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Let there be given a stationary rigid rod; and let its length be l as measured by a measuring-rod which is also stationary. We now imagine the axis of the rod lying along the axis of x of the stationary system of co-ordinates, and that uniform motion of parallel translation with velocity v along the axis of x in the direction of increasing x is then imparted to the rod. We now inquire as to the length of the moving rod, and imagine its length to be ascertained by the following two operations:—(a) The observer moves together with the given measuring-rod and the rod to be measured, and measures the length of the rod directly by superposing the measuring-rod, in just the same way as if all three were at rest. (b) By means of stationary clocks set up in the stationary system and synchronizing in accordance with § 1, the observer ascertains at what points of the stationary system the two ends of the rod to be measured are located at a deﬁnite time. The distance between these two points, measured by the measuring-rod already employed, which in this case is at rest, is also a length which may be designated “the length of the rod.” In accordance with the principle of relativity the length to be discovered by the operation (a)—we will call it “the length of the rod in the moving system”— must be equal to the length l of the stationary rod. The length to be discovered by the operation (b) we will call “the length of the (moving) rod in the stationary system.” This we shall determine on thebasis of our two principles, and we shall ﬁnd that it diﬀers from l.Current kinematics tacitly assumes that the lengths determined by these two operations are precisely equal, or in other words, that a moving rigid body at the epoch t may in geometrical respects be perfectly represented by the same body at rest in a deﬁnite position.We imagine further that at the two ends A and B of the rod, clocks areplaced which synchronize with the clocks of the stationary system, that is to say that their indications correspond at any instant to the “time of the stationary system” at the places where they happen to be. These clocks are thereforesynchronous in the stationary system.”We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks. Let a ray of light depart from A at the time ta, let it be reﬂected at B at the time tB, and reach A again at the time tA′. Taking into consideration the principle of the constancy of the velocity of light we ﬁnd that tB − tA = rAB/c − v and t′A − tB = rAB/c + v where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus ﬁnd that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

Question: what is "c + v" in the preceding quote?

t′A − tB = rAB/c + v

Question: what is "c + v" in the preceding quote? How can c + v even exist when c is supposedly the maximum v in the universe? How, then, did Einstein add the v of the moving rod to c?

Okay Einstein didn't say time dilated by different amounts on the two ends of a moving object with two light beams going in opposite directions, that would mean the clocks were moving at different rates. What he did say is that they would move at the same rate but show different times. He didn't say "lasers", he probably said lanterns but I was just using a modern equivalent which works better because the light is all in one beam. He might not even have used the rocket scenario but it's in videos showing Einstein's theory.

Einstein Fan Club

Looks pretty straight forward to me, the equations are exactly as I showed, a fraction with rAB on top and c+v on the bottom, anyone can easily download the article pdf.

Looks pretty straight forward to me, the equations are exactly as I showed, a fraction with rAB on top and c+v on the bottom

....the time for the beam to go from A to B would be longer than from B to A, nothing ambiguous about it....

Quote from: Centra on 25/01/2022 17:50:16Looks pretty straight forward to me, the equations are exactly as I showed, a fraction with rAB on top and c+v on the bottom, anyone can easily download the article pdf.Seem straight forward to you because you apparently do not understand arithmetic. Your quoted paper does not say c + v. That is like saying that 1\2 + 1 means 1 over 2 + 1.

Actually working it out from the paper changes my mind. It is a poor translation error. rAB is a length, and dividing that by c gives something in units of seconds. You can't add velocity to time since the units don't match, and the math works out when interpreted as rAB / (c ± v).As krypid correctly points out, we're computing the time it takes (in some other frame) for light to travel the length of the moving rod.That said, relativity very much allows valid expressions of speeds higher than c. Given a fast enough ship, I can get to Betelgeuse before I die, which is over 600 light years in maybe 20 years, which is a proper velocity of over 30c, despite my not moving at that velocity relative to any inertial frame. The theory doesn't forbid the expression of such values.

What Einstein was saying with that equation is that if the motion of the moving frame with the rod and clocks was

say, 100,000 km/s that the time for the beam to travel from A to B would be the length of the rod, say 1 km, divided by (300,000 km/s minus 100,000 km/s = 200,000 km/s) = 0.000005 second, and the time from B back to A would be 1 km divided by (300,000 km/s + 100,000 km/s=400,000 km/s) = 0.0000025 second

thus, they would conclude that the two clocks were not synchronized

because if they had been then both ways would have taken the same time, 0.0000033... second.

They would think that clock A was running at half the speed of clock B.

The obvious problem there is that it should in fact have taken the same amount of time both ways, because it should have been exactly the same as if they had been stationary.

Quote from: Centra on 25/01/2022 20:18:07What Einstein was saying with that equation is that if the motion of the moving frame with the rod and clocks wasThe rod is moving relative to the chosen frame. The clocks (four of them?) are not moving in this frame, and are in sync relative to this frame. You didn't explicitly say otherwise (yet), but I want to be clear. It isn't a 'moving frame' except relative to the proper frame of the rod.Quotesay, 100,000 km/s that the time for the beam to travel from A to B would be the length of the rod, say 1 km, divided by (300,000 km/s minus 100,000 km/s = 200,000 km/s) = 0.000005 second, and the time from B back to A would be 1 km divided by (300,000 km/s + 100,000 km/s=400,000 km/s) = 0.0000025 secondRight.Quotethus, they would conclude that the two clocks were not synchronizedThe clocks are presumed synchronized in the frame in question. It really does take light twice as long to go 4/3 km as it does to go 2/3 km.Quotebecause if they had been then both ways would have taken the same time, 0.0000033... second.No, not for a moving rod. That would be true only for a stationary 1 km object, and there's no stationary 1 km object in any frame in your example. The rod has a proper length of about 1.06 km in your example to get the numbers you quote.QuoteThey would think that clock A was running at half the speed of clock B.Nonsense. If that were true, a repeat of the experiment would yield different times to go from one end to the other, not .000005 and .0000025 seconds again. And the measurement can only be done once, after which the rod has passed by the clock. Maybe they can time multiple identical rods going by.QuoteThe obvious problem there is that it should in fact have taken the same amount of time both ways, because it should have been exactly the same as if they had been stationary.Once again, nonsense. The emission and detection events one way are twice as far apart as the emission and detection events the other way. Light cannot take the same time to go two distances, one twice the other.

We imagine further that with each clock there is a moving observer, and that these observers apply to both clocks the criterion established in § 1 for the synchronization of two clocks. Let a ray of light depart from A at the time tA, let it be reﬂected at B at the time tB, and reach A again at the time tA′. Taking into consideration the principle of the constancy of the velocity of light we ﬁnd that tB − tA = rAB/c − v and t′A − tB = rAB/c + v where rAB denotes the length of the moving rod—measured in the stationary system. Observers moving with the moving rod would thus ﬁnd that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous.

So why would the observers moving with the moving rod find the two clocks not to be synchronous?

Does light travel at different speeds in a moving frame, slower in the direction of motion and faster opposite the direction of motion?

That's exactly what Einstein said.

Shine a light the length of a rod and reflect it back, timing the round trip. If there was a 300 km rod with such a clock and mirror at opposite ends, light should take 2 msec to make the round trip if light moved at constant speed in both directions. But if it moved at say 150000 km/sec in one direction and 450000 km/sec in the other, it would take 2msec to move one way and 2/3 sec the other way, for a total of 2 2/3 msec, which would be empirically different than the 2 msec that Galilean relativity demands. At slower speeds the difference would be less, but measurable nevertheless with accurate enough devices, and as has been pointed out above, it has been measured to be frame invariant to at least 17 digits.This falsifies the Newtonian view of absolute time and space that you seem to naively have been pushing.

I know others can't speak for Einstein but everyone who has been posting here seems to support his theories

So your point is that if you were in a frame, say a very long rocket powered boxcar with no windows, on a track, with a 300 km rod with clocks attached to each end, and the boxcar was moving at 150,000 km/s relative to the track, that you could use beams of light between those two clocks to determine that you were in uniform motion rather than being stationary? You do realize that violates the postulate of reciprocity between inertial frames, right?

This falsifies the Newtonian view of absolute time and space that you seem to naively have been pushing.

Observers moving with the moving rod would thus ﬁnd that the twoclocks were not synchronous, while observers in the stationary system woulddeclare the clocks to be synchronous

Quote from: Centra on 26/01/2022 13:48:53So your point is that if you were in a frame, say a very long rocket powered boxcar with no windows, on a track, with a 300 km rod with clocks attached to each end, and the boxcar was moving at 150,000 km/s relative to the track, that you could use beams of light between those two clocks to determine that you were in uniform motion rather than being stationary? You do realize that violates the postulate of reciprocity between inertial frames, right?No, he's saying the opposite. The fact that the speed of light is frame-invariant means you'd always get the same time for a round trip regardless of whether the boxcar is moving or not (if you're in the same frame as the boxcar).