Phys, your rhetorical spin tactics belong in religion or politics. They do NOT belong in a Science forum.
Every time you post, you write some ridiculous distortion that any science minded person sees right through. You are not debating SR at all. You are merely debating “PhysBang vs the World” and hoping to fake out the world.
Either sanely relate what you have to say to the OP or forget it. I’m not going to play rhetoric games with you on a Science forum.
Then do it for the OP scenario and explain how the STATION doesn’t “see” a consistent time dilation for each clock. Don’t forget to use your ∆t.
What would I use ∆t for in the OP?
The flashers move the same distance in the same time as the centered clock “∆x, ∆t”. They have the same dilation from the station clock. Explain why you think the station clock will read t’ of the clock different than it reads t’ from the flasher. Show what the clock would read “tR’” (don’t worry about “4:00”. Use whatever t you want) and what the flasher would read “tF’”. And since t’ for each using ∆t will be the same (tR’ = tF’), explain why the station would not see tR’ as tF’.
But ∆t doesn’t factor into that calculation. It doesn’t matter how far the clocks have moved, or for how long. We can measure the difference between t and t’ for each clock at t=0, regardless of how far they’ve been in inertial frames moving relative to each other.
So I take it that you lack the ability to answer my questions, even the most basic ones about whether or not you used the equation that you meant to use or what equation I got wrong.
As far as I can tell, you are the only one here responding to questions with rhetoric. You are dodging all the questions by merely attacking me.
Carl, the timer moves from -10 to 100 (let’s say). The train clock moves from 0 to 110. That means that their ∆t’ will be the same, correct?
But back at x = -10 and x = 0 (or whatever spacing is needed) we set both the flasher and the clock to be identical (let’s say = 0). So after they have each traveled ∆x = 110, they should still read identically to each other, because they both have the same dilation inside the train (td = 0) and also the same from outside the train (td = ∆t’/∆t). Correct?
But t’ indicates what the station will SEE as the time reading for any timer on the train, correct?
So since both timing devices were set to be identical and they both traveled the same ∆x and have the same ∆t’, the station will calculate the same t’ for each of them. That means that the station will SEE both of them read the same.
t’ indicates the ideal time in a reference frame in which the train is not moving. So if one can see the clocks set to that time, then they see t’.
But according to the station frame, the clocks in the front of the train read differently from the back of the train. If one adjusts the train clocks so that they match the station’s clocks, they no longer match each other according to t’.
Why this happens is explained, not merely stipulated, in some detail in
and in
fourmilab.ch/etexts/einstein/specrel/www/
and in
galileo.phys.virginia.edu/classe … izing.html
…
Phys, Carl should hire you as the Site Librarian so you don’t have to do any actual thinking. As you have been arguing against me, you have been arguing against Einstein.
After the Copenhagen convention, in a letter to Schrodinger, Einstein stated that only Schrodinger understood. He further stated that he “didn’t like what we are doing to Science.” By “we” he was referring to the Quantum Magi who first gave their “perception is reality” speech at Copenhagen. Einstein believed in aether and an absolute frame. But the QM said, “yeah but that is a different aether than what others mean”. They, like you, were using political spin to promote their “perception is reality” ideas, “magic and deception”.
In discussing the relativity of simultaneity, Einstein was talking about a perception issue of light taking longer to get to the non-moving observer from one source than another thus giving the perception of the flashes happening at a different timing. In fact, related to this station-train scenario, he explicitly stated;
In my scenario, the non-moving observer DOES stay at his position “M”. So not only are you misunderstanding Einstein, but also you are directly contradicting him when you claim that the station will not see the events as simultaneous.
To Einstein, “relativity of simultaneity” was merely a perception issue, an illusion that the Quantum Magi wanted to make into “reality”. Since that time, they have distorted Science into the mysticism and religion of Scientism, just as Einstein feared.
Are you high?
I can understand that you don’t really understand what Einstein meant. Try reading Chapter 8 ( bartleby.com/173/8.html ). However, in Chapter 9 Einstein is quite adamant: “Events which are simultaneous with reference to the embankment are not simultaneous with respect to the train, and vice versa (relativity of simultaneity). Every reference-body (co-ordinate system) has its own particular time; unless we are told the reference-body to which the statement of time refers, there is no meaning in a statement of the time of an event.”
To say that he is merely talking about perception is ludicrous.
Funny, I was going to ask you that same question, but I didn’t want to directly imply that you had to be either high or just stupid to make the kind of responses you were making.
I quoted Einstein directly. HE said that there would be no simultaneity issue if the person were stationary at M. And that it was merely an issue of the time it took for the light to travel to the observer that led to the relativity of [perceived] simultaneity.
As I said, you don’t belong on a Science forum.
But also, Einstein didn’t have a clear understanding of time. He described it as merely the turning of hands on a clock. And until just a few years ago, no one else seemed to have understood it very well either. An elite physicist got online (mid 90’s I think) and asked the question, “What is time?” After perhaps 100,000 responses, he declared that Man has no hope. He was right.
I didn’t hear of the question until perhaps around 2005. My first response was “It is merely how many times one thing revolved while another thing moved.” - Very similar to Einstein’s version. But after thinking on it for a while, I published that it is “a measure of relative change”. That is more precisely what it is. You might see that phrase online occasionally. It came from me (so be prepared to hate it too).
I’m tempted to say, “Welcome to Reality”, but I’m sure you haven’t really reached it yet.
“Now before the advent of the theory of relativity it had always tacitly been assumed in physics that the statement of time had an absolute significance, i.e. that it is independent of the state of motion of the body of reference. But we have just seen that this assumption is incompatible with the most natural definition of simultaneity; if we discard this assumption, then the conflict between the law of the propagation of light in vacuo and the principle of relativity (developed in Section VII) disappears.”
In the past, crackpots railed against Einstein and invoked the works of past scientists who had either been misinterpreted of maligned in the service of some kind of conspiracy. Now, it seems, there is a new kind of crackpot, one that wants to return science to the work of Einstein, work that has either been misinterpreted or maligned in the service of some kind of conspiracy.
Neither kind of crackpot seems capable of carrying out the actually intellectual demands of physics. The crackpot gets angrier and angrier the more they are forced to deal with the actual science and the inconsistency in their own position.
Again, you don’t belong in a discussion on Science. As I stated, as did Einstein, “time is a RELATIVE measure”. No one is claiming that it is absolute.
So keep your paranoia and “crackpot” claims to yourself.
I have to wonder what that author meant by;
I have little doubt that he meant, “If we think it was simultaneous, then to our [perceived] reality, it is.” Again, just a perception issue.
"If I think the thunder occurred long after the lightening, then in my frame of reference, it really did." The “relativity of sound”.
And in your case; “If I think that I didn’t say what I just said, then I really didn’t.” The "relativity of speech".
And; “If I think Einstein said something, then in my frame of reference, he really did.”. The “relativity of records”.
“If I think that word means something different, then it really does.” The “relativity of word definitions”.
“If I think that 2+2=3 then in my frame of reference, it really is.” The “relativity of mathematics”.
“If I think that you are insane, then in my frame of reference, you really are.” The “relativity of insanity”.
"If I think that relativity is only relative, then it really is." The “relativity of relativity”.
Dude James,
(edited)
You, Tom Hanks and Sean Penn all have something in common.
Right, but this doesn’t matter. The choice of any distance is arbitrary, and is not used in the Lorentz transformation. The Lorentz transformation operate on one point, transforming the position and time coordinates between two inertial reference frames at motion relative to each other.
We can only set the to be equal in one frame. If the train and the station were moving relative to each other, the clocks can’t be set to all read the same time in both frames.
Your emphasis of the word “see” is troubling. If t is the time in the train’s frame of reference, t’ is the time in the station’s frame of reference.
This difference is not a result of a difference in ∆t, it’s because a difference between t and t’. Simultaneity is not the same in both frames. As I said earlier, if the flasher clocks read 4:00 in one frame, they will read e.g. 3:59 and 4:01 in the other frame. If we add or subtract an equal ∆t or ∆t’ from those readings, it doesn’t change the fact that in one frame they’re simultaneous, and in the other frame they aren’t.
Well, one out of 4 is a start. 
Okay, let’s focus on only this one point of contention until it is resolved {I love Resolution Debating}.
If we had 2 people standing beside the tracks holding synchronous clocks with magnetic feet on them (on the clocks, not the men) when the train came by, and they simultaneously tossed their clocks onto the side of the railcar where the clocks stuck, assuming the train wasn’t made by Tonka, both clocks would very quickly accelerate up to the train’s velocity.
Now the clocks constitute a frame of motion of their own in that they are a constant distance from each other at all times. When they simultaneously touch the train, they simultaneously begin their acceleration up to velocity. They never change the distance from each other and thus never have any time dilation occur between them. Thus on the train, they must still read identically.
If you believe in length dilation, the 2 people can be standing a little closer to each other. That won’t change this concern.
A person standing a little further up the tracks who can read both clocks watches them being tossed. He sees them come in contact with the railcar still reading identically. At the very instant of contact, is there any reason for him to see one instantly become different from the other?
He sees them both accelerate identically and achieving the same ∆x with respect to the tracks as each other (∆x1 and ∆x2). Since they both maintain the same ∆x at all times, they should both experience the same dilation. Thus he sees them both appear to slow down identically. Is there any reason to see one slowing down more than the other? What would be the reasoning?
The person watches the clocks accelerate. For any ∆x as a clock accelerates, we can describe the differential dx and integrate the dx into a complete ∆x.
For any velocity during the constant acceleration a (with “iS{}” == integral sum);
a = v/dt
v = adt = dx/dt
dx = vdt = adtdt
∆x = iS{adtdt}, and
∆v = a*∆t
Both clocks experience the same acceleration, a, and the same dt thus;
∆x1 = ∆x2, and
∆v1 = ∆v2
∆t1 = ∆t2
t’ = L(t – vx/c^2)
∆t’ = L(t1 – vx1/c^2) - L(t0 – vx0/c^2)
= L((t1 – vx1/c^2) – (t0 – vx0/c^2))
= L(t1 – vx1/c^2 – t0 + vx0/c^2)
= L(∆t – (vx1-vx0)/c^2)
= L(∆t – v(∆x)/c^2)
Again, both clocks experience the same affects, thus;
∆t1’ = ∆t2’
So what the person should see for each clock would be;
ClockA = L(∆t – v(∆x)/c^2)
ClockB = L(∆t – v(∆x)/c^2)
Identical readings for each clock.
Hyperbole aside, Religion assumes that everyone is a moron. Science assumes that no one is a moron.
Living in the middle somewhere, in Religion we hate and in Science we debate.
If you can’t follow the math or logic, why bother to read it at all? The religion forum isn’t far away… “Dude”.
In the scenario you describe, the clocks would read equally in the train’s frame, and would read unequally in the station’s frame.
It seems what you’re getting caught up in is the period of acceleration between when the clocks are at rest relative to the station, and when they are moving at constant velocity relative to the station. I haven’t done the math, but I would predict that the change from t to t’ due to acceleration is equal to the momentary difference between t and t’ in the inertial case. What equations are you using to calculate the effect of acceleration on time dilation, and where did you get them? I don’t have a source that presents them straightforwardly, but eyeballing the derivation, I’d expect it to look something like
∫ 1/√(1 – v^2/c^2) (t – vx/c^2) dt
That is, integrating the Lorentz transformation over time. Acceleration in Special Relativity is basically handled as a sum of the differences between inertial frames.
I was merely going through each step of the scenario to cover all of the bases. But I think I see the issue here.
You agree that the clocks would accelerate together and stay in sync with each other, correct?
But then you say that the station will not read them equally. I went through those steps describing what the person on the tracks (or a nearby station) would SEE. He sees each step occurring to both clocks equally.
If t’ = the reading of the clock on the train according to the train (and I agree), then what is the equation for what the station sees as that reading?
t = the station’s time according to the station
t’ = the train’s time according to the train
?? = the train’s time according to the station (what the station “sees”)?
Why would the station see something different than what is there (forgiving the time it might take light to get from one to the other)? How could a hand pointing to the left appear to the station as a hand pointing straight up? Or how could a digital “1” appear to the station as a “3”?
Einstein’s work was all about how long it would take the light to get from one to another and thus leave the perception of different timing.