Thread: Relativity of Simultaneity View Single Post

 Quote by dalespam your language here is a little confused, so i am concerned that your thoughts are also a little confused. When we are talking about the different postulates and conclusions in relativity we are not talking about a "cause and effect" effect relationship, but rather a "logical implication" relationship. A cause and effect relationship involves more than a logical implication relationship, specifically it also implies a temporal ordering where the cause preceeds the effect. There is no such temporal ordering between lc, td, ros, the principle of relativity (por) and the invariance of c (c), so you cannot speak of causes or effects amongst them.
This is one thing that I have trouble with, and it might be down to my lack of understanding of the nuances of the concepts; but my understanding is that that light is physical, and when the speed of light is measured a physical effect is being measured; I'm also of the understanding that LC & TD are physical effects. For that reason I don't understand how we can't speak in terms of cause and effect.

I have not trouble with the idea that the postulates of relativity allow us to reason what should be the case, if the postulates are assumed to be true; and what effects must occur, in order for the assumptions to hold true; but insofar as they relate to the physical world, then I think we have to be able to discuss cause and effect at some point; reasoning alone, as far as I am aware, doesn't cause physical effects (of the sort we are talking about).

 Quote by dalespam So, what we have is properly "implies" and not "causes". The proper way to express this formally is: $(\text{PoR} \cap \text{C})\leftrightarrow (\text{LC} \cap \text{TD} \cap \text{RoS})$ This whole conversation began when you correctly pointed out that $(\text{lc} \cap \text{td})\rightarrow \text{c}$ is false. I responded by correctly pointing out that $(\text{lc} \cap \text{td} \cap \text{ros})\rightarrow \text{c}$ is true and you had neglected ros which is an essential part of sr. You then followed up with the incorrect assertion that $(\text{lc} \cap \text{td})\rightarrow \text{ros}$ to which i gave a counter-exmaple. The rest of the conversation has basically been follow-up to that. I hope this clarifies things.
Apologies, I'm not overly familiar with the formal notation, although I think I understand the above; I just can't write a full response using the correct operators.

It might be worth drawing the distinction, again, between an invariant measurement of c $\text{(mC)}$ and an invariant actual c $\text{(aC)}$; where
$\text{mC}$ does not necessarily imply $\text{aC}$ ; as per the example where two reference frames measure the speed of light to be c, but because the instruments used are of different lengths, the actual speed represented by the measurements is different.

You mentioned that I had neglected RoS, and so had drawn the wrong conclusion; however, RoS is a consequence of the actual speed of light (c) remaining invariant, not necessarily of an invariant measurement of ca 300/000 km/s.

$(\text{PoR} \cap \text{mC})$ does not necessarily imply $(\text{LC} \cap \text{TD} \cap \text{RoS})$

but
$(\text{PoR} \cap \text{mC})\leftrightarrow (\text{LC} \cap \text{TD})$

If we consider, that we can only ever, really, speak about the measurement of c, and not the actual speed", then:
$(\text{PoR} \cap \text{mC} \cap \text{RoS})\leftrightarrow (\text{aC})$

Which leaves us with the formulation:
$(\text{PoR} \cap \text{mC} \cap \text{RoS})\leftrightarrow (\text{LC} \cap \text{TD} \cap \text{RoS})$

which appears to be circular.

 Quote by dalespam No, it is merely a counter example to your claim of post 121 that "lc & td must occur in order for c to be invariant" i.e. $\text{c}\rightarrow (\text{lc} \cap \text{td})$
OK; again, I'm not 100% sure of the relevance of an unrelated transformation. It might be best to just assume that the context is Einseinian relativity.