Larry D

If we are just dealing with Special Relativity, we call the time measured by a clock that is stationary (relative to the frame of reference) "coordinate time."

Do we have an adjective to use when we "level up" to General Relativity and now we put a clock at a stationary point and it is extremely far from any source of gravity?

I realize that when we think about it, we realize this isn't possible realistically (removing all gravity would mean removing all mass and that would take us, our planet, our spaceship, etc. out of the system). I'm thinking though we probably have such a "mythical" variable for the purposes of various calculations.

If I see E = (1/2) m v^2 I immediately call it "kinetic energy".

Is there a similar name for the E in E = m c^2?

I'm collecting together a list of math ideas that are later used in higher science courses. One of the ideas is (a-b) = -(b-a).

The operation is fairly simple.

If a=33 and b=11:

(33-11) = -(11-33)

(22) = -(-22)

22 = 22

I need an official name for this (example: I consider "associative law of addition" to be an official name). I'm in a catch 22 because I can't google this without knowing the name.

I specifically need a name for this idea, not the name of something related or the name of something leading to it.

Thank you.

I'm looking for a technical name or technical term that can be google-searched to find more information.

While working through a problem in Quantum Chemistry, I came across the integral of [1/f(t)] df(t)

The answer, ln f(t) + C

makes sense but this is the first time I've seen the differential df(t).

I can work through it, putting it into more familiar notation by substituting y=f(t) and df(t) = dy. This results in the expected answer.

I'm a little uneasy with this though--is there a name corresponding to this that I can Google search?

I’m having a problem with an equation used in the First Fundamental Theorem of Calculus. I’ve found it in a textbook (author from MIT), on Wikipedia and on several web sites.

F is antiderivative

f is function

F(x) = (integral) (from a to x) f(t) dt

My first thought when I saw this was that it would only be true if a=0. I decided to test it with a simple function f(t) = 2t, a simple range [a,b] = [1,5] and setting the point of interest, x, to be equal to 3 (x=3).

f(t) = 2t

F(x) = t^2

x = 3

F(3) = 9

(integral) (from 1 to 3) 2t dt = t^2 | (from 1 to 3) = 9 - 1 = 8

9 does not equal 8.

Is there a flaw in my logic? (showing me where I went wrong would constitute an answer)

Has this problem been previously acknowledged? (showing an internet reference to this problem would also constitute an answer)

I've been going through the Learning Objectives for AP Physics to review myself

http://apcentral.collegeboard.com/apc/public/repos...

And it crossed my mind, a student who knows everything on all of these could be ready for graduate school.

Your thoughts?

(I could write more to explain what I'm thinking but I'm keeping it brief to allow a wider range of answers--I'm looking for feedback from someone else who has mulled over this thought on their own at some point.)

Several web sites explaining potential energy were found to be writing g as a positive value (+9.8).

It makes sense that the P.E. change is positive if an object is raised a height. However, a student who makes the mistake of writing g = -9.8 will get the problem wrong, as they calculate a negative value for their potential energy change when climbing the ladder of a slide.

Is there a rationalization for changing g to a positive value for this problem -- as opposed to writing

P.E. = - mgh so that g can stay negative and saying work = mgh (and work = -P.E. since work + P.E.= 0)?