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Voting closed: June 02, 2006, 04:16:06 AM

A problem with your "gravity"

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A problem with your "gravity"
« Reply #60 on: June 09, 2006, 01:10:07 PM »
Quote from: "Erasmus"

In any case, I think this latest argument if nothing else ought to convince you that it is relative acceleration and relative velocity that are interesting to us, and not the ill-defined concept of "actual" acceleration.

-Erasmus


But you damn your own theory, for if it only relative acceleration that are interesting to us, than what is the earth accelerating relative to?

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TheEngineer

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A problem with your "gravity"
« Reply #61 on: June 09, 2006, 03:21:49 PM »
Quote from: "Erasmus"


This is essentially the place where I have a problem with your argument.  I am fundamentally opposed to the notion that a treatment of light as a wave involves deformations in some medium.  That is, the "crests" and "troughs" of an electromagnetic wave are not physically identifiable objects.  It certainly seems that your argument requires that there be physically identifiable crests in electromagnetic waves, and that furthermore they arrive one after another at a point whereby some light is passing.  What exactly is it, basically, that is waving in an electromagnetic wave?

I think I can help:

Electromagetism 101 will now begin.

Erasmus, remember when I said that a moving charge creates a magnetic field?  Well it goes both ways.  This is in fact the basis for the electricity that powers our civilization.  As you change the magnetic field in a coil of wire, you get an electric field that moves electrons and produces electricity.  As the mag. field intensifies, it pushes the electrons one way, and when the field gets smaller, it pulls them the other.  This produces the AC current we have delivered to our houses.  It actually doesn't matter which variable you vary, the area of the loop or the strength of the field, the effect is the same.

That said, let's look at light.  A photon is just a packet of energy.  This where the particle/wave dualtity of light comes in.  This packet is just a combination of an electric and a magnetic field.  As the electric field increases, a complimentary magnetic field is produced.  As this mag. field increases, it creates a complimentary electric field.  So as one changes, it affects the other.  By the way, these fields are always perpendicular to each other.  It is called a wave, since when these a ploted together, the graph is a combination of sine waves.

In effect, light is a self propagating wave.  It doesn't need a medium to travel through, it is purely energy and is caused by the changes of the field strengths of it's two components.  

As a side note, the speed of light was first discovered on a chaulkboard, long before it was measured.  Maxwell noted the complimentary nature of the two waves that must make up the light wave and surmised that any wave having these characteristics must have a velocity about equal to 2.99*10^8 m/s.  This was not a variable in his equation, it was fact.  That is why the speed of light is a constant.

I've probably left something out, but hopefully you get the idea.


"I haven't been wrong since 1961, when I thought I made a mistake."
        -- Bob Hudson

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Erasmus

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A problem with your "gravity"
« Reply #62 on: June 09, 2006, 04:49:34 PM »
Quote from: "TheEngineer"
It doesn't need a medium to travel through, it is purely energy and is caused by the changes of the field strengths of it's two components.


Okay, I realize that light does not need a medium.  That's... mostly my point.  Believe me, I have no doubt that wave exhibits both wavelike properties and particle-like properties.

Your description convinces me even more that with light, nothing is "waving", as in, there's no material that moves from side to side (perpendicular to the direction of propagation) or front to back (in the direction of propagation).  I.e. you can't take some light and physically identify a crest or trough; you can see crests and troughs in film exposed by a double slit, but they don't appear to be moving anywhere, let alone in the direction that the wave is moving.

If you can't identify some part of light as a crest or a trough that moves in the direction of wave propagation, then how can it be meaningful to say, "Because the receiver is accelerating, he more wave crests/troughs arrive at his position in a given time interval."?

Also, you still haven't addressed the issue that the transmitter is accelerating away from us, and we are accelerating towards it, and if the effect Doubter describes is real, then they ought to cancel one another out.  i.e., it's relative acceleration that's important, not "absolute" acceleration.

-Erasmus
Why did the chicken cross the Möbius strip?

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Erasmus

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A problem with your "gravity"
« Reply #63 on: June 09, 2006, 04:51:29 PM »
Quote from: "Doubter"
if it only relative acceleration that are interesting to us, than what is the earth accelerating relative to?


Just because acceleration is relative doesn't mean we can't tell that we're accelerating.  All it means is that if somebody else, in a different reference frame, measures our acceleration, he will not get the same result we get when we measure it.

When I said that it's only relative acceleration that's interesting, I was referring only to the frequency shifting phenomenon that you're alleging exists.
Why did the chicken cross the Möbius strip?

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TheEngineer

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A problem with your "gravity"
« Reply #64 on: June 09, 2006, 05:11:44 PM »
Quote from: "Erasmus"


Your description convinces me even more that with light, nothing is "waving", as in, there's no material that moves from side to side (perpendicular to the direction of propagation) or front to back (in the direction of propagation).  I.e. you can't take some light and physically identify a crest or trough; you can see crests and troughs in film exposed by a double slit, but they don't appear to be moving anywhere, let alone in the direction that the wave is moving.


What is actually waving is the electric and mag. fields.  While there is no actual material moving, it is energy that is moving.  Crests and troughs can be identified by their amplitude.  For a pictoral representation, go here: http://en.wikipedia.org/wiki/Electromagnetic_waves

Quote
If you can't identify some part of light as a crest or a trough that moves in the direction of wave propagation, then how can it be meaningful to say, "Because the receiver is accelerating, he more wave crests/troughs arrive at his position in a given time interval."?


Every wave has a maximum amplitude (with very certain exceptions).  The frequency of the wave can be seen by how often the max amplitude is recieved.  The distance between these crests is the wavelength.
Since the speed of light is a constant, it cannot change speed as it travels.  Now if I move towards the source, the speed of light has not changed, I am decreasing the distance between the crests, thereby decreasing the wavelength of the light, blueshifting it.


"I haven't been wrong since 1961, when I thought I made a mistake."
        -- Bob Hudson

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Erasmus

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A problem with your "gravity"
« Reply #65 on: June 09, 2006, 05:27:31 PM »
Quote from: "TheEngineer"
What is actually waving is the electric and mag. fields.  While there is no actual material moving, it is energy that is moving.  Crests and troughs can be identified by their amplitude.  For a pictoral representation, go here: http://en.wikipedia.org/wiki/Electromagnetic_waves


I have seen these representations many times, but I worry that they may be misleading for this particular discussion.

Quote
Since the speed of light is a constant, it cannot change speed as it travels.  Now if I move towards the source, the speed of light has not changed, I am decreasing the distance between the crests, thereby decreasing the wavelength of the light, blueshifting it.


I am aware of relativistic blueshifting.  Maybe you missed the beginning of the discussion, but Doubter is claiming that blue shifting can occur even when the objects in question are immobile with respect to one another.  That is, the receiver is not moving any closer to or farther from the source.

Note that GR predicts a red shift when light travels from a region of high gravitational potential to a region of low potential.
Why did the chicken cross the Möbius strip?

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TheEngineer

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A problem with your "gravity"
« Reply #66 on: June 09, 2006, 05:56:26 PM »
Quote from: "Erasmus"


I am aware of relativistic blueshifting.  Maybe you missed the beginning of the discussion, but Doubter is claiming that blue shifting can occur even when the objects in question are immobile with respect to one another.  That is, the receiver is not moving any closer to or farther from the source.


I guess I did miss it.  When there is no relative movement between the source and reciever, there should be no shifting.


"I haven't been wrong since 1961, when I thought I made a mistake."
        -- Bob Hudson

A problem with your "gravity"
« Reply #67 on: June 12, 2006, 08:47:37 AM »
Quote from: "TheEngineer"
Quote from: "Erasmus"


I am aware of relativistic blueshifting.  Maybe you missed the beginning of the discussion, but Doubter is claiming that blue shifting can occur even when the objects in question are immobile with respect to one another.  That is, the receiver is not moving any closer to or farther from the source.


I guess I did miss it.  When there is no relative movement between the source and reciever, there should be no shifting.


The light is emmited at a specific point and time.  The source is accelerating away, but the light itself is radiating from the point where it was emmited.

The photon once emmitted is no longer subject to the force accelerating the earth and stars, any more than the ball tossed into the air(actually less, the ball is subject to air resistance).

A problem with your "gravity"
« Reply #68 on: June 12, 2006, 12:54:12 PM »
Quote from: "Erasmus"
Quote from: "Doubter"
if it only relative acceleration that are interesting to us, than what is the earth accelerating relative to?


Just because acceleration is relative doesn't mean we can't tell that we're accelerating.  All it means is that if somebody else, in a different reference frame, measures our acceleration, he will not get the same result we get when we measure it.

When I said that it's only relative acceleration that's interesting, I was referring only to the frequency shifting phenomenon that you're alleging exists.


O.K. let me see if I can explain another way:

Part of the confusion is because physicists use MASS in two different ways, and Einstein was particularly guilty of switching his usage depending on his audience.

There is M0  and there is MR .

MR is Relativistic Mass, or Kinetic Mass
M0 is Resting Mass.

When Einstein writes an equation such E=MC2 he is referring to Resting Mass, not Kinetic Mass.  
The Rest Mass of a Photon is 0, the Kinetic Mass of a Photon is 1.  

Photons must always have a speed of light, or else MR would be other than 1.

How does this apply?

Two ways, first Kinetic Mass can never exceed M0C , second even if you do not accept the idea of a M0 that is non-relativistic, you can not increase MR by more than MRC.

Now you suggest that you can continue to accelerate indefinitely because you are outside of any other frame of reference to which you can be compared to the speed of light.  Acceleration in a change in Kinetic energy, and the pseudo gravity experienced is due to the inertia resisting the acceleration.  As you must continue to increase the Kinetic Energy to continue to accelerate you will, as has been noted several places elsewhere, in about 1 year exceed the maximum Kinetic Energy increase.

Or to word it in another way:

Acceleration is relative, but only in reference to your velocity in the past. Therefore the “Light Speed” limit can be applied to the difference between your past velocity and your current.

The Speed of Light…It’s not just a good idea, IT’S THE LAW!

A problem with your "gravity"
« Reply #69 on: June 12, 2006, 12:57:45 PM »
Quote from: "Erasmus"


Note that GR predicts a red shift when light travels from a region of high gravitational potential to a region of low potential.


Yes, which helps show that it is Gravity, not Acceleration that we are affected by.

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Erasmus

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A problem with your "gravity"
« Reply #70 on: June 12, 2006, 01:17:08 PM »
So, it doesn't sound like you're addressing the current topic in this reply, but okay.  

Anyway, I know about relativistic mass.  However, it seems to me that your understanding of the concept is incomplete in that you, like almost everybody who knows a bit about SR, treat the word "relativistic" as meaning "a la Special Relativity" as opposed to "relating to some relative measurements".

Observe:

Quote from: "Doubter"
As you must continue to increase the Kinetic Energy to continue to accelerate you will, as has been noted several places elsewhere, in about 1 year exceed the maximum Kinetic Energy increase.


As I continue to accelerate, as you say, I am always at rest in my own reference frame.  Since all measurements I take are made in my own reference frame, I will always measure myself to be at rest.  Thus I will always measure my mass to be my resting mass (M0 by your notation), and my kinetic energy to be my resting kinetic energy (zero).

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Acceleration is relative, but only in reference to your velocity in the past.


That's incoherent.  Your "velocity in the past" isn't defined, nor is your present velocity, except insofar as they are defined in your reference frame.  In your reference frame, velocity is always zero (in the past as well as the present), so acceleration is always zero, according to your definition.

Quote
The Speed of Light…It’s not just a good idea, IT’S THE LAW!


Funny.  The law is more cryptic than you think.  Can you prove (mathematically, please) that no objects can travel faster than light?
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #71 on: June 12, 2006, 03:48:35 PM »
Quote from: "Erasmus"


That's incoherent.  Your "velocity in the past" isn't defined, nor is your present velocity, except insofar as they are defined in your reference frame.  In your reference frame, velocity is always zero (in the past as well as the present), so acceleration is always zero, according to your definition.


No, just because you write it, does not make it so.  The only way acceleration can be relative is in regards to the object being accelerated.  Whether I move in regards to you has no effect on your kinetic energy.  Your acceleration increases your Kinetic Energy in relation to what it was at a point in the past.  Velocity, Kinetic Energy, and acceleration have no meaning without time.  Since time must enter into the equation there is not reason not to use time as a reference.  Since over time in  accecleration your Kinetic Energy increases, the difference in current and past kinetic energy can describe your acceleration.  Your change in Kinetic energy can also be described by the accumulation of Psuedo Gravity, times the mass of the object.

As I have said several times now, you can't have it both ways, you can not say the you are accelerating to produce your gravitational effect, but not changing your velocity because there is no reference point.  Again, your referece for acceleration is the same as mine for Kinetic energy.



Quote
The Speed of Light…It’s not just a good idea, IT’S THE LAW!


Funny.  The law is more cryptic than you think.  Can you prove (mathematically, please) that no objects can travel faster than light?[/quote]

I don't have to, Einstien did with his Velocity-addition formula:  V = (Vsub1 + Vsub2) /  (1 + (vsub1 * Vsub2)/c**)

Or you can use the calculation for Kinetic Energy:
K = m * c * c * ((1 / (sqrt(1-(v/c)*(v/c)) - 1)

As v approaches c, m approaches infinity, and so does the kinetic energy.

Consider Thermodynamics, since no system is perfect, there would need to be "Waste energy" from this "Dark Matter" propulsion.  As the energy requirements increase to continue the acceleration, the amount of energy "wasted" would also increase, consuming the world.

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Erasmus

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A problem with your "gravity"
« Reply #72 on: June 12, 2006, 04:05:44 PM »
Quote from: "Doubter"
The only way acceleration can be relative is in regards to the object being accelerated.


I give up on trying to figure out what you mean by this.

Quote
Whether I move in regards to you has no effect on your kinetic energy.


Yes, it does.  Energy is relative too.

Quote
As I have said several times now, you can't have it both ways, you can not say the you are accelerating to produce your gravitational effect, but not changing your velocity because there is no reference point.  Again, your referece for acceleration is the same as mine for Kinetic energy.


Look.  How do you measure acceleration?  It's change in velocity over time.  How do you measure velocity?  It's change in position/distance over time.

At some point in time, measure the distance to the ground.  Count to ten.  Measure the distance again.  Same?  Relative velocity is zero, then.  If the relative velocity is zero, the relative kinetic energy is zero.

Quote
The Speed of Light…It’s not just a good idea, IT’S THE LAW!


Funny.  The law is more cryptic than you think.  Can you prove (mathematically, please) that no objects can travel faster than light?[/quote]

Quote
I don't have to, Einstien did with his Velocity-addition formula:  V = (Vsub1 + Vsub2) /  (1 + (vsub1 * Vsub2)/c**)

Or you can use the calculation for Kinetic Energy:
K = m * c * c * ((1 / (sqrt(1-(v/c)*(v/c)) - 1)

As v approaches c, m approaches infinity, and so does the kinetic energy.


I don't see how any of these things prove that no object can travel faster than light.

Quote
Consider Thermodynamics, since no system is perfect, there would need to be "Waste energy" from this "Dark Matter" propulsion.  As the energy requirements increase to continue the acceleration, the amount of energy "wasted" would also increase, consuming the world.


Whatever, we're not talking about problems with the mechanism providing the acceleration for the flat Earth.  Don't change the subject.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #73 on: June 12, 2006, 05:29:59 PM »
Quote from: "Erasmus"
Quote from: "Doubter"
The only way acceleration can be relative is in regards to the object being accelerated.


I give up on trying to figure out what you mean by this.


Try this:
Quote

Acceleration: The rate at which an object's velocity changes with time.  

So any acceleration requires a change of velocity, and time.  

Picture the two us us racing down the road on motorcycles.

I hit the gas, you hit the brakes.  Inertia causes me to pullback on my bike, it cases you to push forward on your handlebars.  We are accelerating in respect to each other and ourselves.

Now same senerio, but you do not brake.  We again are accelerating in respect to each other, but I'm still hanging on to my handle bars, you are not magically pushed forward to you handlebars.  Relative to each other we have accelerated, but within your frame of reference there is no effect, because your Kinetic energy has not changed.

Quote
Whether I move in regards to you has no effect on your kinetic energy.


Yes, it does.  Energy is relative too.

[/quote]

Now you  are are really grasping at straws.  Kinetic energy in not relative, Potential energy is.

If My motion affects your Kinetic energy, then you should be able to determine my motion by your kinetic energy.  The difference between my kinetic energy and yours can determine our relative velocities,  it can be used to determine the effect of an impact between us, and various other things, but it is not relative in the same way that velocity is.

As in the motor bike example, my change of kinetic energy does not effect yours.

Quote

Look.  How do you measure acceleration?  It's change in velocity over time.  How do you measure velocity?  It's change in position/distance over time.

At some point in time, measure the distance to the ground.  Count to ten.  Measure the distance again.  Same?  Relative velocity is zero, then.  If the relative velocity is zero, the relative kinetic energy is zero.

Yes the RELATIVE kinetic energy is zero, but the Kinetic energy is not.
Do the same experiment in a car, when both objects are at the same speed, the same thing happens.  hit that wall ahead of you, and your relative kinetic energy to the wall is what matters.

Quote

I don't see how any of these things prove that no object can travel faster than light.

As v approaches c, m approaches infinity, and so does the kinetic energy.
Since there is finite energy available, you can not increase your kinetic energy infinitely.

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EnragedPenguin

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A problem with your "gravity"
« Reply #74 on: June 12, 2006, 08:15:43 PM »
You guys have me completely befuddled, but I'm trying as ard as I can to figure this whole thing out.


Quote from: "Doubter"
Whether I move in regards to you has no effect on your kinetic energy.

Quote
Yes, it does.  Energy is relative too.



Now you  are are really grasping at straws.  Kinetic energy in not relative, Potential energy is.


Here's my question. If motion is relative (At least, that's what I've always thought) and kinetic energy is energy from motion, shouldn't that mean kinetic energy is relative too?

I'm completely confused now :(
A different world cannot be built by indifferent people.

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Erasmus

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A problem with your "gravity"
« Reply #75 on: June 13, 2006, 12:17:57 AM »
Quote from: "Doubter"
Picture the two us us racing down the road on motorcycles.


Your motorcycle experiment evokes no new understanding in me.

Quote
Now you  are are really grasping at straws.  Kinetic energy in not relative, Potential energy is.


How do you justify this?

Quote
If My motion affects your Kinetic energy, then you should be able to determine my motion by your kinetic energy.


Nono.  Your motion determines your kinetic energy.  I measure your kinetic energy by measuring your speed.  Since speed is relative, kinetic energy must be.

Quote
The difference between my kinetic energy and yours can determine our relative velocities,


Sure, or we could just measure the velocities directly.  What's the point of measuring my own kinetic energy?  I know it's zero.

Quote
it is not relative in the same way that velocity is.


....... wrong.

Quote
Yes the RELATIVE kinetic energy is zero, but the Kinetic energy is not.


Ahh, I see the problem.  You believe that these things have "real" values.  Yeah, you're not talking about relativity anymore; you're talking about some other magic theory that isn't true.

Okay, basically, your understanding of Special Relativity is just plain flawed.  Pretty badly flawed.  Better than most, but flawed.  Don't worry about it; everybody goes through that phase.  Just have a look over some literature on it.  Wikipedia's article is good, if a little dry.  I highly recommend "Modern Physics" by Tipler and Llewellyn.  I'm currently reading through "General Relativity" by Malcolm Ludvigsen... it's pretty hard to follow (he skips  steps in his proofs, overusing the dreaded "clearly" and "obviously") but it's all geometric, which I believe will yield the best intuition.

Basically, you can have a look at the literature, or you can just take my word for it.  I'm happy to try and help you improve your understanding, if you're interested.  Otherwise I don't really think we can communicate on this topic anymore.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #76 on: June 13, 2006, 04:51:25 AM »
i know this exists in older posts but in FE witch makes me keep thinking of iron but back to my point if gravity does not exist then some one explain how the high/low tide works when after all the moon is suposedly just a spot light.?

A problem with your "gravity"
« Reply #77 on: June 13, 2006, 02:33:38 PM »
Quote from: "Erasmus"
Basically, you can have a look at the literature, or you can just take my word for it.  I'm happy to try and help you improve your understanding, if you're interested.  Otherwise I don't really think we can communicate on this topic anymore.


O.K.  let's start with a couple of basic ideas.  Tell me where I've gone wrong.

What is acceleration?  A change in Velocity.
What is Velocity?  Relative movement within a reference frame.

What is the velocity of the Earth? 0
What is the velocity of the earth one year from now? 0
What is the change of velocity of the earth over this period? 0
What is the earth's acceleration during this time? 0

Since there has been no change of velocity to the subject on the earth, there has been no acceleration, and therefore no gravity like force.

In some of the reading I have done on the subject, the theory is that without another frame of reference there can not be acceleration, and there would be no gravity-like effect.  

The opposing idea, which I have tried to represent, but apparently badly, is that you would get the effect because you are increasing kinetic energy.

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Erasmus

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A problem with your "gravity"
« Reply #78 on: June 13, 2006, 04:26:32 PM »
Quote from: "Doubter"
In some of the reading I have done on the subject, the theory is that without another frame of reference there can not be acceleration, and there would be no gravity-like effect.


Not exactly.  The theory is that without making long-distant measurements, you can't tell the difference between gravity and acceleration.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #79 on: June 13, 2006, 07:43:54 PM »
Quote from: "Erasmus"
Quote from: "Doubter"
In some of the reading I have done on the subject, the theory is that without another frame of reference there can not be acceleration, and there would be no gravity-like effect.


Not exactly.  The theory is that without making long-distant measurements, you can't tell the difference between gravity and acceleration.

Again I should have been more clear, when I said "The theory" I wasn't refering directly to to the theory of equivalence.

I will try to dig up the reference.

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Erasmus

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A problem with your "gravity"
« Reply #80 on: June 13, 2006, 08:01:16 PM »
Quote from: "Doubter"
I will try to dig up the reference.


Aha, I see, okay.  That would be interesting.

In that case, what do you think is happening in the "elevator" thought experiments, where you are in an elevator with no windows.  You feel gravity.  It could be that the elevator is accelerating upwards (via cable, or a rocket, or whatever), or it could be that you're in a gravitational field.  According to the theory you're talking about, it would seem that we're forced to conclude that it's a gravitational field, since we can't see any objects in a reference frame other than our own.

That clearly violates the principle of equivalence.  There's no way we should be able to tell.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #81 on: June 14, 2006, 07:50:11 AM »
Quote from: "Erasmus"

 According to the theory you're talking about, it would seem that we're forced to conclude that it's a gravitational field, since we can't see any objects in a reference frame other than our own.

That clearly violates the principle of equivalence.  There's no way we should be able to tell.

Actually the theory says that there would not be any gravity like effect, nor would there be any centripital force without another frame of reference.

That's where part of my problem lies,  Why should an external oberver make a difference to the internal observer who is not aware of it.


You would not argue that an observer traveling outside of the box (offset on the x, y, or z axis) is a valid reference point for making relative observations.  But you seem to argue against a temporally displaced frame of reference.  

An object is travelling at velocity x, where x can be almost anything but with reference to itself is zero.  After 10 seconds of acceleration a 9 meters per second per second  it as you say, is still travelling at 0 in reference to itself, but has increased in velocity to x + 900 meters per second and displaced 450 meters compared to itself ten seconds ago.

The internal observer has experienced the force, but not the displacement.
Again, I don't think you would argue that to an external observer, there is a limit to the acceleration from the external frame of reference, so what happens internally at that point?  Or during your closed box acceleration after youve accelerated for years, and you open a window and can see an external reference point?

I'm also trying to work out relative Kinetic energy.  Consider a particle accelerator.  If I accelerate a particle to 14%C, it's mass increases 1%.  For further acceleration I need to increase the energy accordingly for the new mass, and so on.

Now to the particle, it is still, but the earth is moving, but where is the force coming from to move the earth to 14%C?  Likewise when you increase the earth's mass by 1% due to it's relative velocity, and now there is the consequence of additional gravity.  The particle is in the earths gravitation, and therefore the particle's mass should likewise increase.

New we put the accelerator into Einstein's elevator.  If sitting in a gravity feild, the effect should be the same.  If accelerating, that would have to mean that the relative velocity of the elevator from the frame of the partical has accelerated, which I believe would have to be accounted for with time dialation. Time has slowed down for the particle, so the external acceleration is faster.

Now, if the particle is a radioactive with a know half life, it should be possible to determine it's time difference due to dialation between it and the elevator.  So if the time dialation was 1% and the experimant runs for 100 seconds there should be an apparent difference in age between the accelerated particle (in the particle accelerator) and a like particle that was not accelerated sitting in the elevator.  From the accelerated particle's frame the other particle experienced a time acceleration during the period it was in the accelerator, which would seem like the opposet of what would be expected. (the elevator accelerated to 14%c from the particles frame, but when motion equalizes 1 extra second passed for the elevator).

Now we step outside of the elevator car.  It acclerates to near light speed, it's mass and the mass of everything inside should increase, to the external observer.  So if inside the elevator I hang two objects, and their mass increases, then their gavitational attraction should increace pulling them together.  Yet to the person inside the elevator, there should be no difference. Do the objects pull together or not?

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Erasmus

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A problem with your "gravity"
« Reply #82 on: June 15, 2006, 05:38:27 PM »
I'm not sure I follow you in this most recent post, but I'll try to address it anyway.

Quote from: "Doubter"
That's where part of my problem lies,  Why should an external oberver make a difference to the internal observer who is not aware of it.


Observers don't make a difference to one another in any circumstance (assuming they're not exerting forces on one another, and that they're massless, that is).

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But you seem to argue against a temporally displaced frame of reference.


Here's one of the places that I've really lost you.

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An object is travelling at velocity x, where x can be almost anything but with reference to itself is zero.


There, you're doing it again.  You've totally missed the point of relativity.  Objects don't have a "true velocity" that they travel at.  Also, we don't say "in reference to x", where x is a velocity.

If E is an object -- say, an astronaut -- E can define a reference frame.  It can, for example, put itself at the origin, make x be "to the right", y be "up", and z be "behind".  If E measures E's velocity in this reference frame, he gets zero.

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After 10 seconds of acceleration a 9 meters per second per second


Instead of just throwing that out, let's come up with something a little more undeniably meaningful.

Suppose, for example, Alice is the astronaut in the above paragraph, floating out in deep space.  She's got a jet pack with her, and an accelerometer.  The accelerometer is basically a spring scale with a one-kg mass attached.  You point the spring backwards, fire the jet pack, and read the weight of the mass.  If it's 1 N, you're accelerating at 1 m/s.

So, Alice turns on the jet pack and reads the accelerometer.  It reads 9.8 N for ten seconds.  Then she turns off the jet pack.

Now Alice measures her velocity in her reference frame again.  It's still zero.

Now imagine that Alice was floating next to a gigantic space station, initially at rest with respect to the space station.  Even after she was done accelerating, she would still measure her velocity relative to herself to be zero.  Yes, she could measure her velocity in another reference frame, and she would get a different result.

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Again, I don't think you would argue that to an external observer, there is a limit to the acceleration from the external frame of reference, so what happens internally at that point?


I don't understand this question.

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Now to the particle, it is still, but the earth is moving, but where is the force coming from to move the earth to 14%C?


If the particle can tell that it is accelerating relative to the Earth, then it can conclude that there is some force acting on it.

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The particle is in the earths gravitation, and therefore the particle's mass should likewise increase.


No.  Gravity has no effect on your mass.

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Now we step outside of the elevator car.  It acclerates to near light speed, it's mass and the mass of everything inside should increase, to the external observer.  So if inside the elevator I hang two objects, and their mass increases, then their gavitational attraction should increace pulling them together.  Yet to the person inside the elevator, there should be no difference. Do the objects pull together or not?


The "mass" that would increase to the external observer is the mass  in F=ma.  That is, it appears to the external observer that it takes a greater force to apply a given acceleration to the object.  It doesn't contribute to the gravitational field.  Only invariant mass contributes to the gravitational field.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #83 on: June 16, 2006, 10:48:22 AM »
I'll try to separate ideas to make it a little more clear (and the posts a little smaller).

Quote from: "Erasmus"
I'm not sure I follow you in this most recent post, but I'll try to address it anyway.

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But you seem to argue against a temporally displaced frame of reference.


Here's one of the places that I've really lost you.

Acceleration shows a change in velocity over a period of time.
Likewise the change in velocity, and the amount of force required to achieve that change can be calulated knowing the rate of acceleration, the mass of the object being accelerated and the tim eperiod of acceleration.

While the "Local" frame of reference may still have a velocity of zero.  It's velocity can also be referenced from the frame of it's position in the past.

As the past is simple displacement in the "fourth dimention" which is recognized by Relativity, and Relativity states that you can not have a velocity faster than light if you have mass, than your velocity can not increase beyond C relative to your past.

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Erasmus

  • The Elder Ones
  • 4242
A problem with your "gravity"
« Reply #84 on: June 16, 2006, 11:25:23 AM »
Quote from: "Doubter"
While the "Local" frame of reference may still have a velocity of zero.  It's velocity can also be referenced from the frame of it's position in the past.


How can the velocity be referenced in this way?  How do you measure it?  What do you look at to make the measurement?

And, "just measuring the time and multiplying by the acceleration" isn't a measurement.  It's you, applying a formula that you believe to be lawful.

What do you think about the accelerating astronaut?  Suppose he measures his acceleration to be, say, 0.1c/s, and he leaves his jet pack on for twenty seconds.  What's his velocity at this point?

Also, if you're going to say that it's not possible for him to accelerate at 0.1c/s for  twenty seconds, are you going to suggest that somehow the force of his jetpack decreased over his time so that to his perception, after a while he felt like he wasn't accelerating anymore?  If so, I'd like you to give a formula for what force he feels as a function of time.

Lastly, we picked a point in time when he was not accelerating.  Why is this reference frame preferred to the one that is accelerating with him?  Why, for that matter, is it preferred to some reference frame that existed before?

To explain: Suppose that, from some "past reference frame", our astronaut applies impulse A to accelerate to 0.9c and then stops accelerating.  Pick a reference frame F after this acceleration is done.  The astronaut is at rest in reference frame F now.  Now have him apply impulse A again.  He's now at rest in a frame F', but measured in reference frame F, he is travelling at 0.9c.   Now have him apply impulse A again... see where this is going?  If all that's important is that I'm measuring my velocity "with respect to some reference frame in the past in which I was at rest", then I can always just stop accelerating for a split second, pick the reference frame in which I'm currently at rest, and then apply the same impulse again.  I can make that impulse arbitrarily short.  How -- as measured by the astronaut -- would that be any different from just going at a constant acceleration forever?

Basically, you insist on the existence of preferred reference frames.  The most fundamental premise of relativity is that no such thing exists.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #85 on: June 16, 2006, 11:40:54 AM »
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Now to the particle, it is still, but the earth is moving, but where is the force coming from to move the earth to 14%C?


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If the particle can tell that it is accelerating relative to the Earth, then it can conclude that there is some force acting on it.

So motion is relative but force is not? alice knows the the jetpack is applying force to her, but the station is moving raltive to her.

Doesn't that contradict Relativity, which states the the laws of Physics should be valid for all frames of reference.  If so, then to move an object relative to something, there needs to be force applied to the object.

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The particle is in the earths gravitation, and therefore the particle's mass should likewise increase.


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No.  Gravity has no effect on your mass.


If Gravity is not distinguishable from accleration, then Gravity must affect your relativistic Mass, because acceleration does.

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Now we step outside of the elevator car.  It acclerates to near light speed, it's mass and the mass of everything inside should increase, to the external observer.  So if inside the elevator I hang two objects, and their mass increases, then their gavitational attraction should increace pulling them together.  Yet to the person inside the elevator, there should be no difference. Do the objects pull together or not?


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The "mass" that would increase to the external observer is the mass  in F=ma.  That is, it appears to the external observer that it takes a greater force to apply a given acceleration to the object.  It doesn't contribute to the gravitational field.  Only invariant mass contributes to the gravitational field.

That can not be true.  What is a photon's invariant mass? 0.  If only invariant mass contributes to gravitation, than a photon would not be diverted by gravity.  Also, it is not in accordance with what I've been reading:
Quote from: "http://abyss.uoregon.edu/~js/21st_century_science/lectures/lec07.html"
Whereas the sources of the electric field, the electric charges of particles, have values independent of the state of motion of the instruments by which these charges are measured, the source of the gravitational field, the mass of a particle, varies with the speed of the particle relative to the frame of reference in which it is determined and hence will have different values in different frames of reference. This complicating factor introduces into the task of constructing a relativistic theory of the gravitational field a measure of ambiguity


Equivilence is interesting, but not really true.  We may lack the ability to measure the differences, but theoretically an object under gravity would weigh more at the floor of the elevator than at the ceiling. Sure the difference is only about (6357 * 6357 / 6357.003 * 6357.003) on the surface of a round earth, but it is still there.

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Erasmus

  • The Elder Ones
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A problem with your "gravity"
« Reply #86 on: June 16, 2006, 01:06:34 PM »
Quote from: "Doubter"
That can not be true.  What is a photon's invariant mass? 0.  If only invariant mass contributes to gravitation, than a photon would not be diverted by gravity.


Completely wrong.  You don't have to "contribute to gravitation" in order to fall towards gravitationally "attractive" bodies.  This is because each massive body curves the spacetime around it.  Even if you have no mass, you're still travelling through curved space.


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Also, it is not in accordance with what I've been reading:
Quote from: "http://abyss.uoregon.edu/~js/21st_century_science/lectures/lec07.html"
Whereas the sources of the electric field, the electric charges of particles, have values independent of the state of motion of the instruments by which these charges are measured, the source of the gravitational field, the mass of a particle, varies with the speed of the particle relative to the frame of reference in which it is determined and hence will have different values in different frames of reference. This complicating factor introduces into the task of constructing a relativistic theory of the gravitational field a measure of ambiguity


Okay, I could be wrong about this.

Assuming it's correct however, it implies that some strange things.  If that's the correct interpretation, then the fact is that relativity would seem to predict exactly what you describe -- to the external observer, the objects hanging in the elevator would move towards each other under their mutually increased gravitational attraction, but to the observer in the elevator, they would remain in their original positions.

Either you're okay with that result, or you are forced to conclude that gravitational mass is not relativistic mass.

I'm currently leaning towards the latter.

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Equivilence is interesting, but not really true.  We may lack the ability to measure the differences, but theoretically an object under gravity would weigh more at the floor of the elevator than at the ceiling. Sure the difference is only about (6357 * 6357 / 6357.003 * 6357.003) on the surface of a round earth, but it is still there.


I'm pretty sure we can measure that difference in a sufficiently expensive laboratory :)  However, the principle of equivalence is not refuted by this, since it only talks about local measurements; i.e., ones in which tidal effects can be neglected.

A nice way of thinking about the principal of equivalence is that if spacetime is locally Minkowski, then special relativity applies locally.
Why did the chicken cross the Möbius strip?

A problem with your "gravity"
« Reply #87 on: June 16, 2006, 02:00:26 PM »
Quote from: "Erasmus"
Quote from: "Doubter"
While the "Local" frame of reference may still have a velocity of zero.  It's velocity can also be referenced from the frame of it's position in the past.


What do you think about the accelerating astronaut?  Suppose he measures his acceleration to be, say, 0.1c/s, and he leaves his jet pack on for twenty seconds.  What's his velocity at this point?

Also, if you're going to say that it's not possible for him to accelerate at 0.1c/s for  twenty seconds, are you going to suggest that somehow the force of his jetpack decreased over his time so that to his perception, after a while he felt like he wasn't accelerating anymore?  If so, I'd like you to give a formula for what force he feels as a function of time.

Is that a trick question?  Acceleration is a function of time, the gravity effect of acceleration is a force felt as a function of time.

Also, the math behind relativity account for part of what you describe, to the space station Alice is flying past, her rate of acceleration would gradually decrease as she approaches C.

But on a more practical level, the jet pack has a limit to the velocity of it's exhaust.  Like your car, once you've floored it, you reach a point where it will not accelerate.

Using a rocket whose maximum exhaust velocity is .5C, If I traveled from the hypotetical round earth to the hypothetical sun, assuming your .1C accleration, it would take about 16 minutes to make the trip.  To the pilot, it would be slightly less time due to time dialation, but even to the pilot it would not continue to acclerate at a steady rate, the effectivenss of the thrust decreases as the ship's velocity approaches exhaust velocity.

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To explain: Suppose that, from some "past reference frame", our astronaut applies impulse A to accelerate to 0.9c and then stops accelerating.  Pick a reference frame F after this acceleration is done.  The astronaut is at rest in reference frame F now.  Now have him apply impulse A again.  He's now at rest in a frame F', but measured in reference frame F, he is travelling at 0.9c.

According to relativity, this is not true.  Velocities are not additive in relativistic speeds. To reference frame F you are traveling about .98c not 1.8c.  You are trying to use Newtons laws and apply Relativity.
EvenI know enough of Einstien's theories to know you can't do that.
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Basically, you insist on the existence of preferred reference frames.  The most fundamental premise of relativity is that no such thing exists.

Yet you too are insisting that your perfered frame (the Flat Earth) does not have to comply to any other frame.

I can see how, the FE could experience continued acceleration at 1 G, but could not be part of this universe while doing it.  As this universe is belived to have a finite duration, and perhaps even finite dimentions, you would literally run out of time, as subjective time on your flat earth would slow down as your velocity approached light with respect to the universe.

Now my head is really spinning.  If we are speeding away from the event point of the "Big Bang", we judge 13+ billion years to have passed.  If you were the point in space where the Big Bang happened how much time would have passed?

A problem with your "gravity"
« Reply #88 on: June 16, 2006, 02:11:43 PM »
Quote from: "Erasmus"
Quote from: "Doubter"
That can not be true.  What is a photon's invariant mass? 0.  If only invariant mass contributes to gravitation, than a photon would not be diverted by gravity.


Completely wrong.  You don't have to "contribute to gravitation" in order to fall towards gravitationally "attractive" bodies.  This is because each massive body curves the spacetime around it.  Even if you have no mass, you're still travelling through curved space.


But if gravtational attraction is:
gravity = (mass1 * mass2  * k) / distance2

And Mass1 = 0 then Gravity = 0

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Erasmus

  • The Elder Ones
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A problem with your "gravity"
« Reply #89 on: June 16, 2006, 03:20:16 PM »
Quote from: "Doubter"
But if gravtational attraction is:
gravity = (mass1 * mass2  * k) / distance2

And Mass1 = 0 then Gravity = 0


Valid argument.  However, it turns out that that is not gravitational attraction.  Not in GR, anyway.
Why did the chicken cross the Möbius strip?