Whist this is true, it is too wide a definition. Not all irrational beliefs are religious. I can find all sorts of people that believe in alien abductions, that the government is controlling their thoughts by radio waves, or that the government is controlled by the Bavarian Illuminati. These are not religious beliefs. Schizophrenics can come up with cogent arguments that once you assume the truth of their mania are completely rational. They are simply unable to shake the core axiom - which does make it irrational.
Religions are more focussed. Belief systems, that make specific claims about the nature of existence in a manner that is revealed to the believer as part of the religion. Whilst you can come up with personal religion, involving revealed truths - revealed only to you - this is probably indistinguishable from schizophrenia anyway. In general religion involves the revealing of truths to the believer. Usually revealed in the form of some holy writings. There is a collective belief in these truths that defines the core axioms of the religion. Whilst they are unassailable, and irrational, not every irrational belief is religious.
If you want to characterize any belief about the nature of reality as implicitly religious, you might try, but it isn’t a definition most people would recognise. Certainly painting a belief by the observer that they exit is not something most people would regard as a religious belief. Possibly you are trying to say that it is not rational to use any derivation from the observation of your own existence to claim anything more. Which is an interesting viewpoint. On that, without clearer substantiation might be tarred as being a somewhat religious viewpoint itself.
I think you are misreading what I write. A revealed “truth” is not and has never been considered objectively true to anyone except a believer. It is a technical term. One used in discussing religious beliefs.
This doesn’t work though. If you subtract the time shown on clock B from the time shown on clock C, you have measured the time it takes for light to get from B to C, plus the time it takes for light to get from C to B. This is two may measurement. When you ‘repeat’ the experiment, you will be measuring the time it takes light to get from C to B, plut the time it takes light to get from B to C. It is exactly the same. You are still measuring 2 way trips. Even if the light took longer to go one way than the other, the total time would be the same.
Yes , you are right. I thought it over and came up with a fresh idea. I would like to propose following experiment:Lets have two atomic clocks at points A and B. We can easily measure the time light travels from A to B and back (t2).Now lets send a light pulse from A to B which will start clock B, than t2/2 later lets start clock A. After short (arbitrary) time lets send another light pulse from A to B which will stop clock A and B. Now lets compare the readings on both clocks. If the speed of light is isotropic, the reading on both clocks should be the same.If the speed of light from A to B is faster than from B to A, clock A should indicate more time than B.Am I correct?
Point A fires a laser, waits t2/2 then the clock starts. It waits an arbitrary time (ARB) and fires another laser and the clock stops.
Because the time between the first and second laser being fired is, by definition, t2/2 + ARB, that is always the time between the two lasers arriving at B. You have completely taken the travel time out of the equation.
Clock A will always read the time ARB.
Clock B will always read the time ARB + t2/2 which you already knew.
I’m afraid you are right again. This should teach me a lesson to post something without checking. Thank you for doing it.
However I have another proposition: Lets say we have three clocks: A on the left, B in the middle (we can establish midpoint precisely since two way speed of light is constant) and C on the right. At t0 signal is sent to A and B to synchronise the clocks. So we have three clocks synchronised as follow:
B is at t0, A is at t0-tBA and c is at t0-tBC (tBA is time light travels from B to A). After arbitrary time tARB the signal from A is sent to stop the clocks. We should have the readings of the clocks:at tA: t0-tBA+tARB and at tC: t0-tBC+tAC+tARB. After subtracting tA-tB and noting that tBC=tAB and that tAB+tBA=1/2tACA; tACA is the time light travels from A to C and back we will get the result :tA-tC=2tAC-1/2tACA. I hope I’m right this time.
I’m afraid this is wrong again. But I think this should work:
Lets have two clocks, clock A on the left, D on the right ant two clocks C and D right in the middle. If we send light signals from A to B and from D to C we can synchronise the clocks A and D so the signal from A arrive at B at the same time as signal from D to C (this is quite trivial, so I won’t elaborate). Since both signals arrive at central location at exactly the same time, say at 12.00, they should leave A and D earlier by the amount of time needed to complete one way trip from A to B and from D to C (or B, because time difference between B and C is 0) respectively. So we will have the situation as follows:
tA=tB-tAB; tAB is the time light travels from A to B (tAB=tAD/2)
tD=tB-tDC; tDC is the time light takes to travel from D to C; tDC=tDB=tDA/2
tA=tB-tAD/2
tD=tB-tDA/2
Now if we send the signal at arbitrary time tARB from A to D to stop the clocks at A and D respectively we will have the readings:
tA=tB-tAD/2+tARB
tD=tB-tDA/2+tARB+tAD
lets subtract tA from tD:
td-tA=tAD-tDA/2+tAD/2
tD-tA=tAD-tDA/2+tAD/2 +(tAD/2-tAD/2); tDA+tAD=tADA (time for the light to travel from A to D and back; easy to measure;
so we will have:
tD-tA=2tAD-tADA/2
or explicite expession for the time light travels from A to D:
tAD=(tD-tA+tADA/2)/2
Martin – you may want to check the dates on the posts. Kaltkalt was looking at the wikipedia article as it existed at least two years ago. And they may not even be around this board at this point to respond.
It is not that hard to understand really. If the OP is still around to read this then he might respond, clearly people do read very old threads. If he has moved on then he won’t mind at all, because he won’t see my post. You might call it a local version of the anthropic principle.