View Full Version : SR and the earth, sun, and galaxy.
First of all let me say Hello! as this is my first post. I am a biologist, and have been reading over the very basics of SR for a little while and have a lot of questions on the implications of the theory.
I had been asking my questions primarily at a different forum, one that wasn't directed towards physics so I figured this would be a better place for me to ask my questions and hopefully receive responses that I can comprehend.
This was my last post at the forum, with a lot of the questions I had. We had previously been discussing an example of I was on an airplane traveling at 300mph, and then we talked about if I was on an airplane traveling .865c.
In the first paragraph, if it is not clear, I am wanting to understand what exactly is meant by an "inertial reference frame."
--
Hmm. I think the key to me understanding this is realizing what an "inertial reference frame" is. I know inertia, in the most basic sense, is an objects resistance to movement. So is the inertial reference frame the observer that isn't moving? I mean, the one that isn't traveling at speeds a fraction of c? The one on earth.
I realize the person on earth is the inertial reference frame, but in the advanced calculations, since the earth is rotating around the sun, it is undergoing constant acceleration. Is this ever a factor?
In addition to the above paragraph, could this be what makes us "realize" time as we do? Obviously, a second is a unit of time that we arbitrarily invented, and now it "is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom. (some website from Google)".
But with us rotating around the sun, it seems the sun would be the inertial reference frame (ignoring the wobble from planets) and that we on earth would be undergoing (I guess) constant acceleration. The speed of the earth around the sun is very fast, I'm sure. I'm sure over 30,000 mph. I don't have the data to do the division to see what factor of c that is, but I'm sure it's a decent factor (I'm sure less than .05c though), but much greater than the example of the airplane.
Using what I've learned from you guys here and probably poor logic of my own, I would say the entire earth, and presumably everything on it has aged less than the sun (I know the sun existed before earth). Since the earth (and everything on it) is undergoing the constant acceleration, and the sun, relative to us, is static, the earth would have aged slower than the sun. Would this be a factor in calculating the age of the sun? Or is the age of the sun calculated by other means (fuel left in it)?
But if any of the above has merit, then wouldn't we have to take it back a step further and look at the earth revolving around the sun, which is moving around in our galaxy? And even further, assuming a singularity-type-entity that the big bang came from, our entire galaxy is moving away from the point of the big bang at probably a very fast speed, so we'd age less than the point from which we all came from. Which I guess makes sense, that would make us younger than the universe's existence?
If the Milky Way ages less than the singularity from which we came from, and, uh oh, my mind is locking up I can't see the answer here, but, oh wait no thats wrong. If it were the opposite, that would mean we've been around longer than the universe existed. Hmm. Wait, this last part about our galaxy and the singularity-type-entity we came from really isn't an issue because the singularity exploded, (I'm frowning now), and doesn't exist anymore for us to use as an inertial reference frame.
But as I asked above with more complicated calculations with Einstein's SR, since all the galaxies in the universe are moving away from a central point (or each other? but couldn't you have two moving in somewhat parallel?) could the age differences be calculated using different reference frames?
I'm sure they (you guys, the physicists who know this stuff) have methods for calculating the ages of galaxies, etc. Or for knowing that something is x light years away. I guess my question is if SR theory is used in addition to conventional methods, like to check for differences in ages. I guess this is also all hinged on whether or not the galaxies, our sun, the earth, etc, is moving at speeds which are large enough of a fraction of c to make a difference when calculating ages. It seems to me that even if it's a small part of c, in a time frame of 11 billion years, that it could have an effect. Even my original question/example with a 300 mph airplane would show differences over 11 billion years (although everyone would die from the airplane's disgusting food -- my apologies if anyone is an airplane food chef!)
This is amazing stuff. Really, congratulations to you guys for devoting so much time to studying and understanding this stuff.
Thank you anyone in advance for reading this, and for any reply you may give me. I am enjoying learning the basics of this subject. I know the time you put in a reply may seem to be worthless, but the knowledge that I and others gain is priceless.
--Aychamo
NOTE: I found a figure that says the earth moves about 66,000 mile per hour around the sun. That is 66,000 mi/hr / 60 min/hr / 60 sec/min = 18.33 miles per second. The speed of light is 186,000 miles/second. So the 18.33 (repeating) / 186,000 means the earth is moving at 9.85 * 10^-5c. Pretty insignificant, but still 220 times more significant than an airplane moving at 300 mph. I don't know the math, but over 4.5 billion years that is bound to have some effect. I cannot find figures for the sun's speed in revolving about in the Milky Way (I actually don't like the candy bar).
Originally posted by aychamo
...I know inertia, in the most basic sense, is an objects resistance to movement.
Actually inertia in terms of mass is resistence to "change in" motion. The Newtonian concept of inertia is that things unacted on by ordinary forces remain in constant motion states. In relativity it is that things unacted on by real forces, i.e. four-vector forces, remain in geodesic motion states.
So is the inertial reference frame the observer that isn't moving?
It isn't accelerating.
I mean, the one that isn't traveling at speeds a fraction of c? The one on earth.
The velocity one observer has with respect to another has nothing to do with whether either of them are in inertial states of motion. As long as their velocities are constants and one can neglect gravitation then both of their frames are inertial.
I realize the person on earth is the inertial reference frame, but in the advanced calculations, since the earth is rotating around the sun, it is undergoing constant acceleration. Is this ever a factor?
If your worried about precision to nanoseconds in typical calculations then yes, but in the 0.865c scenarios one usually neglects such small effects.
I don't have the data to do the division to see what factor of c that is, but I'm sure it's a decent factor (I'm sure less than .05c though), but much greater than the example of the airplane.
The airplane clocks weren't compared to a clock held stationary with respect to the sun, but either way the speed of the earth with respect to the sun is 9.9x10^{-5}c, and \gamma is 1.000004.
Using what I've learned from you guys here and probably poor logic of my own, I would say the entire earth, and presumably everything on it has aged less than the sun (I know the sun existed before earth).
Actually compared to the SR time dilation of the Earth the sun's own gravitational time dilation of its own matter is significant.
But if any of the above has merit, then wouldn't we have to take it back a step further and look at the earth revolving around the sun, which is moving around in our galaxy?
These are still small effects.
And even further, assuming a singularity-type-entity that the big bang came from, our entire galaxy is moving away from the point of the big bang at probably a very fast speed, so we'd age less than the point from which we all came from.
The big bang isn't motion of matter away from a center point within our universe. It is an expansion of space itself. There is no one single point within our universe that can be said to be the center of the explosion. The dynamics you want to consider here are general relativistic, not special at all. For example, instead of inertial frames, commoving frames are of interest.
could the age differences be calculated using different reference frames?
The age of the universe refers to the time elapsed for a commoving frame since big bang.
I guess my question is if SR theory is used in addition to conventional methods, like to check for differences in ages.
When it is significant.
The speed of light is 186,000 miles/second. So the 18.33 (repeating) / 186,000 means the earth is moving at 9.85 * 10^-5c. Pretty insignificant, but still 220 times more significant than an airplane moving at 300 mph.
But a comparison wasn't being made to a clock held stationary with respect to the sun.
Originally posted by DW
The age of the universe refers to the time elapsed for a commoving frame since big bang.
I agree, but lately I've wondered if it isnt impossible to define precisely. Comoving frames differ depending on how low or high they are in gravitational potential. I used to imagine there was a universal time (since the bang) that all comoving observers could agree on, now it seems there would at best be only a rough approximate agreement. Or?
Originally posted by DW
The big bang isn't motion of matter away from a center point within our universe. It is an expansion of space itself. There is no one single point within our universe that can be said to be the center of the explosion.
Again I agree, or have gathered the same impression from reading what cosmologists have to say.
The conventional scientific view is that we didnt all come from a point anywhere. The presumed spatial flatness implies that the initial singularity was infinite in spatial extent. So space was already infinite in all directions when it began expanding.
the alternative view, of a finite closed universe begun at a point, has become a minority view---has become marginalized in the past 5 years or so. Maybe some time it will come back in fashion---Ned Wright had something to say about this at his website last time I looked, some tentative evidence of small positive curvature. But the majority view seems to simply be "flat or flat enough for government work". Which means an spatially infinite beginning. Do you have the same impression?
------------------------------
"Actually compared to the SR time dilation of the Earth the sun's own gravitational time dilation of its own matter is significant."
that reminds me, apparently there are or were two time standards: one an imaginary clock at the center of the earth and another at the center of the sun-----no clock on the surface of the earth keeps standard time. Something about rotation. Also the atomic clocks in GPS satellites do not keep surface time but speed up and slow down along the satellites elliptical orbits as their altitudes change (among other things) so their time signals must be corrected using General Relativity (among other considerations). Strange thought, or strange to me anyway.
Marcus: I agree, but lately I've wondered if it isn't impossible to define precisely. Comoving frames differ depending on how low or high they are in gravitational potential. I used to imagine there was a universal time (since the bang) that all comoving observers could agree on, now it seems there would at best be only a rough approximate agreement. Or?
I've always wondered whether the CMBR could be used as a frame of reference for all observers. As the Universe expands with time, the temperature of the CMBR decreases. I suppose that wouldn't be a very practical measure of time as the CMBR wouldn't change significantly over our lifetimes. But it's just a thought. Also, our motion wrt the CMBR has been measured to be around 370 km/s. Can the CMBR be used as an absolute frame of reference with regard to motion?
Originally posted by Jimmy
I've always wondered whether the CMBR could be used as a frame of reference for all observers. As the Universe expands with time, the temperature of the CMBR decreases. I suppose that wouldn't be a very practical measure of time as the CMBR wouldn't change significantly over our lifetimes. But it's just a thought. Also, our motion wrt the CMBR has been measured to be around 370 km/s. Can the CMBR be used as an absolute frame of reference with regard to motion?
Originally posted by Jimmy
Can the CMBR be used as an absolute frame of reference with regard to motion?
Quantum Mechanics will always be within a reduced Dimensional field .
Originally posted by marcus
I agree, but lately I've wondered if it isnt impossible to define precisely. Comoving frames differ depending on how low or high they are in gravitational potential. I used to imagine there was a universal time (since the bang) that all comoving observers could agree on, now it seems there would at best be only a rough approximate agreement. Or?
All comoving frame observers in a Robertson-Walker universe agree on this time. They are not at different "gravitational potentials".
Ned Wright had something to say about this at his website last time I looked, some tentative evidence of small positive curvature. But the majority view seems to simply be "flat or flat enough for government work". Which means an spatially infinite beginning. Do you have the same impression?
I believe it will be found to be closed curved. Right now there isn't a great weight of evidence for which side of asymtoticaly flat it actually is.
Originally posted by DW
All comoving frame observers in a Robertson-Walker universe agree on this time. They are not at different "gravitational potentials".
...
I realize that about a Robertson-Walker universe
but that is where matter is spread out uniformly.
And the moment you have galaxies in the universe it is no longer
exactly R-W.
So a comoving clock somewhere deep in Andromeda's potential-well
will be ticking more slowly than a comoving clock halfway between us and Andromeda.
which will be ticking more slowly than a comoving clock completely out of the Local Group altogether, not near any cluster
once I raised this issue with a gravity expert (i.e. a relativist or GR person) and whether you could slice spacetime cleverly into a foliation that would define a universal time, and he gave me some arguments that you could not-----and that, if I understand what he said, the R-W universal time is just a rough approximation to an ideal which does not exist
it is still a beautiful idea to be at rest with respect to the Hubble flow or, in other words, with respect to the CM Background. and that all those observers at rest wrt Background could agree down on the atomic clock age of the universe, down to the tick.
Originally posted by aychamo
I am a biologist,
In the field of biology, have you ever heard of the term "thermodynamic time"?
aychamo:
The idea that because the Earth is moving (whatever that means), that its time "rate" , (arguing on the ideas of Special Relativity) must be slowed relative to the Sun is erroneous. The opposite is in fact the case.
As a matter of fact there is "point" where a kind of "balance" occurs between the time "rate" decreasing effect of Special Relativity (SR), and the time "rate" increasing effect of "increasing" gravitational potential (ie: General Relativity, GR) as an orbit's (assume circular for simplicity) radius increases.
This balance occurs at a satellite orbit (either natural or man-made satellites, the maths doesn't know any difference) altitude of half the radius of the "central attractor" (eg: this would be the Sun when considering the earth's orbit around it, or the earth, if considering an earth-tied artificial satellite). At this point the time "rate" changes of GR are exactly equal and opposite those attributable to SR. So, for example, an atomic clock in a satellite circularly orbiting the earth at half its radius (around 2,000 miles) would exhibit the same time "rate" as identical atomic clocks at the geographic poles.
See:
Freely Orbiting Satellites (http://gijxixj.home.att.net/Relativity/GrSrTpSatExplns.htm#FreelyOrbitingSatellites)
for more along these lines. (Hope you enjoy the guitar music ;-). It plays fitfully whilst it loads, at least on a slow connection, volume control at TOP of web-page, where you can also mute it when the endless repetition p**s you off too much ;-).
Dennis Revell
russ_watters
Jan6-04, 12:57 PM
Satellites in low earth orbit travel at 16,500mph. At that speed, time dilation in gps satellites is about 7,000 nanoseconds (seven millionths of a second) per day (and then there is the general relativity time dilation which is 45,000 nanoseconds per day in the opposite direction). That works out to about 2.5 seconds per thousand years or just over one year over the life of the universe.
I got the dilation values off the net and I really hate math, so I won't try to calculate the actual values for the earth going around the sun. But it'll be within an order of magnitude or two of what I just gave you. Not real significant on the galactic timescale.
Actually russ, the formula(e) for circular orbits is(are) quite simple:
The ratio of satellite time to planet or star time (actually planet/star time at its geographical poles only), for any circular orbit, is given by:
GAMMA = 1/ SQRT [ 1 + GM/c²{2/R - 3/r} ]
where G = Newton's Gravitational constant, M = mass of the planet/star, c = speed of light, R = planet/star radius, r = radius of orbit.
If this equation hasn't already got a name, why don't I modestly call it the Revell equation. ;-) In terms of the speed v of a circularly orbiting satellite, which speed is fixed by G, M, and r, this becomes:
GAMMA = 1/ SQRT [ 1 + {2h/R - 1}v²/c² ]
here h = satellite altitude (ie: h = r - R).
Btw, GPS satellites do not orbit at 16,500 mph. The formula relating satellite speed to altitude (for circular orbits) is quite simply, and strictly:
v = SQRT [GM/(R + h)], symbols same meanings as given above, so you can work it out for any (circular) situation. Interestingly for h = 0, v² becomes GM/R, which is half of the (escape velocity)², or half of vesc², so the simplest version of expression for GAMMA becomes:
GAMMA = 1/ SQRT [ 1 + {vesc²/c²}{1 - 3R/2r} ]
... unless I made a mishtake somewhere (couldn't be bothered double checking it ;-).
Hi, how are you? I am a programmer, not Physic major. I just read Physic to exercise my bored brain. I did some SR reading, but not GR yet. So, I am really a layman.
I am very interested in this formula regarding time difference between two objects orbiting the same star.
Is there a book that shows the derivation of this formula?
Also, since the object in the Sun does not orbit the Sun, will this formula need to be adjusted to that?
GijXiXj
Jan10-04, 12:45 AM
Sammywu:
The formula/e for circular orbits is/are given in the immediately preceding post to yours. Take your pick but version 2 or 3 are probably most appropriate.
Take ver. 3: GAMMA = 1/ SQRT [ 1 + {v²esc/c²}{1 - 3R/2r} ]. If you know the escape velocity of the star (given, in case you don't, by v²esc = 2GM/R, all symbols usual meanings), then you can apply it in turn to as many orbit radii (r) as you like, the ratios will give the relative plod of time from orbit to orbit. The smaller the value of GAMMA, the more the "time-rate" of a satellite is "speeded up". Each individual GAMMA gives the time plod difference between an individual satellite and the geographical poles of the star/planet. ( and ... duh ... this is a slippery slope I'm sliding down ... )
You could even account for all (reasonably sized circular) orbits by plotting the graph of GAMMA versus r, for any given planet/star, or indeed for a completely imaginary one. ;-)
Well, the proof is from General Relativity, so prepare for a long haul. ;-) The "finale" of this, as far as the information you requested is concerned, can be found as Eq. (392) from Pauli's "Theory of Relativity", Dover paperback edition:
T' = T/SQRT(-g44) = T/SQRT[1 + 2Ø/c²] ... ... (392)
Here Ø is the potential, and in the case you're interested in, it is the gravitational potential. For a circularly orbiting satellite you just add the centrifugal potential of -½v² to the gravitational potential to get the overall potential of (Øgrav - ½v²).
and no doubt, you can find simlilar in many other places, and Einstein's original 1916 paper. But I "like" this equation, very reminiscent of the Special Relativity Lorentz Transformation. For the closest I've found to the "real McCoy", look HERE (http://www.mathpages.com/rr/s6-05/6-05.htm).
Also see HERE (http://gijxixj.home.att.net/Relativity/GrSrTpSatExplns.htm#FreelyOrbitingSatellites) for more, but I should warn you there are huge mistakes there. ;-) (I'm just waiting for a professional relativist to point them out to me. ;-)
If you're not confused, you don't understand the problem.
Sammywu
Jan10-04, 08:20 AM
Before I get too crazy and confused, let me propose an imaginary experiment.
Let me put a space station in a remote space area far away from all gravities. Now a rocket passes by the station; at the event, you can synchronize their clocks, and we will ignite a propeller attched to the rocket pointing perpendicular to the rocket instead of parallel. This propeller will continuously generate a calculated acceleration and make the rocket go thru a circle and back to the station again at the sam speed so that we can synchronize and compare their clocks without any ambiguity again.
You can see we now generate a artificial gravity on this rocket. According to the GR and SR effects you people proposed, will they exactly cancel out each other and their clocks shall match?
Or does the gravity have different effect from a propeller generated artificial gravity?
At the same time, I have proposed a twin paradox that could be realized without gravity and infinite acceleration, which I can not comprehend just as DR. Does this really mean that twin paradox is really meaningless?
Does this click? Please help.
Thanks
Sammywu
Jan10-04, 09:47 AM
Another questions:
Simply when applying this GR effect to some observers, there are three different kind of observers:
1. The one orbiting around the planet or sun in a circular constant speed.
2. The one free falling.
3. The one stands on the surface of the planet and supposed the planet is not spining by itself, this person actually felt two forces: the gravity that pull him down and the ground support that push him up and they cancel each other.
How do their clocks tick differently?
Thanks
Sammywu
Jan10-04, 10:00 AM
Another experiment:
You sent a rocket out of the space station that I put in the space in the previous experiment. Of course, we will give it some push for a certain amount of time until it reaches a certain speed. Let it fly for a certain time in constant speed. Then at one point, we ignite its returning propeller until it reverses its velocity ( I assume velocity is the vector one. If I am wrong, forgive the layman. ) Let it fly for a certain time and then start its bracking propeller so that it will undergo acceleration until it stops at the space station. Again we can synchronize their clocks at two events without ambiguity. Could you tell me what is their clock difference in certain formular? Keep the thought simple, you can either assume you are the clock at the station or or the clock at the rocket and give me your answer as you like. just make it clear what is T and T' in your answer.
Originally posted by Sammywu
Before I get too crazy and confused, let me propose an imaginary experiment.
Let me put a space station in a remote space area far away from all gravities. Now a rocket passes by the station; at the event, you can synchronize their clocks, and we will ignite a propeller attched to the rocket pointing perpendicular to the rocket instead of parallel. This propeller will continuously generate a calculated acceleration and make the rocket go thru a circle and back to the station again at the sam speed so that we can synchronize and compare their clocks without any ambiguity again.
You can see we now generate a artificial gravity on this rocket. According to the GR and SR effects you people proposed, will they exactly cancel out each other and their clocks shall match?
No, the clocks won't match. Since the Rocket from its perpective is always accelerating towards the station it will perceive the station clock as running fast. The station will simply preceive the rocket clock as running slow due to its relative velocity. (The rocket's acceleration will only affect the rocket's clock in as far as it affects the rocket's relative velocity.
Sammywu
Jan10-04, 01:14 PM
Janus, Thank you for your reply. Just one thing I would like to make clear. The rocket passed right by the station and retruned through the station again after a circle. The station is not in the center of the circle. Also, the station is light enough so its gravity shall be neglegible.
If you still maintain your answer, where did the GR effect that proposed by other people go? It seems that only SR is making an effect here.
Thank you very much. Would you might shed some lights to the other two questions?
Originally posted by Sammywu
Janus, Thank you for your reply. Just one thing I would like to make clear. The rocket passed right by the station and retruned through the station again after a circle. The station is not in the center of the circle. Also, the station is light enough so its gravity shall be neglegible.
If you still maintain your answer, where did the GR effect that proposed by other people go? It seems that only SR is making an effect here.
Thank you very much. Would you might shed some lights to the other two questions?
Strictly speaking, you don't need to invoke GR for this problem, just SR as it applies to acceleration.
From the perspective of the station this is simple. you only need to take into account the ship's relative velocity.
From the ship's perspective, dealing with its acceleration is a bit more difficult. (The time dilation the ship measures on the station's clock will vary by not only the postion of the station with respect to the accleration but also its relative distance.)
For instance, while the ship is next to the station, the acceleration of the ship is away from the station like this:
<-------Ship Station.
During this part of the circular path the effect seen by the ship is that the station clock runs slow both due the relative motion and the acceleration the ship experiences. since the distance between ship and station is small, the second effect will be small.
During another part of the path the acceleration of the ship will point towards the station like this:
Ship------->______Station
The relative motion effect will be the same. But now the ship will see the station clock as running fast due to the acceleration. But now that the ship is far away from the station, this effect will be considerable. (Enough to overwhelm the relative motion effect. )
The upshot is that the rocket will measure more total time as passing on the station clock when it come back around.
You really only need to invoke GR if you want to consider the spaceship as being your "at rest" frame during the whole time. Then you have to explain the motion of the station(and the force felt by the passengers of the ship as being due to a gravitational field.
Then you factor in the the gravitational time dilation due to the relative postions of the ship and station in the field.
Sammywu
Jan11-04, 08:35 AM
Janus, Thank you for you reply. I am still not clear. Let me make a wild guess.
The GR effect seems to me is a compensation for the potential energy lost. I need to auume an observer rest at the infinite far away from the start, i.e the center of the gravity source. This observer will run a slowest clock compared with all other clocks cuaght in the vincinity of the gravity, i. e. their potential energy is negative as -GMm/R, refered as EP hereafter. Let's keep it simple for now, assuming them as rest without movement. This GR effect formulate their clock as T/sqrt(1+2EP), knowing EP as negative. This will show that the smaller is R, the higher is the T.
Now lets assume the object in vincinity is moving, then all we need to do is add the SR effect, which is oppsite to the GR figure in sign, back to the object.
If this is correct, we then equate the clock in the infinitely far away to the center of the gravity source. Here is what bother me. How could you simply equate a clock so far to the clock in the gravity center?
Do i do it right? If it's complete baloney, please let me know. If I capture it, please correct me any place I am wrong.
Now, If it somehow captures our mainstream scientist, I will ask another question. Back to the mass-energy equation, my impresion is the mass is really how an outsider measure the internal dynamic of an enclosed confinement ( simply said, box ). Einstein has shown how to derive the mass of photon in the way in SR. Now, gravity potential energy, actually all potential energy, bother me. If I can confine two masses, M and m, and keep them in a certain distance inside that confiement, what will be the mass when we measure it. Could it be M+m-GMm/Rc**2?
Thanks
GijXiXj
Jan11-04, 08:42 AM
Er, Sammyu, isn't what you're talking about exactly the scenario given in Figure 1 (http://gijxixj.home.att.net/Relativity/Image1.gif), from this web-page (http://gijxixj.home.att.net/Relativity/GrSrTpSatExplns.htm)?
Btw, neither propellers (nor jet engines) work in space. ;-)
Janus, your explanation seems awfully complicated. Surely it doesn't really matter that the station is not at the centre of the space-ship's circular path because it's in the same reference frame as the centre of that circular path.
Sammywu
Jan11-04, 11:43 AM
Dear Janus,
When digesting your answers more, I realized that gravity is not the same as a true force in this model.
Because the experiment I gave showed clearly the spaceship has a slower clock than the station, from your answer. Even though from a relative motion point of view, if disreagrding the acceleration, they look symmetric to each other.
While applying gravity and my artifical gravity actually generate two opposite effects. So, gravity is apparently different from the artificial gravity I created with the material-antimaterial propeller, which as I know could be the most efficient propeller thereotically.
The better way to interprete this seems to be taht the gravity is purely a curvature of the space, from lots of people saying, which I had problem to grasp in the past. Now it seems clearer to me what that really means and why Einstein needed to say that. I need sometimes to further digest the idea.
Thank you for tolerating the layman here.
Well, I am still interested in knowing your response to my interpretation of compensation for pontential energy and whether the mass measurement will be different by an outside observer when a box containing two masses in certain distance. If you are willing to tolerate me more. Thanks
Sammywu
Jan12-04, 07:59 AM
Janus,
I think I got closer to the core of the problem.
I modify my experiment to make me comfortable with no ambiguity. My station could have attached a tunnel to the center of the circle. In this way, I am certain the center is at the same reference frame as Gixxie said. I know for you this is unnecessary.
Now, the next thing, I want to talk about is the absoluteness of acceleration. It always bother me when it was said who is feeling the acceleration. Now I found a scientific way without ambiguity to check who is under influence of true acceleration. It's from Einstein any way. I just did not understood it. Simple, whoever see the light bent is under the true accelaration. Most of time, we will not able to see the curve of loght but in an astronomical scale, this effect could become clear.
Back to my experiment, the spaceship will see the light bent. The space staion and the center will not see the light bent.
In a gravity field, the free fall observers and the orbiting observers will not see the light bent because the light is bent together with them. Or, in a different words, all of the three: free falling objects, orbiting objects and light are under the gravity influence.
The objects on the planet, assuming no planet spin, will see light bent, because they are not moving together with light by gavity. So, it has the same clock as the infinite observer, or the center observer in Physicists' textbook. The force acting on them are the electromagnectic forces exercised by the push of the outer electrons in the ground serface. As I remember, somebody has said, all contact push forces actually base on the outer electron push force. Or simply, the ground support force that keep the observers in standing.
The three models: PE compensation, space curvature and absolute acceleration examined by light bent all seem to be able to explain somethings and might have more foundamental implcation embeded.
For one, now we can clearly see gravity is a fake force-- my own term. You people might not agree with that. ElectroMaganetic force seem to be a real force. I am not sure about weak and strong forces.
Does this seem to deviate from the mainstream?
Now what I need to do is putting some math. in, so I can derive a common formula for he GR effect and SR effect in these models. And hopefully that will come out to agree with yours.
Sammywu
Jan12-04, 05:44 PM
Dennis, Sorry, I have misspelled your name. Also, I did not notice you are the same person.
About the propeller, maybe a thruster will be more acceptable to you.
Any way, geting back to this topic. I have to take one statement back. The free falling clock might be different from the orbiting clock. Even though they both see the light straight, so, they are under the influence of the same fraudulent force, the gravity. But the orbiting object has an initial speed and the free falling does not. Let's worry that later.
Now, back to my imginary experiment, the formula between the space station clock Tst and the spaceship's clock Tship shall be Tst=Tship*1/sqrt(1-v^2/c^2). Since only SR is neededin the evaluation of their relationship and Tship has a slower clock. And do not forget, you can compare their clocks after one circle without ambiguity. It's real effect. The person in the spaceship will be younger.
Now, reversing that back to the relationship between the Sun and the Earth, Tearth=Tsun*1/sqrt(1-v^2/c^2). The particles in the Sun shall experience slower clock than the earth clock, assuming there is no spin in both sun and Earth. The Earth is orbiting around the Sun under the fraudulent force. Borrowing your derivation, -v^2/c^2 = 2EP = -GMsun/R, here R is the distance between the Sun and the Earth.
Now, let's look at the spin at the Earth. You probably can assume polar bears are not under any spin effect and need no adjustment compared with the Earth's clock.
My friend supplied me these figures, about 6,000,000m for Earth's radius, 9.8 m/sec2 for gravity experienced in the surface of the Earth. With that figure, I found an object to orbit around earth's surface, you might call that escape velocity, I can't remember why, requires about 7,800 m/sec ( I did the calculation in the office and can't remember the exact figure. Any way, it's high school math. by v^2/g=r.) The spinning speed at the Equator would be very small, like 34 m.sec maybe. Any way, the person on the Earth's surface could be considered standing firm supported by the ground force. This means you can treat that clock just like the Earth clock. I might show you how to adjust that next time.
Now, Let's denote Ts as the clock of a satellite orbiting around the earth at the height of H. By the GR effect I just learned from you, Ts=Tearth*1/sqrt(1-v^2/c^2). Here -v^2/c^2 probably is 2PE which you already derived. 2PE=-2GM/(r+H), here r is the radius of the Earth, not the distance between the Sun and the Earth.
It does not seem to me I need any SR effect, did I miss something?
Thank you for your and Janus's help.
Hi,
I was wondering:
if two objects traveling in a parallel paths near each other, were to approach the speed of light, would the gravitational force attracting the two objects approach infinity? Or would the gravitational force between the two objects remain unchanged? In other words, does the gravitational force of an object increase to infinity as it's velocity approaches the speed of light? I was under the impression that the gravitational force of an object remains unchanged as it approaches the speed of light, otherwise, any object approaching the speed of light would be crushed to a point under it's own gravity, instead of a two dimensional plane under Lorentz contraction. I could be wrong, could someone clarify for me? Thanks!
Best Regards,
Edwin
Sammywu
Jan12-04, 08:18 PM
Edwin,
That's a very interesting question. Maybe I will spend some times thinking about that too.
Any way, the reason I got back here is that I found I got another question answered, probably. The inertial reference frame that bothered me and most people is apparently the frame that light is not bent. Now, we have a physical way to measure which one is the inertial reference frame.
When I thought that through, I will elaborate. Otherwise, for now, it's enough for me. We might be able to discuss whether there is a truely prefered frme or not.
Originally posted by Edwin
Hi,
I was wondering:
if two objects traveling in a parallel paths near each other, were to approach the speed of light, would the gravitational force attracting the two objects approach infinity? Or would the gravitational force between the two objects remain unchanged? In other words, does the gravitational force of an object increase to infinity as it's velocity approaches the speed of light? I was under the impression that the gravitational force of an object remains unchanged as it approaches the speed of light, otherwise, any object approaching the speed of light would be crushed to a point under it's own gravity, instead of a two dimensional plane under Lorentz contraction. I could be wrong, could someone clarify for me? Thanks! If you were sitting on (or in) either object, the other would appear "stationary" - not moving , except towards you. The gravitational force (between your object and the other one) that you would measure would be the same, whether you observed the quasar 3C273 (an arbitrary, distant point of reference) travelling at 0 m/s or 200 million m/s. The usual caveats apply.
Perhaps you are interested in what you would observe if you were a distant observer?
Sammywu
Jan12-04, 08:36 PM
Edwin, My attemp to answer your question might not be authetic. Do not take me as your final answer.
I see your point. Let me elaborate. You need to assume another observer that is static. So, to the observer, their mass were increased, so their gravity attraction shall be increased to the observer. Now, if there was nothing holding them afar. They will be socked to each other, of course. But what will be the clocks and space measured to themselves and the static observer? Note one thing, the distance should be observed unchanged by either the moving observer or static observer, because the length contraction occured on the dimention of movement.
If they are held afar by something, the question will be what force is required to break down the holding architect.
Here, of course we are assuming no acceleration is involved.
Well, I can't get the answer yet. Just trying to clarify your question behind your statements took me some analysis.
Thankyou for your help guys. I am interested both the observations of distant observers as well as the observer on the masses themselves. I have been reading up on relativity, and have not found a specific answer yet, either, but will continue to search. I think it is that General Relativity does not treat gravity as a true force, but describes the affects of gravity in terms of curved space-time and manifolds. Suppose that our two objects traveling near each other are spheres, the idea that the dimensions of the two spherical masses do not contract but in the direction of travel would make me wonder as to whether the gravitational force is homogenous in three spatial dimensions x, y, and z the two masses. What do you think. Would the gravitation forces appear to be homogenous to some third stationary observer on some planet? Would the gravitational force between the two objects appear to be homogenous(equal in strength in all directions) to the observer riding along on one the two objects?
Inquisitively,
Edwin
Sammywu
Jan13-04, 07:51 AM
Edwin, Nereid,
I have no answer for Edwin's question. I can only make his puzzling a little clearer.
There is a hidden paradox in the situation.
Since the distance between the two objects does not contract and the time takes two objects to hit each other is independent upon the Masses. So, the increase of mass measurement by the static observer will be irrelevant. The time taken for the two objects colliding will be the saem by both moving observers and static observer. This contradicts SR's theory of time dilation supposed to be observed by the static observer.
The next case, if we carefully manufacture the holding bar between the two objects so that it can only stands the force of Gm0*m0/R^2. The moving observers shall not see the bar broken. By the static observer, the mass increase will make higher arracting force and will break the holding bar.
Again, we , the nonbelievers, prevail. Albert lost. Take one step back, can we explain the constant light speed phenomenon without Albert's SR? NO. Maybe we shall listen to what Albert wants to say about it. Janus, would you mind representing Albert again?
Another question, if the two objects actually are two electrons, when we add electromagnectic force here. How does that play? I always have problems with Albert's imagination when he says about that infinite current. I tried to decipher that mystery with just two electrons. Did anyones do this already?
Now let me get back my unfinished work on the free falling object in the gravity. It appeared to me, at the initial point, when the free falling object was let go, it shall have a difference in time measurement difference with the pass by orbiting object. With its falling continued, it shall have a different time measurement against the orbiting object. Its velocity ( which one is the scalor one, velocity or speed? ) will increase but less than the orbiting object's velocity until it reach the center of star.
Let me stop here, because i thought of two interesting imaginary experiment.
I can actually extend my imaginary experiment with one more spaceship B, naming the original one A. While the spaceship A passes the station, B will leave the station. With precise control, B will work like the shadow of A on the diameter of the circle that A will move around. If you are familiar with the coircular motion and the oscillation, you know what I am talking about. A and B will meet at the other end of the diameter according to classical theory, another event without ambiguity. Coming back, they will meet with the station agin. What will be the clock difference among the threes?
Another one. making a device like a donut with a connecting tunnel through one of the diameters. With an open end at the joint of the tunnel and the donut, you can shoot an object in with certain speed. Put a comparable significant mass in the middle; now this is a artificial Sun experiment that you can let objects orbiting through the donut and objects falling through the tunnel. You can perform a true comparison of clocks between free falling objects and orbiting objects.
Sammywu
Jan13-04, 08:59 AM
Just one thing. My imaginary two masses in a closed box might not be imaginary. Browsing Physic World, notice a double pulsar being viewed now, that could be an arena for two masses in an enclosed box.
Sammywu
Jan13-04, 07:47 PM
Back to my free fall object issue. My calculation was a litle rusty. I need to correct the calculation of those formula. I wonder why nobody caught it and corrected me. Now, I have to spend long words for just calculation. Ek=(1/2)*mv^2. Ep=-GMm/R. E=Ek+Ep=-(1/2)GMm/R. and v^2=GM/R=-2E/m. Let's denote F=E/m. F is not the potential energy but rather the field, Is it that how you call it? Tst=Tship*(1/sqrt(1-v^2/c^2)). In reverse, Tearth=Tsun*(1/sqrt(1-v^2/c^2))=Tsun*(1/sqrt(1+2F/c^2)).
Please help me to check my math. If I am still wrong, correct it for me.
Back to my experiment, the one with staion timed by Ts, the orbiting ship timed by Ta and the straight moving ship timed by Tb. We already know Ta. The only question is Tb. Let's use Ts as a standard parameter; you can make Tb as a function of Ts. Make integration of Tb along the path with Ts as the parameter. Can any one do the math? I believe this will bring us to the GR effect when we equate Tsun=Ta and Tearth=Tb.
Let's go back to the three models I proposed, The second saying that gravity is a fraudulent force and just the space curvature ( inward though ) is really the same as the third that uses light path as a check for who is under true acceleration. Is it? Maybe not. for now, let's assume they are. The first one will completely match this one, result wise. If yiur calculation showed that they are different, please let me know.
If this proved correct, this seems to be a complete model that we can deduce GR from SR. So GR does not exclude SR, GR is an extention of SR. Without SR, GR has no wher to be born.
This seems to resolve our question. Actually this poses more question. If this is true, does that imply any length dilation and mass measurement issue here in GR? Did GR textbook mention anything like that?
Where does my two stars in one box fit?
Does GR help us to resolve Edwin's paradox?
Sammywu
Jan14-04, 07:43 AM
Back to Edwin's question.
Now, I remember this is really the part of SR's Electrodynamic analysis.
The question actually arised from a phenomenon. Two static free electrons will exercise electric force between them. While two constantly moving electrons ( or current ) will exercise not only electric force but also magnetic force between them. This is a paradox itself. In Newtonian's world, movement is relative. From the view-points of the moving electrons, they are static. There should not be additional force observed.
That's why the question of two electrons came to my mind immediately. I think.
Many people here have studied that part of SR. I think Lawrenzian force came out from that part also. Hope I use the right term. . I can't find my SR book. Anyone want to help Edwin to understand this part. Nereid, David, how about you?
Originally posted by Edwin
Hi,
I was wondering:
if two objects traveling in a parallel paths near each other, were to approach the speed of light, would the gravitational force attracting the two objects approach infinity? Or would the gravitational force between the two objects remain unchanged? In other words, does the gravitational force of an object increase to infinity as it's velocity approaches the speed of light?
Since the second question is easier to answer I''l respond to that one first. Yes. The gravitational force is a function of velocity. The gravitational force on a particle in free fall is given by
G_{\mu} = \frac {1}{2} m g_{\alpha \beta, \mu} v^{\alpha} v^{\beta}
where
g_{\alpha \beta}
are the gravitational potentials, also known as the components of the metric tensor and
g_{\alpha \beta, \mu}
is the partial derivative of the potential with respect to mu. There are ten independant gravitational potentials in GR as opposed to the one in Newtonian gravity. The Newtonian gravitational potential is defined as
\Phi = \frac{c^2}{2} (g_{00}-1)
The mass of the particle with proper mass m0 is given by
m = \gamma m_{0} = m_{0} \frac {dt}{d\tau}
The mass is not only a function of speed but a function of the gravitational potential as well. For detailed defintions and proof of these relations see
http://www.geocities.com/physics_world/gr/grav_force.htm
As the speed increases then the gravitational force goes to infinity. However since the mass increases as well the particle will actually slow down in cases such as a falling towards a black hole or in a uniform gravitational field.
Note: The gravitational force has a relative existance. Einstein's Equivalence Principle states that at any point in spacetime the force my be transformed away at that point. For the falling particle - the particle may be considered to be a rest at the origin of a locally inertial frame. At the origin of this frame the gravitational force is zero. However the tidal force may not be zero as is the case of tidal forces, which cannot be transformed away.
Thankyou everybody for your help. I really appreciate it.
Best Regards,
Edwin
Sammywu
Jan14-04, 04:36 PM
Arcon,
Thank you for the lesson. It's of great help.
As you know, My Physics is just high school grade. It will take me a while to understand all those formulas.
One thing I do'nt know is what is tidal force. I have read it many places. Would you mind explain a little more about it?
Also, if applying this 4-dimention vector formula to my imaginary experiments, does its outcome fit my calculation?
Originally posted by Arcon
The gravitational force on a particle in free fall is given by
G_{\mu} = \frac {1}{2} m g_{\alpha \beta, \mu} v^{\alpha} v^{\beta}
Which you should mention is a fictitious force because it is not a four vector and can be transformed away.
The Newtonian gravitational potential is related to g00 as
\Phi = \frac{c^2}{2} (g_{00}-1)_
Only in a linearized weak field approximation.
The mass of the particle isw hose proper mass is m0 is given by
m = \gamma m_{0} = m_{0} \frac {dt}{d\tau}
This is bad terminology for special relativity and is just plain wrong for arbitrary spacetimes in general relativity. In general the conserved energy parameter is not always even proportional to \frac{dt}{d\tau}
The mass is not only a function of speed but a function of the gravitational potential as well. For detailed defintions and proof of these relations see
http://www.geocities.com/physics_world/gr/grav_force.htm...
This is your own site and is wrong. The word mass of a particle free of a vector potential referrs to the positive root for m in the equation
(mc)^2 = g_{\mu}_{\nu}P^{\mu}P^{\nu}
which is manifestly invariant.
However the tidal force may not be zero as is the case of tidal forces, which cannot be transformed away.
Yes they can. You just can't tranform away both the affine connections and the tides "at the same time" when the gravitational spacetime curvature field described by the Riemann tensor is not zero.
Originally posted by Sammywu
One thing I do'nt know is what is tidal force. I have read it many places. Would you mind explain a little more about it?
The definition of Tidal force is that which causes gravitational tides i.e. tidal forces are what cause ocean tides.
In GR it is defined in hte more precise way as is the difference in the gravitational force as measured in a freelly falling frame of referance.
Also, if applying this 4-dimention vector formula to my imaginary experiments, does its outcome fit my calculation?
Which calculation is that again?
Sammywu
Jan14-04, 05:16 PM
Arcon,
Thank you for the lesson.
There are three simple experiments that could be used to examine your theory:
The experiment I proposed about two spaceships A, B and the station in the remmote space assuming no gravity force.
Another is an experiment that actually similates the movement of free falling bodys and orbiting objects caught by a gravity.
The last one was the two parallel moving electrons.
Once I digested your formulas, maybe I can propose some more complicate experiments for you to demonstrate how your formula can be applied to a problem.
DW, thank you for pointing out it's Arcon's own website.
I guess when you know more, there are more differences that people have to straight out. But, this shall be focused on a basis that we are all looking for the truth.
I have browsed his website. Did you? Any formulas shall be corrected to reflect a current Physics points of view.
Note - I just noticed an error in the influential article The Concept of Mass, Lev B. Okun, Physics Today, June 1989. This is the article which has been highly influential in bad mouthing the concept of relativistic mass. But the problem is that it has several errors in it. There is an expression in this paper for the gravitational force on a moving particle. The equation number is (16) and is on page 34. For the case of radial fall as I did it his expression and my expression are identical.
I have never seen Okun's work in the derivtion of that equation so I didn't know know the meaning of the terms he used. Therefore, before today at least, I was not in a position to comment on it. However. since I took the time recently figure out how he derived this I can now see his errors of which there are two. Okun claims that when v << c the equation reduces to (using my symbols for mass)
F_{g} = \frac {GMm_{0}}{r^{2}}
However that is an invalid statement. That is not the limit. The correct limit is
F_{g} = \frac {GMm}{r^{2}}
The massm, m, this second equation equals
m = \frac {m_{0}}{\sqrt{1 + 2\Phi/c^{2} - \beta^{2}}}
only attains the value m = m0 when
2\Phi/c^{2} = \beta^{2}
For v << c m is not m0. However if v << c and Phi = 0 then m = m0. For example: If the particle is at rest in the field (or moving slowly) then m becomes
m = \frac {m_{0}}{\sqrt{1 + 2\Phi/c^{2}}}
In his article Okun incorreclty implies that Einstein did not use the concept of relativistic mass in Einstein's text The Meaning of Relativity. However Einstein does utilize this expression in a derivation regarding Mach's principle. Notice that that mass is a function of the gravitational potential. Therefore if the particle is in a gravitional field its inertia is altered. For this reason Einstein stated in his text (page 100,102)
(page 100) The inertia of a body must increase when ponderable masses are piled up in its neighboorhood.
(page 102) The inert mass is proportional to 1 + q, and therefore increases when ponderable masses approach the test body
Where Einstein's q is the negative of (gravitational potential)/c2 and is positive. Einstein's relation is an approximation to the exact relation which I gave above. When the field is weak and the particle is movig slowly then
m \approx (1 - \Phi/c^{2})m_{0} = (1 + q)m_{0}
Okun also makes another error in this paper. His formula has an E in it which he claims is energy and his use implies that his E has the value
E = m_{0}c^{2}\frac {dt}{d\tau}
However this is not how the energy for such a particle is defined in general relativity. Okun assumes that the energy E is related to the time component of the 4-momentum P as E = cP0. However this only holds on special relativity. The correct expression in general relativity is P as E = cP0.
Long story short - While Okun't goal was to claim that m can be called E which is the time component of the 4-momentum and doing otherwise causes confusion what he actually did was to prove the opposite. His thinking of m as E has led to make the errors above!!
[:D]
Sammywu
Jan15-04, 08:16 AM
DW, I will be busy for a while. Any way, if you do not agree with Arcon, is that possible that you show how you will handle these problems.
Thanks a lot.
Originally posted by Sammywu
DW, I will be busy for a while. Any way, if you do not agree with Arcon, is that possible that you show how you will handle these problems.
Thanks a lot.
I totally dissagree with pmb's rather Newtonian approach to relativistic problems. As I've mentioned just about every view he has outlined above is wrong in the context of modern general relativity. If you have a specific problem you would like me to show the correct relativistic treatment for just give me the details. Otherwise I'll just give a brief account of the modern relativistic paradigm for gravitation. The equation of motion for general relativity takes a form similar to Newton's second law.
This general relativistic equation is
F^\lambda = \frac{DP^\lambda}{d\tau}
There are 3 important differences between this and Newton's second law. First the vectors represented here contain a fourth element corresponding to a timelike coordinate. Second, the time derivative here is not a derivative with respect to your coordinate frame, but is with respect to time for the "particle" in question. Third this derivative is not an ordinary total derivative, but is a covariant total derivative. It is a sum of an ordinary derivative and an affine connection term which is a description of the curvature of the spacetime coordinates being used. The capitalization of the D in the numerator is not a typo. This force F^\lambda is a real force and is sometimes qualified and called the four vector force because it is a four vector and is different than the ordinary force defined by
f^i = \frac{dP^i}{dt}, and sometimes including f^0 = 0, which is still sometimes used at a basic level in the context of "special" relativity.
The four vector force is real and as such can not be transformed away. In the case that there is only gravitation and no real forces, F^\lambda = 0
and
F^{\lambda};_\mu = 0.
The first is sufficient to guarantee that a frame exists in which the inertial forces also called fictitious forces or pseudo forces including the gravitational force (a sum of terms proportional to affine connections) is "locally" transformed away.
The second is sufficient to guarantee that a frame exists in which the tidal force(a sum of terms proportional to first order derivatives of affine connections) is "locally" transformed away.
The tricky part is that when there is Riemannian spacetime curvature these two things can not be transfomed away going to a "single" frame.
You can transform away one or the other, but when Reimannian spacetime curvature is present at least one of those is always present. In short this is because the Riemann tensor which is the expression for this kind of curvature is a sum of first order derivatives of affine connections and products of the affine connections. A nonzero tensor is nonzero according to every frame, so one set of terms can be transformed away, but never both when this tensor isn't zero.
The Newtonian concept of the gravitational field was an acceleration field. The relativistic analog of the acceleration field is the affine connections. However the affine connections are not tensors and can be transformed away. Therefor by themselves they can not represent a complete real physical entity. For this reason in modern relativity a different quantity all together has become the field concept in gravitation and that is the field of Riemannian spacetime curvature described by the Riemann tensor. This is because when the Riemann tensor is not zero one can never have a frame in which the gravitational force can be "globally" transformed away and when the Riemann tensor is zero one always has a frame according to which the gravitational force is globally transformed away and the fact that the Riemannian spacetime curvature is described by a tensor which can not be transformed away is sufficient that it can be taken to stand alone as a complete real physical entity.
Up to this point I've referred to the gravitational force several times, but as I specifically mentioned it is a fictitious force corresponding to F^\lambda = 0. In fact when doing modern general relativity it is rare that one would even make reference to a "gravitational force". One instead thinks of motion in terms of geodesic motion Vs nongeodesic motion. When there is a real force present the motion is nongeodesic. In the absence of the four vector forces motion is geodesic. This consept replaces Newtons first law of motion. Newton's first law of motion can be expressed as unacted on a particle will have constant motion. The general relativistic law corresponding to this is unacted on a particle will follow geodesic motion. If the four vector force is zero then one can show that the equation of motion for general relativity reduces to what is called the geodesic equation. This equation describes optimal paths between events in a curved four dimentional spacetimes called geodesic paths named after paths between points along the curved surface of the earth that optimise travel distance along the surface. The geodesics of general relativity are also paths of extremal proper time. Consider two events or two different points in space and time. You have an observer that is to start at the location of the first event and to arive at the second. The most time will be recorded by that observers watch if he is to follow the geodesic that connects them. This is the principle of maximal proper time.
Theres a lot more detail I could get into, but I'm getting tired now so I'll just tie it together with the short paragraph following.
The stress energy tensor is the gravitational source for the field of Riemannian spacetime curvature. This curvature determines what forms the differential geometry of spacetime expressed by the metric tensor can take. Things unacted on by four forces then follow geodesics or paths of maximal proper time between events which are paths descriptive of this geometry. Due to the curvalinear nature of the spacetime an observer's coordinates will not be globally rectilinear or inertial which leads to his experiencing of inertial forces which he then calls the gravitational force. Locally transforming this away is as simple as going into free fall.
Sammywu
Jan16-04, 07:40 AM
Arcon, DW, Sorry, I haven't even time to really read your stuff. Just on the train, checking Arcon's formula populated in his website. The first part of Gravitational force.
One question: What is t and tow? Is t the time of the moving observer, or the infinite observer ( the center observer in textbook maybe ), assuming here t and tow are all in a gravity influencing area?
Originally posted by Sammywu
Arcon, DW, Sorry, I haven't even time to really read your stuff. Just on the train, checking Arcon's formula populated in his website. The first part of Gravitational force.
One question: What is t and tow? Is t the time of the moving observer, or the infinite observer ( the center observer in textbook maybe ), assuming here t and tow are all in a gravity influencing area?
t = time as measured by distant observer
Tau = time as measured by wristwatch strapped to falling particle
Arcon
Sammywu
Jan16-04, 02:03 PM
Arcon, What is the 4-velocity? Why did you divide the space measurement by the remote observer by the proper time of the falling obhect?
Originally posted by Sammywu
Arcon, What is the 4-velocity? Why did you divide the space measurement by the remote observer by the proper time of the falling obhect?
Relativity implies that the laws of physics must be expressable in a form which is independant of the coordinate system. I.e. they must be covariant. Such laws may expressed mathematically using tensors. When expressed as tensors then they are said to be manifestly covariant. For example: In Newtonian dynamics the equation
F = \frac {dP}{dt}
is said to be manifestly covariant, i.e. it does not depend on the coordinate system. However the equation
F_{x} = \frac {dP{x}}{dt}
which is one of the components of the first equation, is not manifestly covariant since it is not a vector equation. However it is still covariant since it has the same form in all coordinate systems. I.e. in another, primed, coordinate system
F'_{x} = \frac {dP'{x}}{dt}
A 4-vector is an example of a tensor, i.e. it is said to be a tensor of rank 1. The proper time is also a tensor. It's called a tensor of rank 0, also known as a scalar or an invariant. As such the laws of general relativity must be expressible in the form of tensors. The mathematics of general relativity is of course geometry just as it is in Newtonian dynamics. However in general relativity one deals with differential geometry, the equations of which are often tensor equations. In order to describe motion one starts with 4-vectors and then squeezes out the desired quantities. In this case the 4-vector of velocityh (i.e. 4-velocity) must be obtained by division by a proper time interval rather than a coordinate time interval since the former gives a 4-vector while the later does not.
This is a bit different than one finds in Newtonian dynamics. For example, vectors in Newtonian vectors have components which all have the same physical meaning. However the same cannot be said of 4-vectors. Thus when we have the final result we want to pick off the components to obtain the physical stuff that we're interested in. For example: in special relativity one might desire to work a problem using 4-vectors and yet they might not even be seeking a 4-vector as their goal. They might want to know the the energy of something is and that is a component of a 4-vector. However there is a subtle relationship between components and scalars. Since it takes some space to describe it I wrote it up and posted it in the internet at
http://www.geocities.com/physics_world/ma/invariant.htm
If you have the stomach for it then you might want to take a gander at something I wrote up as an quicky intro to tensors. It's located at
http://www.geocities.com/physics_world/ma/intro_tensor.htm
It describes a few different classes of tensors, namely general tensors, affine tensors, Cartesian tensors and Lorentz tensors.
Creator
Jan16-04, 06:04 PM
Originally posted by Arcon
The Meaning of Relativity. However Einstein does utilize this expression in a derivation regarding Mach's principle. Notice that that mass is a function of the gravitational potential. Therefore if the particle is in a gravitional field its inertia is altered. For this reason Einstein stated in his text (page 100,102)
quote:
--------------------------------------------------------------------------------
(page 100) The inertia of a body must increase when ponderable masses are piled up in its neighboorhood.
(page 102) The inert mass is proportional to 1 + q, and therefore increases when ponderable masses approach the test body
--------------------------------------------------------------------------------
[/B]
Acron;
Though I haven't read Okun's article, I believe you may have incorrectly assumed that Okun was unaware of Einstein's use of 'relativistic' mass.
In reality, he probably simply chose to ignore it. The reason?:
Well, you may be unaware of it but...
The quote you gave above (pg. 100 & 102 from The Meaning of Rel.) is the first of three Machian ideas Einstein initially thought to be consistent with GR.
However, it is well known since then and quite well established (proved in 1962 by Carl Brans) that this first statement is in fact not true of GR.
So although this statement was initially used by Einstein, it has long
ago been recognized as inaccurate due to papers by Carl Brans and others....
In other words, the statment that "The inertia of a body must increase when ponderable masses are piled up in its neighboorhood"
has long ago been shown to be invalid,;
And Okun's statements are simply reflecting his knowledge of that fact.
(I have not the Brans reference ; I'm sure it is locatable).
Creator[:)]
Sammywu
Jan16-04, 06:18 PM
Arcon, David, I can't find my SR book. But I remember it did mention this 4 vector force and momentum, even though I am not sure how to practically use it.
I designed an experiment so as we can see how these four vector momentum and force can be of any pratical use.
Please also list the formula for gamma. I believe that's the contraction factor. But I can not even memorize that formula.
Let's even keep the gravity away from this experiment.
Here, We have one object C whose rest mass is m0. I have two observers A and B. C is moving relative to A in 2/3c northward and B is moving relative to A in 2/3c southward. A , B and C coincide each other at the initial event point.
Now, At the time 0, there is a external force ( not gravity ) acting on C westward. The force is measured as m0*g to C. This force will be acting on C for 10 seconds of C's time.
Actually this experiment will be a little difficult to execute. Any thruster will involve certain mass change. If we put it into an electric field, there will be other issue. Let's just assume this is possible.
Tell me what is the 4 vector force and mometum relative to A, B and C. Will it be fixed or it become a function of tC?
At any time of tC, what is the relationship between tB, tA and tC.
When tC = 10, that is After 10 seconds of C's time, what are tA and tB?
What is the velocity of C relative to A and B at any time of this 10 second interval?
Originally posted by Creator
Acron;
Though I haven't read Okum's article, I believe you may have incorrectly assumed that Okum was unaware of Einstein's use of 'relativistic' mass.
No. He was most definitely unaware of it. He told me that in e-mail. I.e. I was curious as to his comments given Einstein's text so I e-mailed him and asked him. Until I mentioned it to him he was unaware of it.
In reality, he probably simply chose to ignore it. The reason?:
Well, you may be unaware of it but...
The quote you gave above (pg. 100 & 102 from The Meaning of Rel.) is the first of three Machian ideas Einstein initially thought to be consistent with GR. However, it is well known since then and quite well established (proved in 1962 by Carl Brans) that this first statement is in fact not true of GR.
That is incorrect. The statement of Einstein's is by no means false and has never been proven wrong. What you're most likely thinking about is a different use of the term "mass"
Okun refered to Einstein's text in order to prove that Einstein did not use the idea in his book. The problem was that he didn't look closely enough. I guess if you don't expect to find something then its not surprising that you don't find it.
But please note - Einstein is by no means wrong. That statement is by no means wrong. If it was wrong then Einstein's GR would be wrong since all Einstein assumed there was that the quantity M = m_o*dt/dT (m_o = proper mass) was a function of the gravitational potentials (note - for a coordinate system in which the metric is time-orthogonal, i.e. g_0k = 0, metric its a function only of Phi = (g_oo - 1)c2/2) as well as a function of speed of the particle.
Thank you for your response though. If you find that reference to the 1962 article by Carl Brans please pass it on. Thanks.
Sammywu
Jan16-04, 07:19 PM
Arcon, DW,
I am digesting your statements. Arcon seems to indicate that the object under gravity does lose mass. DW, I like your explanation of geodisc and ficticious force.
I know I have seen that strees energy tensor somewhere. I might ask question related to it later. Any way, need some times to digest these. I also read your formula in another thread. I like it. But I still have problems reading it. The denotation is still not clear enough. I think that's common to you, but they are not to us, the laymen. If you can make clear which is what mesaured by whom, I will catch them quicker. Sorry, DW, My previous sample experiment was for Arcon and you. Trying to locate where exactly your arguments are.
I wrote David because somebody else was in my mind.
Sammywu
Jan16-04, 07:53 PM
DW, Would you mind elaborate a little more on the inertial force? What is it? How to calculate it? take a simple example is good enough. Also, are the inertial force and gravity the only two known ficticious forces?
Gravity is caused by the timespace curvature. How will the inertial force been considered? Does it appear as part of the stress energy tensor?
Sammywu
Jan16-04, 08:03 PM
DW, I always thought Reimannian spacetime curvature is a description of gravity. From what you said, it's not. What is it then?
Sammywu
Jan16-04, 08:08 PM
DW, Is the Riemann tensor the stress energy tensor? How does it look like?
Originally posted by Sammywu
I know I have seen that strees energy tensor somewhere. I might ask question related to it later.
When you're ready then see
http://www.geocities.com/physics_world/sr/mass_tensor.htm
I call it the mass tensor in that page but the mass tensor and the energy-momentum tensor differ only by the multiplicative constant c2. The reason for the namd "mass tensor" was only to make the point that the energy-momentum tensor is not a name that is fundamental to that tensor and that since mass is related to energy then one can speak of mass instead of energy. That fundamental fact seems to be lost on most students lately.
But the energy-momentum tensor is not something that is only used in GR. It's used in SR as well.
An inertial force is a force which is not present in an inertial frame of reference. For example: Suppose you're in an inertial frame of reference in a spaceship. A particle moving in a straight line at constant velocity in an inertial is said to be a free-particle. If the ship starts to accelerate then that same particle will now be an accelerating particle as viewed from your frame of reference. It behaves exactly as if it was in free-fall in a uniform gravitational field. That's Einstein's Equivalence Principle. Newton would have called that a fictitious force (hence dw is Newtonian in this respect). However Einstein would call it a real force. The reason Einstein refered to inertial forces as real is that they have the same nature as the gravitational force which Einstein considered to he real - almost by definition. In fact he did call it a real force. Many physicists do today as well.
Here's one such example: From Newtonian Mechanics, A.P. French, The M.I.T. Introductory Physics Series, W.W. Norton Pub. , (1971) , page 499.
From the standpoint of an observer in the accelerating frame, the inertial force is actually present. If one took steps to keep an object "at rest" in S', by tying it down with springs, these springs would be observed to elongate or contract in such a way as to provide a counteracting force to balance the inertial force. To describe such force as "fictitious" is therefore somewhat misleading. One would like to have some convenient label that distinguishes inertial forces from forces that arise from true physical interactions, and the term "psuedo-force" is often used. Even this, however, does not do justice to such forces experienced by someone who is actually in the accelerating frame of reference. Probably the original, strictly technical name, "inertial force," which is free of any questionable overtones, remains the best description.
The Coriolis force is a real force as well. As Einstein explained in the February 17, 1921 issue of Nature
Can gravitation and inertia be identical? This question leads directly to the General Theory of Relativity. Is it not possible for me to regard the earth as free from rotation, if I conceive of the centrifugal force, which acts on all bodies at rest relatively to the earth, as being a "real" gravitational field of gravitation, or part of such a field? If this idea can be carried out, then we shall have proved in very truth the identity of gravitation and inertia. For the same property which is regarded as inertia from the point of view of a system not taking part of the rotation can be interpreted as gravitation when considered with respect to a system that shares this rotation. According to Newton, this interpretation is impossible, because in Newton's theory there is no "real" field of the "Coriolis-field" type. But perhaps Newton's law of field could be replaced by another that fits in with the field which holds with respect to a "rotating" system of co-ordiantes? My conviction of the identity of inertial and gravitational mass aroused within me the feeling of absolute confidence in the correctness of this interpretation.
Sammywu
Jan16-04, 08:43 PM
Arcon,
Your statements are very interesting. I gradually figured that your arguments with DW seems to focus on whether the gravity mass is the same as inertial mass and the measured mass is a gravity mass.
Any way, I need some times to read them.
DW, you seem to believe that the inertial mass described by E/c^2 not the same as gravity mass. In one place, you said that g=GM/R^2 and the M is the rest mass. By the way, I am confused, proper time is the time of the mover's own time. How about proper mass? Straightly, it seems to be the rest mass. But you seem to say that's the E/c^2. Please help me with the terms. . Thanks.
Sammywu
Jan16-04, 09:08 PM
DW, Arcon, If the argument focused on whether gravity mass is the same as inertial mass. I would like to ask another question.
If you noticed, some people are trying to capture a light in a confinement. Let's say if I am able to keep some photons bouncing inside a confinement, will this confinement exihibit any gravity from the mass by E/c^2 of the lights inside the confinement to the objects nearby?
What will your predict?
If I know a little bit about laser, maybe we can do some experiemnts with lasers.
Originally posted by Sammywu
Your statements are very interesting.
We have Einstein to thank for that. :-)
By the way, the terms proper mass and rest mass are synonyms. I choose to use the term proper mass because the term rest mass can easily be interpreted to mean the relativistic mass when v = 0. However that is not always true since relativistic mass is also a function of the gravitational potential so you there can be a difference. I.e. for v = 0 relativistic mass, m, and proper mass, m0 are related as
m = \frac {m_0}{\sqrt{1 + 2\Phi/c^{2} }}
Let's say if I am able to keep some photons bouncing inside a confinement, will this confinement exihibit any gravity from the mass by E/c^2 of the lights inside the confinement to the objects nearby?
Yes. But even a straight beam of light can generate a gravitational field sicne it has energy and energy has mass. John Archibald Wheeler once showed that it's possible for light was able to form an object and act as a gravitating body even with no confining walls. That object is called a geon. Its unstable though so it wouldn't last long.
Sammywu
Jan17-04, 07:06 AM
Arcon, DW,
How about this one?
Since higher temperature means a group of molecules moving in higher average speed against us. Will they be measured with higher inertial mass and gravity mass?
Sammywu
Jan17-04, 08:16 AM
DW, Arcon,
How about these two:
When an electron in a atom absorbed a photon and elevated to a higher state, will the atom show higher inertial mass and gavity mass?
If I spin a top to very high spin speed, not only now we have angular mometum also will it be measured with higher mI and mG?
Sammywu - You do understand that there are two commonly used terms in relativity regarding mass right? These term terms are relativisitic mass and proper mass. In all cases when I use the term "mass" it means "relativisitic mass. It is in this sense of the term that I'm answering your questions. In the present case it makes no difference.
Originally posted by Sammywu
Since higher temperature means a group of molecules moving in higher average speed against us. Will they be measured with higher inertial mass and gravity mass?
Yes.
When an electron in a atom absorbed a photon and elevated to a higher state, will the atom show higher inertial mass and gavity mass?
Yes.
[quote]
If I spin a top to very high spin speed, not only now we have angular mometum also will it be measured with higher mI and mG?
Yes. In fact I worked out a similar question this past week regarding a rotating cylinder. See
http://www.geocities.com/physics_world/sr/rotating_cylinder.htm
Sammywu
Jan17-04, 10:00 AM
Arcon, DW, Actually I like the term proper mass. It's better than rest mass. Put it simple, there is no such thing as rest mass. An electron in an atom has different mass from a free electron as we already show, if Arcon is right and DW agreed. The elctron has a spin of 1/2, so its true rest-rest mass shall be smaller than our published figure if you can stop its spin. This is my stupid thought. Do not bother that much.
DW, can we show that stress energy tensor or the mass energe tensor changed due to the spin of the top with your modern SR? Arcon, when I have time, I will browse your explanation of spinning of cylinders.
Thanks.
Sammywu
Jan17-04, 01:59 PM
DW, Arcon, About relativistic mass, I can't say what I agree. Let's examine another case. Let two objects A and B with the same mass m, moving toward each other in 2/3c relative to me, the observer. The threes are all in its own inertial reference frames. Agreed?
Now, two objects collide and lump together as one. What is the mass when they lump as one? What were they before lumped together, relative to me, A or B?
Can we apply this to the phenomenon of an electron and a positron lumped together as a gamma photon for a short period?
Is there a term. for apparent mass? Is it the same as relativistic mass?
By the way, I think DW does have good intention in puclishing the modern relativity formulae in another thread. I think we can show our difference in thinking and theory. Actually DW shall make his formulae even more clearer. Discussion of how they were derived will be even better.
Thanks
Originally posted by Sammywu
DW, Arcon, About relativistic mass, I can't say what I agree. Let's examine another case. Let two objects A and B with the same mass m, moving toward each other in 2/3c relative to me, the observer. The threes are all in its own inertial reference frames. Agreed?
Im sorry but I don't understand what you mean by "The threes". Please clarify. Do you mean you and the other two particles are each in an inertial frame of referance?
Now, two objects collide and lump together as one. What is the mass when they lump as one?
Let the proper mass of each particle before collision be m0. Let the proper mass of the final lump be M0. The mass of each particle before collision is
m = \gamma m_{0}
The total mass, mtotal, before the collision is the sum of the two masses
m_{total} = 2\gamma m_{0}
Since this is a closed system mass is conserved and therefore
M_{0} = m_{total} = 2\gamma m_{0}
Can we apply this to the phenomenon of an electron and a positron lumped together as a gamma photon for a short period?
No. If you did then the momentum of the system would not be conserved. Look at this from the viewpoint of the zero momentum frame of reference. When there were the electron and positron then there was no momentum. When these particles appear and only one photon appears there will now be momentum since one single photon always has momentum. But you can have them anihilate and yield two photons moving in opposite directions as obeserved in the zero momentum frame of reference.
Sammywu
Jan17-04, 05:31 PM
Arcon, Yes. When I say the three, I mean looking from the three different objectss points of view. Can we show the 4 vector momentum conservation and energy conservation from the three different observers.
About the other statement, I oversimplified that, there needs to be a subvelocity 2/3c in both particles toward the screen. so, their true velocity will be c*sqrt(8/9) with an angle toward the middle of teh seceren.
Sorry about that.
Sammywu
Jan18-04, 08:13 AM
Arcon, Any way this went far away from our original topic. Let's try to resolve my question about a free falling object in a gravity field in your equations.
Your gamma is dt/dt'. Since dt/dt' is a not constant here, this gamma is not constant either. OK, I stuck here and don't know how to continue. Want to help me. Thanks.
My object is try to derive the clock difference when the object returns to the initial falling point in your formulae.
Sammywu
Jan18-04, 10:25 AM
Arcon, This is the test case for you in case you do not know what test I was up to.
Making a device like a donut with a connecting tunnel through one of the diameters. With an open end at the joint of the tunnel and the donut, you can shoot an object in with certain speed. Put a comparable significant mass in the middle; now this is a artificial Sun experiment that you can let objects orbiting through the donut and objects falling through the tunnel. You can perform a true comparison of clocks between free falling objects and orbiting objects.
YOu can assume the middle mass is a evenly densed sphere of M0 with a radius of RM. Apparently your formula will only deal with detail gravity density.
Sammywu - Please read your PM
Arcon
Originally posted by Sammywu
You can perform a true comparison of clocks between free falling objects and orbiting objects.
Orbiting objects are free falling objects.
Originally posted by Janus
Orbiting objects are free falling objects.
Good point. Even objects moving away from the Earth, even at escape velocity, are in free-fall.
Sammywu
Jan18-04, 07:32 PM
Arcon, Janus, Well. We can agree on that. But here I am just trying to understand how Arcon's formulae shall be applied to resolve a problem. My free falling object refers to the one that was held by some thing at one end of the tunnel before the orbiting object was shot into the donut. If you like you can assume there is a clock at this initial point and compare the moving object( the oscillating object from a Newtonian point-of-view. ) to this static clock. You don't even need to bother a comparison between the obiting object, hereafter as A, and the object in question here, hereafter as B.
I have pointed out that since the relative velocity will be changing, the time dilation factor is not a constant rather a function of Ts ( for Time of static ) or Tb.
My poitn is let's test Arcon's formulae to a case and just want to see how it can be used to solve a practical problem. If a Physic theory is just a whole bunch of formulae but unable to predict or soleve some problems, what is the use of these formulae?
I tried to skip a few evaluation in Arcon's formulae and come to a point that Fex=0, so f(total)= 0+G. Now Arcon' (21) is a denotation I am not familiar with. How would you put the G here as a function of r or Tb?
Sammywu
Jan18-04, 07:34 PM
DW, By the way, who is pmb, when you said you do not agree with his Newtonian's approach?
Thanks
Sammywu
Jan20-04, 05:06 PM
Arcon,
Back to your Gravtitational Force, I Think I could decipher some of your notations now. Your v with either alpha on the upper or lower right is the velocity in vector form but in a vertical or horizon writing.
What is the reverse L? Is it a matrix, or it denotes a matrix calculation?
Is it correct?
Originally posted by Sammywu
Arcon,
Back to your Gravtitational Force, I Think I could decipher some of your notations now. Your v with either alpha on the upper or lower right is the velocity in vector form but in a vertical or horizon writing.
What is the reverse L? Is it a matrix, or it denotes a matrix calculation?
Is it correct?
I'm not sure which symbols you're refering to. Please post the equation number. Thanks
Arcon
Sammywu
Jan20-04, 07:30 PM
Arcon,
In equation (1), you describe the 4-force to be big D differential of 4-momentum to the proper time. Then it equals to the sum of two components. The first part is a formula of sum of partial differential multiplied by the dx(b)/d(tau). The second part is a product of a so-called connexion and the two different forms of 4-momentum and 4-vector. The product of 4-momentum and the 4-vector shall be a 4x4 matrix. The conection is also a 4x4 matrix. their matrix product shall be a 4x4 matrix. But this special symbol must mean you sum a row of the 4x4 matrix to make a subcomponet of the vector.
In equation (4), you said the total of the two parts is the external force.
maybe I shall buy a modern SR textbook to decipher your notation here.
Originally posted by Sammywu
In equation (1), you describe the 4-force to be big D differential of 4-momentum to the proper time.
Yes. That is the absolute derivative (aka derivative along the curve) of the covariant 4-momentum.
Then it equals to the sum of two components. The first part is a formula of sum of partial differential multiplied by the dx(b)/d(tau).
Yes. Notice that the chain rule is used, i.e.
\frac {\partial P_{\mu}}{\partial x^{\beta}} \frac {dx^{\beta}}{d\tau} = \frac {dP_{\mu}}{d\tau}}
The second part is a product of a so-called connexion and the two different forms of 4-momentum and 4-vector. The product of 4-momentum and the 4-vector shall be a 4x4 matrix. The conection is also a 4x4 matrix. their matrix product shall be a 4x4 matrix. But this special symbol must mean you sum a row of the 4x4 matrix to make a subcomponet of the vector.
If you wish to view that last part in matrix form then note that the affine connection has 3 indices and as such is not a 4x4 matrix. A 4x4 matrix has 16 components. The affine connection has 4x4x4 = 64 components. If you want to form a matrix equation then view the connetion as a 4x4 matrix which is a function of "u" (mu). Then note that you have to rearrange it to read
P_{\alpha} \Gamma^{\alpha}_{\mu \beta} U^{\beta}
This then reads in matrix form
P\Gamma(\mu)U
In equation (4), you said the total of the two parts is the external force.
Eq.(4) has on the left side dP/dt while on the right side it has Fexternal + Gravitational Force
maybe I shall buy a modern SR textbook to decipher your notation here.
An SR text won't have this material. You need a text either on GR or on differential geometry. There is a nice text by D'Inverno that I like. It's called Introducing Einstein's Relativity, Ray D'Inverno, Oxford Univ. Press, (1992)
I'm working on writing tutorials at the moment. Here is one on an intro to tensors in general
http://www.geocities.com/physics_world/ma/intro_tensor.htm
This one gives you an idea of the geometrical meaning of the Christoffel symbols (affine connection)
Arcon
Sammywu
Jan20-04, 09:14 PM
Arcon, Thanks. You did not comment on my part of connexion. was the guess correct?
Before I can get to a book store and buy a modern SR book, I can only limp along your formulae.
By the way, in my experiment of two objects hiting each other in 2/3c relative to me and lumping together, where is the inertial energy?
Sammywu
Jan20-04, 09:16 PM
Arcon, Thanks. I just saw your reply of connexion. I was trying to guess thatis a 4X4X4 matrix or different things. You definitely pointed me to correct path.
Originally posted by Sammywu
Arcon, Thanks. You did not comment on my part of connexion. was the guess correct?
Which part was that? All I saw was that you said it was in the second part and was a 4x4 matrix. The gamma part is the connection but it is not a 4x4 matrix.
By the way, in my experiment of two objects hiting each other in 2/3c relative to me and lumping together, where is the inertial energy?
Did you read the PM message I sent you?
Sammywu
Jan21-04, 08:31 AM
Arcon, I saw your complete response after I wrote that response. So don't bother with it.
What do you mean by the PM mark?
Sammywu
Jan21-04, 08:34 AM
Arcon, I am sorry. What do you mean by the PM message?
Did you send me an Email?
Sammywu
Jan21-04, 08:38 AM
Arcon, I am reading your "Introduction to Tensors". I believe that will be very helpful.
Thanks
Originally posted by Sammywu
Arcon, I am sorry. What do you mean by the PM message?
Did you send me an Email?
Take a look below this post where you normally click on "quote" to respond. To the left there are other buttons to click. One says "pm" - when you click that you can send the person a "Private Message (PM)"
To read PM's take a look at the top of this web page. There is a button that says "user cp" which stands for "User Control Panel"
I sent you a few PM's. I guess that you don't know about this function. Click on PM and you can see what I sent you.
Arcon
Sammywu
Jan21-04, 12:33 PM
Arcon, I finished your tensor introduction. A contravariant is a matrix whose values will change with the coordinates we use. A covariant is a matrix whose value will not change with the coordinates.
Your equation (31) might have messed up between j and k. Either k shall be j or k shall be j.
In equation (16), shall it be rank 1 or rank zero?
Hopefully I am right.
The 4-momentum and 4-force shall all be covariant.
How about the connexion? Is it also a covariant?
Sammywu
Jan21-04, 12:47 PM
Arcon, I am sorry. Take back part of the first statement. An invariant is a matrix or vector that won't change with any coordinates we use. A product of covariant and its own unit cubicle will not change with any coordinates we use.
Thanks
Sammywu
Jan21-04, 01:23 PM
Acon, In equation (31), is it a covariant tensor? Or, a tensor is a covariant? Your numerators and denumerators in the fraction of the partial differentials probably are in reversed places.
Thanks
Originally posted by Sammywu
A contravariant is a matrix whose values will change with the coordinates we use. A covariant is a matrix whose value will not change with the coordinates.
To be precise - a covariant vector is a geometrical object whose components transform from one coordinate system to another as shown in the notes. The components may be represented using matrix notation.
Your equation (31) might have messed up between j and k. Either k shall be j or k shall be j.
Thanks. It seems I did make an error. I'll look into it and correct it if neccesary. All those little damn numbers can look confusing huh?
In equation (16), shall it be rank 1 or rank zero?
A tensor of rank 1. There is one index so the rank is one.
The 4-momentum and 4-force shall all be covariant.
There are both covariant and contravariant forms of all vectors. A covariant vector is said to the the dual of a contravariant vector since there is a one to one relationship between the two.
How about the connexion? Is it also a covariant?
The term "covariant" as used there refers to a type of tensor. Since the affine connection is not a tensor the term does not apply.
Originally posted by Sammywu
Acon, In equation (31), is it a covariant tensor? Or, a tensor is a covariant? Your numerators and denumerators in the fraction of the partial differentials probably are in reversed places.
Thanks
That is called a mixed tensor since it is covariant with respect to some indices and contravariant with respect to other indices.
A tensor is said to be covariant if and only if all indices are covariant.
A tensor is said to be contravariant if and only if all indices are contravariant.
Sammywu
Jan21-04, 07:23 PM
Arcon, Reading your article "Invariant", most important concept is the metric tensor. It's not a constant matrix. Its value will change with the coordinates we choose. Take (ct, x, y, z) as the coordinate of an inertial frame ( or flat spacetime ), it shall be the matrix with the diagonal value as (-1,1,1,1) and the rest subcomponent is 0.
Let me denote it as Gs.
In a gravity field, with (cTf, Rf, yf, zf ) as coordinates ( make Rf for the x ) for a free fall observer, it shall be Gs as well, since the free fall observer will see straight light beam ( Something seem to need to be adjusted here, it must be related to its initial field potential if this person was held steady in the gravity field before released. ) When we replaced Rf, yf, zf with R, y,z , the coordinate observed by the static infinite observer, what would this metric matrix look like? hmm, I need to think.
Sammywu
Jan22-04, 07:40 AM
Actually, I found I have to redo some thinkings done for the two imaginary experiment I brought in to examine the SR and GR effect.
The orbiting object, the free fall object released from a point inside the gravity field and the standing person in the gravity field might all see a different degree of light bent.
I have to assume a far remote observer who is in a tre flat spacetime and a free fall object coming from this far remote place: in other words, EP=0. These two seem to have the same clock and they will see all light beams straight.
Sammywu
Jan22-04, 08:48 AM
In order to make the pont clearer, I would like to build a closed box, julst like the elevator in Einstein's elvator, but much bigger. I will put three holes, to let light thru, on one sidewall in different altitude, or height. On the opposite wall, I will mark the corresponding spots at the opposite wall. I will lock myself at the same latitude as the middle hole. In a true inertial frame, no gavity, all three light beams will hit the corresponding spots. When I put the closed box in a gravity field, either free fall or standing, all of three light beams will hit different spots to reflect different acceleration at different altitude.
To me, the observer, all the light beams will probably look straight, because light beams are the only perception for guiding a straight line. Only by the spots shift we can tell the lights actually bend because this box was built in a environemnt without gravity.
Sammywu
Jan23-04, 12:45 PM
Actually, I found it interesting to draw how the light beams will travel in this box in different environments: far away, free fall from far away, free fall released from one point in the gravity field, standing rest in the gravity field.
My gut feeling is that the standing rest observer in the gravity field has the same clock as the clock released from this point. Though I am wondering how I can prove that.
I also got this formula from the site that Marcus showed me at another thread.
This formula describes the clock standing at distance r from the earth center and a geocentric latitude A.
(delta)v/v= [ V(r,A) - I^2*r^2*cosA^2/2 - ( V(a1,0) - I^2a1^2/2)]/c^2
delta)v/v is the fractional frequency, a way to show clock difference in fraction, I believe. No time to check the detail.
V(r,a) is the gravitational potntial.
I: Earth's angular rotation rate.
a1: Earth's quatorial radius.
Sammywu
Jan23-04, 12:59 PM
Just make it clearer.
(delta)v/v= [ V(r,A) - I^2*r^2*cosA^2/2 - ( V(a1,0) - I^2*a1^2/2)]/c^2
Something was wrong here, or the document did not make it clear. This delta fraction is definitely compared with a clock at the sea level high in equator. If you replace r as a1 and A as 0, you will see this turned to 0. Even though the document claimed it's compared to earth centered clock. Maybe I did not carefully read the document.
Sammywu
Jan23-04, 04:58 PM
Arcon, Janus,
I am confused. This document apparently was written by some authoritative sources.
But its model contradicts what I thougt here.
Its formula actually says that the higher you are, ao as the higher your field potential ( Since -GM/R is negative, the higher you are, the smaller absolute field potentail you have, the higher field potential after you apply the negative sign.), the faster your clock will run.
It also has a sample to make that clear.
By my model, it will be reversed. The higher we are, we will be more close to the infitely far away clock, which is the slowest clock.
This is really true clock difference, not a measurement issue. See the two experiments I proposed.
Please help. What's wrong with my model?
Sammywu
Jan23-04, 06:35 PM
Janus, Arcon,
I checked a little further. There is a 1976 rocket experiment that proved the GR time dilation effect. This further bothered me. Doesn't this contradict to the twin paradox? We put energy to push this rocket up and let it fall back going thru certain brake accelaration. Its clock turned out faster. Isn't this a twin paradox experiment?
So, is the twin paradox just some balloney? It seemed to show that the astronauts sent out will be older rather younger.
How do we reconcile these two theory?
russ_watters
Jan24-04, 12:15 AM
Originally posted by Sammywu
I checked a little further. There is a 1976 rocket experiment that proved the GR time dilation effect. This further bothered me. Doesn't this contradict to the twin paradox? We put energy to push this rocket up and let it fall back going thru certain brake accelaration. Its clock turned out faster. Isn't this a twin paradox experiment?
So, is the twin paradox just some balloney? It seemed to show that the astronauts sent out will be older rather younger.
How do we reconcile these two theory? There are quite a number of experiments that confirm the SR and GR time dilation effects. They involve clocks on the ground compared to clocks on towers, clocks in space, clocks in planes, etc. The GPS system is my favorite example: the satellites are launched with their clocks calibrated to run slower than identical clocks on earth. When they reach orbit, they stay synchronized with their twins on earth.
The twins paradox is a mental exercise. It only seems like a paradox. If it were a real paradox, SR and GR would be invalid.
Originally posted by Sammywu
Gravity is caused by the timespace curvature. How will the inertial force been considered? Does it appear as part of the stress energy tensor?
Please note that spacetime curvature is not a cause of gravity. The term spacetime curvature is merely a geometric analogy of purposes of description and should not be taken as a literal truth and should, by no means, not be thought of as a cause. It is an unfortunate truth that this geometric interpretation has become predominant in general relativity because it makes people stop thinking. Especially since spacetime curvature is not always present when there is a gravitational field present.
Regarding this analogy Steven Weigberg comments on this very point in his GR text, which is a main staple of any GRist. From page 147
... the geometric interpretation of the theory of gravitation has dwindled to a mere analogy, which lingers in our language in terms like "metric, "affine connection," and "curvature," but is not otherwise very useful.
[...]
(The reader should be warned that these views are heterodox and would meet with objections from many general relativists).
Good ole Weinberg! Good stuff as is expected from a Nobel Laureate [:)]
Sammywu
Jan24-04, 08:08 AM
Russ, Arcon, Janus,
So do we all agreed that an astronaut going out of the Earth and made a trip back to the Earth will be older rather than younger than his twin brother? He most likely will be making trips inside gravity fields of some major masses: the Earth, the Sun, and the Galaxy.
Sammywu
Jan24-04, 08:26 AM
Russ, Arcon, Janus,
Were any known experiments doone to check a clock that was brougt to the international space station and back to the Earth by astronauts?
That shall confirm this faster clock on higher ground also. Correct?
Originally posted by Sammywu
Russ, Arcon, Janus,
Were any known experiments doone to check a clock that was brougt to the international space station and back to the Earth by astronauts?
That shall confirm this faster clock on higher ground also. Correct?
Sammywu - I've asked you a direct question and you've refused to answer it. Until you have the courtesy to respond then I will no longer respond to any further questions. I asked if you got my PM message I was met with stone silence. If you got it and don't acknowledge it and the content I'm going to assume that this is something you will always do and therefore not respond to anything else from you
Sammywu
Jan24-04, 01:26 PM
Arcon,
I am sorry. I have read your PM messages, after you told me how to. I assumed that you should know that. I don't remember there was any other questions you asked me.
Did I skip other questions you asked me? Please let me know which one.
You can send me another PM.
If you refered to the question of relativistic mass, I would say the concept of relativistic mass is useful. I tends to agree with it more and more now. But I still have questions on how gravity and EM energy will be included.
By the way, I am still reading your gravity force document, I found it is very useful to me. But I still have problems truely able to apply it or totally comprehend it.
That's why I came back to revisit my concept of GR effect.
Any way, I tried another model to resolve this FR effect.
Assumptions:
1. an object on its geodisc will run fastest clock because it is not affected by any true external forces.
2. An object free fall from far remote runs a clock in sync with the far remote clock because it is on its own geodisc. Let's denote its time as T for this paragraph.
3. When this object passes by a standing object, the standing object has a slower clock because it undergoes the support force, an external force.
This will derive a GR effect if 1/2*v^2=-FP; FP stands for field potential and equals to -GM/R.
This model will match the formula I got from this document.
Just one problem, if applying m*c^2-m0*c^2=-EP, 1/2*v^2 will not be the same as -FP.
Let me threw this problem aside, and check the model against the three observers I have mentioned:
1. For a standing clock supported by a ground support force, its time will be T*SQRT(1-2GM/(R*c^2)).
2. For a free fall clock with initial speed 0, it will depend on where this clock was released. It's on its own geodisc, but related to where it was originally released. Its clock is the same as the standing clock where it was released.
3. For an orbiting clock, its clock seems to be T*(1-2GM/(R*c^2)+v^2/c^2).
4. In general any objects in the gravity field with certain speed v, will follow the same formula used for item 3. It does nto matter where the speed direction is. This at first bothered me, but I realized in the free fall object from far remote will pass thru the midtunnel, assuming we build a midtunnel that it can fly thru, and fly away from the mass center in a decreasing speed v toward far remote. So, the direction of its speed apparently does not matter.
This model match most of experiments that Russ mentioned and the formula published in this authoritative document.
Originally posted by Sammywu
Arcon,
I am sorry. I have read your PM messages, after you told me how to. I assumed that you should know that. I don't remember there was any other questions you asked me.
Okay. I just wanted to make sure. I explained that I was very busy lately and that I don't find that I have the time to respond/read all of what you're posting. I didn't want you to get the impression that I was ignoring you or being rude. Some people never read e-mail/PM and thus I had no way of knowing if you read it or were one of those people. Thanks for clarifying. Much appreciated. I can see that you are not one of those people.
Arcon
Sammywu
Jan24-04, 02:01 PM
Arcon,
Thanks. I did read that. I really appreciate your help.
Sammywu
Jan24-04, 04:47 PM
Just try a sanity check on this GR effect as the Earth to the Sun.
-FP=(6.67*10^-11)*(2.0*10^30)/(150*10^9)= 1*10^10 m^2/sec^2.
-2FP/c^2=2*10^10/(9*10^16)=0.22*10^-6
sqrt(1-2FP/c^2) will be close to 1.
So, even if a spaceship escapes out of the gravity field of the Sun, the clock speed won't go too fast.
Try escaping from the Earth's surface relative to outside of earth's gravity.
-FP=(6.67*10^-11)*(6.0*10^24)/(6.4*10^6)=6.8*10^7
-2FP/c^2=15.6*10^7/(9*10^16)=1.7*10^(-9)
sqrt(1-2FP/c^2) is even smaller than the one in the Sun's gravity field.
Can't find Milky way's data.
Sammywu
Jan24-04, 05:10 PM
Just correct my item 3 formula.
T*SQRT(1-2GM/(R*c^2)-v^2/c^2).
There is an incorrect sign. Another possible formula is more likely
T*SQRT(1-2GM/(R*c^2))/SQRT(1-v^2/c^2).
So, for a free fall from far remote with initial speed=0, the denominator and numerator will be the same.
Sammywu
Jan24-04, 05:34 PM
Try Sun surface's clock:
-FP=(6.67*10^-11)*(2.0*10^30)/(0.7*10^9)= 19*10^10 m^2/sec^2.
-2FP/c^2=38*10^10/(9*10^16)=4*10^-6
sqrt(1-2FP/c^2) will be close to 1.
This is trying to verify David's claim that muon's apparent life expectancy inflation is due to rather GR effect than SR effect. This small difference between the Sun's surface's clock and the Earth's clock will not produce the significant apparent time inflation in the muon phenomenon.
Sammywu - I've just completed one item that I was working on. See
http://www.geocities.com/physics_world/gr/geodesic_deviation.htm
See Eq. (22). It represents the tidal acceleration of two particles in free-fall. That is to say that if there are two nearby particles in free-fall then that eqations tells you what the relative accelerations are. That relative accelertion cannot be transformed away, i.e. it exists in all frames of referance. Notice that its velocity dependant!!
Arcon
Sammywu
Jan25-04, 02:23 PM
Arcon,
Thanks. I will definitely take a look.
I just finished first part of your gravitational force. Trying too make clear.
t: proper time of a far remote observer.
tau: proper time of any objects , hereafter denoted as O, in the gravity.
bolded v: velocity of the object O from the far remote observer's view.
bolded r: Cartesian coondinate from far remote observer's view.
m: the apparent ( or relativistic ) mass of object O.
Spacetime event define all events in far remote observer's coordiante.
4-Velocity: O's 4-velocity as defined as the change from far remote observer's view but derivative of O's proper time. As we mentioned, it is a covariant.
The v(a) here is a 4-vector from of bold v adding time parameter.
Upper a or lower a seems to just denote a different form of the vector, either in horizotal or vertical.
4-Momentum and 4-force: the covariant form of the momentum and force . Just like 4-velocity.
Absolute Derivative is the subcomponent of 4-force. 4-force will be the total of the external force plus the adjustment from the connexion.
++++++++++++++++++++++++++++++++++++++++++++++++++ +++
Another book I happened to read said the connexion is the partitial differentials of g(ij), the geometric attribute or the metric unit.
It also denote the form in a way that upper subscripts are the inverse of the lower scripts.
Just FYI.
++++++++++++++++++++++++++++++++++++++++++++++++++ ++++
Sammywu
Jan25-04, 03:38 PM
Arcon,
Finshed your (1) thru (7) in the "Gravitational Force'.
Correct one item in the last paragraph:
Absolute Derivative is the subcomponent of 4-force. 4-force will be the total of the external force minus the adjustment from the connexion.
From (1) Thru (7) you show the total force from the far remote observer's view as 3-vector is the space componet of the 4-force/dilation factor plus G.
G is the space componet of the product of relativistic mass , the velocity of O measured by far remote oberver ( hereafter denoted OF ), connexion and O'svelocity again.
I need to think about your statments in explanation of (4) and (5).
Sammywu
Jan25-04, 04:42 PM
Arcon,
By the way, you do not always need to respond to me unless you see it necessary to correct me.
Let me do a sanity check for myself. Let me assume a free fall object in the gravitional feld; its velocity is not constant from OF's view and so 4-U and 4-P are not constant either. dP(mu)/d(tau) will not be zero. F(mu) shall be zero because the affine connection adjustment l cancel out the dP(mu)/d(tau) in that the object is on its own geodisc.
Also, in an inertial frame, shall the affine connection be the zero 4x4x4 matrix?
Thanks
Sammywu
Jan25-04, 06:18 PM
Arcon,
Following thru (14b), Your statements using Equivalence Principle to explain the gravity is "the force resulting entirely fromobserving the particles... frame" is crucial.
Up to (14b), you show the comparison between Loretz force against how the same concept to be applied to gravity. Even though I still had no idea how you do (11), the "more familar" form to me is unfamiliar, unfortunatelly. I will read the your appendix to see whether I got it there.
Any way, here you show gravitational charge as m, the relativistic mass, not the proper mass.
Sammywu
Jan26-04, 08:23 AM
Arcon,
The mainpart that I have no idea is where the Christoffel'symbol is from. I can only assume that's true. I can follow you thru (16) now.
Also, I realized that your purpose is to prove the gravity mass is the relativistic mass.
I tried to use what you show me here to derive GR effect. It seems that you already assume GR and SR effct.
I also noticed your two threads about mass tensors.
Any way, remember I said in my model of GR effect, there is a problem. If mc^2-m0^c2=-EP=GMm0/R, then 1/2v^2 will not equal to -phi( I denote -FP). Now, if GMm0/R^2 needs to be adjusted to GMm/R^2, or with your mass tensor, will that correct this problem?
Originally posted by Sammywu
Arcon,
The mainpart that I have no idea is where the Christoffel'symbol is from. I can only assume that's true. I can follow you thru (16) now.
There is a discussion of the meaning and derivation here -
http://www.geocities.com/physics_world/ma/chris_sym.htm
Also, I realized that your purpose is to prove the gravity mass is the relativistic mass.
No. That is not my purpose. It's just something that has always been true in general relativity. But there has been so much misinformation being passed around the internet on this topic that I've made things clear on my web site for those who really want to learn it correctly. Mind you - this was done close to 100 years ago by Einstein. It's nothing new by any means. That equation is in Einstein's text The Meaning or Relativity. Problem is that people don't know that it its there. This misinformation was started in an article by Lev Okun on the concept of mass when he failed to comment on this. But in all fairness to Okun he didn't know Einstein did that in Einstein's book. The problem was that Okun made it appear as if Einstein never used relativistic mass and that's a false claim.
Any way, remember I said in my model of GR effect, there is a problem. If mc^2-m0^c2=-EP=GMm0/R, then 1/2v^2 will not equal to -phi( I denote -FP). Now, if GMm0/R^2 needs to be adjusted to GMm/R^2, or with your mass tensor, will that correct this problem?
You wrote so much that I was unable to read it all. But I don't know what EP is. The quantity "mc^2-m0^c2" is Kinetic energy so I take it that your EP is potential energy. What has happened is that you're using invalid approximations. I'll be getting to the correct one soon.
Sammywu
Jan28-04, 08:17 AM
Arcon, Thanks. I am trying to gather some more info. from Sean M. Carroll's " Lecture Notes on General Relativity". Hopefully, that will resolve some questions about a few paradoxes I found in the GR effects.
Sammywu
Jan28-04, 09:40 AM
I just read about this "Gravitational redshift". I never paid too much attention about redshift because I though that is always purely measurement issues from whose views.
But this seems to be related to the GR time dilation effect.
I discussed it with my friend. So he asked this question. If a light beam was shot from the tower down to the groud, the ground observer shall se a blueshift instead of redshift. Is this right?
Sammywu
Jan28-04, 10:03 AM
I followed through that Gravitational Redshift actually leads to the algorithm of GR effect. Do a sanity check here.
If I continuously send a light beam to the tower for one second: wave length * number of wave packet = 1 second. The tower observer will see longer wave length * number of wave packet which means more than one second. So, the tower's clock needs to be faster than mine.
Sammywu
Jan28-04, 12:55 PM
The analogy between two elevators in the same frame and the tower and the ground in the Earth could be extended to add another elevator by the side of the higher elevator but with lower acceleration; this shall give a comparison between the two objects in the gravitational field but different acceleration.
Sammywu
Jan28-04, 01:51 PM
This analogy can explain the difference between two standing clocks at different alttudes. Note the elevator's acceleration is upward just in reverse to gravity. It's like the supporting forces for standing clocks.
This shall not straightly applied to any free fall or orbiting objects.
The only thing certain here is the initial clock of a free fall object shall be the same as the standing clock at where it was released.
Note if I released a clock to free fall, it will oscillate between two points in the imaginary experiment I proposed as digging a tunnel through a significant mass. We shall be able to compare its clock against many standing clock along the tunnel. Since it is on its own geodesic, its clock shall be faster than the standing clock which is always at its original release point. We can actually compare their clocks without ambiguity when the free fall clock oscillate back to the point. No need for redshift or anything else. True clock comparison.
Sammywu
Jan29-04, 09:41 AM
My impressions from the first 70 pages of Sean's lecture notes.
1. Event space is a manifold.
2. At each point of the event space, there is a metric tensor, which shall be viewed an unique one at this point from all objects' worldlines that pass thru this event point. Even though all objects have their own time-space coordinate, the metric tensor is the same one, so the metric tensor seen from different coordinates follows the rule of tensor transformation equation by different coordiantes.
3. All objects moving around this event space will leave a world-line like a time-like curve in the manifold.
4. The connexion is 4 4x4 matrixes, which is related to metric unit in the EQ (3.21), that is needed to make partial derivatives to become tensors.
5. In the page 26 thru 30, the 4-velocity is not the covariant 4-velocity.
Sammywu
Jan30-04, 09:25 AM
I still see a flaw in this clock formula used to adjust GPS clock. This formula basically adjust the clock based on GR effect and then on SR effect.
Since there are experiments proving the GR effect is real, I think this effect is true without doubt. But applying the SR effect right after the GR effect actually conflicts with the underlying assumption that Gravity needs to be traeted differently from a regular external force.
SR effect is a relative effect, not absolute effect. GR effect is an absolute effect because the two clocks are relatively fixed.
When I compare a clock Ta standing at the GPS's orbit, ( we can build a very high tower ), while the GPS is orbiting around the earth, its clock Tg shows a SR difference from Ta. After the GPS makes a circle, the clock can be compared. The point here is SR effect is a relative effect, Tg is slower than Ta from Ta's view, Ta is slower Tg from Tg's view. Which one will be truely slower after a circle?
Gravity does not do any work on the GPS. ( Is this correct? ) The supporting force to the standing clock does not do any work either.
The GPS is on its own geodesic. Ta is the one pushed off its geodesic.
From an analogy of this to a sample experiment created as artificial gravity, the one on the artificial gravity shall have a slower clock. This will implys the standing clock has a slower clock.
This is different from the formula applied to GPS clock setting.
Apparently I must be wrong.
Before I can further provide a calculation from this tensors and SR/GR, I thought of some analogies that show these interesting facts and underlying implications:
Let us send two rockets to height H from the earcth surface. One we will gradually curve it so as it will evently orbit the Erach at H. Once it gain its height and its escape velocity, it will orbit the Earth without further energy expenditure. Another one will be sent staright up and kept afloat by continuously pump out energy to gain the exact acceleration to counter the gravity GMm/R^2.
We will see very clear that it's actually more difficult to keep a rocket standing than orbiting at the same height.
Another one is the one we send straight out to the same hwight and let it free fall back to the Earth. We can see this one expends the least energy.
A side finding: This seems to imply that by holding us at the Earth's surface, The Earth has to expend some energy continuously.
My first try to resolve a SR/GR clock question with this modern SR theory.
The easiest question is the spaceship flying in a circle about a statis observer in a flat spacetime.
The most important issue here seems to choose a stable reference frame. It's very difficult to use the spaceship's reference frame because it actually change with time.
Using the flat spacetime as the reference frame, (t,x,y,z) for the world line of the spaceship, (t, R*cos(2*3.14*t/T), Rsin(2*3.14*t/T), 0) denotes the path using t as the parameter.
Replacing t by t', the proper time of the spaceship, d(t,x,y,z)/dt'= dt/dt'*(1, -2*3.14*R/T*sin(2*3.14*t/T), 2*3.14*R/T*cos(2*3.14*t/T), 0). dt/dt' can be assumed to be a constant as 1/sqrt(1-v^2/c^2) since the spaceship is kept in constant velocity.
When integrating this from t =0 thru T or t'= 0 thru T*sqrt(1-v^2/c^2), we will get T'=T*sqrt(1-v^2/c^2).
The whole process actually is redundent. When dt/dt' = sqrt*(1-v^2/c^2) was determined, the outcome is basically completely based on this.
From the spaceship's view, by SR, dt'/dt shall be also sqrt*(1-v^2-c^2).
The only reason this at last shows that spaceship has a slower clock is because the flatspace has a referencible coordinates.
Sammywu
Feb13-04, 08:10 AM
By people's help from here, especially Arcon ( pmb_phy ), I read a few documents on tensor algebra and modern SR/GR, I got enought to come back to solve this porblem, even though I still do not understnd all the informations contained in thes documents.
1). We need a stable coordinates to start with. The assumed coordinate is (T(lab), X, Y Z ). T(lab) is also T(world) or T(coordinate). You can see Okun's paper about Gravity Redshift for a better description of it.
2). In this coordinate, g(00) , the 00 component of metric tensor, is -(1+2Phi). I perfer to write it as -(1-2|Phi|), since Phi is negative. In this formula, we can easier see 1-2|Phi| is smaller than 1.
3). For a standing clock in the gravitational potential Phi, Its time can be denoted as T(stand). We can establish another coordinate as (T(stand), X, Y Z ).
4). T(stand) is the same as a clock at thecomoving ( in this case, velocity as zero ) local inertial frame, denoted as T(loc1), to the order of the first differentials.
You can see Okun's paper for better description of this frame. I personally think T(stand) and T(loc1) does have difference when we integrate them over the world line, but to the order of second differentials, ignored by most treatment and for now.
T(loc1) is really the clock held at the same area and just released to free fall. Just at this moment dT(loc1) is the same as dT(stand).
When the object falled to a diffrent area and with aspeed not zero any more, its ticking rate might be different.
5). Based on ds^2=g(ij)dx(i)dx(j), dT(loc1) is the ds and dT(lab) is the dx(0). dx(1), dx(2) and dx(3) are all zero because the clock never move. -(dT(Loc1)^2)=-(1-2|Phi|)*dT(lab)^2. So dT(Loc1)=sqrt(1-2|Phi|)dT(lab).
6). dT(stand) is very close to dT(loc1). So dT(stand)=sqrt(1-2|Phi|)dT(lab). The deeper you are in the gravitational field, the higher is 2|Phi|. dT(stand) is smaller and it means T(stand) is slower.
Once we established T(stand)'s relationship with T(lab), we can now use a different coordinates (T(lab), X, Y, Z). This cordinates can be unchangely describing the orbit of an orbiter. Now lets see how to calculate the T(orbit), the time of an orbiter around the center mass.
7). First, we can establish another T(loc2) for T(orbit), T(loc2) is the free fall object going on the same speed as the orbiter. So, the orbiter is in the local inertal frame. T(loc2)=T(orbit), no doubt.
8). From T(loc1)'s view, T(loc2)=T(loc1)*gamma. Here beta is v/c and gamma is sqrt(1-beta^2).
Why don't we look from T(loc2)'s view? It's the most important question. The coordinates we can establish with clear relationship between time and space is the coordinate (T(stand), X, Y Z ). The rotating of the orbiter makes it diffecult to relate X,Y,Z to its T(orbit).
The selection of the coordinate again determines how the resulit will be. Anyway, I will think more on this.
9) T(orbit)=T(loc2)=gamma*T(loc1)=gamma*T(stand)=gamm a*sqrt(1-2|Phi|)*T(lab). So, the orbiter is slower than a standing clock.
Note all I did above, I simplified that taking T as c*T. The coordinates established assuming c=1.
My question is:
How much confidence will we have about the existence of T(lab) and the relationship established between T(lab), X, Y and Z? T(lab) is the same as the far remote obserevr free of gravity I have imagined in the first place.
Sammywu
Feb14-04, 09:06 PM
Using EQ (3.48) in the "Lecture Notes on General Relativity" written by Sean M. Carroll, with the same coordinate labeled (T(lab), X, Y, Z ) as before, g(00)=-(1+2Phi), g(11), g(22) and g(33) are close to 1. T(orbit)=integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-R^2), where T is the lab. time for the orbiter to make a circle. R is the raius of the orbit. The item R^2 comes from dX/dA^2+dY/dA^2, where dA is the angular velocity of the orbiter. X=R*cosA and Y=R*sinA, so dX/dA=R*(-sinA) and dY/dA=R*cosA.
Any way, 2Pie*R=T*beta, where beta=v/c as SR people know. R=T*beta/2Pie.
So, T(orbit)= integration from 0 to 2Pie of SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2).
SQRT((1-2|Phi|)*(T/2Pie)^2-(T/2Pie)^2*beta^2)= (T/2Pie)*SQRT(1-2|Phi|-bate^2).
T(orbit)=T*SQRT(1-2|Phi|-beta^2). Note T=T(lab), so
T(orbit)=T(lab)*SQRT(1-2|Phi|-beta^2).
We already know T(stand)=T(lab)*SQRT(1-2|Phi|).
This shows a match with what we know as GPS time formula.
Also this shall match the H&K experiment.
Note the formula can be applied to an airplane flying around the Earth with smaller amount of v than the so-called escape velocity that drives an orbiter, where the airplane will need to rely on air pressure beneath the wing to support it from not falling down.
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