Re: the basis of relativity

Ken S. Tucker wrote:
Tom Roberts wrote: 
The actual value of the Action does not matter -- one can add an
arbitrary constant to it without changing any physics.

I find no evidence to support your statement,

Then you haven't looked. This is an elementary theorem in variational calculus.


check
Weinberg pg 360-1, it's not there, provide a ref please.

In many ways Weinberg's book is merely a superficial treatment of the mathematics of GR. You need more, better textbooks. <shrug>


	Indeed, there is a much larger group of functions one
	can add without changing any physics....


Yes, but action is not *usually* considered in that
group.

OF COURSE NOT! Please look up there and note that this is a group of functions that are ADDED TO the action. So of course the action itself is not a member of the group. Sheesh!


Just examine adding an arbituary constant to
Plancks "h", such as -h, the physics will change :-).

This is utterly and completely unrelated to this discussion, which is of a _classical_ theory (GR).


And, as the previous posts pointed out repeatedly: R^a_bcd can be
nonzero but R is zero (because the terms in the contractions sum to
zero), and this _ALWAYS_ happens in a vacuum region.


The problem is: If R=0 then constant g_uv's may be
used to describe that field over a finite region.


Aparently you have not been paying attention.

As I have said before: If R^a_bcd is nonzero in a region, then there are no coordianates for which the {g_uv} are constant. And yet if this is a vacuum region then necessarily both R_uv and R are zero.


Hence R=0 means R^a_bcd=0, that's a fact. 

No, it's wrong. Once again you got the implication backwards:

	R^a_bcd = 0 implies R=0
	But R can be 0 even when some of the {R^a_bcd} are nonzero,
	such as in a vacuum region.


Study metrics where R=0, i.e. No Curvature, in that
case we can surely use constant g_uv's right?


No.

Consider the Schwarzschild metric in the region r>2M. R=0 but yet Riemann is nonzero, and there is no set of coordinates for which the {g_uv} are constant.

You _REALLY_ need to pay attention! This has been mentioned at least twice.

In differential geometry, "no curvature" means the RIEMANN CURVATURE TENSOR is zero (not, as you claim, that the Ricci scalar is zero).


Tell us what volume we
should use to calculate the density and hence vacuum
using mass/volume=density and why.

Select a point in spacetime at which you wish to know the density. Select any small 3-volume surrounding that point (contained in the 3-surface of constant time for the time you are interested in), total the mass/energy in that volume and divide by the volume. Now take the limit as that volume goes to zero, always containing the point in question. The result is the density at that point.

As I keep telling you, you _REALLY_ need to learn the basics.


Please define
vacuum (aka density) or ref as we should use it in GR.


I have done so at least twice before in this thread:

	In GR, a vacuum region has T^uv = 0 throughout.

BTW "vacuum" is not at all "aka density".


> [... more gobbledygook too tiresome to respond to]


Don't expect me to respond until you learn to pay attention....


Tom Roberts	tjroberts@xxxxxxxxxx
.

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