Place is a word that also means locates it, says it is at etc. Did you want to actually want to deal with the exact and specific math I presented yes or no.
SR Issue
- Thread starter chinglu
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Place is a word that also means locates it, says it is at etc. Did you want to actually want to deal with the exact and specific math I presented yes or no.
1) Link one shows I believe in using mathematical logic. I am not sure what you are trying to prove, but it is true.
2) Link two shows a link that describes evolution. If you disagree with what I posted from the link indicate why.
3) Your third links shows where I proved a case where t'=t for a light pulse emitted from the origins. If you can prove this relationship is false, have at it. You are going to lose.
You sure do keep a log on me all the way back to 2010, LOL.
I did not say this. Perhaps you should start by explaining in detail what words I used for you to come to such a preposterous conclusion.
I said C' and M (corresponding to lines h and f, respectively) "are only co-located at one event in all of space-time, event P" (event P happens to be the only event common to lines f and h). Since we are working with events that have $$y=z=y'=z'=0$$ we only have two degrees of freedom and so any two lines which are distinct and not parallel must meet at some zero-dimensional locus -- event P.
Look, you say that M' and M both see lightning located on "the positive x-axis". Doesn't M' have a positive x'-axis? If not, how can M and M' be in relative motion?chinglu said:
Yea, exactly what I am saying. But, you said the predicate C' and M are co-located is not universally true. Now prove it. Otherwise, retract your claim.
No, I never said "see". I use the term calculate.
And, they both have a positive x-axis. What is your point?
Ah, so M' doesn't see the same lightning in two places, but calculates the locations, and one of these is "wrong"?chinglu said:
Well, you keep implying there is only one, "the" positive x-axis. You don't distinguish the axes for M and M' when you present your "exact and specific" math.
1) I am not sure about your first part. The LP in M' claims the lightning is at one place along the positive x-axis of M' and ROS claims the lightning is at a completely different location along the positive x-axis of the M' frame if the condition "C' and M are co-located" is true.
2) Yes, there is one x-axis., It is common to both frames.
Wow.
Ok, your second response indicates we have a problem. In what sense can the x axis of a moving frame be "common to" the x axis of a stationary frame?
They can be parallel, but the moving frame takes its own axis with it, surely?
Ahem. From the OP, C' moves with respect to M with constant velocity v > 0. Therefore at all times prior to the co-location event, C' is a non-zero distance away from M and at all times after the co-location event, C' is also a non-zero distance away from M. Thus the co-location event is a point-like event of zero duration and zero spatial extent.
It follows directly from the definition of constant non-zero relative motion that two things that at one time are in the same place at all other times are in distinct places.
This is not the least bit controversial, if words are given their common meaning.
This is true for any frame, including Σ where M is at rest or Σ' where C' is at rest. No special definition of "time" is needed, because the duration of the co-location event is zero.
Technically, there is no part of special relativity used for this proof, so it is valid in classical kinematics as well.
Likewise it follows from geometry, algebra, or first-order logic.
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The author of the OP has used confusing notation. Rather than treating the four orthogonal axes as lines in space-time that meet at the origin event O, the author often thinks only of the 3 spatial axes (and of infinite duration) which meet at world-line f in frame Σ and world-line g in frame Σ'.
The x spatial axis is described as every event with $$y=z=0$$ while the x' spatial axis is described as every event with $$y'=z'=0$$, so both are the same 2-D locus of space-time events and are the same (imaginary) object. In fact, all events (O, P, Q, R) and all space-time lines (f, g, h, k, j and ℓ) are contained in this two-dimensional sheet. So the OP' author's limitation of thinking in only three spatial dimensions is inherently confusing.
'sigh'
I guess you have to learn to think Minkowski. I'm still struggling with how that diagram I keep posting (here it is again)
. .. indicates that the Lorentz transformation acts symmetrically since both t' and X' have the same angle wrt c = v. But then each frame sees the other as time dilated and length contracted, which sounds like an asymmetry. It's as if there's a direct relation between t and t' (they are equivalent) where t' has the dilation and this corresponds to the geometry in the diagram, but between X and X' this relation is inverted, although they are also equivalent spaces.
Yeah, and a Euclidean version of this diagram would have Cartesian coordinates, and no time axis.
I guess you have to learn to think Minkowski. I'm still struggling with how that diagram I keep posting (here it is again)
. .. indicates that the Lorentz transformation acts symmetrically since both t' and X' have the same angle wrt c = v. But then each frame sees the other as time dilated and length contracted, which sounds like an asymmetry. It's as if there's a direct relation between t and t' (they are equivalent) where t' has the dilation and this corresponds to the geometry in the diagram, but between X and X' this relation is inverted, although they are also equivalent spaces.
Yeah, and a Euclidean version of this diagram would have Cartesian coordinates, and no time axis.
The Lorentz transformation can be thought of a hyperbolic analogue of rotation (points move on hyperbolas, not circles).
Alternately it can be thought of stretching one diagonal by a factor of $$\sqrt{\frac{c + v}{c - v}}$$ and shrinking the other diagonal by the same factor.
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Alternately it can be thought of stretching one diagonal by a factor of $$\sqrt{\frac{c + v}{c - v}}$$ and shrinking the other diagonal by the same factor.
I really appreciate your ability to make worded arguments this clear and concise, and that you often refrain from using your vast mathematical skills (which I often can't follow).
Ditto. He shares mathematical physics along with the 'clear and concise' worded arguments. In detail. I'm always learning something when I read rpenner.
We don't have a problem. The x-axis is common to both frames. The markings are different frame to frame, but it is not a problem. That is basic SR.
You see, the lightning is in the positive x-direction for both frames from the origins.
We are in agreement.
AGREED
Wrong.
The light postulate is used to decide the position on the positive x-axis based in the M' frame if C' and M are co-located.
On the other hand, in the M frame if C' and M are co-located, the light postulate is used in the M frame to decide the M frame coordinates of the lightning. Then, LT is used to translate that coordinate to the M' frame.
All of this is pure SR.
You failed again.
Wrong.
The OP proves if C' and M are co-located, the lightning is at $$(d',0,0,d'/c)$$ in the M' frame. That is a fact of the light postulate and not subject to debate.
Yet, the M frame claims if if C' and M are co-located, the lightning is at $$(d'(1-v/c),0,0,d'(1-v/c)/c)$$ in the M' frame. SR gets the answer wrong period.
Now, some can say, that is ROS and indeed, it is. But, the answer is wrong period.
Ok (???). Well suppose I'm standing somewhere and "looking" in my x direction. Suppose also that someone else is walking past me in the same direction, x. How does this situation make this x direction "common to both frames"?chinglu said:
For me, the x direction and the direction the other person is walking remain the same (we can assume this person knows how to walk in a straight line), but can I really claim my x axis (an abstract straight line extending away from me) is the same as the walking person's? Suppose the person walking is also bouncing a basketball as they move, so to them the ball is moving straight up and down, and not moving along their x axis, whereas the ball does move along my x axis. How can they be the "same axis"?
To keep thinking that they are "the same" despite evidence of motion "looking different" in either frame is tantamount to insanity or something. You could I suppose, also revise what "the same" means in the context of axes or directions of motion (or something).
Ok, Galileo got locked up by the Pope, but later on we all learned the Pope was the crazy one (or maybe he was just protecting his precious Church and its dogma). Or something. Something, something . . . something else, yada yada . . .
Neddy Bate
Valued Senior Member
You can eliminate the apparent contradiction by using more careful wording:
At the time when C' and M are co-located, frame M' finds the light flash to be located at $$(x', y', z', t') = (d'(1-v/c),0,0,d'(1-v/c)/c)$$.
At the time when C' and M are co-located, frame M finds the light flash to be located at $$(x', y', z', t') = (d',0,0,d'/c)$$.
SR is fully equipped to handle two different frames disagreeing on which events are simultaneous. That is relativity of simultaneity (ROS).
It would be inconsistent with nature if SR claimed the light flash is at two different M' locations at the same time, but that is not what SR claims.
What SR really claims is that the light flash is at two different M' locations at two different times. That is not inconsistent with nature. The light is always moving at c, so at two different times, the light must be at two different M' locations.