New look at gravity..

Alright, as this thread has already given me a ‘Headache’ thinking about it. Let it continue ……..

Isn’t it true that Gravity whatever it is ‘Seems’ to operate FASTER than Light ?

As everyone knows, when you look up at the Sky at SOL, it is not really there but,
has moved almost one of it’s diameters and, the light has not yet reached us .

I read somewhere, that experiments have shown, that the Earth is being ‘pulled’
towards where the Sun “Actually IS “ in the sky, rather than where it ‘appears’ to be.

Otherwise, how could the gravity between widely separated Galaxies affect one another
if it only operated at the speed of light ?

I’m expecting my headache will increase in intensity with the forthcoming answers :-)

Regards,

Bill

I wouldn't panic about the speed of gravity. Try thinking of it in the same way we do about light. Let's assume you are comfortable with light having a speed. That means if I switch on a torch over here there is some delay in transit before Bill over there sees the flash. Gravity works just like that too, but invisibly so for our human senses. [ imagine if we were blind to light but not to gravity, we would know all about the motion of large bodies way away from here, but bump into everything in the nearby acre!!! :-) ]

So if I am running along flashing my torch on and off but still shining towards Bill, then he will see these flashes arrive from where I was with the relevant click of the switch of the torch. So I could be performing all sorts of manouevres with light signaling going on, possibly in a quite complex manner, but always Bill will be behind the play due to the delay in light transit.

NOW here comes the tricky bit!! When I say at some time, this delay, that moment etc .... from whose point of view is this? Mine? Bill's? Martin's? Because of delay, then now here isn't generally going to be now over there.

Probably the simplest, and generally accepted way, to relate times at different places is a bit like an internet ping. Send a light impulse from A to B, which encodes the time at A, 'immediately' send it back from B to A, also encoding the time at B when the impulse arrived. At A you divide the round trip time ( as measured by a clock at A ) by two and compare with what B said the time was, and give an adjusting offset back to B.

[ So if I send '12:00pm' to B from A, B sends me back '12.01pm' that I receive at 12.04pm, then I deduce B is 'running behind' one minute - they should have said '12:02pm' - and advise them to roll their clock forward one minute. ]

Don't worry too much about the detail of synchronising, the point is that one can have clocks marching together like this. You don't have to, indeed one could set up a system of separated clocks in some other way.

As far as planet orbits go it's a bit like this:

- there's a delay due to gravity having a speed limited to that of light.

- so yes, right now the Earth is responding to where the Sun was about 8 minutes ago.

- we couldn't have known that 'beforehand' anyway as nothing travels faster than light ( that we are aware of )

- so Earth and all other planets and bodies are behaving according to a 'time delayed recording'

- clearly this is worse the farther out you are, so Pluto I think is about a half day behind the Sun's position compared to Mercury which is a matter of seconds.

- but gravity is inverse square with distance. Thus if the delay is proportional to distance but the magnitude is inverse square, the nett effect is ( more-or-less ) like inverse distance. So Mercury is still the most affected by this. Pluto may have to wait longer but the effect is way, way smaller proportionally.

- this is why Newtonian calculations ( ie. assuming no delay and infinite speed of propagation of influences ) only gave ( about one hundred years ago ) significant disparity with observations. The small discrepancy was completely adjusted for by the use of General Relativity - which does account for delay.

As regards close binary systems it is often stated that there is considerable 'self energy in the gravitational field'. This is a way of saying that with the large forces close in ( because the neutron stars are massive and nearby ), plus the high speed of movement, have created a considerable 'queue' of influence that is in propagation between the two neutron stars at any given moment. As we like things like conservation of energy and momentum to be true everywhere and always then we need to 'give' the field the amounts that have 'left' one neutron star but haven't yet been 'received' by the other. It's one strategy for keeping the accounting straight with all these delays going on. :-) :-)

Cheers, Mike.

Sorry, Martin :-) :-)

You've assumed the Sun doesn't move. You will get a nett increase in angular momentum if you assume the Sun stays still and planet's behave according to retarded time. However you've missed that the Sun responds to where the Earth was 8 minutes ago - and that preserves action and reaction - so you don't then get a force component tangential to the line between. Hence angular momentum is conserved.

Ditto :-)

The Earth doesn't revolve around the Sun! They both revolve around a common centre of mass, and that centre is one focus of the relevant ellipse ( but with perturbations etc ).

Ditto^2 :-)

More ditto .... the 'back reaction' or 'radiative damping' is well covered by Einstein's GR equations ... which makes them fiendish to solve.

Cheers, Mike.

Except for information relayed between entangled particles. That is instantaneous, right?

Well Zippy gets back first, as he left at the same time as Slowpoke and was going faster.
If light takes 8 years for a round trip, and Zippy goes at 0.9c then his time is 8/0.9 ~ 8.89 years ( time = distance/velocity )
For Slowpoke the time is 8/0.1 = 80 years

[ we haven't done any modern relativity yet. Just applied what we mean by 'speed is v' in a given frame, in this case we have chosen Static's rest frame to compare light, Zippy and Slowpoke going there and back to Alpha Centauri. Newton and Gallileo would be quite happy up til this point. If there was an object going some speed relative to Slowpoke or Zippy and we wanted to know what the object's speed relative to Static was, then we'd be doing modern relativity. ]

So we have to determine the age differences upon return now. The way I will present it is as follows, with :

ds = a small amount of spacetime
dt = a small amount of time
dx = a small amount of distance in the x direction
dy = a small amount of distance in the y direction
dz = a small amount of distance in the z direction

then :

ds^2 = dt^2 - ( dx^2 + dy^2 + dz^2 )

I am using velocity units such that c = 1. I'll stay in the frame where Static is at rest, thus any dt's, dx's, dy's and dz's are obtained from a co-ordinate system not moving with respect to Static. So Static has velocity 0.0, Zippy 0.9, Slowpoke 0.1 and light 1.0 on this scale.

It turns out ( I won't prove this ) that ds can be considered as the time that elapses as experienced by one that travels along path indicated by dt, dx, dy and dz. So instead of ds let's use dT where T is proper time. Thus dT is a little bit of time passing on the path indicated by dt, dx, dy and dz. Now we have :

dT^2 = dt^2 - ( dx^2 + dy^2 + dz^2 )

so let's align our x axis along the Earth-Alpha Centauri axis, with light and our intrepid travelers not moving off that line. So dy = 0 and dz = 0 :

dT^2 = dt^2 - dx^2

thus dividing thru both sides by dt^2 :

dT^2/dt^2 = dt^2/dt^2 - dx^2/dt^2

(dT/dt)^2 = 1 - (dx/dt)^2

So what is dT/dt? It is the rate that time passes along a spacetime path in our reference frame. Let v = dx/dt :

(dT/dt)^2 = 1 - v^2

dT/dt = SQRT[1 - v^2]

Look familiar? Let's do some cases :

Static, v = 0 thus dT/dt = SQRT[1 - 0^2] = 1. So time passes for Static at the same rate that .... time is passing for Static. Phew, that's a relief. :-)

Light, v = 1 thus dT/dt = SQRT[1 - 1^2] = 0. So time doesn't pass for light. No biggie.

Zippy, v = 0.9 thus dT/dt = SQRT[1 - (0.9)^2] = SQRT[1 - 0.81] = SQRT[0.19] ~ 0.436

Slowpoke, v = 0.1 thus dT/dt = SQRT[1 - (0.1)^2] = SQRT[1 - 0.01] = SQRT[0.99] ~ 0.995

This gives when Zippy returns at (8/0.9) years his age being (8/0.9) * 0.436 ~ 3.87 years older than when he left.

For Slowpoke returning at 80 years the age is 80 * 0.995 ~ 79.6 years older than when he left.

Supposing they all met for a coffee on Slowpoke's eventual return, what are their age differences then? Static will be 80 years older, Slowpoke will be 79.6 years older ( or 0.4 years younger that Static ), as for Zippy his extra age is 3.87 + ( 80 - 8.89 ) = 74.98 years ( or 5.02 years younger that Static ). While Zippy 'gained' biological time by coming back sooner he still had to wait around for Slowpoke to come back to have that cup of coffee. Here's an illustrating diagram ( don't take this as to exact scale ):

[ I used the word's 'instantly' and 'immediately' in the problem description so we'd just have straight world lines and I wouldn't have to stuff about integrating curves of non-constant relative velocity. Implicitly I have integrated the dT's etc but hey, a straight line is a boring integral! ]

The worldline of Static is the time axis, as is the parts of the worldlines of Zippy and Slowpoke after they returned ( or before they went ). You always travel in time, it's the movement in space that you have the choice of.

Cheers, Mike.

( edit ) dt^2 - ( dx^2 + dy^2 + dz^2 ) nearly looks like some sort of Pythagorus's Theorem wotchamacallit. It's the minus sign that spoils it. The spacetime we are dealing with is not Euclidean but that of Minkowski ( or hyperbolic ). If we were to change our description to another frame, moving with respect to Static, then a sort of rotation is involved but it would look like this :

the new frame being the primed symbols. Note that this preserves the speed of light as always being c regardless of frame. With a greater relative speed between frames the 'scissors close' and the time' and distance' axes fold onto the light line!

( edit ) I've only done SR here. With GR the dt's, dx's, dy's and dz's from one frame to another kerfuddle each other without mercy! You have a 4x4 matrix relating two frames, totalling 16 entries, but symmetry implies that 6 are redundant/degenerate, so you're left with 10 independent ones. Ah, go figure ...... :-)

( edit ) And then there's the fuss about time travel. Going into the future is no problem, albeit maybe at different positive rates. So if velocity exceeds c? Then dT^2 is going to be negative and thus dT is what ????? Not a real number alas .... :-)