- #106
yuiop
- 3,962
- 20
Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration and that is exactly the opposite to the generally perceived wisdom.
kev said:Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration .
and that is exactly the opposite to the generally perceived wisdom
kev said:Are you saying that the proper acceleration is:
[tex]\frac{d\Phi}{dr'} = \frac{d\Phi}{dr/\sqrt{(1-2M/R)}} = \frac{GM}{r^2}\left(1-\frac{2GM}{R}\right)^{-1/2} [/tex]
(assuming [tex]dr/dr' = \sqrt{(1-2M/R)} [/tex] where dr/dr' is the ratio of coordinate length to local (proper) length in the radial direction)?
starthaus said:You still have a mistake:
[tex]\frac{d\Phi}{dr'} = \frac{d\Phi}{dr/\sqrt{(1-2M/r)}} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1/2} [/tex]
You can't get anything right.
kev said:There is no error in my calculation above.
kev said:LOL. Great coming from someone who made a complete hash of their calculations in this thread https://www.physicsforums.com/showthread.php?t=403978&page=5 and when I clearly pointed out your errors in #71 you continued arguing the mistake was mine until finally admitting your error in #81.
There is no error in my calculation above, unless you are making a petty reference to inconsistent use of explicitly stating G and sometimes using G=1. In that case you have made the a similar error of explicitly stating c^2 in the final term and using implied c^2 =1 in the middle term. We are all big boys here and occasionally lapse between using explicit c or G and sometimes using units of G=c=1. Get over it. We all do it. I suspect this is yet another red herring to distract us from a major blunder in your work.
starthaus said:Nah, you are mixing [tex]R[/tex] with [tex]r[/tex]. There is no [tex]R[/tex]. But worse, you have no clue what you have calculated, even after I took you through the derivation step by step.
kev said:Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:
[tex] \gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}} [/tex][tex]a_0 = a \gamma^3[/tex]
kev said:The important issue is why your equation for coordinate acceleration is so wrong.
So are you claiming coordinate acceleration is greater than proper acceleration or not? Your response here seems to agree with kev's statement.starthaus said:You don't know what you are doing. It is really simple:kev said:Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration .
[tex]a_{proper}=\frac{d\Phi}{dr'}=\frac{d\Phi}{dr}*\frac{dr}{dr'}[/tex]
[tex]a_{coordinate}=\frac{d\Phi}{dr}[/tex]
If [tex]\frac{dr}{dr'}=\sqrt{1-\frac{2GM}{rc^2}[/tex] by what factor do the two accelerations differ?
Al68 said:So are you claiming coordinate acceleration is greater than proper acceleration or not? Your response here seems to agree with kev's statement.
starthaus said:Because you don't know what you are doing, even after someone leads you through the derivation step by step.
You really need to take a break from numerology and reflect on what I explained to you in posts 107 and 114.
Starthaus, I must say that this is the part that troubles me most about the potential approach. It is indeed more elegant, but this field approximation step makes it very suspicious to me. Particularly in light of the results for the rotating reference frame where we found that obtaining the correct answer with the potential approach depended critically on whether you were using a strong-field approximation or a weak-field approximation. We should have been able to tell, from the beginning, that we needed to use the strong-field approximation.kev said:I can do a derivation based on the exact Schwarzschild solution that gives equations that agree exactly with equations in #1 and disagree significantly with your conclusion based on a field aproximation.
DaleSpam said:Starthaus, I must say that this is the part that troubles me most about the potential approach. It is indeed more elegant, but this field approximation step makes it very suspicious to me. Particularly in light of the results for the rotating reference frame where we found that obtaining the correct answer with the potential approach depended critically on whether you were using a strong-field approximation or a weak-field approximation. We should have been able to tell, from the beginning, that we needed to use the strong-field approximation.
Can you explain more about that step? When is the weak field approximation safe,
under what conditions is the strong field approximation required, and when would even the strong field approximation fail?
Without that information it seems that we have to do the brute-force approach anyway, just to check if the field approximations were good.
kev said:I can do a derivation based on the exact Schwarzschild solution that gives equations that agree exactly with equations in #1
and disagree significantly with your conclusion based on a field aproximation.
Thanks, this is helpful. I am looking at the whole chapter:starthaus said:Now, to the justification of the derivation based on the strong field:
6. Rindler gives the justification for this approach in chapter 9.6.
7. Rindler uses it for deriving proper acceleration for rotation in 9.7.
DaleSpam said:Thanks, this is helpful. I am looking at the whole chapter:
http://books.google.com/books?id=Mu...a&dq=rindler 9.6&pg=PA183#v=onepage&q&f=false
Post 1 is just a collection of statements and assumptions requesting input from other members of this forum. Whether it is a mess or not, depends on whether the statements are true or not and that remains to be determined.starthaus said:First things first.
1. Do you agree that post 1 of the OP is a mess?
I never claimed that #1 was a formal derivation. (See accusation 1.)starthaus said:2. That the results are put in by hand, none of them is derived?
We agree on the equation for proper acceleration. I disgree about your claim that the rest of the equations are wrong.starthaus said:3. That the only correct formula is the one for proper acceleration [itex]a_0=\frac{GM}{r^2}(1-\frac{GM}{rc^2})^{-1/2}[/itex]?.
If the statement's in #1 are correct, it does not really matter how I obtained them, but for the record they are substantially based on the mathpages relativity website which is usually fairly reliable and rigorous.starthaus said:5. Since kev did not derive the above for proper acceleration until very late under my guidance, it isn't clear if he din't simply lucked out in the OP.
From the above it is clear that Rindler has not directly derived the coordinate gravitational acceleration and so the incorrect equation you have obtained for coordinate acceleration, is your own hash up rather than Rindler's.starthaus said:Now, to the justification of the derivation based on the strong field:
6. Rindler gives the justification for this approach in chapter 9.6.
7. Rindler uses it for deriving proper acceleration for rotation in 9.7.
8. Rindler uses it for deriving proper acceleration in a gravitational field in 11.2.
starthaus said:9. What troubled me was something different, the fact that he uses the approach by equating the strong field metric
[tex]ds^2=e^{2 \Phi/c^2} dt^2-...[/tex]
with the Newtonian approximation for the weak field [tex]ds^2=(1-\frac{2m}{r})dt^2-...[/tex]. I wrote to him about this (in conjunction to criticizing his circular derivation of the equations of accelerated motion in 3.7). He (Rindler) got very defensive and said (textually) that he prefers simpler proofs to the more rigorous ones, so I let the issue drop.
You seem to think that coordinate force is obtained by multiplying the coordinate acceleration by the rest mass. That is simply wrong in both SR and GR. It is very easy to demonstrate that your assumption is false in SR.starthaus said:That's too bad, if you end up buying Rindler, you will find out that, contrary to your post 1 (and to your incorrect claims above),
[tex]\vec{f} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1} [/tex]
is indeed the coordinate force per unit mass. You can multiply by [tex]m_0[/tex] all by yourself. Besides, you have now the results derived rather than put in by hand as you did in post 1.
This is just all wrong. See my comments above. It is the strong field approximation of potential substituted into the Schwarzschild metric. It is not a combination of weak and strong field approximations. Does Rindler really give the exact equation you have quoted above for coordinate acceleration or is it just a botched extrapolation of what he said, by you?starthaus said:...
Both fields are required (see point 9 , above).
kev said:Post 1 is just a collection of statements and assumptions requesting input from other members of this forum. Whether it is a mess or not, depends on whether the statements are true or not and that remains to be determined.
I never claimed that #1 was a formal derivation. (See accusation 1.)
We agree on the equation for proper acceleration. I disgree about your claim that the rest of the equations are wrong.
If the statement's in #1 are correct,
it does not really matter how I obtained them,
kev said:Allow me to clear up some of your misconceptions. The metric [tex]ds^2=(1-\frac{2m}{r})dt^2-...[/tex] is not an approximation. It is the exact Schwarzschild solution (with units of c=1). The weak field approximation is the assumption that the gravitational potential is given by [tex] \Phi = GM/r [/tex] (same as the Newtonian potential) and strong field approximation is the assumption that the potential is given by [tex] \Phi = (1/2)c^2 \log(1-2GM/(rc^2)) [/tex]
starthaus said:[tex]ds^2=e^{2\Phi/c^2}dt^2-...[/tex]
is approximated in the weak field by:
[tex]ds^2=(1+\frac{2\Phi}{c^2})dt^2-...[/tex]
(simple Taylor expansion)
Now, substitute [tex]\Phi=-mc^2/r[/tex] above.
starthaus said:Start with the Schwarzschild solution in the weak field approximation:
[tex](cd\tau)^2=(1+\frac{2\Phi}{c^2})(cdt)^2+(1+\frac{2\Phi}{c^2})^{-1}(dr)^2+...[/tex]
...
...
At the Earth surface :
[tex]\Phi_1=-\frac{GM}{R}[/tex]
starthaus said:...
How about the blunder:
[tex]a_0=a\gamma^3[/tex] ?
kev said:I concede that
[tex]ds^2= (1+\frac{2\Phi}{c^2})dt^2 - (1+\frac{2\Phi}{c^2})^{-1}dr^2 - ...[/tex]
approximates to the Newtonian gravitational field when (1-GM/r) is approximately equal to unity and when [tex]\Phi=-GMc^2/r[/tex] is substituted. It is also independent of the value or units used for c.
I was thrown off course by this contradictory post https://www.physicsforums.com/showpost.php?p=2730381&postcount=81 by you in another thread:
The result of the statements you make in the above quote is that the weak field approximation is:
[tex](cd\tau)^2=(1-\frac{2GM}{Rc^2})(cdt)^2+(1-\frac{2GM}{Rc^2})^{-1}(dr)^2+...[/tex]
which is not correct.
Assume radial motion only and assume dr/dt is small so that
[tex]d\tau \approx dt[/tex]
That is not a mistake. I still maintain the equations in #1 are correct. I can admit and correct my mistakes as I have done in this post. I learn and move on. You should try doing the same. It makes for much better progress all round.
kev said:...
Assume radial motion only and assume dr/dt is small so that
[tex]d\tau \approx dt[/tex]
starthaus said:But this is a horrible hack that is not true. You know very well the value of the raitio [tex]d\tau/dt=\sqrt{1-2m/r}[/tex]
pepeherborn said:Why don' you two guys do the following: You must arrive to Newtons law of Gravitation in the weak field also. Suppose two spherical masses of 1 and 2 kg at a distance of 1m.For example made of iron with density 7.9 g/cm^3.
Write on the left hand side :
m1 = 2 kg
m2 = 1 kg
r1 = 1m
G = 6,67 *10^-11*m^3/(kg*s^2)
F = G* m1*m2/r^2
F = 1.334*10^-10 N
Now write down on the right hand side every single step you need to calculate to get this single number in General relativity including conversions for G = c = 1 etc. So this would settle this discussion (hopefully!)
kev said:kev said:...
Assume radial motion only and assume dr/dt is small so that
[tex]d\tau \approx dt[/tex]I know that, but approximation solutions always involve "horrible hacks" and should be used with caution. If you do not like hacks, then you should probably avoid derivations based on the weak or strong field approximations. They start with failure built in.
Physics and numerology are different disciplines. What you are doing isn't physics. I suggest that you invest in some good books , I recommended quite a few.(Rindler,Moller)There is no notion of any distinction between proper time and coordinate time in Newtonian equations and from that point of view all Newtonian equations are a horrible hack as you put it.
While true this is not a valid argument for hacking GR. There are perfectly correct solutions to the problem that don't involve hacks. You have been given at least two in this thread, I don't see the point in your continuing to produce hacks.The formula you base you derivation on, of
[tex]\vec{F}=-grad\Phi[/tex]
is true in Newtonian physics and happens to also be true for proper acceleration and force in relativity, but it is not true for coordinate acceleration without modification.
Are you reading these things in any book or are you simply making them up?
kev said:Assume radial motion only and assume dr/dt is small so that
[tex]d\tau \approx dt[/tex]
[tex]dt^2 = (1-\frac{2GM}{r})dt^2 - (1+\frac{2GM}{r})^{-1}dr^2 [/tex]
[tex]\frac{dr}{dt} = \sqrt{ \frac{4G^2M^2}{r^2}-\frac{2GM}{r} }[/tex]
[tex]\frac{d^2r}{dt^2} = \frac{d(dr/dt)}{dt} = \frac{d(dr/dt)}{dr}\frac{dr}{dt}= \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3} [/tex]
In the weak field (1-2GM/r) is approximately unity so:
[tex]\frac{d^2r}{dt^2} = \frac{GM}{r^2}(1-2GM/r) - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} - \frac{2G^2M^2}{r^3} \approx \frac{GM}{r^2} (1-2GM/r) \approx \frac{GM}{r^2}[/tex]
starthaus said:If the above were correct (it isn't) and since we have already established that :
[tex]a_0= \frac{GM}{r^2}(1-2GM/(rc^2))^{-1/2}[/tex]
it becomes clearly apparent that you can't have
[tex]a_0=a\gamma^3[/tex]
So, is the your coordinate acceleration expression false or is the relationship between proper and coordinate acceleration false?
Hint: they are both false.
kev said:Now dr/dt is differentiated again:
[tex]\frac{d^2r}{dt^2} \;=\; \frac{d(dr/dt))}{dr}\frac{dr}{dt}\;=\; -\frac{m}{r^2}\left(\frac{3\alpha^2}{k^2}-2\alpha \right )[/tex]
For a particle falling from infinity, k has the value 1, but for a particle at rest at a given radial coordinate, [tex]k = \sqrt{\alpha} .[/tex]
(See https://www.physicsforums.com/showpost.php?p=2714572&postcount=4 for how the value of k is obtained).
Substituting [tex]k=\sqrt{\alpha}[/tex] gives the coordinate acceleration of a particle at its apogee at r as:
Nope, coupled with [tex]k=\sqrt{\alpha}[/tex] (see your previous post) that would makekev said:
Substitute [tex]k/\alpha[/tex] for dt/ds: (See previous post).
[tex]\frac{dr}{ds} = \sqrt{k^2 - \alpha}[/tex]
Differentiate again with respect to ds:
[tex]\frac{d^2r}{ds^2} = \frac{d(dr/ds)}{dr} \frac{dr}{ds} = -\frac{m}{r^2}[/tex]
Why is that bad? Shouldn't [tex]\frac{dr}{ds} = 0[/tex] be true for a particle at apogee?starthaus said:Nope, coupled with [tex]k=\sqrt{\alpha}[/tex] (see your previous post) that would make
[tex]\frac{dr}{ds} = 0[/tex]
This is really bad.
You are quite right Al. I think the objection that Starthaus is making is that this implies the next step is differentiating zero with respect to s. The clever part of Prof Brown's derivation is that by using the constant k he can get around this and carry out the differentiation. If Starthaus was as good at calculus as he makes out he is, he would realize there is no problem with the step taken by K. Brown. You only have to read Prof Brown's website to realize he very good at calculus and physics.starthaus said:Nope, coupled with [tex]k=\sqrt{\alpha}[/tex] (see your previous post) that would make
[tex]\frac{dr}{ds} = 0[/tex]
This is really bad.
Al68 said:Why is that bad? Shouldn't [tex]\frac{dr}{ds} = 0[/tex] be true for a particle at apogee?
Not trying to butt in, just trying to follow along.