Galilean relativity and acceleration

AndyVo

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.

Jimmy

In General Relativity, a freely falling object is equivalent to an object moving with a constant velocity in flat space-time. 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 space-time.

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/s²- the latter scenario being equivalent to being on a ship in flat space-time 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 space-time. The curvature of space-time is caused by massive objects.

One caveat: If the object is large enough, or is falling toward a very massive object- a black hole for example, tidal forces become significant and the equivalence between the two scenarios break.

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

Nugatory

You're doing just fine here, as this is the basic insight that led Einstein to realize that free-fall in a gravitational field is no different than floating free in the absence of any gravity - they're both inertial frames. In general relativity, gravity isn't a force and it doesn't produce acceleration.

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

AndyVo

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 :).

ghwellsjr

You're proposing a gravitational tsunami. What the physicists wouldn't give for such an occurrence. They're building very delicate instruments to detect extremely small gravity waves. Of course you would be able to detect the gravity from the sudden creation of new mass. But don't worry, it's not going to happen.

DiracPool

Actually, you WOULD experience a force, a tidal force squeezing you together. Kind of like someone walking up behind you and pushing your shoulders together. Your free fall acceleration wouldn't be detected for the reasons outlined by the other commentators, but depending on the density and distance of this "presto" star, you would feel that tidal effect more or less. If that star were a black hole not so far away, that pushing on your shoulders would soon spaghettify you in a way that the force was certainly felt.

AndyVo

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.

haushofer

The equivalence principle also holds in Newtonian gravity. There, any time-dependent 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 time-dependent rotation cannot pretend he/she is in a gravitational field, while in GR this would be possible (locally).

PeterDonis

Strictly speaking, this can't happen: it would violate energy conservation. But you could imagine the closed box starting out far away from all other objects, but happening to have a trajectory that will take it close to a star. The effect would be basically the same, and everything the other posters have said would still apply.

harrylin

In addition to the other responses, you will of course be able to detect that you are falling towards that star if you look out of the window; the box is not anymore approximately an inertial frame (in the usual sense of "Galilean" reference system).

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

pervect

The one thing that makes the box not an inertial frame if you look far enough outside it is the tidal forces. In the absence of rotation of the box, the presence of tidal forces is both necessary and sufficient to show that you're not in an inertial frame. (With a sophisticated enough definition of tidal force you might be able to eliminate the remark about the box rotating, but if you just naively look for differences in acceleration on two ends of a rigid bar, you could conceiviable mislabel rotation of the box as a tidal force.)

"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 non-zero curvature tensor, which means GR.

harrylin

I meant "looking" literal: surely optical effects can show it (aberration? Doppler? ..)

ghwellsjr

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.