
#1
Nov812, 09:27 PM

P: 3

Hello all. I'm a long time reader and a first time poster. I should start by saying that I am not a physicists or a physics student and am studying it merely out of curiosity so please forgive any ambiguous terms I may use that are not standard.
I came up with a sort of thought experiment to do with classical or special relativity. It is my understanding that to an observer in a closed environment such as a train there is no experiment that can betray uniform velocity. However acceleration can be detected. This makes intuitive sense when we think about how it feels to travel in an actual train. But consider an observer in a closed box in outer space. Say that all of a sudden a star were to manifest itself from nothing some distance from this box. Suddenly there will be a force imparted on this box from the gravity of the star and it will be accelerated towards the star. Is there anyway to detect this occurrence? That is, does the has the box's inertial frame of reference change? Intuitively it seems that this isn't detectable since everything inside the box is being accelerated at the same rate. (Doesn't free fall feel the same as 0 gravity?) Yet relativity states that acceleration is detectable. I'm pretty certain my confusion stems from a fundamental misunderstanding of relativity but I'm not sure where I've gone wrong. Any help would be greatly appreciated. 



#2
Nov812, 10:29 PM

P: 633

In General Relativity, a freely falling object is equivalent to an object moving with a constant velocity in flat spacetime. Both are considered inertial.
In your sealed box, you wouldn't be able to tell the difference between falling toward a star and moving with a constant velocity in flat spacetime. Relative to the star, the falling object is accelerating but it isn't proper acceleration. If you had an accelerometer while freely falling, it would read zero. In contrast, standing on the surface of the earth, you are rest relative to the ground but your instrument would measure an acceleration of 9.8m/s^{2} the latter scenario being equivalent to being on a ship in flat spacetime firing its rockets and accelerating. In GR, gravity isn't a force and objects aren't really attracted to each other; they are merely following curved paths (geodesics) through spacetime. The curvature of spacetime is caused by massive objects. Tidal forces still exist for a small object falling toward the earth or the sun, but are so small they can be ignored. Here are a couple lectures that may help: http://www.youtube.com/watch?v=VXDoFB8uQRQ http://www.youtube.com/watch?v=rsvOZX2J_SY 



#3
Nov812, 10:32 PM

Sci Advisor
Thanks
P: 2,951

Search this forum and the net for "equivalence principle" for more details. 



#4
Nov812, 10:45 PM

P: 3

Galilean relativity and acceleration
All right. Thanks for the quick and informative replies. It's starting to make a bit more sense. I'll look up the "equivalence principle" for more information :).




#5
Nov812, 10:53 PM

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#6
Nov812, 11:55 PM

P: 492





#7
Nov912, 12:08 AM

P: 3

Haha yes that makes sense. So you would feel it but not in the sense that you feel you are accelerating. Interesting stuff, for me at least since it's all fairly new.




#8
Nov912, 03:31 AM

Sci Advisor
P: 869

The equivalence principle also holds in Newtonian gravity. There, any timedependent acceleration can be rewritten as a gravitational field. The big difference with GR is that in GR any coordinate transformation is equivalent to some gravitational field. So e.g., in Newtonian gravity an observer with a timedependent rotation cannot pretend he/she is in a gravitational field, while in GR this would be possible (locally).




#9
Nov912, 10:17 AM

Physics
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#10
Nov1012, 03:31 AM

P: 3,178

Also a disambiguation: in post #3 "inertial frame" includes gravitation with inertia  thus a "freefallframe". 



#11
Nov1012, 05:58 AM

Emeritus
Sci Advisor
P: 7,436

"Looking out the window" just gives you a more sensitive way to measure the tidal force  a tidal force that has little effect over the cramped inside of the box might be much more obvious when integrated over a longer distance. So I'd say that "looking out the window" doesn't really tell you all that much, it's mostly just a matter of even a small tidal force becoming noticeable if you integrate it over a long enough distance. To present the formal argument, the presence of a true tidal force shows that there is geodesic deviation, and the geodesic deviation demonstrates that there is a nonzero curvature tensor, which means GR. 



#12
Nov1012, 07:00 AM

P: 3,178





#13
Nov1012, 09:17 AM

PF Gold
P: 4,524

Why are we talking about a window in the box? The whole idea of the box is to prevent you from seeing or looking outside of it. It's not suppose to be a structure that you make measurements from. You're supposed to imagine that you are freely floating inside the box which is what will happen in reality.



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