danielj

Thank you, thats all very helpful. Would you mind expanding the above statement please? Cosmological Principle? at sufficiently large scales, the universe appears as homogeneous and isotropic Copied from Mordreds thread ‘Cosmological Principle’ there I go with a guess again 🫣

My take away from this is that I am lacking logical thinking skills, when it comes to physics. In that, I can’t make assumptions and expect to get the right answer. Understanding the question is more important than the answer I think. thank you

I see, yes. Thank you. I get caught in the details, not always helpful to me.

Thank you, much appreciated I’m just trying visualise some of the mechanics of this, so would it also be true to say that if I (tried) to take a ‘snap-shot’ measurement, with any combination of inertial/accelerating reference frames that in reality there is no way to take a snap-shot, as what was the time frame of the snap shot (by that phrase I mean a static measurement) and no matter how small the time frame, change will have occurred in one or more dimensions? The questions should get less stupid as I progress, but I can’t promise So they were above you if you used the ground as your reference frame, but not if you used the rotor itself? Your brain used the ground? reading this back I see there being potentially many different frames of reference, equally valid?

Would you mind just expanding that a little for me please? I can see that human perception skews reality somewhat, but I can’t work out that bit for myself. thank you

The relativistic effects of being in an accelerated frame of reference can be derived from special relativity. That is, acceleration doesn't produce relativistic effects separate from that of velocity. However, the relativistic effects of an accelerated frame of reference are nevertheless different from that of relative velocity. Specifically, clocks that are below you are slower, and clocks that are above you are faster, with the amount by which the clocks are slower or faster depending on the distance of the clock from your location in your accelerated frame of reference. Thank you , that is a very useful explanation that I can (just) grasp.

It does. Sorry, I didn’t mean it the way it came across. Obviously there would be an effect of time slowing at an accelerated rate from the (let’s say) constant acceleration. It was more the ‘are they equivalent’. I guess they are from your other answers and from the answer that time would slow for both observers from the others point of view. thank you

Could you clarify for me please: the statement ‘(the faster you go, the more slowly you age)’, quoted from World Treasury, T.Ferris. If it were me travelling faster ( and faster) would time be passing more slowly for me? (Would I feel it) or only from the viewpoint of some one else not in the same location taking measurements? Or, if I were taking measurements of them, would I record time speeding up for them. Do you see what I’m getting at? also, what would the difference be between me in an accelerating rocket going away from a planet, and measuring from the rocket, and standing on the planet and measuring the rocket receding into the distance? Are they equivalent? Simple as you can please thank you Or in other words, does the fact that the rocket is accelerating have a bearing on the measurement (the dilation effect)? I’m rambling a bit , I know but just trying to think it through. One question leads to another! does the orbiting planet also use energy (as the rocket does to accelerate) to stay in orbit, rather than just fly off in a straight line into space?

Thank you

Could anyone tell me what the proper name and meaning of the mathematical symbol that is a simple arrow pointing to the right please? for instance a2=c2-b2 arrow here a=square root c2-b2

No, I think I still got that wrong. Got a bit confused there somewhere along the line, swapping the n still wouldn’t give that answer. I’ll start writing it down, tough for me to do it in my head.

Ah, yes, got it now. And when I used 1 for n, I think I accidentally swapped the n to the wrong side. That’s cleared that mess up!, thank you

Is the zero degrees with 1 and 2 correct? And if so, what does that mean? does it prove that there are no regular polygons with 1 or 2 sides?

(can you say why?) From the free pdf, a Polygon is defined in Greek as ‘many-angle’, and with one straight line there are no angles or vertices, and with two straight lines there is one internal and one external. Also, one or two straight lines cannot define an area. Also, to try it out, I gave n a value of 1 and then 2 and in both cases it gives an answer of an internal angle of 0, which is nonsense I think? I can understand your teaching, it just takes me awhile and several reads through to digest it, thank you. I understand reasonably well x,y and z axes, positive and negative as I have to occasionally use a laser cnc machine at work. I can visualise it.

Could you confirm (or not) my understanding of the following equation please? I know it’s very simple, but that’s where I am. Formula to calculate interior angle of a regular polygon. Formula for Interior angle of regular polygon=180 degrees(1-2/n) interior angle of regular polygon Square=180degrees(1-2/4) 4 because there are 4 sides. Can you use the vertices instead? 1-2/4=1/2 1/2 x 180=90 interior angle of regular polygon square=90 degrees obviously I know that is the correct answer, but the 1 made me think. I didn’t know why there was a 1 in the equation, I guess that the 1 represents the ‘whole’ or ‘sum of all angles’ ?not sure how to describe that. also, if there is no gap between symbols, eg: 180(1-2/n), does that always mean multiply? thanks