Special Relativity Contains Massive Error

As v approaches c, L approaches zero. Therefore, as v approaches c, the distance between that spaceship and every point in the universe also approaches zero.Now, the other part of your question was how much time each photon thinks has passed. The answer to this question is a little different, as relativity doesn't really apply to photons, because they don't have a reference frame. This is one of the ways in which relativity is different from classical mechanics.One way to think about the "photon's frame" is that it is the frame of the universe, in which all distances are zero. Since all distances are zero, there is no way to measure the time
  • #1
Tempest
Suppose an atom that is at rest in an inertial reference frame simultaneously emits two identical photons in exactly the opposite direction. For all intent and purpose consider the radius of the atom as vanishingly small. Now consider things after one second has passed in this 'atomic frame'.

According to the theory of SR, any photon in an inertial reference frame must have speed c=299792458 meters per second in that frame. Thus, after one second has passed, the distance each photon is away from the atom, in the atomic frame absolutely has to be 299792458 meters. Let us denote this distance by D.

Now consider things from the point of view of a coordinate system whose origin is one of the photons. In this system, the speed of this photon is zero, and the atom is moving (lets say to the left) at speed c, and the other photon is moving to the left at a speed greater than c.

Now, either the distance the atom is away from the photon in this photonic frame has been Lorentz contracted or not. Assume it has been contracted. Let d denote the distance the atom and photon are separated by in the photonic frame, after 1 second has passed in the atomic frame.

The exact amount of contraction is given by:

d = D(1-v^2/c^2)^1/2 = D(1-c^2/c^2)^1/2 = 0

Hence, after one second has passed in the atomic frame, the distance between the two photons in the photonic frame is zero. Hence, in the photonic frame, the two photons are not in relative motion, which is impossible (that would only be true if they were emitted in the same direction, and they were emitted in opposite directions). Thus, the Galilean transformations hold when we switch from the atomic frame to the photonic frame.

Now, consider a third reference frame F3, which was initially moving away from the atom (to the right) at a constant speed S, and such that after one second had passed in the atomic frame, photon 1 was located at the origin of F3.

We know that in the atomic frame, that after one second has passed photon one traveled 299792458 meters. Suppose that S = 1 meter per second. Thus, F3 is moving away from the atom at a constant speed of one meter per second.

And after one second has passed in the atomic frame, we have stipulated that the photon is at the origin of F3. Therefore, at the moment the photon's were emitted, the origin of F3 was located 299792457 meters away from the atom, in the atom's frame. It would then follow that after one second had passed in the atomic frame, that the origin of F3 would be 299792458 meters away from the atom, and hence the photon would be at the origin of F3 as stipulated, so everything is fine.

Hence in the atomic frame (atom at rest), we have an event which begins when the origin of F3 is 299792457 meters away from the atom, and ends when the origin of F3 is 299792458 meters away from the atom, and this event takes 1 second in the atomic frame.

(work in progress)
 
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  • #2
If the atom thinks 1 second has passed, how much time does each photon think has passed?
 
  • #3
Perhaps you should have titled this thread: "DOES SR contain a massive error?"

The answer is that, no, it does not.
Originally posted by Tempest
Now consider things from the point of view of a coordinate system whose origin is one of the photons. In this system, the speed of this photon is zero, and the atom is moving (lets say to the left) at speed c, and the other photon is moving to the left at a speed greater than c.
You've chosen to use a coordinate system which is singular. In other words, for a photon, time effectively does not exist. The distance between itself and any other point in the universe is length-contracted to zero, and the time it takes to get anywhere in the universe is also zero.

In other words, in the photon's rest frame, the atom, the other photon, and your Aunt Sally are all at the same distance: zero. And they always will be. Thus, there is no paradox, or error, or whatever you'd like to call it.

I can understand that it can be a little challenging to learn and intuitively understand special relativity -- but you need to think a little more before trying to convince us that it's wrong.

I should also note that this sort of discussion is permitted on physicsforums.com only in the Theory Development forum (which is a subforum of the General Physics forum).

- Warren
 
  • #4
Originally posted by russ_watters
If the atom thinks 1 second has passed, how much time does each photon think has passed?
A very pertinent question...

- Warren
 
  • #5


Originally posted by chroot
Perhaps you should have titled this thread: "DOES SR contain a massive error?"


In other words, in the photon's rest frame, the atom, the other photon, and your Aunt Sally are all at the same distance: zero. And they always will be. Thus, there is no paradox, or error, or whatever you'd like to call it.


What do you mean they are all "at the same distance: zero"?
 
  • #6
Originally posted by russ_watters
If the atom thinks 1 second has passed, how much time does each photon think has passed?

Answer: 1 second
 
  • #7


What do you mean they are all "at the same distance: zero"?
What I mean is that if you were the captain of a hypothetical starship capable of traveling at c (it's hypothetical, of course, because a starship necessarily has mass), you would be able to travel to any point in the universe in exactly zero time. This means that every point in the universe appears to be zero distance from you.

In more mathematical terms, length contraction can be expressed as:

[tex]L = \frac{L_0}{\gamma} = L_0 \sqrt{1 - \frac{v^2}{c^2}}[/tex]

where [itex]L_0[/itex] is the distance between two objects when you're not moving at all relative to them, and [itex]L[/itex] is their distance when you are traveling at some velocity [itex]v[/tex]. As you can see,

[tex]\lim_{v \rightarrow c} L = 0[/tex]

In other words, when you're going c, everything in the universe is zero distance from you.

- Warren
 
  • #8
Originally posted by Tempest
Answer: 1 second
That would be an incorrect answer, Tempest. The answer is zero. Do you know how to use the Lorentz transformations to demonstrate time dilation?

- Warren
 
  • #9
It was shown that the galilean transformations hold when swtiching from the atomic frame to the photonic frame, since the event took one second in the atomic frame, it also took one second in the photonic frame. Do you mean derive the time dilation formula from the Lorentz coordinate transformations? Probably I do.
 
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  • #10
Originally posted by Tempest
It was shown that the galilean transformations hold when swtiching from the atomic frame to the photonic frame, since the event took one second in the atomic frame, it also took one second in the photonic frame.
The Galilean transformation never holds. It is a low-velocity approximation of the Lorentz transformation.

Surely, if you're going to be dealing with things going at the speed of light, you're no longer able to use a low-velocity approximation.

- Warren
 
  • #11
If the galilean transformation never holds, then the distance the two photons are separated after one second has passed in the atomic frame is zero, hence the two photons are not in relative motion, hence the two photons are traveling in the same direction, contrary to the stipulation that they were emitted in opposite directions. Hence, the galilean transformation sometimes holds.
 
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  • #12
Originally posted by Tempest
If the galilean transformation never holds, then the distance the two photons are separated after one second has passed in the atomic frame is zero, hence the two photons are not in relative motion, hence the two photons are traveling in the same direction, contrary to the stipulation that they were emitted in opposite directions. Hence, the galilean transformation sometimes holds.
Nope. Sorry. The very concept of 'relative motion' really just has no meaning for photons, since they don't experience distance (or time).

And no, the Galilean transformation never, ever holds. Special relativity replaces it with a generalization, the Lorentz transform.

You're welcome to say "Special relativity has a massive error if I assume the Galilean transformation holds," but that's not very useful -- since the Galilean transformation never holds in special relativity.

- Warren
 
  • #13
Originally posted by chroot
A very pertinent question...

- Warren
Hey, I'm not completely useless outside the engineering forums.

Tempest, the very title of your thread should set off a blinking red light in the back of your mind when you read it back to yourself. If SR did contain a massive error, don't you think it would have been noticed before? The implications of what you are saying are so fundamental (it really is a massive error you are suggesting) that the theory would fall apart completely and all the technology we use that depends on it (GPS for example) would not work.

You need to check to see if your warning light is working properly.
 
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  • #14
originally posted by Tempest
According to the theory of SR, any photon in an inertial reference frame must have speed
c=299792458 meters per second in that frame.


Originally posted by Tempest
Now consider things from the point of view of a coordinate system whose origin is one of the photons. In this system, the speed of this photon is zero, and the atom is moving (lets say to the left) at speed c, and the other photon is moving to the left at a speed greater than c.


Doesn't it occur to you that those two statements are contradictory? The only way you can assert the second is to deny the first. In other words your whole argument is "If special relativity is wrong, then it contains an error"!
 
  • #15
Originally posted by russ_watters
Hey, I'm not completely useless outside the engineering forums.

Tempest, the very title of your thread should set off a blinking red light in the back of your mind when you read it back to yourself. If SR did contain a massive error, don't you think it would have been noticed before?

Naaaah.

The implications of what you are saying are so fundamental (it really is a massive error you are suggesting) that the theory would fall apart completely and all the technology we use that depends on it (GPS for example) would not work.

GPS would still work, because it does work AND SR is wrong.



You need to check to see if your warning light is working properly.

Yeah, it went on the moment I read your post.
 
  • #16
Originally posted by HallsofIvy
Doesn't it occur to you that those two statements are contradictory? The only way you can assert the second is to deny the first. In other words your whole argument is "If special relativity is wrong, then it contains an error"!

I) IF the speed of light is c in any inertial reference frame then the photonic frame isn't an inertial reference frame.

II) If in some inertial reference frames the speed of light isn't c then the photonic frame is an inertial reference frame.



So if SR doesn't contain a massive error then the photonic frame is a non-inertial reference frame.

Hence, if the photonic frame is an inertial reference frame, then SR contains a massive error. :)
 
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  • #17
Originally posted by Tempest
Hence, if the photonic frame is an inertial reference frame, then SR contains a massive error. :)
Correct logic, but the premise is invalid -- a frame centered on a photon is not inertial.

- Warren
 
  • #18
Originally posted by chroot
Correct logic, but the premise is invalid -- a frame centered on a photon is not inertial.

- Warren



How sure are you?
 
  • #19
Originally posted by Tempest
How sure are you?
110%. Do you understand the definition of an inertial frame?

- Warren
 
  • #20
Originally posted by chroot
110%. Do you understand the definition of an inertial frame?

- Warren

Refresh my memory, something about a frame which isn't accelerating if I do recall. Only problem with that, is acceleration is relative.

I am thinking more along the lines of a frame in which all three of Newton's laws are true.
 
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  • #21
Originally posted by Tempest
Only problem with that, is acceleration is relative.
Sorry, acceleration is not relative. Let's say you're flying about in a spaceship. You can build a device (an accelerometer) which can measure your spaceship's acceleration without reference to anything outside the ship. You can close all the windows, and the accelerometer will keep working.

On the other hand, it's impossible to build a speedometer without looking out the window. If you're cruising along at uniform velocity (any uniform velocity), there's no way an experiment inside your spaceship can detect the motion when the windows are closed.

- Warren
 
  • #22
Originally posted by Tempest
Yeah, it went on the moment I read your post.
My tone may have been a little sarcastic, but I was serious - when you find what you think is a major flaw in a major theory of science, you should start looking for the flaw in your logic/understanding that leads you to think the theory has a flaw. Believe me I understand that can be difficult - I argued with my teachers in school all the time, but they were almost always right.
 
  • #23
Originally posted by chroot
Sorry, acceleration is not relative. Let's say you're flying about in a spaceship. You can build a device (an accelerometer) which can measure your spaceship's acceleration without reference to anything outside the ship. You can close all the windows, and the accelerometer will keep working.

On the other hand, it's impossible to build a speedometer without looking out the window. If you're cruising along at uniform velocity (any uniform velocity), there's no way an experiment inside your spaceship can detect the motion when the windows are closed.

- Warren

The two devices you are referring to, have nothing to do with the concept of 'relative'. Acceleration has meaning only with respect to a coordinate system. So consider a coordinate system whose center of mass is the center of mass of a uniformly accelerating spaceship. Now, consider a particle moving in the frame. The coordinates of the particle are varying in time. The particle will not travel in a straight line at a constant speed. The particle will be accelerating. Using the mathematical definition of acceleration, we can write the forumla for it mathematically, and there must be an answer, so that we cannot say that the particle isn't accelerating in the coordinate system. Thus, in this coordinate system, the particle IS accelerating. However, we should not infer that a force external to the particle is the cause of the acceleration, because the ship is a non-inertial reference frame.

Suppose the particle is not subjected to any force, and that the force on the spaceship is F. Let M denote the mass of the spaceship, the acceleration of the spaceship relative to the particle is a=F/M, but this is also the acceleration of the particle with respect to the ship.

The point is, the mathematical definition of acceleration makes no reference to whether or not the coordinate system is inertial. Hence, there will be an answer for what the acceleration of the particle is, even if that acceleration is computed in a non-inertial frame. So you cannot say that the particle isn't accelerating with respect to the spaceship.

So the argument really goes like this. There is a definition of acceleration in a coordinate system. This definition doesn't care whether the coordinate system is inertial or not. So there will be a nonzero value for the acceleration of the particle in the rocket's frame. Thus, you cannot say that the acceleration of the particle is undefined in the rocket's frame.

What you do not want to do, is conclude that the particle is being subjected to force F. So while the systems are kinematically equivalent, they are not dynamically equivalent. Or in other words, the two systems are not equivalent. This is usually expressed by saying the rocket frame is non-inertial, and the particle frame is inertial.

Suppose that you come up with a value of a for the particle's acceleration in the rocket frame, and that the particles mass is m. If you blindly apply Newton's second law, you will conclude that the particle is being subjected to a force equal to ma, but this is wrong. But observers in the rocket would know that their rocket is being subjected to a force F. Now in a coordinate system at rest with respect to the particle, the acceleration of the ship is a. Let the mass of the ship be M, and so by Newton's second law F = Ma. Now the observers in the ship would know their mass M, and so could compute their acceleration relative to the particle, and they will get a = M/F. And so, when they plot the particles position in their frame in time, and they see that its equivalent to M/F, they will then conclude that the particle isn't being subjected to a force.

The point then, is that acceleration must be relative, because the mathematical definition of it is not limited to coordinate systems which are inertial, but two systems which are accelerating with respect to each other, are not equivalent. Something about them isn't relative, but rather is absolute. And there are many ways of saying this. The simplest is to say that one system is absolutely an inertial reference frame, and the other system is absolutely not an inertial reference frame.
 
  • #24
Tempest: At 12-17- 2003 07:52 AM you asked:
I am not really sure I have the concept of an inertial reference frame down, can anyone help me?
and then at 12-17- 2003 04:36 PM you proclaimed:
Special Relativity Contains Massive Error
This sets off a big blinking red light in my mind.
 
  • #25
Originally posted by Redfern
The two devices you are referring to, have nothing to do with the concept of 'relative'.
Sorry redfern, the word 'relative' already has a precise meaning in relativity theory. If you don't like it, you'll have to complain to someone else. I don't care what you think it should mean.

The rest of your post is bunk. The concept has nothing at all to do with coordinate systems, which exist independently (if you choose to use one at all -- some mathematicians prefer not to).

- Warren
 
  • #26
Originally posted by Tempest
Now, either the distance the atom is away from the photon in this photonic frame has been Lorentz contracted or not. Assume it has been contracted.

Psst, in a 1907 paper, Einstein admitted that the geometric shape of a body does not really “contract” due only to “relative motion”. So if we have your frame in which the two photons are emitted be about 186000 miles long, in each direction from the atom that emits the photons, the length of that frame doesn’t really “shrivel up” just because some other frame is moving relative to it.

He said it this way:

Einstein said in 1907
“The shape of a body in the sense indicated we will call its ‘geometrical shape’. The latter obviously does not depend on the state of motion of a reference frame.”

In Lorentz’s 1904 relativity theory, his “contraction” was due to a mass moving through an ether and being contracted by that motion, by means of the mass somehow “feeling” a resistance to its motion through the ether and contracting as a result, much like if we hold a balloon inside a moving car and it has a round shape, then we hold it outside the window of the car and the air resistance changes its shape. But just “relative motion” alone, between two masses, can not cause a “length contraction” in either of the masses.
 
  • #27


Originally posted by David
Psst, in a 1907 paper, Einstein admitted that the geometric shape of a body does not really “contract” due only to “relative motion”. So if we have your frame in which the two photons are emitted be about 186000 miles long, in each direction from the atom that emits the photons, the length of that frame doesn’t really “shrivel up” just because some other frame is moving relative to it.

He said it this way:



In Lorentz’s 1904 relativity theory, his “contraction” was due to a mass moving through an ether and being contracted by that motion, by means of the mass somehow “feeling” a resistance to its motion through the ether and contracting as a result, much like if we hold a balloon inside a moving car and it has a round shape, then we hold it outside the window of the car and the air resistance changes its shape. But just “relative motion” alone, between two masses, can not cause a “length contraction” in either of the masses.

David, it seems you have misunderstood the mathematics of relativity. There is something called the Lorentz transformations, and these transformations have to do with how the coordinates of one system transform into another system in relative motion to it.

Now, as a result of these transformations, a ruler which runs from zero to one meter in one coordinate system, will NOT have that length in a coordinate system in relative motion, so you are misunderstanding an implication of the theory of relativity. You are misunderstanding it, because you don't understand what the Lorentz transformations say. One of the consequences of the Lorentz transformations is that the length of the ruler is a function of the relative speed v, and the speed of light c. This means that a ruler which is one mile long in a frame at rest with respect to the ruler, will be shorter in a frame in which that ruler is moving. The length the ruler will be in a frame in which it is moving is given by:

[tex] L = L_0 \sqrt{1-v^2/c^2} [/tex]

So as you can see, if the ruler is accelerating in some inertial frame, the closer its speed gets to c, the shorter it gets, and when its speed actually reaches c, then its length in that frame is equal to zero, contrary to the fact that it is a ruler, and by definition a ruler is a one dimensional entity. Hence, no ruler can be accelerated to the speed of light in an inertial frame.

Go study the Lorentz transformations. I can give you a link to a site if you need one. I would also suggest trying to derive the contraction formula, and the time dilation formula from the transformations. It's a good mathematical excercise. Respond to me if you have any questions about this.
 
  • #28
Tempest,

This post has really got me thinking. Things are very clear from the atomic frame. Two photons are simultaneously emitted from an atom which is at rest in an inertial reference system, after one second, they are both 299792458 meters away from the center of mass of the atom, by one of the postulates of relativity.

Then you do something interesting, you switch to a reference frame traveling with one of the photons, so that in that frame, the photon is at rest (lets say at the origin). But clearly, by the relativity postulate, a photon cannot be at rest in an inertial reference frame, since its speed must be equal to c in an inertial reference frame. So if the theory of relativity doesn't contradict, the only possible conclusion is that this frame isn't an inertial reference frame.

That being said, if you can now show that the frame in question MUST be an inertial reference frame, you would likely be onto something. And so this is why you have gotten me thinking. Here is what would have to be done. You would have to prove that if X is an inertial reference frame, and Y is a reference frame whose origin is moving at a constant speed relative to the origin of X, and whose axes aren't rotating in frame X, then Y is an inertial reference frame.

Then, because the photon is moving at a constant speed in the atomic frame, it would follow by a theorem not yet proven, that a frame in which the photon is at rest, is an inertial reference frame. And since one of the consequences of the special theory is that in any inertial reference frame the speed of any photon is c, you would have accomplished something.

So this brings the attention to inertial reference frames. If we simply define an inertial reference frame as a reference frame which isn't accelerating relative to any other inertial reference frame, your proof would be complete, but your definition would be circular.

So then you need a definition which avoids circularity.

Now, a well known consequence of the special theory, is that mass (like length and time) is relative. The formula used by Einstein was:

[tex] M = \frac{m}{\sqrt{1-v^2/c^2}} [/tex]

So since we are still operating under the assumption that the relativity postulate is true, we have access to the above equation, and I was thinking of trying to use it in a definition of an inertial reference frame, other than a reference frame which isn't accelerating. Let me know your thoughts

Also I am wondering if you can use the fact that in the 'photonic' frame the photon's rest mass is equal to zero.

I was sort of thinking like this

Suppose that a body of mass M is at rest in the atomic system. It would then follow that its mass M is equal to its rest mass m, since v=0. And similarly for all other bodies at rest in the atomic frame. And for anybody which is moving in the atomic frame, the mass of that object isn't equal to its rest mass, but rather is something greater, given by the formula above. Now, for photons in the atomic frame, we have division by zero error from that formula, unless the rest mass of a photon is zero. So a consequence of the theory of relativity, is that in any reference frame in which a photon is at rest, that photon MUST have a mass of zero. And we already know that relativity implies that there are no inertial reference frames in which a photon is at rest, though in your 'photonic' frame the photon is clearly 'at rest', from which we must conclude that the photonic frame is non-inertial. But here is what gets me.

Because there were two photons emitted from the atom, the atom wasn't subjected to a net force. Now from in the photonic system, the atom is following Newton's law of inertia, from which it does follow that the photonic system is an inertial system. You are onto something.

In other words, the atom never accelerated relative to the photon, and it was subject to no net force, and so the proper inference from the photonic system, is that the photonic system is an inertial reference frame.

Thank you


P.S. I just started wondering if a clear contradiction to quantum physics is visible. We know that in the atomic frame, the energy of the photon is given by E = hf, so I am wondering if this can be used. It would at least show the two theories are incompatible. IN that photonic frame the rest mass of the photon is zero.

[tex] Mc^2 = hf = hc/\lambda [/tex]

From which it follows that

[tex] Mc = h/\lambda [/tex]

Since c is greater than zero, we can divide boths sides by c to obtain

[tex] M \lambda = h/c [/tex]

And so we have:

[tex] \frac{m \lambda}{\sqrt{1-v^2/c^2}} = h/c [/tex]

And I think the wavelength is relative too. Hence there is a relationship:

[tex] \lambda = \lambda_0 \sqrt{1-v^2/c^2} [/tex]

Hence we have:

[tex] m \lambda_0 = h/c [/tex]

and since m =0 we arrive at:

[tex] 0 = h/c [/tex]

Which is a clear contradiction. This at least shows quantum physics and relativity to be at odds. Let me know what you think.
 
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  • #29


Originally posted by StarThrower
David, it seems you have misunderstood the mathematics of relativity. There is something called the Lorentz transformations, and these transformations have to do with how the coordinates of one system transform into another system in relative motion to it.

Now, as a result of these transformations, a ruler which runs from zero to one meter in one coordinate system, will NOT have that length in a coordinate system in relative motion, so you are misunderstanding an implication of the theory of relativity. You are misunderstanding it, because you don't understand what the Lorentz transformations say. One of the consequences of the Lorentz transformations is that the length of the ruler is a function of the relative speed v, and the speed of light c. This means that a ruler which is one mile long in a frame at rest with respect to the ruler, will be shorter in a frame in which that ruler is moving. The length the ruler will be in a frame in which it is moving is given by:

[tex] L = L_0 \sqrt{1-v^2/c^2} [/tex]



I have copies of Lorentz’s 1895 and 1904 relativity papers. He introduced his Lorentz Transformation in 1895 as a way to try to explain the results of the Michelson Morley experiment.

He specifically said in his 1895 paper:

Lorentz said in 1895:

”Thus one would have to imagine that the motion of a solid body (such as a brass rod or the stone disc employed in the later experiments) though the resting ether exerts upon the dimensions of that body an influence which varies according to the orientation of the body with respect to the direction of motion.”

Just “relative motion” can not “exert an influence” on a body in any manner.

Sorry, but rulers don’t “contract” just due to “relative motion”, and Einstein agreed in 1907.

A ruler in “relative motion” to some distant ruler, doesn’t know it’s “moving”, and thus feels no force compelling it to shrink. You need to study more about the various relativity theories.

Originally posted by StarThrower
So as you can see, if the ruler is accelerating in some inertial frame, the closer its speed gets to c...

You can’t jump from a “relatively moving” ruler to an “accelerating ruler” just like that. Those are two different things.

And there is no such thing as something “accelerating in some inertial frame”.
 
  • #30
Originally posted by StarThrower
But clearly, by the relativity postulate, a photon cannot be at rest in an inertial reference frame, since its speed must be equal to c in an inertial reference frame.

I've generally found that nature doesn't always obey the ancient postulates of mortal men.

Actually, many astronomers today say that distant 1-c galaxies are moving away from the Earth at the relative speed of 1-c, and, therefore, when they emit a photon aimed in our direction, that photon is moving at “0” velocity in our direction. The photon later speeds up relative to us. See the Davis-Lineweaver paper to learn about this new theory.

It’s time to move on in physics and astronomy and move into the 21st Century.
 
  • #31
Tempest

You can let your atom's frame be absolute relative to the ether. Then galilean transformations hold, and time dilation effects in that frame are zero.

When you put yourself in a moving frame F3 (moving at 1 m/s) then you must take into account a small time dilation effect, but it can be ignored and the galilean transformation still holds (on the assumption sr is wrong).

In the photon frame the effect of time dilation is infinite and so an observer cannot make any measurements.

Wisp theory http://www.kevin.harkess.btinternet.co.uk has transformations (sections 7.11 and 7.11.1) that might help you out. It looks at this from both absolute and relative frames of reference. The theory does not use the Lorentz - FitzGerald contraction; it treats events as having absolute simultaneity, and treats distances as invariant.

The theory does not support sr's claim that the speed of light is constant, although it does show that much of the predictions of sr are correct.

Your ideas will lead you to thinking that the speed of light cannot be constant and you will find it near impossible to convince anyone that this is so.
 
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  • #32


Originally posted by chroot

In other words, when you're going c, everything in the universe is zero distance from you.

- Warren
It quite amazes me.If every thing is zero distant from the photon when it travels at c then if it has to go anywhere then it's already there even without moving.
It's not all, but then photon is already every where.

Does it confirms that the light transmition is instaneous from the photon's point of view?

My purpose is a good debate. I'll jepardise my membership for it if I have to!
 
  • #33


Originally posted by David
I have copies of Lorentz’s 1895 and 1904 relativity papers. He introduced his Lorentz Transformation in 1895 as a way to try to explain the results of the Michelson Morley experiment.

He specifically said in his 1895 paper:



Just “relative motion” can not “exert an influence” on a body in any manner.

Sorry, but rulers don’t “contract” just due to “relative motion”, and Einstein agreed in 1907.

A ruler in “relative motion” to some distant ruler, doesn’t know it’s “moving”, and thus feels no force compelling it to shrink. You need to study more about the various relativity theories.



You can’t jump from a “relatively moving” ruler to an “accelerating ruler” just like that. Those are two different things.

And there is no such thing as something “accelerating in some inertial frame”.

David, the ether concept was discarded, and the transformations are now interpreted as transformation of coordinates from one reference frame to another. It is completely irrelevant what their formulator intended them to mean. The mathematics drive the conclusions. That math discards the notion of ether.

What Lorentz felt was responsible for the shrinking is totally irrelevant to the mathematics of relativity. The transformations are coordinate transformations. You can derive the formula for length contraction and time dilation from using a very simple argument, which will prove to you that the length of a moving ruler shrinks, because it is in motion in some inertial reference frame.

No jump is made, you perform the necessary mathematics. You differentiate velocity with respect to time.

Acceleration = dv/dt. Nothing to it.

And of course there is such a thing as accelerating in an inertial reference frame.

An inertial reference frame is a reference frame in which all three of Newton's laws are satisfied. One of those laws is Newton's third law, which states that

F12 = -F21

The symbol F is for force, and force is equal to mass times acceleration. Hence, objects in inertial reference frames can accelerate in those frames. You are making a conceptual error.
 
Last edited:
  • #34
Originally posted by David
I've generally found that nature doesn't always obey the ancient postulates of mortal men.

Actually, many astronomers today say that distant 1-c galaxies are moving away from the Earth at the relative speed of 1-c, and, therefore, when they emit a photon aimed in our direction, that photon is moving at “0” velocity in our direction. The photon later speeds up relative to us. See the Davis-Lineweaver paper to learn about this new theory.

It’s time to move on in physics and astronomy and move into the 21st Century.

David,

You left out the remainder of the quote:

"So if the theory of relativity doesn't contradict, the only possible conclusion is that this frame isn't an inertial reference frame."

The logic in that thread was flawless, there was nothing to question at all.
 
  • #35


Originally posted by StarThrower
from using a very simple argument, which will prove to you that the length of a moving ruler shrinks, because it is in motion in some inertial reference frame.

Einstein basically retracted his “the moving ruler shrinks” hypothesis in 1907, when he said:

“The shape of a body in the sense indicated we will call its ‘geometrical shape’. The latter obviously does not depend on the state of motion of a reference frame.”

Just “relative motion” alone can not cause any ruler in the universe to “shrink”. Some type of “force” or changing force must be placed upon the ruler to get it to shrink. However, I know of quite a lot of people who do believe a “relatively moving” ruler “shrinks”, because they’ve heard this rumor so many times.
 
<h2>1. What is "Special Relativity Contains Massive Error"?</h2><p>"Special Relativity Contains Massive Error" is a claim made by some scientists that the theory of special relativity, proposed by Albert Einstein, contains a fundamental mistake or error that invalidates the entire theory.</p><h2>2. What is special relativity and why is it important?</h2><p>Special relativity is a theory that explains the relationship between space and time, and how they are affected by the motion of objects. It is important because it has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds, which is crucial for our understanding of the universe.</p><h2>3. What evidence supports the claim that special relativity contains an error?</h2><p>There is currently no scientific evidence that supports the claim that special relativity contains a massive error. The theory has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds. Any claims of an error in the theory would need to be backed up by solid evidence and rigorous scientific testing.</p><h2>4. Who is making the claim that special relativity contains an error?</h2><p>The claim that special relativity contains a massive error is made by a small group of scientists who have not been able to provide any substantial evidence to support their claim. The majority of the scientific community continues to accept and use special relativity as a valid and accurate theory.</p><h2>5. How does the scientific community respond to the claim that special relativity contains an error?</h2><p>The scientific community responds to this claim with skepticism and requires rigorous evidence to support any new theories or claims. The majority of scientists continue to use and accept special relativity as a valid theory, as it has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds.</p>

1. What is "Special Relativity Contains Massive Error"?

"Special Relativity Contains Massive Error" is a claim made by some scientists that the theory of special relativity, proposed by Albert Einstein, contains a fundamental mistake or error that invalidates the entire theory.

2. What is special relativity and why is it important?

Special relativity is a theory that explains the relationship between space and time, and how they are affected by the motion of objects. It is important because it has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds, which is crucial for our understanding of the universe.

3. What evidence supports the claim that special relativity contains an error?

There is currently no scientific evidence that supports the claim that special relativity contains a massive error. The theory has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds. Any claims of an error in the theory would need to be backed up by solid evidence and rigorous scientific testing.

4. Who is making the claim that special relativity contains an error?

The claim that special relativity contains a massive error is made by a small group of scientists who have not been able to provide any substantial evidence to support their claim. The majority of the scientific community continues to accept and use special relativity as a valid and accurate theory.

5. How does the scientific community respond to the claim that special relativity contains an error?

The scientific community responds to this claim with skepticism and requires rigorous evidence to support any new theories or claims. The majority of scientists continue to use and accept special relativity as a valid theory, as it has been extensively tested and has been shown to accurately predict the behavior of objects at high speeds.

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