Relativity and Cosmology

I agree that in flat space the ratio between the circumference of a circle and its diameter is π;

circumference/diameter=π

But is that true for curved space as well? If the ratio between circumference and diameter of a circle in curved space is less than π, a smaller circumference should be expected. The faster the velocity, the more curved the space, the smaller the ratio C/D for that curved space, just as contraction increases with velocity.

Here is an interesting discussion concerning the ratio of C/D in curved space;

http://mathforum.org/library/drmath/view/55198.html

I'm not sure on this and wouldn't mind feedback.

Hi Roger,

You wrote: "circumference/diameter=π ... But is that true for curved space as well? If the ratio between circumference and diameter of a circle in curved space is less than π, a smaller circumference should be expected. The faster the velocity, the more curved the space ...."

There is no space- or space-time curvature applicable in this rigid disk rotating in flat space-time. Velocity per se does not curve space-time. It is a simple velocity time dilation problem caused by clock synchronization offsets - there is no real reduction in the circumference of the disk. In a real-world disk, the circumferences will actually increase due to stretching of the disk material under centrifugal force.

What you referred to above is curved space-time in the presence of gravitating matter, where circumferences of circles do become less than πd. This is simply the consequence of our inability to measure the 'real' radius - we measure the radius along a curved line, while the circumference is effectively measured in a 'flat plane'. The article that you linked to also talks about curved space-time.

Regards, Jorrie

I guess I was thinking gravitation and acceleration were indistinguishable and that spinning disk feels a constant acceleration since it's direction is constantly changing. Would this not be the case?

Hi Roger,

What you were thinking is a common misconception. Only inside a vanishingly small lab would that be the case. As I said in the OP above, "Invoking the equivalence principle and then pretend that the perimeter clock sits in a gravitational field is no good, although you may have seen this before. Acceleration per se does not affect atomic clocks."

The reason for this is perhaps more readily understood by means of a few equations (as I know is your preferential method!) Consider the equations for gravitational acceleration and gravitational time dilation:

Gravitational time dilation is given by the ratio: dτ/dt = sqrt[1-2GM/(rc²)],

where dτ is the propertime interval of a clock sitting momentarily stationary at a radial distance r from an isolated mass M and dt is the coordinate time interval, far away from this mass.

The equation for gravitational acceleration is: a = -GM/r², showing that gravitational time dilation and gravitational acceleration cannot be directly proportional.

If you play around with different mass to radial distances ratios (M/r), you will find that there is no correlation between gravitational time dilation and gravitational acceleration. For this reason, the acceleration (and the equivalence principle) cannot be invoked to explain the measured time difference between a stationary observer and an observer riding on the edge of the disk.

I hope it helps; otherwise, shout and we will try again...

Regards, Jorrie

Jorrie,

Sorry, I've been gone a while. Reading what you've said above, I'm sure your probably right, but I'm not seeing it (I keep thinking the fact that pi isn't constant but radially dependent in a curved space raises doubts in my head). I'd rather just do the math and see it doesn't work myself. It's hard to do the math and I need your help. I think that although this may be an act of futility, I might learn an awful lot about curved space, so I'll ask you to indulge (and help me).

Lets look at a disk spinning counterclockwise at speed ω with radius R. We'll concern ourselves with a particle at the edge of the disc. It wants to go straight, but it keeps getting turned. Now in the real world the molecule is being turned by electrostatic forces, but its perfectly ok to say its a central potential that is accellerating the particle towards the center of the disk with acceleration;

a=v²/R

Where v is the tangental velocity of the particle and R is the radius of the disk. To calculate the curvature of space due to the central potential causing the radial acceleration (a=v²/R), I think we just divide the acceleration by c² (Like you did on page 67 of your book).

k=a/c² = v²/Rc²

r = Rc²/v²

Where k is the curvature of space and r is the radius of curvature of the curved space. The formula at least makes at least a little sense because it's saying that the smaller the tangential velocity v, the larger the radius of curvature, the flatter the space for a given section of the curved space.

Now what I'm interested in is how pi changes as a function of space curvature π(r) for a fixed radial distance R of the particle. We know that for V=0, k=0 and π(0)=3.1415....

Now this is where I'm stuck. Is the radius of curvature I get above the radius of curvature of a sphere, or a hypersphere?

If its a sphere, then for v=c, the curved space would be a sphere of radius R centered at the center of the disk. In that case π would be equal to 2 (I got this value of Pi from the bottom of this webpage) and thus using;

Circumference = π(0) x d = 2Rπ(0) = 4R

The circumference would be equal to 4R instead of the 6.28 R of a circle in flat space, meaning that the circumferal distance has decreased by about a third. This is unacceptable since the circumference should go to zero as v (tangental velocity) approaches c due to the Lorentz Factor;

What I need to know is how to calculate the ratio of a circumference to diameter of a circle at the equator of a hypersphere, and it better be getting small and approaching zero otherwise there is no connection between Lorentz Contraction and changing Pi on a spherical surface. I have no idea how to calculate Pi on the equator of a hypersphere.

Every particle on circumference is not moving in linear direction but changing the direction. Does contraction at high velocity take place only in case on linear velocity?

Hi nandan, you wrote: "Does contraction at high velocity take place only in case on linear velocity?"

No, not at all, but - contraction is only an artifact of how we measure things using rulers and synchronized clocks. There is no physical contraction at high velocity, whether linear on not!

If you read all the posts on "Paradoxes of Relativity", you will find the underlying principle as simply the different definitions of simultaneity in different inertial reference frames.

Regards, Jorrie

Dear Jorrie,

Please inform if matter can exist in the same form at the velocities approaching C, Light Velocity. I feel at very high velocities matter will get disintegrated into subatomic particle. If you find my querry too preliminary type just excuse me for that.

Nandan

Hi Guest, you wrote: "I feel at very high velocities matter will get disintegrated into subatomic particle. If you find my query too preliminary type just excuse me for that."

Why would you think so? There is no theoretical or observational reason to believe that ordinary matter cannot approach a relative velocity of c and stay intact. The operative word is relative. If a subatomic particle whizzes past you at 0.99c, you are in effect moving at -0.99c in the particle's frame of reference - and you surely do not disintegrate!

True, it is not possible for us here on Earth to accelerated massive (atomic) matter to near light speed, but in astronomy there are observational evidence that black holes can sling-shot massive objects outward at near light-speed (relative to the black hole) and they seem to remain intact.

Jorrie

Does Frame of reference exist at velocity near C. You mentioned my relative velocity reference to subatomic particle at 0.99c. Does frame of reference exist for subatomic particle at that speed.

Are the massive objects slinged out of Black Hole are near light speed when observed or their velocity is reverse calculated at the time of emergence from black hole. Sub atomic particles after slowing down may obtain observable matter like happend after big bang.

nandan

Hi nandan, you asked: "Does Frame of reference exist at velocity near C?"

Sure it does. I would not go as far as to say there is a frame of reference at exactly c, but as near as you like, it exists!

Massive objects are not "slinged out of Black Hole are near light speed", but rather slingshots around a black hole. Nothing escapes from a black hole, not even Hawking radiation - it is formed outside of the hole's event horizon.

You also wrote: "Sub atomic particles after slowing down may obtain observable matter like happened after big bang."

Remember that sub-atomic particles are just ordinary electrons, protons, neutrons etc, i.e., they are observable matter. After the BB, it was not the 'slowing down' of the particles that made them to form atoms, it was the decrease of the temperature that did the trick.

Regards, Jorrie