Gravity and length

[timo]

I know I used the layman answer' date=' but I need help to understand exactly what "space time curve" means. [/quote']

Technically speaking it means that even in a local geodesic system (=the closest you get to special relativity) the second derivative of the metric does not vanish (in SR the metric is usually constant = diag(1, -1, -1, -1)).

 

I am learning Special Relativity now. I know the boost of the spacetime diagram. Does a 'boost' transformation exist for gravity?

In the form SR is usually presented you are limited to Lorentz transformations which the boosts are part of. With a little generalization you can go to arbitrary coordinate transformations. If you extend this to non-flat spacetimes you´re at GR.

 

To really understand what "spacetime is curved" means you´d best read up on differential geometry. Keeping the surface of a sphere in mind as a good example of a curved surface might help. The big key to understanding GR is to realize that there exists a vector space of velocities, momenta, ... at each point of spacetime while spacetime itself (position) is not a vector space.

 

@Halucigenia: The problem is that r is just a parameter with little physical meaning. So by comparing ds with dr you compare a physical quantity (lenght) with an arbitrary parameter. In this case, the parameter is chosen to equal a lenght if no spacetime curvature due to mass was there so one could argue that this comparison makes a bit of sense. I also thought about this example earlier on but in the end you are comparing two different spacetimes which makes argumentation a bit vague, I think.

[□h=-16πT]

After extensive googleing I found the length contraction in a gravitational feild formula for GR here:-

Length Contraction Formula

 

Here it is in LaTeX -

[math]ds=\left(1-\frac{2GM}{c^2r}\right)^{-1/2} dr[/math]

 

Not that I understand most of the math used for GR myself' date=' but I think this looks simple enough so that if you plugged the figures in it would give the difference in length of a standard measuring rod under the influence of a gravitational field.[/quote']

 

Yes, that's correct. That is only the infinitesimal radial length though, a total length requires one to intergrate over that between two radii.

[Jacques]

Thanks for the equation!

Is it valid for non radial length?

It is the same as the proper time equation!

 

Time and space contract in sync.

 

I found an estimate of the mass of the universe made from the cosmic microwave background radiation data. They estimated the density of the universe to [math]3*10^3^0 g/cm^3[/math]and computed a mass of [math]3 * 10^5^2 kg[/math] for the visible universe (14 billions years). But in many others threads I readed that there is no center for the bigbang so I cannot use these equations since the r variable doesn't exist. There is no center to mesure r .

I think, I need an equation who use only density. Something not vectorial. The effect of mass density on spacetime.

[timo]

They estimated the density of the universe to [math]3*10^3^0 g/cm^3[/math] [...']

[math] 3*10^{-30} g/cm^3 [/math]

 

But in many others threads I readed that there is no center for the bigbang so I cannot use these equations since the r variable doesn't exist. There is no center to mesure r .

You could just take any point as r=0, that´s not the problem. The prob is that the equation you want to use is only valid for a special metric in a special coordinate system. You could as well try to solve the Kepler problem by saying "hey, I know gravitational force in Newtonian mechanics is F = m*g, g = (0, 0, -9.81)m/s²".

 

I think, I need an equation who use only density. Something not vectorial. The effect of mass density on spacetime.

Try "Robertson-Walker metric" and "Friedmann universes" for cosmological predictions of GR. What do you mean by "an equation which only uses density" ?

EDIT: Oh, and the equations for the effect of mass on spacetime are called "Einstein equations".

[□h=-16πT]

I think, I need an equation who use only density. Something not vectorial. The effect of mass density on spacetime.

 

You need to solve Einstein's field equations for that one.

[Bill Wolfe]

GR and SR (and if you don't know what those stand for you probably won't be able to follow the rest of this. . .) are being seriously challenged by the new generation of physicists.

 

The quick and simple answer to the original question is that gravity IS curved space. And yes, let's say that from a light-year away you have measured an object and have an accurate result. If a very massive gravity well (like a black hole) were to approach your object to the point where it was experiencing tremendous gravitational acceleration and you were able to measure that object again, it would appear to 'stretch' toward the gravity well.

 

However, if you were floating in space next to it, it would not appear to have stretched at all. Why? Because you, too would be in the area of curved space and so would whatever instrument you were using to measure with.

 

The specifics of your question involves a discussion of the Higgs Boson and the nature of matter, but the answer is both yes and no. YES--if you are far enough away that you and your measuring instruments are not effected by the curved space that we call gravity. And NO, if you and your instruments are too close.

 

Sorry, but that's the best answer there is. If you have any problems with it, feel free to talk to the designer.

[Halucigenia]

Yes, that's correct. That is only the infinitesimal radial length though, a total length requires one to intergrate over that between two radii.

Yes, I think that is what I meant, the two radii / difference in length, being between the start of the measuring rod and the end of the measuring rod, in the direction of the gravitational field:-

1. without the influence of gravity and

2. with the influence of gravity

 

Anyone want to try plugging in the figures for :-

1. at an infinite distance away from the Earth and

2. on the Earth's surface,

for a 1 meter measuring rod?

 

It's a long time since I ever integrated anything.

[swansont]

GR and SR (and if you don't know what those stand for you probably won't be able to follow the rest of this. . .) are being seriously challenged by the new generation of physicists.

 

 

They are?

[Halucigenia]

If a very massive gravity well (like a black hole) were to approach your object to the point where it was experiencing tremendous gravitational acceleration and you were able to measure that object again' date=' it would appear to 'stretch' toward the gravity well.

[/quote'] I thought we were discussing length contraction due to gravity in GR, not spaghettification due to tidal forces under tremendous gravitational acceleration.

However, if you were floating in space next to it, it would not appear to have stretched at all.

I don't think that this is right, spaghettification is not a relativistic effect, you would actually stretch. Though your measuring device would stretch as well, you would notice it's length/width ratio changing. Also I think that you would notice yourself stretching at these tremendous gravitational accelerations, it might hurt a bit . Spaghettification