A pair of lightlike events would have different spacetime intervals in different frames if the spacetime interval is defined as Δs2 = Δt2 + Δx2.
Correct, the invariant would be zero but the timespace interval would not be zero. It just happens that, in lightlike cases, the time it takes light...
I have read Taylor and Wheeler's Spacetime Physics, I have solved many of the problems and I do appreciate the usefulness of the quantity (Δ t)2 − ( Δ x )2
This is my point:
The spacetime interval, as currently expressed, is not the same as Pythagorean theorem. It is typically implied as a necessary consequence of the geometry and that is misleading. It is a consequence of the expression chosen to define the interval (an expression that is...
All my calculations are based on using the same frame.
... I defined Δs2 = Δt2 + Δx2. Then I showed that the same invariant can be arrived at with this definition by reference to a lightlike pair of events which shares one of the timelike pair of events.
That is all very well. However, I did provide an invariant which is the same as the existing invariant. A symbol can always be created for the invariant as I expressed it. It seems to me that the interpretation of that symbol could stand for something consistent with the common understanding...
With regard to special relativity…
Whenever, I come across the spacetime interval, written like this, say, (Δs)2 = (Δt)2 – (Δx)2 – (Δy)2 – (Δz)2 , it is as if it has to be that way. However, it seems to me it is this way by definition and does not have to be so. Sometimes, it seems to be...
This is the way I think about it. I think it is less abstract and I hope it is correct.
The required accuracy is that the measuring instruments are not able to detect the effects of gravity. Taylor and Wheeler are saying that at the particular location (relative to the Earth's surface) and...
That is a moot point. "the number three is blue" is wrong on the face of it. However, it depends on context. We try to avoid context by defining things as precisely as we can. If it is necessarily meaningless then it is wrong in the sense that it has no application.
I am happy with that.
Thank you, but I cannot make sense of that, except as a theoretical formality. That is, I can make sense of saying that it is undefined, but not sense in saying that it is neither right nor wrong.