Is Lorentz transformation absolute

[Chrisc]

An observer at the mid point between two frames A and B that are moving toward her, measures each of their speeds to be 0.75c. According to SR, the Lorentz transformation will determine the speed of A and B measured by the other to be less than 1.5c. (in fact less than c)

Being aware of the effects of SR, the observer at the mid point will also be aware that their measurement of the speed of A and B is subject to the Lorentz transformation and therefore is a measure of some speed greater than 0.75c contracted to 0.75c by the Lorentz effect. After accounting for the Lorentz factor that gave their measurement for A and B they find the speed of each is greater than 0.75c.
Since the observer at the mid point is not necessary in the measurements made by A and B, should the observer calculate the speed of A and B as measured by each as a Lorentz transformation of 0.75c, or the greater, untransformed speed?
Why?

 

[Doc Al]

An observer at the mid point between two frames A and B that are moving toward her, measures each of their speeds to be 0.75c. According to SR, the Lorentz transformation will determine the speed of A and B measured by the other to be less than 1.5c. (in fact less than c)
As in this case the speed of all three frames are a measure of relative motion, the observer between A and B has a relative speed toward A and B of 0.375c.

I thought you just said that A and B are moving with a speed of 0.75c with respect to that midpoint observer? Where do you get the speed of 0.375c?

 

[Chrisc]

Sorry Doc Al, I left that line in from a previous version of the question. Thanks for mentioning it.

 

[Doc Al]

Being aware of the effects of SR, the observer at the mid point will also be aware that their measurement of the speed of A and B is subject to the Lorentz transformation and therefore is a measure of some speed greater than 0.75c contracted to 0.75c by the Lorentz effect.

What do you mean here? Let's call that midpoint observer M. Both A and B are moving at speed 0.75c with respect to M.

After accounting for the Lorentz factor that gave their measurement for A and B they find the speed of each is greater than 0.75c.

What do you mean by "after accounting for the Lorentz factor"?

All measurements of the speed of A with respect to M are made using distances (and times) as measured by M. So there's no need for a "Lorentz factor".

 

[Chrisc]

Let me restate the question using your suggestion of M as my frame.
If I (M) calculate the speed of A as measured by B, I must invoke the Lorentz transformation as it affects measurements made by B. I must also invoke it to calculate the speed of B as measured by A as it affects the measurements made by A. If I calculate the speed of M (my frame) as measured by B, I invoke the Lorentz transformation as it affects the measurements made by B. If B calculates the speed of B as measured by M, they invoke the Lorentz transformation as it affects the measurements made by M (me).
Now, being fully aware of the effects of SR on all measurements of constant linear motion, how do I consider the initial v of B or A, input into the Lorentz factor and applied to the Lorentz transformation of the measurements made by M(me) on the axis of motion and rate of time, is "valid" if I by the definition of relative motion, my(M's) measurement of v must already be affected by the Lorentz transformation?

 

[DavidWhitbeck]

The statement of the problem doesn't make sense to me. The role of the observer is to make the measurements. If the observer is not measuring anything, why is that observer there? Furthermore, the observer doesn't measure the speed of reference frames, the observer defines the reference frame.

The correct language should be in the form of observer A in reference frame A measures position and clock settings of these events as follows... observer B in reference frame B measures position and clock settings of these events as follows...

I wouldn't start jumping into Lorentz factors until you identify the invariant spacetime interval. And also the simplest way to clear up a paradox in SR is to draw a spacetime diagram, the picture itself if drawn with care will usually illustrate clearly the resolution to the problem.

 

[Doc Al]

Let me restate the question using your suggestion of M as my frame.
If I (M) calculate the speed of A as measured by B, I must invoke the Lorentz transformation as it affects measurements made by B.

The reason why you need to use the Lorentz transformations (the relativistic addition of velocity formula, in this case) is because the only data you have to work with are measurements made in frame M. In order to transform the speed measurement of A with respect to M into a speed measurement of A with respect to B, you need to use the relativistic addition of velocity formula. It doesn't "affect the measurements"--it allows you to predict them. If B directly measured the speed of A, he'd of course get the same answer--without the need to use any Lorentz transformations.

I must also invoke it to calculate the speed of B as measured by A as it affects the measurements made by A. If I calculate the speed of M (my frame) as measured by B, I invoke the Lorentz transformation as it affects the measurements made by B.

The same "addition of velocity" formula applies here.

If B calculates the speed of B as measured by M, they invoke the Lorentz transformation as it affects the measurements made by M (me).

No. Since we are given the speed of B with respect to M, no calculation is needed to deduce the speed of M with respect to B. It's the same! (Opposite direction, of course.)

Now, being fully aware of the effects of SR on all measurements of constant linear motion, how do I consider the initial v of B or A, input into the Lorentz factor and applied to the Lorentz transformation of the measurements made by M(me) on the axis of motion and rate of time, is "valid" if I by the definition of relative motion, my(M's) measurement of v must already be affected by the Lorentz transformation?

You lost me on that last sentence. But read my comments above and see if you still have a question.

 

[Histspec]

Let me restate the question using your suggestion of M as my frame.
If I (M) calculate the speed of A as measured by B, I must invoke the Lorentz transformation as it affects measurements made by B. I must also invoke it to calculate the speed of B as measured by A as it affects the measurements made by A.

Yes, if B measures the speed of A, und you want to now the speed of A in a third reference frame like M, you must apply the Lorentz-Transformation (or better the "relativistic velocity addition rule").

If I calculate the speed of M (my frame) as measured by B, I invoke the Lorentz transformation as it affects the measurements made by B. If B calculates the speed of B as measured by M, they invoke the Lorentz transformation as it affects the measurements made by M (me).

No. That is another situation which is completely different, because the relative speed between two observers does not change. So if B measures that the speed of M is -0.75c, than M measures that the speed of B is 0.75c. Only the sign +- of the velocity is changing and no Lorentz-transform is needed.

Now, being fully aware of the effects of SR on all measurements of constant linear motion, how do I consider the initial v of B or A, input into the Lorentz factor and applied to the Lorentz transformation of the measurements made by M(me) on the axis of motion and rate of time, is "valid" if I by the definition of relative motion, my(M's) measurement of v must already be affected by the Lorentz transformation?

There is no problem - see the explanation above.

 

[Chrisc]

The principle of relativity removes the property absolute rest from the laws of mechanics. If M and B have a relative speed of 0.75c, neither can say they are at rest or in motion. Such a claim by either would be a nonsensical statement without an absolute frame of rest from which to make a measurement. The property rest is then a relative measure. If the equations of mechanics are upheld in a frame, the coordinates of the frame hold the property rest. This is the key, the frame is not at rest in the classical sense, the frame holds the property rest with respect to the equations of mechanics being upheld in (accurately predicting) the measured motion (mechanics) of bodies.
To say a frame is at rest or in constant linear motion is nonsensical except with respect to another frame. There is no test of mechanics that will distinguish rest from constant linear motion, because there "is" no such distinction.
So with this line of reasoning, let me restate the question.
I am in the frame M speeding toward B at 0.75c. A is also speeding toward M in the same direction at 0.75c and therefore speeding toward B at 1.5c. But B does not measure the speed of A to be 1.5c, since B's measurements are contracted and time is dilated with respect to M, B will always measure the speed of A to be less than c.
If your scratching your head because I said M was speeding toward B, but then I attributed the contraction and dilation to the measurements made by B, you will see my point.
How does M, between A and B, both speeding toward M at 0.75c, reason and predict the measurement of the speed of A as made by B will be less that c due to the contraction and dilation suffered at B, when the same reasoning would require the contraction and dilation should have been suffered at M and not B in the previous example?
If you're thinking it is simply a matter of relative motion, you are stating my original question. Why does M not prime the measurement of the speed of B with the Lorentz transformation before considering the speed of A as measured by B?

 

[Histspec]

The principle of relativity removes the property absolute rest from the laws of mechanics. If M and B have a relative speed of 0.75c, neither can say they are at rest or in motion. Such a claim by either would be a nonsensical statement without an absolute frame of rest from which to make a measurement. The property rest is then a relative measure. If the equations of mechanics are upheld in a frame, the coordinates of the frame hold the property rest. This is the key, the frame is not at rest in the classical sense, the frame holds the property rest with respect to the equations of mechanics being upheld in (accurately predicting) the measured motion (mechanics) of bodies.
To say a frame is at rest or in constant linear motion is nonsensical except with respect to another frame. There is no test of mechanics that will distinguish rest from constant linear motion, because there "is" no such distinction.

It seems that you have got problems to understand the relativity principle and the definition of reference frames. Of course the observer in M can say he's at rest an B is in motion; and also B can say he's at rest and M is in motion. And because both can say this, the relativity principle is fulfilled. This is as true for the Galilei-transformation (without electrodynamics) as it is true for the Lorentz-transformation. So, when we say A or B or M are at rest, we mean that we measure those conditions with instruments (rods and clocks) which are at rest in specific inertial reference frames. So let us discuss your example more precisely:

I am in the frame M speeding toward B at 0.75c. A is also speeding toward M in the same direction at 0.75c and therefore speeding toward B at 1.5c. But B does not measure the speed of A to be 1.5c, since B's measurements are contracted and time is dilated with respect to M, B will always measure the speed of A to be less than c.
If your scratching your head because I said M was speeding toward B, but then I attributed the contraction and dilation to the measurements made by B, you will see my point.
How does M, between A and B, both speeding toward M at 0.75c, reason and predict the measurement of the speed of A as made by B will be less that c due to the contraction and dilation suffered at B, when the same reasoning would require the contraction and dilation should have been suffered at M and not B in the previous example?
If you're thinking it is simply a matter of relative motion, you are stating my original question. Why does M not prime the measurement of the speed of B with the Lorentz transformation before considering the speed of A as measured by B?

To define the inertial reference frames, let us state: A is resting in M1; B is resting in M2; C is resting in M3.

a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c, that the speed of A is -0.75c, and that the speed of B is 0.75c. Therefore in M3 the relative speed between A and B is 1.5c. (Of course, 1.5c is not a signal velocity and therefore there is no violation of special relativity).

b) Using rods and clocks which are at rest in M1 we measure that the speed of C is 0.75c, that the speed of A is 0c, and that the speed of B (using relativistic velocity addition) is 0.96c. Therefore in M1 the relative speed between A and B is 0.96c.

c) Using rods and clocks which are at rest in M2 we measure that that the speed of C is -0.75c, that the speed of A (using relativistic velocity addition) is -0.96c, and that the speed of B is 0c. Therefore in M2 the relative speed between A and B is 0.96c.

That's all.

 

[Doc Al]

I am in the frame M speeding toward B at 0.75c.

If you are in the frame M, then B is speeding towards you. (You are at rest with respect to yourself.)

A is also speeding toward M in the same direction at 0.75c and therefore speeding toward B at 1.5c.

No, A is not speeding toward B at 1.5c--meaning: The speed of A as measured by B (and vice versa) is not 1.5c.

But B does not measure the speed of A to be 1.5c, since B's measurements are contracted and time is dilated with respect to M, B will always measure the speed of A to be less than c.

Since M is moving with respect to B, B will have to use the Lorentz transformations in order to interpret any measurements of A made by M.

If your scratching your head because I said M was speeding toward B, but then I attributed the contraction and dilation to the measurements made by B, you will see my point.

Not yet.

How does M, between A and B, both speeding toward M at 0.75c, reason and predict the measurement of the speed of A as made by B will be less that c due to the contraction and dilation suffered at B, when the same reasoning would require the contraction and dilation should have been suffered at M and not B in the previous example?

I don't see your reasoning at all. M does not require any "transformations" to interpret data that he has directly measured: The speeds of A and B with respect to M are given in the problem! Of course, anyone (in any frame, not just M's) who wishes to calculate--based on the given data--the speed of B as measured by A will have to use the "addition of velocities" formula. So?

If you're thinking it is simply a matter of relative motion, you are stating my original question. Why does M not prime the measurement of the speed of B with the Lorentz transformation before considering the speed of A as measured by B?

I don't see your point at all. If A moves at 0.75c with respect to M, then M moves at 0.75c with respect to A.

 

[russ_watters]

The principle of relativity removes the property absolute rest from the laws of mechanics.

Yes.

If M and B have a relative speed of 0.75c, neither can say they are at rest or in motion. Such a claim by either would be a nonsensical statement without an absolute frame of rest from which to make a measurement.

You have that precisely backwards. Knowing and accepting that there is no absolute rest or motion means either can be said to be at rest or in motion.

It seems like you make the statement that there is no absolute rest/motion, but still won't let the idea go.

The property rest is then a relative measure.

Yes. As is motion.

 

[Chrisc]

I think we're getting down to the point of confusion or disagreement.
Russ, you claim I have it "precisely backwards" and yet you claim motion and rest are relative measures.
Let me state what I think you are saying in that it may help make more clear what I am saying.
You say "Knowing and accepting that there is no absolute rest or motion means either (frame) can be said to be at rest or in motion".
I agree with that statement as far as it goes, but by itself it clarifies nothing. But take it as a true statement about a specific frame (which you seem unwilling to do) and you will find - I (M) am in motion and B is at rest. This is a valid observation, it does not change the quantitative values of dimension expressed by the Lorentz transformation between me (M) and B.
Doc Al, you claim "You are at rest with respect to yourself". By this I'm guessing you mean that relative motion must be a quantitative property of change you assign to the observed frame instead of your own frame. If that is what you mean, I don't know why you would make this distinction. If you could prove this distinction, you would be proving the existence of an "absolute" frame of rest. If you want me to disprove it I would only suggest that if you are at rest with respect to yourself then you must according to you and Russ, be in motion with respect to yourself. It becomes clear that "with respect to yourself" is "self-referential" and of no meaning at all.
When I state the properties of a frame, I make the assumption the frame is, or extends from an internal (note that is internal, not inertial) observer. This means there is no concept of "with respect to yourself" that is not "with respect to your frame of reference". Your frame cannot "claim" to be at rest or motion except with respect to another frame, but it "always" possesses the property rest (or constant linear motion) with respect to the equations of mechanics.
Once another frame is identified the property rest and motion only serve to establish a quantitative transformation of the dimensions from one frame to the other.
Before I go on to the specific question again, can you verify or clarify what I've said in terms of what you are arguing against.

 

[Doc Al]

If you think you can have a non-zero speed with respect to yourself, I don't think there's much to talk about and little point in continuing.

 

[Chrisc]

Doc Al, I don't.
It is now clear I don't even know what you mean by the phrase "I am at rest with respect to myself".
Please explain.

 

[matheinste]

Hello Chrisc.

From #10 by Histspec.

------------------------------------------------------------------------------------

To define the inertial reference frames, let us state: A is resting in M1; B is resting in M2; C is resting in M3.

a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c, that the speed of A is -0.75c, and that the speed of B is 0.75c. Therefore in M3 the relative speed between A and B is 1.5c. (Of course, 1.5c is not a signal velocity and therefore there is no violation of special relativity).

b) Using rods and clocks which are at rest in M1 we measure that the speed of C is 0.75c, that the speed of A is 0c, and that the speed of B (using relativistic velocity addition) is 0.96c. Therefore in M1 the relative speed between A and B is 0.96c.

c) Using rods and clocks which are at rest in M2 we measure that that the speed of C is -0.75c, that the speed of A (using relativistic velocity addition) is -0.96c, and that the speed of B is 0c. Therefore in M2 the relative speed between A and B is 0.96c.

That's all.

--------------------------------------------------------------------------------------

I don't think you could wish for a more logically presented explanation.

Matheinste.

 

[Doc Al]

Doc Al, I don't.

You don't what?

 

[Chrisc]

I do not "think I can have non-zero speed with respect to myself."
As I said, I don't know what you mean by this.

 

[Chrisc]

matheinste, I think you are missing the point.
You and Histspec begin your "logic" with:
"a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c."
With respect to what?

 

[matheinste]

Hello Chrisc.

Are there more than one of you ?

Matheinste.

 

[Chrisc]

Hello matheinste.
No, well... not unless you count the alien on my shoulder.
But don't tell Doc Al or anyone else about him, they wouldn't understand.

Your point is not lost, but you are assuming the laws of physics distinguish rest from constant linear motion. If you can state on the strength of your measurements that your frame is at rest, and in that I mean as distinguished from constant linear motion, please tell me how.

 

[Histspec]

matheinste, I think you are missing the point.
You and Histspec begin your "logic" with:
"a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c."
With respect to what?

Really, you need to learn to work with reference frames:
a) M3 is an inertial reference frame (i.e. some sort of coordinate system).
b) We have many clocks and rods, which are at rest in respect to M3.
c) Then we have many other frames (including M1, M2). In those frames we have many clocks and rods as well. Those instruments are at rest in respect to M1 or M2.
d) And with all these instruments we measure the velocities of the A, B and C. So please read my post (#10) again.

 

[Doc Al]

Your point is not lost, but you are assuming the laws of physics distinguish rest from constant linear motion. If you can state on the strength of your measurements that your frame is at rest, and in that I mean as distinguished from constant linear motion, please tell me how.

All we are saying is that it is at rest with respect to itself. Of course it's in constant linear motion with respect to some other frame. No one is proposing a way to distinguish rest from constant linear motion in any absolute sense. That would be quite antithetical to relativity!

 

[matheinste]

Hello again.

We cannot say, or determine whether a frame is at rest. I do'nt believe it even has a meaning. We can however say if we, or anything else, is at rest ( or moving ) relative to a given inertial frame.

If you are at rest in an inertial frame you are at rest relative to all other objects at rest in that frame. Therefore, you being one of the objects at rest in that frame, are at rest relative to youself.

Matheinste.

 

[Chrisc]

Doc Al, you've said this ("at rest with respect to itself") a few times now and I have asked you to explain it. If this is what I am missing, I would be grateful for an answer.
Let me set the stage and you finish it.
You are in free space (no gravitation) with any number of identical measuring rods and identical clocks. You measure out 100 meters on three perpendicular axis x, y, z.
Now you will do some tests to show me this frame is at rest?

 

[paw]

...begin your "logic" with:
"a) Using rods and clocks which are at rest in M3 we measure that the speed of C is 0c."
With respect to what?...

With respect to M3!

There are an infinite number of inertial reference frames. We are free to pick the one that is most useful to us. For the purposes of the system you are trying to analyse the frame where C is at rest is the most logical frame to use. It eliminates any motion on the part of C. Therefore, by definition, C is not moving with respect to M3.

The logical frame to use for analysing A's motion is M1, the frame where A is at rest and the logical frame to analyse B's motion is M2 where B is at rest. The results of these choices have already been given correctly.

 

[Doc Al]

Doc Al, you've said this ("at rest with respect to itself") a few times now and I have asked you to explain it. If this is what I am missing, I would be grateful for an answer.
Let me set the stage and you finish it.
You are in free space (no gravitation) with any number of identical measuring rods and identical clocks. You measure out 100 meters on three perpendicular axis x, y, z.
Now you will do some tests to show me this frame is at rest?

No tests are required! It's true "by definition" of what it means to be a reference frame.

You are in a rocketship traveling 0.6c with respect to the earth. What's the speed of the rocketship in the frame of the rocketship? Zero of course. (How can it be anything else?) That's all I'm saying.

 

[Chrisc]

matheinste, thank you. That is what everyone seems to be missing.
If your frame is at rest it is only a statement about the relative motion of another frame.
Likewise, if your frame is in motion it is at rest relative to all other frames of identical motion.
If only one or two or some number of frames you reference, have identical motion to yours, you are all at rest with respect to each other. So, is your frame at rest or in motion?
It is irrelevant, nonsensical and of no use, except in determining the motion between three frames of different motion. All three will measure dimension differently and to predict what one will measure of another you must make the first translation from your frame to theirs.

 

[matheinste]

Hello Chrisc.

------matheinste, thank you. That is what everyone seems to be missing.-----

I don't think they were missing the point I thought that was what most people were trying to tell you and i was merely agreeing with them. I am sorry but i am a little lost now. Probably my misreading somewhere.

Mateinste.

 

[DrGreg]

Doc Al, you've said this ("at rest with respect to itself") a few times now and I have asked you to explain it. If this is what I am missing, I would be grateful for an answer.

As has been said before, it's meaningless to say simply "A is at rest" (without qualification). But it is not meaningless to say "A is at rest with respect to B". All it means is that the distance between A and B, measured by B, remains constant over time. (Or to be technically accurate I should say "displacement" rather than "distance"; B moving in a circle around A doesn't count.)

With that definition, it should be obvious that A is at rest with respect to A. The distance between A and A, measured by A, remains constant over time, and is zero.

 

[Chrisc]

You're saying the origin of a frame is at rest with respect to itself?
I can't even imagine such a statement in physics.

I think I stated the question so poorly from the beginning it has led to clarification rather than answers.
I will start a new thread so as not to confuse anyone else with this one.

 

[Mettra]

You're saying the origin of a frame is at rest with respect to itself?
I can't even imagine such a statement in physics.

This is what the guy is trying to say:

Forget this problem you have posed. Imagine now that you are an observer. You're not experiencing any acceleration (this is SR after all). Therefore, you can plop down a coordinate system and call yourself the origin.

Since you are an observer, there is NO REASON for you to move relative to your origin. It would be entirely NONSENSICAL for an observer to move with respect to his origin. All that does is complicate matters - it adds another frame of reference to the problem.

Now let's consider a simple example problem to demonstrate what I'm going on about.

You are floating in space one day, and you see a comet moving. You think to yourself that you'd like very much to measure it's velocity and so on. So you plop down an x-y coordinate system so that the comet is moving in your x-y plane.

Now, you are observer A, and the comet is observer B. What is the comet's speed? You can check using your x-y coordinates. If there is another guy on the comet, he can also check your speed with respect to him. He has plopped down his own x'-y' coordinates to see how fast you are moving (with respect to him, obviously).

Now let's say that you accidentally hit the rocket booster on your suit for .1 seconds. What does the guy on the comet see? He sees you accelerate very slightly and then end up with a different uniform velocity. He can then measure it just fine in his coordinate system.

You, on the other hand, notice something odd about the comet's movement. It has suddenly changed its velocity (note that I'm ignoring the acceleration to simplify the problem, it isn't kosher). Then you notice that the x-y coordinates that you plopped down earlier are all wrong! They're moving! So you pick them up and plop them down again in their proper place with the origin at your feet.

----------
Measuring something with a moving coordinate system is nonsensical.

If you roll a ball across a room and want to measure its velocity, you don't string up a coordinate system from the ceiling and then fly it towards the ball so that the ball and coordinate system will intercept each other. You paint the coordinate system on the floor and you stand on the floor. You don't run past the coordinate system when the ball rolls along it either - all that silly stuff just makes it difficult to measure it.

Of course you can include an arbitrary number of reference frames and always end up with the correct answers, but there's no reason to overcomplicate a simple problem other than as a theoretical exercise.

 

[DavidWhitbeck]

It's like asking what is the velocity of the center-of-mass in the center-of-mass frame. The correct answer would be "it's zero by definition." I don't see what's so hard to grasp about being stationary with respect to one self. I appear to be stationary in my reference frame while I sit in front of my laptop, but relative to an observer in space, I appear to be moving since I'm in a frame co-rotating with the Earth. Just because that observer noticed that I am moving relative to him did not invalidate my own observation that I'm stationary relative to myself.

 

[Chrisc]

Mettra, you have just explained the "nonsensical" use of the phrase "at rest with respect to yourself"
It is nonsensical because the only alternative (if you even want to contemplate such an extraneous complication of reason) is that you are "not" at rest with respect to yourself. This would mean you are now moving with respect to yourself. Motion and rest are relative measures. You can no more be at rest with respect to yourself than you can be in motion with respect to yourself.

DavidWhitbeck, substitute "relatively, or relative to" for the phrase "with respect to" and you will find the statement now reads: "I am at rest relative to myself". Rest and motion are relative measures. To measure yourself relative to yourself in either state would not only require two of you, both with infinitesimally small rulers, but both of you would simultaneously prove the existence of two frames.
I am not unaware of what you and Mettra are trying to say, but you are misstating, and/or misunderstanding the principle of relativity. There is nothing "one" thing can have, possesses or exhibit relative to "itself"..

 

[DavidWhitbeck]

No Chris, that's simply wrong. If you can't a priori assume reflexivity then relative velocity would be ill-defined. What you wrote sounds less like a reasoned argument, and more like a paradox that you got yourself stuck in.

 

[Chrisc]

That's an interesting comment.
I never shy away from an "interesting" philosophical discussion, but you've stumped me here.
Are you implying that the principle of relativity should stand the test of reflexivity in so much as anything relative to anything else must also be relative to itself?

 

[matheinste]

Hello Chrisc.

------There is nothing "one" thing can have, possesses or exhibit relative to "itself"..-----

How about equality? Or being in the same place etc.

Matheinste.

 

[Mettra]

That's an interesting comment.
I never shy away from an "interesting" philosophical discussion, but you've stumped me here.
Are you implying that the principle of relativity should stand the test of reflexivity in so much as anything relative to anything else must also be relative to itself?

With all due respect, if you want to have an inevitably meaningless semantic argument, you should choose an audience other than physicists. You asked several questions, and others have responded in order to try to be on the same wavelength with you, even providing examples and so on.

However, with each progression, you have cherry-picked a particular wording or phrase and taken it to mean something other than what is immediately obvious in context with the rest of the posts (and all in the greater context of physics). As you can see, this leads us nowhere. If you have a physical question, you can ask it here. If you have a question about the physical ramifications of some realm of physics, there are knowledgeable people here that are willing to help you.

Even though we're not all top-notch English-major-grade psychology-mastering communicators, we make up for these failings by using everyday words along with words that mean something specific in a physics context. Try to work with others here in that regime. These aren't the philosophy forums, after all.

 

[Chrisc]

Mettra, I understand your frustration, I have been the victim of too many trolls in my time.
But I think if you read the whole thread more carefully you will see you've misjudged me.
The content of the posts on this thread are not meaningless and certainly not semantics.
Of all the physical sciences, the buck stops at physics. All other physical sciences
reach the bottom end of their realm of expertise and hand off the torch to the next.
Physics does not have such a privilege. Below the fundamentals of physics is philosophy
which is why it plays such a large role in theoretical physics.
If you are interested, read the short introduction to Carlo Rovelli's "Unfinished revolution"
http://fr.arxiv.org/abs/gr-qc/0604045
Below is a quote from same.

"The search for a quantum theory of gravity raises once more old questions
such as: What is space? What is time? What is the meaning of “moving”? Is motion to be defined
with respect to objects or with respect to space? And also: What is causality? What is the role of the
observer in physics? Questions of this kind have played a central role in periods of major advances
in physics. For instance, they played a central role for Einstein, Heisenberg, and Bohr. But also for
Descartes, Galileo, Newton and their contemporaries, as well as for Faraday and Maxwell.
Today some physicists view this manner of posing problems as “too philosophical”. Many physicists
of the second half of the twentieth century, indeed, have viewed questions of this nature as irrelevant.
This view was appropriate for the problems they were facing. When the basics are clear and the issue
is problem-solving within a given conceptual scheme, there is no reason to worry about foundations: a
pragmatic approach is the most effective one. Today the kind of difficulties that fundamental physics
faces have changed. To understand quantum spacetime, physics has to return, once more, to those
foundational questions."

 

[DavidWhitbeck]

Mettra is right, you set up straw man arguments in all of your replies. Would you stop being argumentative and show enough respect to reply to the heart of what others say?

You don't impress anybody by quoting Rovelli, he wasn't defending your pontifications, he was defending the necessity of having mathematical physicists concerned with foundational issues of quantum gravity.

I'm going to avoid your threads in the future, and I hope that others do the same.

 

[phyti]

Today some physicists view this manner of posing problems as “too philosophical”.

...and still earn a Ph.D!

 

[Chrisc]

matheinste, I am not ignoring your last post, I've just been considering the value of continuing this thread.
As I said in post # 10, this thread has little to do with my original question, but after considering the nature of the posts it would be too easy to just walk away from them. So I'll attempt to answer yours with all the others as they are all arguing against the same point.

These forums are host to a great number of people of great diversity from around the world. Langauge is going to be an issue from time to time. But language should not be used as an excuse or intentionally misused in any serious discussion of physics.
The people that disagree with me in this thread appear to to take one of two positions:
1) those that have not thought through the concepts and dismiss the argument as trivial, semantics or simply argumentative intentions.
2) those that actually think through the concepts and disagree with the physical application.

To all of you who are of the first opinion, I can only suggest you think a little longer. Dismissing concepts such as these will only limit your opportunity to be or understand the next Einstein or Newton. I thought quoting Rovelli in my last post would make this point, but clearly some of those who dismiss the concepts discussed here are just as happy to dismiss the evidence. That is unfortunate not only for them but possibly for physics as a well as no one knows who will start the next revolution. It may be someone that dropped out of school, someone with PhD, or a lowly patent clerk.

To all of you who are of the second opinion, thank you for putting in the effort to express logic, reason and examples. It's easy to dismiss or abandon your conclusions in the face of intimidation or ridicule, it's much harder to persevere and demand reason. If through reasoned argument, I change your mind or you change mine, we have both accomplished something worth the time invested.
So for those of you still as certain of your argument as I am of mine, I'll try again.

The core of this argument is the physical meaning of the concept of relativity. I have taken exception to "physics" accepting a concept that has no "physical" meaning and one that only serves to confuse a basic principle (relativity) which is too fundamentally important to let it suffer from language or be trivialized "as" semantics.

The specific error has been expressed as: an observer is at rest relative to himself.
If something is said to be relative, it is relative to something else. A person can be tall relative to their age, heavy relative to their height, or any similar comparative, dependancy or conjugation of two or more properties or attributes. A person cannot be tall relative to their height, old relative to their age, or possesses any other property expressed as a comparison to itself.
It is easy to understand how this is taken a semantics, but it is crucial to understand how this concept of relativity applies to the laws of physics. A person as an observer is a frame of reference even if not clearly geometrically defined. A person or frame of reference cannot possesses a property that is qualified or quantified relative to itself. i.e. a frame of reference cannot be large relative to its size, or fast relative to its speed. Similarly a frame of reference cannot be at rest relative to its state of motion, therefore: a frame of reference cannot be at rest relative to itself.
The only posts that have made any clarification of this statement are those that attempt to make the statement a reference to a person or frame within a frame. In this sense a person may be at rest relative to another frame that encompasses their frame. But in this case there are still two distinctly different frames which allow for the comparison of one relative to other. This is not the same as one being relative to itself.

 

[DrGreg]

The specific error has been expressed as: an observer is at rest relative to himself.
If something is said to be relative, it is relative to something else. A person can be tall relative to their age, heavy relative to their height, or any similar comparative, dependancy or conjugation of two or more properties or attributes. A person cannot be tall relative to their height, old relative to their age, or possesses any other property expressed as a comparison to itself.

By your logic, then, because 2 \neq 2 is nonsensical, then 2 = 2 is nonsensical too!

 

[Dale]

The specific error has been expressed as: an observer is at rest relative to himself.

In the context of SR, the phrase "X is at rest relative to Y" is clearly understood to mean "X is at rest in the rest frame of Y". So "an observer is at rest relative to himself" simply means "an observer is at rest in his own rest frame".

Such a statement is trivially true, but it is not ambiguous or ill-formed in any way.

 

[paw]

A person as an observer is a frame of reference

I think this may be the root of your problem in understanding what people are saying to you. A person (observer) is not a frame of reference. An observer is just that; an observer. A frame of reference is a coordinate system with certain properties that the observer uses in order to analyse a physical system. The observer can most definitely be at rest with respect to the chosen frame.

As I mentioned in a earlier post (which you didn't comment on) there are an infinite number of possible frames even when we restrict their properties to those that are inertial. The observer is free to choose any of those frames to analyse the physical system in SR. Since the observer can choose any of an infinite number of frames it only makes sense to choose one that simplifies the analysis. Therefore we usually choose a frame in which the observer is at rest.

 

[Chrisc]

No, DrGreg, you are exchanging a property for a quantity and relativity for equality which is not the argument. This is not a question of mathematical value or equivalence.

DaleSpam, Yes, your first statement is true as I and others have stated. I have no argument with specifying an observer at rest in a frame. But as for your second comment that this is not ambiguous in any way, that is what has become the ambiguity and specifically in error.
From post # 11 on, the idea of an observer at rest in a frame, has become "a frame at rest with respect to itself."
If that was just a momentary laps of reason (as I have had many times) I would not be still posting in this thread. When I asked in #15 for an explanation, the thread took a turn where a number of people insisted the principle of relativity refers to the property of a frame ...(fine)...with respect to itself...(NOT FINE).

paw, what I said above is the root of my problem. You're right, an observer is not a frame, but this method is commonly used to avoid the kind of confusion seen in this thread, If you define an observer "in" a frame as opposed to the frame itself, you then must define the observer at rest in the frame. To be at rest in a frame requires all the concepts and constructs of SR necessary to define the frame of the observer with respect to the frame they are in, as is necessary to then define that frame with respect to another. If an observer is not at rest relative to the frame they are in ... what's the point of putting them in the frame? That is why we use the term - frame of "reference".

 

[Dale]

DaleSpam, ... But as for your second comment that this is not ambiguous in any way, that is what has become the ambiguity and specifically in error.

Sorry, I just don't see any ambiguity whatsoever here. All of the following equivalent statements are well-defined and trivially true:

"an observer is at rest relative to himself"
"an observer is at rest in his own rest frame"
"the velocity of an observer is zero in the frame where the observer's velocity is zero"

Do you see any ambiguity or confusion in any of the following three concepts:

1) The velocity of reference frame A wrt reference frame B
(The first time derivative of the transformation equation from B to A)

2) The velocity of object A wrt reference frame B
(the first time derivative of the position of A in reference frame B)

3) The velocity of object A wrt object B
(the first time derivative of the position of A in the reference frame where the first time derivative of the position of B is 0 - a.k.a. the rest frame of B)

In all cases if A=B the velocity is trivially 0 and is not ambiguous or poorly-defined in any way.

 

[phyti]

It seems to be semantics.
Motion is defined as the change of position of one object relative to that of another. The definition cannot apply to a single object (the observer), therefore the statement
'the observer is at rest relative to himself' is meaningless.
An inertial frame by it's definition cannot contain any motion.
If you want to start with first principles, begin with the definitions.

The observer/(detection device) is a frame of reference,it's what the relativity theory is all about, the transformation of one observers description to that of another. He could be in a spacesuit floating in remote space, there is no need of a physical 'frame'.

 

[Dale]

'the observer is at rest relative to himself' is meaningless.

No it isn't. It means "the first time derivative of the position the observer is 0 in the reference frame where the first time derivative of the position of the observer is 0". It is trivial, not meaningless.

I have a hard time believing that you and chrisc really want to turn this ridiculous assertion into an argument. It is simply stunning to me that you seem to have so much trouble with such a trivial matter after it has been explained to you.

You both do realize that 0+0=0, right? And that d/dt 0 = 0 also, right?

 

[paw]

...but this method is commonly used to avoid the kind of confusion seen in this thread...

Your definition of a 'frame of reference' is clearly different to what I was taught. I'd respectfully submit that your definition is the one that's causing the confusion. I guess you can use whatever definition you like but then don't be surprised when you get contradictory results.