Recent content by baffledMatt

No, because this is the definition of P(\vec{x},t). But that doesn't mean it has any physical relevance. But we have discussed this at some length somewhere. The problem is that Dick refuses to acknowledge when he is making assumptions, leading him to believe that he has a perfectly logical...

No. Just a sense of "been there, done that". Arguing with Dr Dick is like hitting your head against a brick wall which is mocking you because it thinks it's made of paper. Matt

You should try reading "A new kind of science" by Stephen Wolfram. He has the same kind of idea that these simple models (called cellular automota) can be used to think about pretty much everything in the universe. Just one word of warning, if you read his book you may be fooled into...

You took the words right out of my mouth. I think there is a tendency for people to read something like And then going on about how imagination is more important in science. If you're a genius like Einstein then fair enough, but unfortunately the rest of us mortals have to slog it out...

Why should we expect this according to SR? Have you done the calculations...? Oh, sorry geistkiesel, I forgot you don't like math. Matt

That's an interesting question. Does anybody know the answer? I would guess that to make it meaningful you have to specify the order of the limits, ie something like m/(1-v/c) -> something. Can someone make sense of this? Or at least tell me whether this is just nonsense :wink: Matt

Just to add to what pmb_phy said, we use mgh much of the time in calculations because it is a good approximation to the change in the gravitational potential (given by U = -GMm/r) for small distances. i.e. when h << r. Matt

Blimey, well I hope it wasn't directed at me. Well geistkiesel, I'm afraid that I've had enough of your attitude so I'll leave you to wallow in a pit of your own, well, whatever it is you are wallowing in at this moment in time. Believe what you like, ignore everyone and everything around...

No, you have no idea what the laws of physics are so you are not qualified to make this statement. You will find the midpoint, but each observer will measure a different distance to that midpoint. Why is this so difficult for you to understand? Go ahead, you'll only be shooting yourself...

Well, a look in my handy little book of physics formulas tells me that the speed of sound in an ideal gas is: v = \left( \frac{\gamma R T}{\mu}\right)^{1/2} where R is the molar gas constant, T is the temperature, \gamma is the ratio of heat capacities and \mu is the mean molecular mass...

Politeness would be a good start. Being civil is not a sign of weakness. And if you stop with all the vulgarity people might be willing to listen more carefully to what you are saying. Matt

Well, we can't help what you personally think about yourself, but that still doesn't warrant all this hostility. Besides, if they have been rude to you surely it would be a better show of character not to stoop to their level? Matt

Southampton eh? A splendid part of the world if you don't mind my saying so. I have a veritably sound aquaintance who originates from a vicinity not far from there and I must confess that they are most agreeable. Matt

No, by specifying v=1 and dt=1 you are setting the time and velocity. distance is velocity multiplied by time, ergo you have also specified distance. Ipso facto your subsequent analysis is erroneous. I know that they were simultaneous in the stationary frame because that's how the...

What an absurd suggestion. Educated man, educated! jimmy_p: Born in Essex (I am not proud of my heritage), raised near Cambridge and University in London - Kensington you know. I think that would put me safely in the 'soft southerner' category. Matt