Recent content by haushofer

Never mind, solved it. This was serious confusion :-p

I guess my confusion is this: why would the hotter sphere cool down a bit by absorbing radiation from the colder sphere?

I think every teacher knows that words are really important in physics. Physics is also about interpretation, or "ontology", or "concepts" as you put it. And that involves wording, which shapes our understanding of everything, including physics. Ultimately, we don't just use math to understand...

My 2 cents: if you're being pushed in your chair because the train accelerates, it's a frame-independent fact that an engine makes the train accelerating. That's why we call this force "real". Your body with inertia just resists this change of speed. In your frame you describe this as a force...

@kuruman I only see now your comment about absorption, emission and minus signs. But isn't the whole point that if (say) sphere 1 absorbs power from sphere 2, eventually this extra energy is reemitted? That's why I choose a plus sign for c.

Maybe I should write $$P_1'=P_1+c \cdot P_2 \ \ \ , \ \ \ P_2'=P_2+c \cdot P_1$$ to find the new equilibrium, so without the accents on the right hand sides. I'll look at it more closely tomorrow, but clearly I'm misunderstanding something fundamental here.

Positive; it's a ratio of surfaces. But I agree this is the tricky equation. I reasoned that sphere 1 receives (hence the + sign) an additional radiation power of ##c \cdot P_2'## from sphere 2, which eventually after reaching equilibrium again has to be reemited. So it's temperature rises...

Dear all, in an encounter of an infamous claim by Gerlich and Tscheuschner that the Greenhouse effect is inconsistent with the 2nd law of thermodynamics I came to a simple thought experiment which I wanted to share with you to check my understanding and brush up my knowledge. The thought...

Yes.

It's turtles all the way down.

I don't understand what that equation says.

1) No, not necessarily; look e.g. at the definition of the covariant derivative. A partial derivative is not a tensor under gct's, and neither is the connection, but the inhomogeneous terms of both cancel out such that their sum is a tensor. 2) Yes, as long as you add/subtract tensors of the...

So, to go back to the OP, if this confuses people, they should be careful and distinguish between \delta(g_{\mu\nu}) and (\delta g)_{\mu\nu}. But I guess this is also confusing, since g is also used for the determinant of the metric. But I don't see why this raises the question whether these...

@anuttarasammyak: By the way, the answer is already given here below. Yes. I guess that one could be pedantic and distinguish between \delta (V_{\mu}) and (\delta V)_{\mu}. The first one means lowering an index and then vary, while the second one means vary first and then lower an index on...

I don't understand that comment. The validation of the minus-sign is because you consider the variation of the very metric tensor itself, not just some arbitrary tensor. I also don't understand this comment: You say that the metric components g_{\mu\nu} don't constitute components of a...