Gravitational force and acceleration in General Relativity.

yuiop

If you are refering to this thread http://www.physicsforums.com/showthr...397403&page=11 then it clear that atyy, George Jones and Dalespam showed you were wrong and no one supported your argument. If fact Dalespam gave a lengthy, rigorous and very impressive demonstration that you were wrong starting at #165 of that thread and ending at #169.

Actually most of the stuff you say is not incorrect in itself, but for some reason you always seem to be talking about a different subject to the subject of thread and what everyone else is talking about. You are doing that in this thread too.

yuiop

You can if you are lucky and it looks like I was.

Actually it wasn't luck. I just refer to conclusions that are rigorously derived by other people that know what they are doing.

starthaus

I knew you were going to go there. I don't know why you would do that since there were so many calculus errors that you made in that thread. As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree. If you want to discuss more on the subject of transformation of centripetal acceleration in SR, we can do it in the appropriate thread.

LukeD

Right, this is the radial acceleration of our test particle as measured against the coordinate r. But this isn't the acceleration measured by our test particle. It's what we'd get if we parallel transported the acceleration vector all the way to infinity and then asked the observer at infinity how much acceleration he would measure if he had that acceleration vector.

To get the acceleration measured by our observer at r, you need to measure the acceleration vector in the orthonormal basis used by an observer at r. So the observed acceleration is
[tex]\frac{d^2 x^\alpha}{d\tau^2} \cdot \hat e^r[/tex]
Where [tex]|\hat e^r|^2=1[/tex] and [tex]\frac{d x^a}{d\tau} \cdot \hat e^r = 0[/tex]. e^r is directed toward larger r value.

It is very important to remember that formula for the dot product involves the metric.
Because our observer is at constant r, just like our observer at infinity, we can choose [tex]\hat e^r[/tex] to be in the same direction as the observer @ infinity's radial vector, but
[tex]\hat e^r[/tex] has a different length
(remember, a length of a vector is only defined at a point in spacetime. If we put both of these vectors at the same point, then e^r is bigger by a factor of (1-2M/R)^(-1/2))

yuiop

In that thread it took over 150 posts and a lot of bickering to play the "guess what Starthaus is thinking" game when you could have simply stated what you were thinking in one or two posts and now it seems you want to play that game again in this thread.

So far, in our new "guess what Starthaus is thinking" game, I have guessed in #15 that you are thinking that the relationship between the proper time and the coordinate time of a stationary particle is given by:

[tex]\frac{dt}{dS} = \frac{k}{1-2m/r}[/tex]

and you have not denied it. Right there is a mistake in your thinking.

DaleSpam

We certainly can go back and do this again if you wish. My derivation and my results are correct, the "error" you pointed out comes straight from the reference (Gron) that you agreed to use, and your error lies in not knowing what proper acceleration is.

LukeD

If you got different answers by the 2 methods, then you were doing something wrong. I've checked kev's original answer with a few different sources as well as with my own work, and he made no mistakes at all. I also checked your formulas (the few that you wrote), and they were going in the right direction. You never got very far with them though, and your later comments suggest to me that you probably don't know the correct way to proceed.

In particular, you mention dr/ds as being the proper speed, but this is incorrect. The observer at r has to deal with length contraction in addition to the time dilation that you handled. He measures a dr'/ds, not dr/ds.
Working with 4-vectors instead of the individual coordinates leads to a much cleaner treatment that allows you to transform between observers' measurements easily.

Attacking someone and insisting that they are wrong just because they didn't show their work (which he said he would if you asked) usually doesn't get you a nice response...

Anyway, the question's been cleared up..

starthaus

It is not my problem that you had so much trouble with getting a few basic derivatives correct <shrug>. So many errors in your math makes it so painful teaching you.

starthaus

I can add a very nice derivation from Moller that supports my point and contradicts yours to the one from Rindler.


Nothing productive will come out of this.

starthaus

How do you "manufacture" dr' out of dr? In a rigorous way, not by putting in a scaling factor by hand?


Agreed but all you have are the equations of motion that I derived correctly.


There is no work to be shown, the different expressions were put in by hand. You do not know but this is a repeat of another thread on the same exact subject where kev did the same thing, put in the results by hand. It took a painful 150 posts to guide him through a proper derivation.

yuiop

I admitted in that thread that I am not a mathematician and my calculus is lousy. Maybe you get a kick out of getting me to say that again and again. Your calculus might be good, but you are lousy at understanding the physical interpretation of the equations.

Thanks Luke

starthaus

Physics is about deriving results, not putting them in by hand . For that, you need to know math. As to my physical interpretation, it comes through the ability to derive results rather than cobbling them from websites.

DaleSpam

Well, at least we agree on that.

If you want to back up this claim then you are more than welcome to do so and I will be more than glad to point out once again that you have derived a coordinate acceleration instead of the proper acceleration.

starthaus

Sure, we'll go over it again. Should be fun, once we are done, you an also take up your claim with Moller.

yuiop

It took 150 posts to figure out that you were talking about some form of coordinate acceleration when everyone else was talking about proper acceleration.

Your derivation concluded that proper centrifugal acceleration is zero and in another place you claimed proper centrifugal acceleration is defined as [itex]a_0 = a\gamma[/itex] and in another place you got [itex]a_0 = a[/itex]. You never did arrive at the conclusion that everyone else arrived at, that proper centrifugal acceleration is [itex]a_0 = a\gamma^2[/itex]. So can you leave out the condescending tone?

DaleSpam

I just had a kind of flash-back from 2nd grade. "I'll tell my Moller on you".

starthaus

No, no :lol:. You'll have to tell Moller, not me :lol: