I did no such thing. I pointed out one of the (at least) two ways that your thought experiment was flawed. The idea that a flawed thought experiment disproves the equivalence principle is simply not true. Please stop misrepresenting my point.
Gravitational acceleration is just acceleration, too. The point is that both observers experience inequivalent accelerations, and thus, the equivalence principle implies that their situation is inequivalent. For that matter, one could easily remove AndyL’s point about your thought experiment: have the chest caught by a contraption, say a spring or a rubber band, that decelerates it with the required 1g (or rather, in such a way that the chest’s occupant experiences 1g), then releases it to the ground when its relative motion to it is zero. This requires some fiddling, but I think it ought to be possible. Any way, it’s not the important point; that still is that the moving observer needs to change reference frames in a way the stationary one doesn’t when he wants to return to Earth, thus, both situations are physically inequivalent.
And by the way, if you regard the gravitational field as ether, then what you’re calling ether, physicists call spacetime, and you’re just using different names to talk about the same concepts, and ultimately will end up with the same theories (providing you did everything correctly).
So far Tomh has obviously failed to set-up a situation where the two observers experince ‘equivalent’ accelerations, another objection to his scenario is that the two observers start and finish in spatially seperated postions so any comparison between the two depends on the method used to synchronise clocks.
However, both these objections are fixable by altering the scenario in several ways. First of all I’ll explain these fixes (which unfortunately ends up with amore complex sceanrio than Tomh envisaged) and then in the next post I’ll explain why the equivalence principle still stays intact.
Two observers A and A’ (both of mass m) start at rest on the surface of the Earth. A will be the stay at home observer (though he will have to move to duplicate the acceleration patterns experinced by A’, as we will see) and A’ will move out to a far off point and return back to the Earth. I will divide each observer’s proper times T and T’ in to 6 intervals to (hopefully) make it easier to follow what is going on.
One other thing to note is that I won’t treat the Earth as a physical barrier, so do not worry that both the observers will travel through the Earth several times. Also when I talk about forces, I will be purely talking about non-gravitational forces.
T[sub]1[/sub]: A is subjected to a force of 1mg acting against the gravity of the Earth’s surface which keeps him in a stationary position on the Earth’s surface.
**
T’[sub]1[/sub]**: A’ is also subjected to a force of 1mg, but this acts downwards towards the Earth’s centre causing him to travel (with initial acceleration of 2g) through the centre of the Earth and continue through the other side. We will assume that T’[sub]1[/sub] is long enough for A’ to get a significant distance from the Earth and travel at relativstic speeds relative to the Earth that is a signifcant fraction of c.
T[sub]1[/sub] = T’[sub]1[/sub] (this is just a statement)
T[sub]2[/sub]: A is subject to no force and so goes in to free fall through the centre of the Earth until he reaches the other side of the Earth where he stops.
T’[sub]2[/sub]: A’ is also subject to no force and drifts in space far from Earth.
T[sub]2[/sub] = T’[sub]2[/sub]
T[sub]3[/sub]: A is subject to a force of 1mg that keeps him stationary in his new (antipodal) postion on the Earth’s surface.
T’[sub]3[/sub]: A’ is subject to a force of 1mg in the opposite direction to his direction of motion until he is at rest relative to the faraway Earth.
T[sub]3[/sub] = T’[sub]3[/sub]
T[sub]4[/sub]: A continues to be subjected to same force as in T[sub]3[/sub]
T’[sub]4[/sub]: A’ continues to be subject to the same force as in T’[sub]3[/sub] until he is in the same postion relative to the Earth as he was at the start of interval T’[sub]3[/sub]
T[sub]4[/sub] = T’[sub]4[/sub]
T[sub]5[/sub]: A goes in to free fall again and travels thorugh the centre of the Earth and arrives back at his intial postion.
T’[sub]5[/sub]: A’ is subject to no force and drifts towards the Earth, by symmetry, arriving at the same postion relative to the Earth he was at the start of interval T’[sub]2[/sub]
T[sub]5[/sub] = T’[sub]5[/sub] (note: that these two proper times are equal by symmetry where as the other proper times have been equal simply because we have stated they are equal)
T[sub]6[/sub] A is now back in his intial psotion and experinces the same force holding him there as in interval T[sub]1[/sub] he maintains this postion and eagerly awaits the return of A’ where they compare clocks.
T’[sub]6[/sub]: A’ is subject to a force of 1mg as in T’[sub]1[/sub] which will cause him ] and arrive back at his starting postion where he compares clocks with A.
The question is: does T[sub]6[/sub] = T’[sub]6[/sub]???
The answer is: no, they are not equal T[sub]6[/sub] will be significantly longer than T’[sub]6[/sub] (this follows from arguments made previously in the thread). Other than the compartive lengths of T[sub]6[/sub] and T’[sub]6[/sub] (which is what we’ve used to measure the time dialtion between them) there is no experiment either one could’ve performed locally to distinguish themselves from each other, yet there is still an objective time dilation between them!
N.B. it may see that the time dialtion effect occurs entirely on the sixth legs of the observers’ journeys, but it is just the way I’ve set it up so that whether the sixth legs of the their journeys are equal or not becomes the measure of whether there is a time dialtion between them or not. If we were to arrive at a figure for the difference between T[sub]6[/sub] and T’[sub]6[/sub] we would have to take the whole journeys of both observers in to account.
The basic flaw in the analysis that the above somehow disporves the equivalance principle is that it involves the comparison of the experinces of two different observers, who may start and finish locally, but are not local to each other for the length of their journeys. The equivalnce principle applies locally only, once you take the ‘local’ part out you’re no longer analysing the equivalnce principle.
To illustrate this, you can describe an entire spacetime using the differences between various observers (and this is frequently just the way it is done more or less). If you use 3 free-falling observers to perform a parallel transport, that tells you whether or not the spacetime is curved. Simlairly in this situation, just comapring their results A and A’ can conclude that they live in a curved spacetime, though they won’t be able to conclude much else.
The assymmetry between A and A’ arrives from how the curvature of spacetime assymetrically ‘influences’ their worldlines.
Next question then. Does it predict E[sub]k[/sub] = mc^2 (1 - γ) like SR does? Or explain it in any way?
Naita asked
“How does your ideas deal with the results of Michelson-Morley?”
I forgot to credit the original author, who was Tom Van Flandern. It was he who said :-
All the facts that seem to require special or general relativity can be more simply explained by assuming an ether that corresponds to the local gravitational field. Michelson found no “ether wind,” or fringe shift, because of course the Earth’s gravitational field moves forward with the Earth. As for the bending of starlight near the Sun, the confirmation of general relativity that made Einstein world-famous, it is easily explained given a non-uniform light medium.
Einstein did not mention “local”. To make sure, I just re-read chapter XX. The last sentence above is therefore wrong. If you think about it, the chest is accelerating, so cannot stay “local”.
“So far Tomh has obviously failed to set-up a situation where the two observers experince ‘equivalent’ accelerations, another objection to his scenario is that the two observers start and finish in spatially seperated postions so any comparison between the two depends on the method used to synchronise clocks.”
As mentioned above, this thought experiment was originally set up by Einstein, who said that the two observers were experiencing equivalent accelerations, and “proved” the EQ.
TAMOP then goes on to change the experiment, asks a question about that one, and then answers his own question.
Andy L pointed out a way that the man in the chest on Earth can tell that he is in a gravitational field and not an accelerated field…
“For the box on Earth, the accelerometer reads 1g throughout, except for the brief period when the box is somehow made weightless (and there are problems with that idea - after a minute of free fall, the box would be moving very fast relative to the Earth, so how do you stop it without extreme accelerations). The gyroscopes keep the accelerometer pointing in the same direction even as the whole box is rotated 360 degrees around the accelerometer.
For the box in space, the first half of the trip looks just the same as what happens on Earth. But when the box is rotated 180 degrees, the gyroscopes detect that, and register the fact that the acceleration is now negative 1 gee for the rest of the trip - a pretty significant asymmetry.”
That is a direct unedited quote from an earlier posting. A later one is this :- “The idea that a flawed thought experiment disproves the equivalence principle is simply not true. Please stop misrepresenting my point.”
Your thought experiment is an amended version of mine, which in turn is an amended version of Einstein’s. Where is the flaw? The ingredients are there as Einstein set them out - a chest in 1G caused by gravity, and a chest in 1G caused by acceleration.
You pointed out a way to distinguish the two from the assymmetry. Now which is it, can you tell them apart, or can’t you?
I checked out your link, I have been there before. I think this is the part you are pointing me to :-
“It’s useful to think of the problem in terms of relativistic mass, since this is what each rocket motor “feels” as it strives to maintain a 1g acceleration or deceleration. The relativistic mass of each traveller’s rocket is continually decreasing throughout their trip (since it’s being converted to exhaust energy). It turns out that at the half way point, Laurel’s total relativistic mass (for fuel plus payload) is about 28m, and from here until the trip’s end, this relativistic mass only decreases by a tiny amount, so that Laurel’s rocket needs to do very little work. So at the halfway point his fuel:payload ratio turns out to be about 1.”
Why use the term “relativistic mass”? There is no velocity or motion of any kind “except perhaps vibration”, between the rocket and its motor. The rocket motor “feels” the mass of the rocket sure enough, but as the motor and the rocket are in the same FR, v = 0, and this is its rest mass. This mass is decreasing as the fuel is burned off, so the thrust of the motor is gradually reduced throughout the journey to maintain the 1G. As for “the mass only decreases by a tiny amount”, this depends on the definition of “tiny”. The amount of fuel used per unit of time will decrease as the total mass is decreasing and the motor thrust is being reduced, but especially on the outward leg will not be tiny.
There is a small green face in my preview, I did not put it there, and I cannot get rid of it. Not my fault.
You claimed that you were setting up a situation in which the observers couldn’t be distinguished - but you were wrong. You added a detectable reversal of acceleration, and you added a period of weightlessness that was inconsistent with the previous claim one chest was stationary on the Earth. The original thought experiment by Einstein was valid, until you added those extraneous asymmetric factors.
It’s relativistic mass in that paragraph, because it’s dealing with the situation as observed from a frame of reference that is not moving with the rocket. You can translate everything into the frame of reference of the rocket if you wish. Then the rocket doesn’t change mass, the distance the rocket travels is contracted, etc.
I should add that Einstein’s thought experiment did not prove the equivalence principle - various experiments had done so (and continue to do so). Einstein formulated his thought experiment to clarify the consequences of the equivalence principle.
Can you promise not to mention what Einstein did or didn’t say? It’s entirely irrelevant, a matter of historical interest only (FTR thoguh by enclosing his observers in boxes Einstein clearly meant to restrict them to local observations., even if that term had not yet entered common physics parlance).
“Local” in this sense just means what is occurring in the immediate vicinity of the observer, all observers whether accelerating or non-accelerating have a local frame of reference.
The thought experiment of Einstein is about two generic observers, who did not compare each others observations directly. They are assumed to be completely seperated in space and do not compare clocks.
I changed the experiment, because you had failed to acheive what you set out to acheive.
We only want to compare the localobservations of each observer in a fairly generic way, by either setting up a method to synchronise clocks nor by demanding they start and finish in the same spot we go outside of the relam of the local.
If you are having problems understanding the equivalance principle, try this re-statement of it: In a vanishingly small volume, the gravitational forces and inertial (pseudo-)forces a body experinces are both directly proportional to it’s mass.
That’s just Newtonian dynamics and Newtonian gravitation, the point being that the equivalence principle isn’t really unique to general relativity, Einstein took his cue from Newtonain gravity and d’Alembert’s principle.
Try this Tomh:
Let’s say I resort to extremely underhand tatics to demonstarte to you the equivalance principle and drug and kidnap you.
You wake up in small box-like room made of an impenetrable material completely sealed off from the outside world.
The first thing you see a note that I’ve left you saying that is on a rocket ship in deep space. But you notice that the gravity seems to be operating in the room in the same way it does on Earth. I.e. your feet ‘stick’ to the floor as normal, if you drop anything it falls to the floor with an acceleration of 1g, etc, so your a little dubious of my note.
I’ve been quite generous and I’ve fitted the room out with equipment that allows you to make any conceivable measurement you like within the room, though given the impenetrable nature of the room you can’t measure or communicate with anything outside of the room. Additionally the room is too small for any variation in the Earth’s gravitational field within it to be detectable.
How can you tell if the box is standing on the Earth’s surface or if you are accelerating in deep space at 1g?
For one thing, he stands fairly alone claiming it’s a more simple explanation. And I highly doubt it can be used to explain the timing observations of the moons of the solar system without some heavy “epicycles”. Our observations of the moons of Jupiter and Saturn (and others) are consistent with the same constant light speed throughout the solar system.
And again, how does ether theories account for the relativistic formula for kinetic energy?
[quote=“tomh4040, post:187, topic:388750”]
Einstein did not mention “local”. To make sure, I just re-read chapter XX. The last sentence above is therefore wrong.
[QUOTE]
Whether or not Einstein stated it clearly in your book is of no importance; it’s clear that the equivalence principle is valid only locally. Just look at the first sentence in the wiki entry, which talks about
(bolding mine)
‘Staying in one place’ is not what locality means in this context, and it’s an ill-defined concept anyway. Rather, ‘locally’ means something like ‘in a sufficiently small region’ – if you turn towards bigger regions, tidal effects always make it possible to distinguish between a gravitational field and an accelerated frame of reference.
Yeah. General relativity is confirmed by the measurements of Mercury’s orbit, the decay of neutron star orbits measured over decades (Hulse–Taylor pulsar - Wikipedia), the time dilation experiments I mentioned in an earlier post, the results of the Gravity Probe B (Gravity Probe B - Wikipedia), and various other experiments, so any theory that supplants GR is going to have to a) match its predictions for all of the above and b) make some further predictions that differ from GR and can be tested. It’s a high hurdle. For SR, it’s even worse for competing theories, since SR is built into quantum electrodynamics, which is probably the most confirmed theory in existence.
I have been accused of ignoring posts, so I looked back to see which I missed. Yours is one which I will comment on, but if I don’t reply it is usually because the answer is so obvious. What is perhaps not so obvious in yours is that gamma contains v : gamma = 1/sqrt(1 - (v/c)^2) .
Using the formula above, for a rocket with an integral motor, v = 0 . Therefore gamma = 1.
Ek = mc^2 (1 -1) = mc^2 X 0
Ek = 0
By having the accelerating force removed for eg 60 seconds and the chest rotated as previously shown. This can be done for the chest on the rocket very easily. It can be done on Earth only with a great deal of difficulty, which is why this is a thought experiment. As Andy L pointed out, after doing this manouver, the G forces would be extreme for the man on Earth, and negligible for the man in the chest. The man in the chest will dust off his knees and stand up, the man on Earth would be strawberry jam.
In case you have not read it, this is from “A Quiestion of Time” by Hans Zweig :
Beginning in Chapter 20 of his 1917 Book on Relativity Theory Einstein uses a common sense thought experiment to demonstrate the equivalence of the force of gravity and other, more common known forces, which produce acceleration. His analysis involves a man standing in a box, in the absence of gravitational force, but with a hook and rope attached to the top of the box and an imaginary engine accelerating the box upwards. The man’s kinesthetic experience is the same as though the box were standing on earth and gravity were anchoring his feet to the ground. This compelling image is the source of his belief that gravity has the same effect as acceleration produced by any other means.
The facts in the experiment are correct - the conclusion is wrong. To bring this point home an alternative thought experiment is offered which stresses the distinction between the force of gravity and other forces with which we are familiar, such as the force of an engine pulling and accelerating a train, or the powder in a gun which accelerates a bullet to produce its muzzle velocity.
To distinguish the force of gravity from such other forces consider an idealized experiment in which a train is moving along an embankment on a planet on which the force of gravity is negligible. In one case we let an engine accelerate the train. In a second case we imagine a large body ahead of the train which attracts the train due to its gravitational pull. We can also imagine this second case as a train falling, or racing, to earth.
If the train were in uniform motion then it would be valid to compare a walk forward on the train with a laser firing a pulse of light, or a gun shooting a bullet from the rear of the train in the direction of the train’s motion. The velocity of the walker, the bullet or the photon remains constant relative to the velocity of the train. But if the train is accelerating because of the engine pulling it this is no longer true. In that case the walker, at each step, is in touch with the instantaneous velocity of the train, so that his walk can remain essentially constant with respect to the instantaneous velocity of the accelerating train. But the bullet or the laser beam do not remain in contact with the train so their velocity will decrease relative to that of the accelerating train as time passes.
On the other hand, if the train were falling towards earth, or pulled forward by a large gravitational mass, the acceleration would be due to gravity and the bullet fired from the gun (and possibly the laser light) would also be subject to the continuing force of gravity so the velocity relative to that of the train would be constant as is the case for the walker. This differentiates the case of gravitational acceleration from the force producing acceleration that acts only on the train.
What is clear is that we have a distinction between a specific force operating on a specific body and a general force, the force of gravity, operating in an undifferentiated way on all bodies, objects, possibly even photons. Since the effects are different in the two cases we cannot claim that the force of gravity is the same as any other force that produces acceleration. The validity of General Relativity, like that of Special Relativity, can therefore be questioned.
That’s not an answer to my question. That the kinetic energy is zero for something with zero velocity is blindingly obvious.
The aforementioned formula for kinetic energy is derived from special relativity and experimentally verified. It can be shown mathematically to approximate 1/2 mv^2 at low velocities. Unless you can show that your alternative to special relativity is compatible with or improves this formula there is really no reason to discuss it.
However I do not expect you to come up with a rational answer since you still haven’t grasped what’s wrong with your understanding of the equivalence principle. Of course he’ll know he’s in an accelerated box if he can turn it around or change the acceleration, but that’s not what the EP is about. It’s entirely and exclusively about a situation where you’re in a small space, with no knowledge of or control over anything outside said space, or the acceleration or orientation of that space. Assuming there’d be no way to tell in that situation whether it’s 1g of acceleration or 1g of gravity that keeps you on the floor, leads to certain conclusions about the behaviour of light in gravity, all of which have been confirmed by experiment.
Your inability to grasp this, or to see how an ether linked to gravity fields is incompatible with basic astronomical observations in the solar system along with hundreds of other experiments and observations, or ability to show that contrary to known science all those experiments and observations are actually compatible with that idea, makes me believe it’s not worth it to continue participating in this thread.
Once again, you answer that someone can tell the difference only if you do different things to each chest - for one chest, you remove acceleration and then add it again, for the other chest you remove acceleration, and then apply a huge amount of acceleration. It’s not surprising that if you do different things to each chest, the person on the inside can tell the difference.
However, I should note that it is possible to correct this flaw in your thought experiment (with an idea introduced upthread by someone else). If you drill a hole in the earth connecting two points on its surface that are at the same latitude (or at opposite latitudes), evacuate the atmosphere out of the hole, and arrange for the sides of the hole to be frictionless, an object dropped into the hole will be in free fall, and will return to the original location at zero velocity in 85 minutes or so. You could make the free-fall portion of the space chest 85 minutes, also, so that both chests have identical free-fall experiences.
However, this does not correct the fatal flaw in your thought experiment, which is that the chest in space is accelerating in the opposite direction after the free-fall period than relative to the acceleration before the free-fall period - and this reversal in acceleration does not occur before and after the free-fall period on the chest on Earth.
Since your attempt to disprove time dilation is only valid if both chests have identical experiences, once again (even after our best efforts to help you correct your errors) it fails
[quote=“naita, post:197, topic:388750”]
That’s not an answer to my question. That the kinetic energy is zero for something with zero velocity is blindingly obvious.QUOTE]
It is also blindingly obvious that a rocket with its own internal motor will not undergo a mass increase. There is therefore no reason for it not to accelerate past light speed.
[Quote Originally Posted by naita
For one thing, he stands fairly alone claiming it’s a more simple explanation. And I highly doubt it can be used to explain the timing observations of the moons of the solar system without some heavy “epicycles”. Our observations of the moons of Jupiter and Saturn (and others) are consistent with the same constant light speed throughout the solar system…[Quote]
You are not correct in saying I stand alone, or if you are referring to TVF, that he stood alone. Our group is growing, yours is shrinking. Of course you are correct in saying “Our observations of the moons of Jupiter and Saturn (and others) are consistent with the same constant light speed throughout the solar system…” That speed is constant relative to the Sun, slightly modified by its passage through our atmosphere.
[Quote, Andy L]
“Once again, you answer that someone can tell the difference only if you do different things to each chest - for one chest, you remove acceleration and then add it again, for the other chest you remove acceleration, and then apply a huge amount of acceleration.”
The huge amount of acceleration to the Earth chest was not applied by me, it was a consequence of the chest being in a gravitational field as opposed to an accelerated field.