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Author Topic:   what is the big bang and how do i understand it?
JustinC
Member (Idle past 4872 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 82 of 122 (241989)
09-09-2005 8:36 PM


Cavediver,I have a question about GR and didn't want to start a new thread, so I'll post it here. I'm sure you can answer it easily.
In GR, the curveture of spacetime, expressed as excess radius of a sphere, is given by the formula:
Radius excess= (GM)/(3c^2)
(This is according to Feynman's "Six Not So Easy Pieces", though in the preface it's noted that he leaves out the sum of the pressures in this statement.)
Then, it's stated that this excess radius will be independent of frames of reference. How is this so, since mass is dependent on the frame of reference?
My guess is that although the frames of reference will get a different mass, they are using different coordinate systems, and in their coordinate system they will get the same excess radius. Is this correct?

Replies to this message:
 Message 83 by cavediver, posted 09-11-2005 1:55 PM JustinC has replied

  
JustinC
Member (Idle past 4872 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 84 of 122 (242283)
09-11-2005 3:02 PM
Reply to: Message 83 by cavediver
09-11-2005 1:55 PM


quote:
Easy, it's not! This is the big misunderstanding with SR. Velocities do not add. Naively: we see a spacecraft (mass M, velocity V<~c) exert thrust F for time T. The spacecraft deltaV is F/M x T, so we expect to see new velocity = V + F/M x T. We don't because veocities don't add. We see something smaller than this. The false conclusion is that M has increased.
I think you've been saying this all along, but I'm just beginning to understand it. So let me see if I have a good grasp of time dilation, lenght contraction, and mass increase.
These concepts are the result of only looking at the 3D components of a 4D vector. Whereas length of an object shoots off to zero as an obsever sees the object increase in speed, this is only a result of just looking at the 3D vectors. In spacetime, the term (c^2)(t^2) will compenstate for lenght going to zero.
Also, is relativistic mass just a result of maintaining Newton's laws? Because, as you said, the only way to interpreted the nonadditive velocities in terms of Newton's laws would be to suppose the mass has increased.

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 Message 83 by cavediver, posted 09-11-2005 1:55 PM cavediver has replied

Replies to this message:
 Message 87 by cavediver, posted 09-11-2005 6:37 PM JustinC has replied

  
JustinC
Member (Idle past 4872 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 91 of 122 (242380)
09-12-2005 1:12 AM
Reply to: Message 87 by cavediver
09-11-2005 6:37 PM


quote:
Look like you've got it!
Yep, I almost convinced myself that I understood it also. Then I read pmb's posts at the SciForum:
Relativistic Mass vs. Rest Mass | Sciforums
He seems to be saying that the strength of a gravitational field is dependent on one's reference frame, since mass increases with speed.
Consequently I've been reading alot about mass, and there seems to be some dispute about whether one should use relativistic or proper mass when explaining GR and SR. Is this true?
How is it false to conclude that mass has increased since it seems to have more inertia the faster it goes? Even if one uses the energy-momentum equation:
E^2 - p^2=mo^2
It still seems that as you rotate in spacetime, you will increase the E and p. And since p=mv, m would eventually have to increase.

This message is a reply to:
 Message 87 by cavediver, posted 09-11-2005 6:37 PM cavediver has replied

Replies to this message:
 Message 92 by cavediver, posted 09-12-2005 5:07 AM JustinC has replied
 Message 93 by cavediver, posted 09-12-2005 5:21 AM JustinC has replied

  
JustinC
Member (Idle past 4872 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 94 of 122 (242612)
09-12-2005 2:36 PM
Reply to: Message 92 by cavediver
09-12-2005 5:07 AM


How does one become a crank? There seem to be so many of them boobytrapping my education. I guess I'll read up on Ricci curvature, though I'm sure you are correct.

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 Message 92 by cavediver, posted 09-12-2005 5:07 AM cavediver has not replied

  
JustinC
Member (Idle past 4872 days)
Posts: 624
From: Pittsburgh, PA, USA
Joined: 07-21-2003


Message 96 of 122 (242676)
09-12-2005 4:59 PM
Reply to: Message 93 by cavediver
09-12-2005 5:21 AM


quote:
Well, E and p do increase as they are not scalars but you have Mo remaining invariant as it is a scalar. p=mv is not a relativistically sensible expression. neither p, m, nor v are scalars here.
Would the relativistically sensible expression be:
p=(lambda)[(mo)(v)]
and some people interpret (lambda)(mo) to be a mass increase, when it could just as easily be interpreted as an adjustment factor of momentum in whole?
Can all experiments which are interpreted in terms of mass increase be interpreted in terms of momentum increases?

This message is a reply to:
 Message 93 by cavediver, posted 09-12-2005 5:21 AM cavediver has not replied

  
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