# Are inertial forces "real"

Discussion in 'Physics & Math' started by pmb, Nov 22, 2013.

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## Are inertial forces real

Poll closed Nov 29, 2013.

6 vote(s)
75.0%

0 vote(s)
0.0%

2 vote(s)
25.0%
1. ### TachBannedBanned

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Yes, obviously.

Uniform gravitational field does not have zero (intrinsic) curvature, the Riemann tensor is not null, so where did you get the idea above?

3. ### PeteIt's not rocket surgeryRegistered Senior Member

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You said it in post 18:

5. ### TachBannedBanned

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That was a mistake, I wanted to emphasize that "0 spacetime" is meaningless , the correct expression is "zero intrinsic curvature", I copied the reference to "uniform gravitational field" from pmb's post by mistake. The previous post (17) is quite clear on this subject. Hope that this clarifies it.

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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Thank you.
How do you determine the Riemann tensor of a uniform gravitational field?
(I'm maybe asking too much. I should bite the bullet and start learning some differential geometry for myself.)

8. ### PeteIt's not rocket surgeryRegistered Senior Member

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Reflecting on this, isn't it implied by the equivalence principle?
A uniformly accelerating reference frame in flat spacetime is equivalent to a uniform gravitational field, right?
Conversely, a free-falling reference frame in a uniform gravitational field is indistinguishable from an inertial reference frame in flat space, right?

Changing your reference frame can't make intrinsic curvature appear or disappear, can it?

9. ### TachBannedBanned

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$ds^2=(1+gz/c^2)^2(cdt)^2-dx^2-dy^2-dz^2$

Use the Bianchi identities to construct the tensor (or you can calculate the connection coefficients directly, from definition).

Yes, you should.

Last edited: Nov 26, 2013
10. ### PeteIt's not rocket surgeryRegistered Senior Member

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That's more than I can do. What is your end result?

11. ### pmbBannedBanned

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extrinsic spacetime curvature has meaning onlyt in that it doesn't exist. There would have to be a fifth dimension into which the spacetime could curve.

yes.

The spacetime is the same in that they have the same geometry, i.e. flat spacetime. They're different in that observations made by observers at rest in such a field are different than those made by freely falling observers.

12. ### PeteIt's not rocket surgeryRegistered Senior Member

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Thanks.
Is it ever useful to model a higher-dimensional space in which spacetime is embedded with some extrinsic curvature?
(Or does that not even make sense?)

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14. ### PeteIt's not rocket surgeryRegistered Senior Member

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I should change that. Final exams were yesterday!
Until the release of results collapses the function, my state is something like:
$c_1 \mid \textit{doctor} \rangle + c_2 \mid \textit{med student} \rangle$

15. ### brucepValued Senior Member

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4,098
I really don't care what tach has to say about anything. I don't feel as though I can trust what he says because he frequently chooses bad physics for the sake of creating adversarial relationships. There's a difference between being wrong on a subject and choosing to be wrong on a subject. Anyway I could be wrong with what I said. I find it helpful to think about the experimental side of the physics. So what kind of relativistic experiment is conducted in an approximate uniform gravitational field. My first thought was CERN and particle experiments. This is the laboratory frame of SR where 'most' local experimental measurements disregard any effects due to gravity because they're meaningless to 'most' results. Not all just 'most'. Counter with the GPS and Gravity Probe-B. So CERN is sitting at r_shell on the surface of earth where any delta g_shell is so small as to be meaningless to the experimental results. I shouldn't have to explain why we choose to use the mathematics of SR in the local proper frame where the experimental measurements are made. Why the spacetime is Minkowski and which metric measures this spacetime. If the curvature at boundary of the uniform gravitational field was required to be 0 we'd need to account for any > 0. That's my practical approach. I could be missing something and I would like to hear other analysis. I think pmb linked some discussion at google groups worth reading. I'm interested in any further discussion not initiated by a troll.

Thanks
brucep

Last edited: Nov 24, 2013
16. ### TachBannedBanned

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The reason for ignoring the gravitational effects is because the (Earth) field is weak, not because it is uniform.
As an aside, the Earth gravitational field is not even close to being uniform.

17. ### TrappedBannedBanned

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So... Pete asked kindly that there be no fighting and whilst you are attempting to thread some physics into this post, it's pretty much threaded in insults directed towards Tach.

We all know why this is, it's because you were shown to be posting bad physics. Then is it a matter of delusion you appear to believe that Tach is the one posting bad science, yet he has even put his effort into this thread correcting some mistakes yet again you fail to listen.

Maybe moderators should be concentrating on the motives of material posted like Pete had concentrated on me, when my thread was hijacked with insults, accusations and bad science!

18. ### PeteIt's not rocket surgeryRegistered Senior Member

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OK, back on topic now people. Further rabbl can be pm'd, or posted in open government, or something. Just not here.

19. ### UndefinedBannedBanned

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Hi pmb.

No worries, mate. I try to read fairly and in context, and without (as some used to do in the bad old days) waiting to 'pounce' on trivialities in order to manufacture any excuse for 'ego-tripping' etc.

I try to clarify possible typos etc. to be sure I understand correctly before proceeding on the substantive points.

Cheers.

20. ### UndefinedBannedBanned

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Hi Pete.

Isn't that what 'Brane/String theorists do?

Good luck in your final exams results, mate!

21. ### pmbBannedBanned

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Then you're okay in my book.

22. ### TrappedBannedBanned

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A really strongly correlated subject to the OP is the matter of weak relativistic equivalence.

All inertial forces are due to motion in gravitational fields - to hit the real nail on the head, is to just speak about gravity (and which folds acceleration and curvature) in the context of inertial pseudoforces. It appears through equivalence, that inertia and matter really is the same thing even without invoking the Higgs Mechanism responsible for provided matter to particles. A really nice simple mathematical theorem proves this. I suppose this demonstration helps to distinguish the relationship between the inertial properties of matter and that with gravitating masses

Mass and force is classically related as

$F = Ma$

The weak equivalence is found then by starting with the definition of two masses with acceleration

$F_{XY} = M_Xa_X$

and

$F_{YX} = M_Ya_Y$

such that

$F_{XY} = -F_{YX}$

and

$\frac{M_X}{M_Y} = -\frac{a_Y}{a_X}$

plugging this into our equivalence we get

$\frac{M_g}{M_i} = \frac{a_i}{a_g}$

This is a well-known statement. The ratio of the gravitational mass and the inertial mass appears to be theoretically equivalent to their respective accelerations. A psuedoforce appears to be anything associated to the dynamics of accelerations and curvatures.

23. ### pmbBannedBanned

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I don't know of any.