more than 1 century out of date when you do the same for time dilation.

And yes, a clock is time dilated in any reference frame in which it is moving. Note how easy it is to specify wrt what the relative quantities are measured.

Well, maybe you are correct. Can you provide the math using the clock frame and prove the observer frame is time dilated compared to the clock frame?

We are using light by the way as the test as is for WIKI.

 

[chinglu1998]

I think you're getting sidetracked by the labels "moving" and "rest", which here refer to the observed speed of the light clock itself. Better to just call them the "unprimed" and "primed" frames. Δt is the time for one cycle of the 'light clock' in a frame in which the light clock is at rest. Δt' is the time for that same cycle as measured from a frame in which that 'light clock' is moving. So, the infamous 'time dilation' maxim that "moving clocks run slow" is exactly what is described by the given equation Δt' = γΔt. All is well.

(As seen in the primed frame, the light clock is a moving clock.)

Alright. Primed and unprimed is fine with me.

What if the unprimed frame is taken as stationary? What results do you get?

Or, maybe say, the unprimed frame is the context.

 

[chinglu1998]

What does "times of each frame" mean? You can only compare the times between a specific pair of events, comparing "time" without specifying events is totally meaningless. What events do you want to look at? If we look at two events on the worldline of an object at rest in the unprimed frame, then the time between them in each frame is given by t' = t γ. If we pick two events on the worldline of an object at rest in the primed frame, the time between them in each frame is given by t = t' γ. If that doesn't answer your question you'll have to be more specific about what you're asking. But please don't avoid my own question, which was asking what you meant by "time expanded" in this comment:


Obviously light that is sent by a light source moving relative to the observer will be aberrated in the observer frame, but beyond this I have no idea what you mean by "use" light aberration. Again, you'll have to be more specific.

"Get" for what specific quantity? Again, are you asking about the time between some events in the clock frame?

Let us assume you take the context of the frame with the clock. The start event is the light emission and the end event is the light reaching y=L.

Please calculate t' and t.

 

[Doc Al]

What if the unprimed frame is taken as stationary? What results do you get?

Not sure what you mean, since the light clock is at rest in the unprimed frame. But if there were another clock at rest in the primed frame, then observers in the unprimed frame would see it as running slow. In that case, the 'time dilation' formula would be Δt = γΔt'. (As seen from the unprimed frame, that second light clock is a moving clock.)

 

[JesseM]

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?

No, because the LT doesn't say anything about being restricted to events which are all on the same "slice" of a light cone, you can transform a set of events that all happened at different times in your frame.

In the stationary frame, the light sphere is a sphere.

If you pick events on the light cone that all happened at the same time in the frame you label "stationary", then their positions form a sphere. If you pick a bunch of events that happened at different times in this frame (and I'm not talking about doing a LT, I'm saying you're free to pick as your initial data a set of events which are non-simultaneous in whatever frame you start out with) then their positions may form some other shape like an ellipsoid. If you pick all events on the worldlines of the light beams at all possible times in this frame, they form a 4D cone.

Similarly, if your initial set of events was such that when you transform into the "moving frame", you get a bunch of events that are simultaneous in the moving frame, then their positions form a sphere. If your events in the moving frame are non-simultaneous, then they may form some other shape like an ellipsoid. And if you are looking at all events on the worldlines of the light beam in the moving frame, they form a 4D cone. So, I still can't make any sense of your distinction between "Euclidean space" in the stationary frame and "Minkowski space" in the moving one, still seems like a totally incoherent distinction.

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.

I have no idea what you mean by "sees a sphere". If you think of yourself as an actual physical observer at rest in some frame (as opposed to adopting the omniscient perspective of someone reading a problem in a textbook), then you understand that you can't actually visually "see" a set of simultaneous events in your frame at a single moment, right? Since you are at different distances from different points in space, what you see visually at a single moment will be light from a bunch of events at different times in your frame. Statements about what was happening at a single t-coordinate can only be made in retrospect, like if in 2010 I receive a signal from an event E1 10 light-years away in my frame, and in 2020 I receive a signal from an event E2 20 light-years away in my frame, and I conclude retroactively that they both happened simultaneously at the t-coordinate of 2000 in my frame. So the "light sphere" is every bit as much of a retroactive reconstruction as the "light cone", both involve charting the coordinates of a bunch of events that I didn't become aware of until various later times.

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.

"Stationary" is meaningless unless understood to mean "stationary relative to" some object or frame. Certainly an observer (or any other object) is stationary relative to their own frame, but moving relative to other objects and frames.

That observer does not move and thinks all other objects move.

In their own frame yes, but the observer is perfectly capable of understanding that they would be seen as "moving" in other frames, unless they are an idiot who doesn't understand the LT.

Just look at LT and that will help you udnerstand this fact.

What fact? What aspect of the LT will "help me understand"? It would certainly be helpful if you would give some actual math rather than these endless incoherent verbal arguments.

 

[Dale]

What if the unprimed frame is taken as stationary? What results do you get?

Do you have some sort of diagnosed learning deficit? Stationary wrt what?

 

[chinglu1998]

Not sure what you mean, since the light clock is at rest in the unprimed frame. But if there were another clock at rest in the primed frame, then observers in the unprimed frame would see it as running slow. In that case, the 'time dilation' formula would be Δt = γΔt'. (As seen from the unprimed frame, that second light clock is a moving clock.)

Δt = γΔt'.

Can you please show this by including light aberration just like WIKI did and following the same style argument? Thanks.

 

[chinglu1998]

Do you have some sort of diagnosed learning deficit? Stationary wrt what?

Stationary with respect to the observer in WIKI. Does WIKI have something else in the problem?

 

[JesseM]

Let us assume you take the context of the frame with the clock. The start event is the light emission and the end event is the light reaching y=L.

Please calculate t' and t.

I already told you that "If we look at two events on the worldline of an object at rest in the unprimed frame, then the time between them in each frame is given by t' = t γ." I suppose since one event is on the worldline of the bottom mirror and the other event is on the worldline of the top mirror this doesn't precisely fit the sentence I wrote, but since both events are at the same position on the x-axis, and the x-axis is the axis of relative motion between the two frames, it still works out the same. If you need a derivation for this I could give you one, but please first do me the courtesy of answering the question about what you meant by "time expanded" in this comment, as I have asked twice now:

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.

 

[JesseM]

I am fine with your typing above. But, now take the next step as I have.

Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good.2 In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the ``stationary system.''
http://www.fourmilab.ch/etexts/einstein/specrel/www/

In my view, this is cruciel in understanding SR. When you take a frame as stationary, your measurements are Euclidian and your physics is Newtonian. The other frame when mapped with LT is not.

Any Newtonian equation that holds in one frame holds in all of them. As usual, you base your statements on some weird notion of "Minkowski space" that no actual physicists use (as opposed to Minkowski spacetime, which we all understand), but none of your attempts to justify the notion that one frame involves "Euclidean space" and the others involve "Minkowski space" make the slightest bit of sense to me--see my recent comments in post #195.

So, this is not a simple verbal distinction, but a logical one.

You may have the confused belief that it's more than simply a verbal distinction, but please stop using Einstein's use of the word "stationary" to support your belief, since he never introduces any nonsensical notion that Newtonian/Euclidean laws only work in the "stationary" system (this would contradict the first postulate of relativity in section 2, which says 'The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems of co-ordinates in uniform translatory motion'), and he explicitly says that his purpose in introducing the word "stationary" is "to distinguish this system of co-ordinates verbally from others".