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Thread: A Flaw of General Relativity, a Fix, and Cosmological...

  1. #1 A Flaw of General Relativity, a Fix, and Cosmological... 
    Forum Freshman Zanket's Avatar
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    Abstract: A flaw of general relativity is exposed and is shown to source from a misapplication of the equivalence principle, the theory’s core postulate. A replacement for the Schwarzschild metric is simply derived. (The vast majority of experimental tests of general relativity have been tests of the Schwarzschild metric.) The new metric is shown to be confirmed to all significant digits by experiments of the three classical tests of general relativity. The predictions of the new metric are shown to diverge from those of the Schwarzschild metric as gravity strengthens. The cosmological implications explain some observations simpler than do alternative explanations.

    A Flaw of General Relativity, a Fix, and Cosmological Implications


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    What?


    "Consider the daffodil. And while your doing that, I'll be over here, going through your stuff."
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    Nobody here is willing to challenge my claim?
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    Let two relativistic rockets approach each other. Thanks to length contraction, in the respective crews’ frames the other rocket can approach at an effective velocity arbitrarily greater than c, shrinking the distance between them arbitrarily fast while they move at an actual velocity less than c relative to each other. The equivalence principle lets the crews suppose that the rockets stand unpowered on the respective surfaces of two objects free-falling toward each other in a uniform gravitational field. Any gravitational field is locally uniform. Then in the respective frames of two objects free-falling toward each other, the distance between them can shrink arbitrarily fast while they move at less than c relative to each other.
    Thanks to length contraction... I have a problem with this, zanket.
    'Length's' don't contract, time dilates. You state that relativistic rockets
    can shrink the distance between them. Let's do an example.

    To make the example easier to calculate, let's assume the rockets are
    two light seconds apart and have a relative velocity of .866c according
    to a stationary observer at infinity. Thats 599,584,916 meters apart according to the observer at infinity. Now, what do you claim the crews
    of the relativistic rockets will measure the 'contracted' distance as? You
    state earlier that the meter 'shrinks'. The two crews would still measure
    599,584,916 of the shorter meters between them, correct? If the crews
    transverse this distance in ONE of the observer-at-infinity's seconds, their
    calculated velocity will be 1.732c. If the crews transverse the same
    599,584,916 short meters in their own frame, how long will it take according to their clocks? Now consider the example with dilated clocks
    instead of contracted meters.
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    Quote Originally Posted by 2inquisitive
    'Length's' don't contract, time dilates.
    Lengths do contract. Length contraction and time dilation go hand-in-hand in special relativity.

    The two crews would still measure 599,584,916 of the shorter meters between them, correct?
    That the meters are shorter for them (to use your way of putting it) is length contraction.
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    take a close look at the Michelson-Morley experiment and Feynman's Six Not So Easy Pieces
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    Quote Originally Posted by Zanket
    Nobody here is willing to challenge my claim?
    Your claim has already been disproven. Third post on this page. I recommend you read VERY CAREFULLY this time.
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    Quote Originally Posted by Aer
    Your claim has already been disproven. Third post on this page. I recommend you read VERY CAREFULLY this time.
    All your issues were subsequently resolved in my favor. Raising an issue does not win an argument.
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    Quote Originally Posted by Zanket
    Quote Originally Posted by 2inquisitive
    'Length's' don't contract, time dilates.
    Lengths do contract. Length contraction and time dilation go hand-in-hand in special relativity.

    The two crews would still measure 599,584,916 of the shorter meters between them, correct?
    That the meters are shorter for them (to use your way of putting it) is length contraction.
    "The two crews would still measure
    599,584,916 of the shorter meters between them, correct? If the crews
    transverse this distance in ONE of the observer-at-infinity's seconds, their
    calculated velocity will be 1.732c. If the crews transverse the same
    599,584,916 short meters in their own frame, how long will it take according to their clocks? "
    You never did answer this question, zanket. How much time will tick off
    on the crew's clock in their rest frame while transversing the 599,584,916
    meters? I will show you the only correct way for the crew to arrive at the
    destination in a relatively quicker time.
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    Quote Originally Posted by Zanket
    All your issues were subsequently resolved in my favor. Raising an issue does not win an argument.
    Hah! None of the issues have been "resolved" in your favor. I've said all along that you are using Newton's equation for velocity when the Special Relativity equation for velocity is required. You cannot support your claim that you do not use Newton's equation for velocity as shown here

    Stop spreading your filth on the internet.
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  12. #11  
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    Quote Originally Posted by 2inquisitive
    "The two crews would still measure
    599,584,916 of the shorter meters between them, correct? If the crews
    transverse this distance in ONE of the observer-at-infinity's seconds, their
    calculated velocity will be 1.732c. If the crews transverse the same
    599,584,916 short meters in their own frame, how long will it take according to their clocks? "
    You never did answer this question, zanket. How much time will tick off
    on the crew's clock in their rest frame while transversing the 599,584,916
    meters? I will show you the only correct way for the crew to arrive at the
    destination in a relatively quicker time.
    Regardless of your personal opinion on what physics "actually says", Special relativity requires both time dilation and length contraction.
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    Quote Originally Posted by 2inquisitive
    You never did answer this question, zanket.
    OK, what you challenge in the paper is just basic SR length contraction. I’ll go through your example:

    Quote Originally Posted by 2inquisitive
    To make the example easier to calculate, let's assume the rockets are
    two light seconds apart and have a relative velocity of .866c according
    to a stationary observer at infinity. Thats 599,584,916 meters apart according to the observer at infinity.
    This is confusing. When you say “let’s assume the rockets are two light seconds apart,” I don’t know in what frame that is until you say “Thats 599,584,916 meters apart according to the observer at infinity”. And “the observer at infinity” is misleading. This observer need not be at infinity, or a proxy for that, as is often the case in examples involving gravity. This observer can be any third observer. And why translate “two light seconds” as “599,584,916 meters”? Why not just stick with “two light seconds”? That’s easier.

    To make the example easier to calculate, let's assume the rockets are two light seconds apart as observed by their shared gantry (they launched from the same gantry) and have a relative velocity of 0.866c according to the gantry.

    Quote Originally Posted by 2inquisitive
    Now, what do you claim the crews
    of the relativistic rockets will measure the 'contracted' distance as?
    A specific value cannot be determined, because it’s indeterminate what each rocket’s velocity is relative to the gantry. One rocket could be stationary with respect to the gantry (it came to a full stop relative to the gantry, and now is “accelerating” at a = 0) and the other rocket could be moving at 0.866c relative to them. Or each rocket could be moving at 0.866c / 2 relative to the gantry. You need to specify more info to make the question answerable. The best that can be done is to state how they’d calculate it, as I do in my next paragraph:

    Quote Originally Posted by 2inquisitive
    You
    state earlier that the meter 'shrinks'. The two crews would still measure
    599,584,916 of the shorter meters between them, correct?
    Yes, they would respectively measure two light seconds contracted to a percentage given by eq. 9.13, where v is their velocity relative to gantry. Were the distance marked out by metersticks that each measure one meter of length in the gantry’s frame, they’d measure the same number of metersticks, each contracted in length to a percentage given by eq. 9.13.

    Quote Originally Posted by 2inquisitive
    If the crews
    transverse this distance in ONE of the observer-at-infinity's seconds, their
    calculated velocity will be 1.732c.
    Their velocity relative to the gantry, or the other rocket? You have three observers here, you need to be more specific. In any case, velocity is always less than c in relativity theory—there is no mathematical way to get a velocity >= c. And the velocity of the rockets relative to each other in the gantry’s frame is a given, so you’re starting a new example here; you can’t calculate a velocity that’s already been given. So the example becomes too confusing to work with at this point. I have to stop.

    If you can fix the example as I noted, I’ll go through it again, if you are not trying to disprove SR here—if that’s what you’re doing, you should do that in another thread. My paper assumes that SR is valid.

    If you agree that the crew can traverse a galaxy in their lifetime without exceeding c relative to the galaxy, as SR says, then you implicitly agree to length contraction, for there is no way that thousands of light years of distance can be crossed in a lifetime at less than c. The distance traversed must be less in light years than the maximum years of a human lifetime.

    Quote Originally Posted by Aer
    Stop spreading your filth on the internet.
    Now that's irony!
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    Quote Originally Posted by Zanket
    If you agree that the crew can traverse a galaxy in their lifetime ....
    I, for one, would not agree to that.

    Where is the evidence that this can be done?
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    Quote Originally Posted by Guitarist
    Quote Originally Posted by Zanket
    If you agree that the crew can traverse a galaxy in their lifetime ....
    I, for one, would not agree to that.

    Where is the evidence that this can be done?
    Books on SR explain this. So does The Relativistic Rocket. The evidence is the experimental confirmation of SR.
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    If the galaxy is 100,000ly across, and you could travel at 0.9999999c (that's seven 9's) your shipboard time would indicate something on the order of 45 years.

    According to SR and the Lorentz transforms.
    Huh?
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    Quote Originally Posted by superluminal
    If the galaxy is 100,000ly across, and you could travel at 0.9999999c (that's seven 9's) your shipboard time would indicate something on the order of 45 years.
    Check this out – As noted in section 2, that’s an effective velocity of (100000 / 45) = 2222c. Plug 2222 into eq. 2.2 returns (2222 / sqrt(1 + 2222^2)) = 0.9999999c.

    Or, plug 0.9999999c into eq. 2.1, to get an effective velocity of (0.9999999 / sqrt(1 – 0.9999999^2)) = 2236c. Then (100000 / 2236) = 45 years.
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    Amazing how that works, isn't it?
    Huh?
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    Quote Originally Posted by superluminal
    Amazing how that works, isn't it?
    So, are you taking a jab at zanket's meaningless "effective velocity" or are you agreeing with the conclusions of his paper - notably, -General Relativity is wrong!-
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    Quote Originally Posted by superluminal
    Amazing how that works, isn't it?
    The cool thing is, the book Exploring Black Holes has the example that converts to years, but goes through something like six steps, using some approximations.
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    Aer:

    So, are you taking a jab at zanket's meaningless "effective velocity" or are you agreeing with the conclusions of his paper - notably, -General Relativity is wrong!-
    Not agreeing or disagreeing with anything! Just noting how you can arrive at the same answer in multiple ways.

    I told MacM and others that I'm not debating the correctness/incorrectness of relativity anymore. I'll poke into things to help clarify the way SR or GR works (if I can!).

    Is his 'effective velocity' thingy meaningless? I haven't looked at it in any detail.

    P.S. IMHO relativity is completely correct to the limits of current measurement technology, based on hundreds of experimental results in both SR and GR. Therefore, by default, I think everyones proofs to the contrary are wrong. Publish in Phys. Letters or Phys. Rev. D and get tons of experimental support for your proof's, then we'll talk... :wink:
    Huh?
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  22. #21  
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    Quote Originally Posted by Zanket
    Quote Originally Posted by Aer
    Stop spreading your filth on the internet.
    Now that's irony!
    You thought I might not see this since you hid it in a response to 2inquisitive - sorry pal.

    Why did you not respond to the point of my thread? Here it is again so you can have a second go at it:

    Quote Originally Posted by Aer
    I've said all along that you are using Newton's equation for velocity when the Special Relativity equation for velocity is required. You cannot support your claim that you do not use Newton's equation for velocity as shown here
    Now as sciforums is down, I'll reproduce it here:

    Quote Originally Posted by Zanket on sciforums
    The derivation is right in the paper so I elaborate only on the pertinent part here: Eq. 2.4 is a relativistic equation, derived by rearranging the relativistic rocket equations.
    veff = a * t [2.4] note:derived from relativity equations

    Quote Originally Posted by Zanket on sciforums
    It shows that Newtonian velocity (v in Newton’s equations) represents effective velocity, not actual velocity.
    v = a * t [2.3] note:Newton's equation for velocity

    Quote Originally Posted by Zanket on sciforums
    Then v in eq. 3.1 must be veff, as it is in eq. 3.2, the one you post above.
    v = sqrt(R / r) [3.1] note:this is a Newton equation

    veff = sqrt(R / r) [3.2]

    Quote Originally Posted by Zanket on sciforums
    See, no invalid equation was used. I refer to eq. 2.3,

    v = a * t [2.3]

    Quote Originally Posted by Zanket on sciforums
    but only to note how it could be changed to become a valid relativistic equation;
    Ha ha! Was it really that hard to show you used Newton's equation. Look at equation 2.3 - that is Newton's equation for velocity. Let me repeat. Your derivation as you stated above is as follows mathematically (which I note, is a lot simplier than your explainations above).

    veff = a * t ... (your relativity equation 2.4)
    veff = v ... (from Newton's equation, v = a * t - equation 2.3)
    veff = sqrt(R / r) (from Newton's equation v = sqrt(R / r) - equation 3.1)
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    Quote Originally Posted by superluminal
    Is his 'effective velocity' thingy meaningless? I haven't looked at it in any detail.
    As a velocity, it is meaningless. He is taking the length of the universe from one frame (let's call it the rest frame) and the time duration from another frame (let's call it the moving frame) and dividing the two. That is veff = rest-length / time-dilated. If you think that is a velocity, you are terribly mistaken. But as you have pointed out, it explains how someone will travel very vast distances in one frame, but only age slightly - but that is not a velocity - you may as well call it an aging factor.
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    Zanket,

    To avoid problems, why are you using any newtonian form of equation for velocity or anything? Einstein works at all velocities. Why not just use the big E from the get go? Or am I missing something?
    Huh?
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    If you think that is a velocity, you are terribly mistaken
    I certainly do not.

    Again, I don't know why you would not just use the Lorentz transforms in the correct frames and work in the relativistic world (the real world) from the start.
    Huh?
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    Quote Originally Posted by superluminal
    Is his 'effective velocity' thingy meaningless? I haven't looked at it in any detail.
    You can see for yourself, in the paper, that it’s just a different way of expressing an actual velocity. An effective velocity equals a specific actual velocity. One benefit of it, as demonstrated above, is that it makes story problems easier.

    Quote Originally Posted by Aer
    You thought I might not see this since you hid it in a response to 2inquisitive - sorry pal.
    The mod here does not want consecutive posts by the same poster.

    Quote Originally Posted by Aer
    Here it is again so you can have a second go at it:
    Asked and answered. You’re a heckler, as anyone can see. So long as the points you make are spurious, I will ignore you and let others draw their own conclusions.
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    The mod here does not want consecutive posts by the same poster.

    Oops! Did I do a bad thing?
    Huh?
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    Quote Originally Posted by superluminal
    Again, I don't know why you would not just use the Lorentz transforms in the correct frames and work in the relativistic world (the real world) from the start.
    The entire premise of his paper as Pete pointed out quite awhile ago is to take an equation from Newton and apply it in Relativity. I know this is retarded which is why I keep telling zanket his paper is retarded as well.
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    Quote Originally Posted by superluminal
    To avoid problems, why are you using any newtonian form of equation for velocity or anything? Einstein works at all velocities. Why not just use the big E from the get go? Or am I missing something?
    I can't be expected to respond your questions regarding Aer's interpretation of the paper, right? It seems that's what you're doing here. Why not go direct to the source and frame your question from that?
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    Ahh well. You see why I stopped debating the validity of arguments against relativity?
    Huh?
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    Quote Originally Posted by Zanket
    The mod here does not want consecutive posts by the same poster.
    Perhaps they don't want abusive consecutive posts which makes sense, but merely replying to different people is not abusive.

    Quote Originally Posted by Zanket
    Asked and answered. You’re a heckler, as anyone can see. So long as the points you make are spurious, I will ignore you and let others draw their own conclusions.
    You think you answered the question? No one supports you, even Pete has given up on your position.

    Here is the last response you gave to the point I made against your paper:

    Quote Originally Posted by Zanket on sciforums
    You are claiming here that simply refering to an invalid equation, without using it in a way that depends on it being valid, is invalid.
    And my answer was: Your statement "You are claiming here that simply refering to an invalid equation, without using it in a way that depends on it being valid, is invalid. " is so asinine that it deserves repeating. Read it! I never said any such thing - I said you are using an invalid equation, not "refering" to it, what is the point of "refering" to an equation in a derivation without "using" it? Are you having trouble with such simple definitions for English words?

    Now let's analyze this world you are in which you say, you can derive something by refering to an invalid equation without using it in anyway that requires it to be valid:

    I am going to derive for you the escape velocity. For brevity, I am not going to explain variables that are already commonly defined - look them up for yourself.

    A given object has a potential energy, P, defined as:

    P = GmM/r

    Moving with respect to the body it wishes to escape from, it has a kinetic energy, K, defined as:

    K = 1/2 m v^2

    From Aer's repertoire of randomly generated equations:

    T = (pi - 2.8)^m / M

    I refer to the above equation for T for the mere purpose of saying it is silly and in order to get the escape velocity, Vesc, the kinetic energy must balance the potential energy:

    P = K

    GmM/r = 1/2 m Vesc^2

    Therefore,

    Vesc = √ (2MG/r)

    Thus I derived the escape velocity with the added reference to my equation for T which, I might point out, is an invalid equation for temperature.
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    Zanket:

    I can't be expected to respond your questions regarding Aer's interpretation of the paper, right? It seems that's what you're doing here. Why not go direct to the source and frame your question from that?
    Sorry. You are correct.

    Again, I'm not interested in analyzing arguments against relativity. You may be completely correct. GR could be flawed. I'd love to read about it in SA. My only interest is in doing my best to see that relativity, as it is currently formulated, and given my limited understanding, is not abused too badly.
    Huh?
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    Quote Originally Posted by superluminal
    Ahh well. You see why I stopped debating the validity of arguments against relativity?
    I avoid that too, for SR anyway!

    FWIW, I’m not using Newton’s equations in a way that depends on their validity. In fact, I’m depending upon the invalidity of those equations. Aer is basically saying that even this is disallowed. His argument against the “effective velocity” coined in the paper amounts to an argument against algebraic substitution.

    Quote Originally Posted by superluminal
    Again, I'm not interested in analyzing arguments against relativity. You may be completely correct. GR could be flawed. I'd love to read about it in SA.
    What’s SA?
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    Sorry. Scientific American.
    Huh?
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    Zanket,

    If you could post a condensed, simple version of the sequence of algebraic operations in question, I could look at it if you wish. I'm not being lazy, it's just that I have a hard time pulling out the relevant information from detailed papers and descriptions.

    If you don't want to, that's fine. No problem.
    Huh?
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    Quote Originally Posted by superluminal
    Sorry. Scientific American.
    You certainly won’t see it there, unless maybe Hawking comes up with it! The current state of science is that almost nobody wants to hear an argument against relativity—they’ve heard it all before, and I don’t blame them. Few would read past the title of a paper like this, unless it was authored by Hawking or maybe one of a few others. The best I can hope for is to get the paper vetted by forumites. But there we have Aer... here he comes to pounce on this comment!
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  37. #36  
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    Quote Originally Posted by Zanket
    FWIW, I’m not using Newton’s equations in a way that depends on their validity.
    Wrong.

    Quote Originally Posted by Zanket
    In fact, I’m depending upon the invalidity of those equations.
    Here is your derivation:

    veff = a * t [2.4]

    "Eq. 2.4 is a relativistic equation, derived by rearranging the relativistic rocket equations"

    v = a * t [2.3]

    "It shows that Newtonian velocity [equation 2.3] (v in Newton’s equations) represents effective velocity, not actual velocity."

    v = sqrt(R / r) [3.1]

    "Then v in eq. 3.1 must be veff, as it is in eq. 3.2, the one you post above."

    veff = sqrt(R / r) [3.2]

    All sentences in quotes are exact quotes from zanket himself.

    Answer this: HOW DO YOU GO FROM EQUAITON 2.4 TO EQUATION 3.2 WITHOUT USING EQUATION 2.3? You must be forgetting that v = a * t is an approximation to the velocity that special relativity predicts. Your veff is not an approximation to the velocity that special relativity predicts. In fact for v = .999999999c your veff goes to infinity.



    Quote Originally Posted by Zanket
    Aer is basically saying that even this is disallowed.
    If by "this" you mean using Newton's equation v = a * t, then yes, I say "this" is not allowed.


    Quote Originally Posted by Zanket
    His argument against the “effective velocity” coined in the paper amounts to an argument against algebraic substitution.
    My argument was and always has been that it is not a "velocity". Even superluminal pointed that out to you.

    Quote Originally Posted by superluminal
    Zanket,

    If you could post a condensed, simple version of the sequence of algebraic operations in question, I could look at it if you wish. I'm not being lazy, it's just that I have a hard time pulling out the relevant information from detailed papers and descriptions.
    There they are, up there 8)

    Quote Originally Posted by Zanket
    You certainly won’t see it there, unless maybe Hawking comes up with it! The current state of science is that almost nobody wants to hear an argument against relativity—they’ve heard it all before, and I don’t blame them. Few would read past the title of a paper like this, unless it was authored by Hawking or maybe one of a few others. The best I can hope for is to get the paper vetted by forumites. But there we have Aer... here he comes to pounce on this comment!
    Damn right! You failed to comprehend what superluminal told you. He wants to see experimental evidence, not your nonsensical brand of physics.
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    Zanket,

    Maybe this has been asked before, but is there an experimental descrepancy that prompted you to write this paper?

    Usually, if theory is being validated by experiment, scientists seek to test it to ever finer accuracy, for therein lie the seeds of new physics. But without an obvious experimental descrepancy, posing a new theory is pointless.

    This is why scientists will not read past the title of certain papers. If the paper is claiming to fix a problem that dosen't exist, why should they?
    Huh?
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    Quote Originally Posted by superluminal
    This is why scientists will not read past the title of certain papers. If the paper is claiming to fix a problem that dosen't exist, why should they?
    Zanket went to go hide for awhile, he usually comes back after he's had a little bit of time to think in solitude - thereby regaining his disillusionment.
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    If you both agree that this is the sticking point:

    Here is your derivation:

    veff = a * t [2.4]

    "Eq. 2.4 is a relativistic equation, derived by rearranging the relativistic rocket equations"
    I find this a curious statement.

    If you accelerate for 100yrs at 1g then, by relativity, your velocity may be on the order of 0.9999c.

    If you use this equation, your velocity will be something like 103.0c!

    So using the results from eq 2.4, even by substitution, will lead to misleading results.

    A proof of this would be to, instead of substituting algebraically and calculating the final result (eq 3.2?), use real values and calculate at each explicit step and see if the results make sense at each step. All equations in a derivation must be valid. At any step you should be able to calculate a valid intermediate result.

    103.0c is not a valid result in this universe and would seem to invalidate the entire chain of reasoning.

    What have I missed?
    Huh?
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    So, Aer. If this is what you are arguing, then I must agree with you.
    Huh?
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    Quote Originally Posted by superluminal
    What have I missed?
    What you have missed is that v<sub>eff</sub> is not an actual velocity as Aer misrepresents it to be. See below.

    Quote Originally Posted by superluminal
    If you could post a condensed, simple version of the sequence of algebraic operations in question, I could look at it if you wish.
    As simple as I think the paper already is, I’ll take you up on that.

    I define:

    v<sub>eff</sub> ≡ v / sqrt(1 - v^2) [eq. 2.1]

    I call v<sub>eff</sub> “effective velocity” in the paper, taking care to distinguish it from an actual velocity denoted by v. Then I show that, by rearranging the equations at The Relativistic Rocket (which are also in my section 9), that:

    v<sub>eff</sub> = a * t [eq. 2.4]

    (This is a relativistic equation, no Newton used yet.) Eq. 2.4 can then of course be expanded to:

    v / sqrt(1 - v^2) = a * t

    Again, you can derive this just by rearranging the relativistic rocket equations.

    Now, Newton’s equation is:

    v = a * t [eq. 2.3 - Newton]

    Here the variables a and t have the same meaning as for the relativistic rocket equations. (The relativistic equations require acceleration to be measured more strictly than Newtonian mechanics does, but in an otherwise compatible way.) Notice that the right-hand side of eq. 2.4 is the same as the right-hand side of eq. 2.3. Eq. 2.4 is a valid, relativistic equation. Then a * t returns not actual velocity, as Newton thought, but rather effective velocity, or v / sqrt(1 - v^2) if you like.

    The paper carries on from here, but this is a good place to stop, for by this point Aer would be having a tizzy fit. He thinks the “effective velocity” is somehow bogus, even though it’s just a descriptive name for a new symbol, and he thinks that this analysis uses Newton’s equation (eq. 2.3) invalidly. Notice that I did not even need to mention Newton. I could just have said that a * t returns effective velocity, or v / sqrt(1 - v^2) if you like.

    Do you see anything wrong with that?
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    Ok Zanket. Let me see...
    Huh?
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    Quote Originally Posted by superluminal
    Ok Zanket. Let me see...
    I wish I could be here to watch this play out Unfortunately I won't be back until tomorrow.
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    Quote Originally Posted by superluminal
    Maybe this has been asked before, but is there an experimental descrepancy that prompted you to write this paper?
    No.

    Usually, if theory is being validated by experiment, scientists seek to test it to ever finer accuracy, for therein lie the seeds of new physics. But without an obvious experimental descrepancy, posing a new theory is pointless.
    A discrepancy need not be revealed by the experiments. It can be revealed in the theory itself. The paper shows a flaw of GR, then fixes it. The paper shows that GR uses its equivalence principle inconsistently. And regardless, an alternate theory that matches the generally accepted theory for experiments done to date is not pointless--it's a theory with equal stature.

    This is why scientists will not read past the title of certain papers. If the paper is claiming to fix a problem that dosen't exist, why should they?
    That’s not the case here. The title starts with “A Flaw of General Relativity, a Fix...” It is clear that the paper purports to show a problem and then fix the problem. Still, they would not get past the first five words.
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    Zanket:

    I define:

    veff == v / sqrt(1 - v^2) [eq. 2.1]

    I call veff “effective velocity” in the paper, taking care to distinguish it...
    Ok. The first problem I see is that your definition is invalid.

    For unit analysis, if v is in terms of m/s (c), then you have:

    veff = c/sqrt(1-c^2) which yields c/c (or m/s / m/s) which is unitless. You may say it is 'not a real velocity' but the units still have to be consistent.

    And v * gamma in relativity is meaningless since it is selfreferential, i.e. gamma is itself dependent on v.

    Further:

    v / sqrt(1 - v^2) = a * t

    is setting a unitless quantity equal to a value in m/s (or 'c') and is meaningless.
    Huh?
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    Quote Originally Posted by superluminal
    Ok. The first problem I see is that your definition is invalid.

    For unit analysis, if v is in terms of m/s (c), then you have:

    veff = c/sqrt(1-c^2) which yields c/c (or m/s / m/s) which is unitless. You may say it is 'not a real velocity' but the units still have to be consistent.
    I thought we covered this already on the other forum. I didn’t copy my units into the post. Can you look at the paper for that? v is unitless in the paper, and in the other forum I showed that this is okay.

    From Exploring Black Holes:
    Measure distance s and time lapse t in the same unit. For example, a spaceship travels half a light-year of distance during one year of time; its speed is then 0.5 year / year and the units cancel. As another example, if an elementary particle moves 0.7 meter in one meter of light-travel time its speed is 0.7. Hence the speed v has no units. In this book the symbol v represents the speed of an object as a fraction of the speed of light.
    Then v<sub>eff</sub> is also unitless.

    I have to go for now. I’ll respond to the rest later.
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    Zanket:

    and in the other forum I showed that this is okay.
    I was probably not paying close attention at that point.

    Anyway, how can:

    v / sqrt(1 - v^2) = a * t

    be 'okay' ???

    You can't set unitless definitions equal to unit-rich entities! They can be included in an equation:

    a1 * t1 * K = a2 * t2

    (K = v / sqrt(1 - v^2) ... a unitless definition) and you still have m/s = m/s.

    How about 4.56 = a*t? Nonsense, right? Units on both sides of an EQUATION must always match! Basic algebra. Yes?

    4.56 m/s = a * t is fine!

    But 3.6c/c = 3.6 = a*t is not!

    The results of a * t must be a real velocity!
    Huh?
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    Quote Originally Posted by superluminal
    Anyway, how can:

    v / sqrt(1 - v^2) = a * t

    be 'okay' ???
    Both sides of the equation are unitless. The quote above from the book explains that units of ly / yr are unitless. In units of ly (light years) for distance and yr (years) for time, the variable a is in units of ly / yr^2 and t is in units of yr. Then the right-hand side is (ly / yr^2) * yr = ly / yr, unitless, the same as the left-hand side. Both sides return the same fraction of c, where c = 1. (And c is also unitless when set = 1.)

    Quote Originally Posted by superluminal
    And v * gamma in relativity is meaningless since it is selfreferential, i.e. gamma is itself dependent on v.
    It is equivalent to a * t. If it is meaningless, then so is a * t, right? The equation above is derived from a manipulation of the relativistic rocket equations, hence valid, assuming my derivation (in section 10.1) is valid. And note how I used it in your story problem above. How did you calculate the result as 45 years? Was it as easy as I did it? Section 2 in the paper explains its meaning. But you need not subscribe to that—you need not believe it has meaning. It need not have meaning to be valid, just as one side of any equation in a manipulation of equations need not have meaning.
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    The quote above from the book explains that units of ly / yr are unitless
    I can't find this on the website. And I can't see how that's possible.

    1ly = 9.4605284 × 10^15 meters

    yr = years

    1 ly/yr = c = 9.4605284 × 10^15 meters / year.

    I'm just trying to use basic mathematical reasoning here.
    Huh?
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    Both sides of the equation are unitless.
    Wait.

    Do I read a * t as "acceleration time time"? In other words m/s<sup>2</sup> x s = m/s

    How is this unitless?
    Huh?
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    Quote Originally Posted by superluminal
    I can't find this on the website.
    The quote is above in my Wed Jul 27, 2005 3:58 pm post.

    And I can't see how that's possible.
    For more info you can also google for “no units” and “speed of light” (those phrases in double quotes).
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    Quote Originally Posted by superluminal
    Both sides of the equation are unitless.
    Wait.

    Do I read a * t as "acceleration time time"? In other words m/s<sup>2</sup> x s = m/s

    How is this unitless?
    It's unitless when the units are ly / yr, or some other units in which c = 1.

    Think of it this way: v = 0.5c is unambiguous, despite no units.
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    Zanket,

    I freely admit to bieng lost. Could you pleas show me the unit analysis that demonstrates this to be unitless?

    v / sqrt(1 - v<sup>2</sup>) = a * t

    The left is clearly unitless. My analysis reveals that if you use ly/yr on the right you get:

    ly/yr<sup>2</sup> * yr = ly/yr, a real velocity.

    Help.

    What unit do you choose for a and t such that they cancel to yield a unitless quantity?
    Huh?
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    Quote Originally Posted by superluminal
    Could you pleas show me the unit analysis that demonstrates this to be unitless?
    Our posts sailed past each other. Look above, especially at the quote from the book in my Wed Jul 27, 2005 3:58 pm post.
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    Zanket

    Look above, especially at the quote from the book in my Wed Jul 27, 2005 3:58 pm post.
    Not trying to be difficult, but for the life of me, I look in the website (The Relativistic Rocket, right?), and I can't find the quote. And I don't see it in your post.

    Could you just cut and paste it please?
    Huh?
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    Sorry: I'm on US eastern time. Ignore the last few posts.

    I'm studying this quote.
    Huh?
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    Quote Originally Posted by superluminal
    Sorry: I'm on US eastern time.
    Isn't PST the only time zone? :-D
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    Quote:
    From Exploring Black Holes:

    Measure distance s and time lapse t in the same unit. For example, a spaceship travels half a light-year of distance during one year of time; its speed is then 0.5 year / year and the units cancel. As another example, if an elementary particle moves 0.7 meter in one meter of light-travel time its speed is 0.7. Hence the speed v has no units. In this book the symbol v represents the speed of an object as a fraction of the speed of light.
    I'm sorry but this reads as nonsense to me. I know very well that Taylor and Wheeler are renouned physicists.

    Is there some preceeding context missing from this quote?

    The statement:

    "Measure distance s and time lapse t in the same unit" alone is gibberish to me. Is it not to you also?

    "a spaceship travels half a light-year of distance during one year of time;"

    Yields for me, 0.5ly/yr NOT 0.5yr/yr????!!!!

    A ly is a pure unit of distance. It can be interpreted in no other way.

    "a spaceship travels half a bobalink of distance during one year of time;"

    Yields for me, 0.5bobalinks/yr

    If I said a train travels 10m during 1 second of time, how would you translate that?

    10m/s.

    I'm totally confused.
    Huh?
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    Isn't PST the only time zone?
    Oh yeah! Forgot. Sorry.
    Huh?
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    Quote Originally Posted by superluminal
    I'm sorry but this reads as nonsense to me. I know very well that Taylor and Wheeler are renouned physicists.

    Is there some preceeding context missing from this quote?
    Alas, there is not. I wish I could explain it better. It has to do with a light year, say, being a year of light travel time. To me it makes sense because, v = 0.5c, say, is unambiguous, despite no units. Units can’t matter if it’s unambiguous. I refer you to the google search on [“no units” “speed of light”] where you can find links that explain this further.

    I’ve gotten used to working with these units, no units that is, and the equations do work out correctly. On the other forum I showed that whether I used the units in the paper or m / s, I got equivalent answers. And using the units in the paper is easier. For example, it allows me to strip out c from the relativistic rocket equations, since c = 1.
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    Well I understand v = 0.5c very well. The units are 'c'. I googled on your suggestions but didn't get much of use. Maybe I'll try later.

    My head hurts.
    Huh?
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    Ah HA!

    http://quantumrelativity.calsci.com/.../Chapter3.html

    Excerpt:

    From now on, we'll agree that we're going to use the same units for time and space. For example, as we've already seen, one foot of distance equals one nanosecond at the speed of light, so if we say a foot of time we mean the same thing as if we say a nanosecond. A meter of time is about 3 nanoseconds. Velocity, distance per time, is now meters per meter, so velocity has no dimensions. The speed of light is now just 1 with no units.
    My head still hurts, but I feel better.
    Huh?
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    Sorry I didn't see this sooner, I could have saved you a headache. Check wikipedia under "geometrized units". Fairly standard pratice for GR.
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    In your paper, section 9:

    γ = sqrt(1 + (a * t)^2) [9.10]


    The Lorentz factor. A time dilation or length contraction factor. Ranges between unity and a limit of infinity:

    γ = 1 / sqrt(1 - v^2)

    Are these gammas supposed to be equivalent?
    Huh?
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    Thanks kevinalm. Yea, I could have used that!
    Huh?
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    Quote Originally Posted by Zanket
    This is why scientists will not read past the title of certain papers. If the paper is claiming to fix a problem that dosen't exist, why should they?
    ... Still, they would not get past the first five words.
    Actually they would read the first five words and already be able to disprove your theory.

    Quote Originally Posted by superluminal
    Quote Originally Posted by Zanket
    I define:

    veff == v / sqrt(1 - v^2) [eq. 2.1]

    I call veff “effective velocity” in the paper, taking care to distinguish it...
    Ok. The first problem I see is that your definition is invalid.

    Zanket took c out of every one of his equations which is why they sometimes appear retarded. He made the requirement that c=1 - so sometimes v is v and othertimes v is v/c but the c just isn't written because it is necessarily 1. This is bad practice for any "generalized theory" but whatever, I resolved this issue on sciforums - though I still wouldn't recommend this use but zanket can't see why it is retarded so I'll leave it at that. Most notable reason why it is retarded is because you are restricted to using units of ly/yr or light-second/second or light-decade/decade etc. meters/second would be an invalid unit in these equations.

    Quote Originally Posted by Zanket
    Quote Originally Posted by superluminal
    Anyway, how can:

    v / sqrt(1 - v^2) = a * t

    be 'okay' ???
    Both sides of the equation are unitless. The quote above from the book explains that units of ly / yr are unitless.
    !!! This is most certainly a retardation of whatever book your are refering to is saying. ly / yr is unitless. Wow! ly is a unit of length, yr is a unit of time.

    Quote Originally Posted by Zanket
    In units of ly (light years) for distance and yr (years) for time, the variable a is in units of ly / yr^2 and t is in units of yr. Then the right-hand side is (ly / yr^2) * yr = ly / yr, unitless, the same as the left-hand side.
    There is no hope for you.

    Quote Originally Posted by Zanket
    Both sides return the same fraction of c, where c = 1. (And c is also unitless when set = 1.)
    I realize you are using "geometric units" (i.e. every time interval is interpreted as the distance travelled by light during that given time interval) but that does not make anything unitless.

    It would be the same if I required c = 299,792,458. I could say that it could be in units of "m/s" or "kilo-meter/kilo-seconds" or "mega-meter/mega-seconds". That doesn't make c "unitless". They probably refer to c as a unitless value that can have any range of units as shown above and you obvious make a retardation out of this statement by saying v, veff, c and whatever else is "unitless".

    Quote Originally Posted by Zanket
    Quote Originally Posted by superluminal
    Both sides of the equation are unitless.
    Wait.

    Do I read a * t as "acceleration time time"? In other words m/s<sup>2</sup> x s = m/s

    How is this unitless?
    It's unitless when the units are ly / yr, or some other units in which c = 1.

    Think of it this way: v = 0.5c is unambiguous, despite no units.
    Regardless of the fact that you set c=1, there are units imbedded in c whether it be a "ly/yr" or "light-second/second" or whatever you may what to use.


    Quote Originally Posted by Zanket
    Alas, there is not. I wish I could explain it better.
    Your problem is you are explaining it wrong.





    Quote Originally Posted by Zanket
    It has to do with a light year, say, being a year of light travel time. To me it makes sense because, v = 0.5c, say, is unambiguous, despite no units.
    There you go again, spouting out nonsense. The units of v are imbedded in c. Just because you see no units doesn't mean there are no units. Please try the following: place your hands over your eyes. Is your monitor still there?

    Quote Originally Posted by Zanket
    I’ve gotten used to working with these units, no units that is,
    Yet you still have no idea what you are talking about. Impressive feat of ignorance.



    Quote Originally Posted by superluminal
    Well I understand v = 0.5c very well. The units are 'c'.
    You do not know the relief you have just brought to me. Thanks superluminal - finally an intelligent statement.
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    Aer,

    I have great difficulty following Zanket's math. My previous post shows two instances of a definiton for gamma on the same page. Do you understand this? The same variable is used (γ) with two equations that are not equivalent.

    EDIT: Ok. I see that one is for a rocket with constant positive acceleration, and the other is for uniform motion.
    Huh?
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    Ok, since all that has been discussed are issues of semantics, let's move on to the real issue:

    Quote Originally Posted by Zanket
    Quote Originally Posted by superluminal
    What have I missed?
    What you have missed is that v<sub>eff</sub> is not an actual velocity as Aer misrepresents it to be. See below.
    I was the one who pointed out to you that "effective velocity" was a misnomer because it is not a velocity (read: actual velocity as you would think of it in your language). It may be more appriopriately labeled an "aging factor" as I alluded to earlier.


    Quote Originally Posted by Zanket
    Quote Originally Posted by superluminal
    If you could post a condensed, simple version of the sequence of algebraic operations in question, I could look at it if you wish.
    As simple as I think the paper already is, I’ll take you up on that.

    I define:

    v<sub>eff</sub> ≡ v / sqrt(1 - v^2) [eq. 2.1]

    I call v<sub>eff</sub> “effective velocity” in the paper, taking care to distinguish it from an actual velocity denoted by v. Then I show that, by rearranging the equations at The Relativistic Rocket (which are also in my section 9), that:

    v<sub>eff</sub> = a * t [eq. 2.4]

    (This is a relativistic equation, no Newton used yet.) Eq. 2.4 can then of course be expanded to:

    v / sqrt(1 - v^2) = a * t

    Again, you can derive this just by rearranging the relativistic rocket equations.
    Everything is ok so far, but note, c=1 so that is why you don't see v / sqrt(1 - (v/c)^2)


    Quote Originally Posted by Zanket
    Now, Newton’s equation is:

    v = a * t [eq. 2.3 - Newton]

    Here the variables a and t have the same meaning as for the relativistic rocket equations. (The relativistic equations require acceleration to be measured more strictly than Newtonian mechanics does, but in an otherwise compatible way.) Notice that the right-hand side of eq. 2.4 is the same as the right-hand side of eq. 2.3. Eq. 2.4 is a valid, relativistic equation. Then a * t returns not actual velocity, as Newton thought, but rather effective velocity, or v / sqrt(1 - v^2) if you like.
    This explaination shows lack of intelligence.

    "measured more strictly than Newtonian mechanics does" Do you wish to elaborate?

    "Notice that the right-hand side of eq. 2.4 is the same as the right-hand side of eq. 2.3" - Yep, they sure are, so now you are going to tell me that v = veff. Wait a minute - that is too stupid. So you are going to beat around the bush but say the exact same thing.

    "Then a * t returns not actual velocity, as Newton thought, but rather effective velocity" - Yep I was right. The only problem is, you actually believe this is not the same as the stupid statement above.


    Quote Originally Posted by Zanket
    The paper carries on from here, but this is a good place to stop, for by this point Aer would be having a tizzy fit.
    And appropriately so!


    Quote Originally Posted by Zanket
    He thinks the “effective velocity” is somehow bogus,
    In what context? As a velocity, "effective velocity" is bogus - yes.


    Quote Originally Posted by Zanket
    even though it’s just a descriptive name for a new symbol,
    Not velocity I might add.

    Quote Originally Posted by Zanket
    and he thinks that this analysis uses Newton’s equation (eq. 2.3) invalidly.
    Why even bring up Newton's equation if you don't use it!

    Quote Originally Posted by Zanket
    Notice that I did not even need to mention Newton. I could just have said that a * t returns effective velocity, or v / sqrt(1 - v^2) if you like.
    how do you explain your equation: veff = sqrt(R/r) - you cannot get this without assuming a * t returns actual velocity as well.



    Quote Originally Posted by Zanket
    Do you see anything wrong with that?
    Yes.
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    v / sqrt(1 - v^2) = a * t
    How does one evaluate this expression?

    If I try to solve for a I get:

    a = v / sqrt(1 - v^2) / t

    The top portion is unitless. t has some units. If I divide UNITLESS by t (in units of seconds say) I get a = 1/s. If a is an acceleration it must be the derivative wrt time of velocity (dv/dt).

    Unless someone can explain how [v / sqrt(1 - v^2) / t] is a rate of change of velocity wrt time, i.e. dist/time<sup>2</sup> (and remember, the top is unitless), then I am lost.

    Does this make any sense to anyone?
    Huh?
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    Quote Originally Posted by superluminal

    v / sqrt(1 - v^2) = a * t
    How does one evaluate this expression?

    If I try to solve for a I get:

    a = v / sqrt(1 - v^2) / t

    The top portion is unitless. t has some units. If I divide UNITLESS by t (in units of seconds say) I get a = 1/s. If a is an acceleration it must be the derivative wrt time of velocity (dv/dt).

    Unless someone can explain how [v / sqrt(1 - v^2) / t] is a rate of change of velocity wrt time, i.e. dist/time<sup>2</sup> (and remember, the top is unitless), then I am lost.

    Does this make any sense to anyone?
    Did you see my post where I said:

    Zanket took c out of every one of his equations which is why they sometimes appear retarded. He made the requirement that c=1 - so sometimes v is v and othertimes v is v/c but the c just isn't written because it is necessarily 1. This is bad practice for any "generalized theory" but whatever, I resolved this issue on sciforums - though I still wouldn't recommend this use but zanket can't see why it is retarded so I'll leave it at that. Most notable reason why it is retarded is because you are restricted to using units of ly/yr or light-second/second or light-decade/decade etc. meters/second would be an invalid unit in these equations.


    Basically you have to guess when v is v and when v is v/c. Fun, isn't it?
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    But even if v is v or v/c what sense does that expression make? Does he not say that a and t are acceleration and time?
    Huh?
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    Quote Originally Posted by superluminal
    But even if v is v or v/c what sense does that expression make? Does he not say that a and t are acceleration and time?
    I'll rewrite the equation in terms of unit quantities:

    a = v / sqrt(1 - v^2) / t

    becomes:

    (ly/yr^2) = (ly/yr) / sqrt(1 - (unitless)^2) / yr

    ly/yr^2 = ly/yr^2

    So as you can see, the first v is v, while the second v is v/c.
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    One thing is for sure. Zanket should rewrite his paper to make the equations more conventional otherwise he could have discovered the secret to Life, The Universe, and Everything, and we won't be able to figure it out.
    Huh?
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    Well shit! That's completely useless! Two different v's? They need to be different variables!

    That makes absolute hash out of everything if true.
    Huh?
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    Quote Originally Posted by superluminal
    Well shit! That's completely useless! Two different v's? They need to be different variables!

    That makes absolute hash out of everything if true.
    He defends it by requiring c=1 which I argued with him to no end so I just gave up and pointed to the bigger flaws.
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    Well that's just arbitrarily changing the nature of the whole expression. And since this relationship is fundamental to the whole thing, it just blows up from there. No wonder I was confused.

    He needs to rewrite the paper without geometrized entities.
    Huh?
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    Geometrized unit system
    Quote Originally Posted by wikipedia
    In geometric units, every time interval is interpreted as the distance travelled by light during that given time interval. That is, one second is interpreted as one light second, so time has the geometric units of length. This is dimensionally consistent with the notion that, according to the kinematical laws of special relativity, time and distance are on an equal footing.
    According to this, it really is valid for velocity to be unitless.
    Acceleration would have units of L<sup>-1</sup> (or T<sup>-1</sup>, since L and T are equivalent in this system).

    Pretty freaky, but workable I think.

    But, it gets even freakier:
    Quote Originally Posted by wikipedia
    Energy and momentum are interpreted as components of the four-momentum vector, and mass is the magnitude of this vector, so in geometric units these must all have the dimension of length. We can convert a mass expressed in kilograms to the equivalent mass expressed in meters by multiplying by the conversion factor G/c2. For example, the Sun's mass of 2.0×1030 kg in SI units is equivalent to 1.5 km. This is half the Schwarzschild radius of a one solar mass black hole. All other conversion factors can be worked out by combining these two.
    So the only unit in the geometrized unit system is powers of L!

    It immediately follows in this system that c really is identical to 1 - no units, and that that v/c and v are identical and unitless.
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    But this does not mean you can arbitrarily assign things. Acceleration still has to have a derivative relationship with velocity wrt time (t<sup>-2</sup>) right?

    Edit: nevermind. I see your point.
    Huh?
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    Quote Originally Posted by Pete
    Geometrized unit system
    Quote Originally Posted by wikipedia
    In geometric units, every time interval is interpreted as the distance travelled by light during that given time interval. That is, one second is interpreted as one light second, so time has the geometric units of length. This is dimensionally consistent with the notion that, according to the kinematical laws of special relativity, time and distance are on an equal footing.
    According to this, it really is valid for velocity to be unitless.
    Acceleration would have units of L<sup>-1</sup> (or T<sup>-1</sup>, since L and T are equivalent in this system).

    Pretty freaky, but workable I think.

    But, it gets even freakier:
    Quote Originally Posted by wikipedia
    Energy and momentum are interpreted as components of the four-momentum vector, and mass is the magnitude of this vector, so in geometric units these must all have the dimension of length. We can convert a mass expressed in kilograms to the equivalent mass expressed in meters by multiplying by the conversion factor G/c2. For example, the Sun's mass of 2.0×1030 kg in SI units is equivalent to 1.5 km. This is half the Schwarzschild radius of a one solar mass black hole. All other conversion factors can be worked out by combining these two.
    So the only unit in the geometrized unit system is powers of L!
    Not that freaky, it all depends on the speed of light as it is constant in all inertial reference frames, they are just replacing "light-year" with "year" and placing some significance to it. Then miniscule brains like zanket read it and think that velocity is unitless. Please don't tell me what non-dimensionalizing is - only a non-dimensional number is unitless, any experience in the real world (i.e. engineering would probably help you understand the difference).
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    Quote Originally Posted by superluminal
    But this does not mean you can arbitrarily assign things. Acceleration still has to have a derivative relationship with velocity wrt time (t<sup>-2</sup>) right?
    Exactly, and it is all dependent on the consistency of the speed of light. It is a stupid definition that has no practicality.
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    He needs to rewrite the paper without geometrized entities.
    I don't know if he needs to, but...

    zanket, do you think that you could rework your paper in the standard LMT unit system? Would it make a difference to the conclusions?

    superluminal / Aer,
    Are you able to work through zanket's paper using the geometrized unit system? If not, why not?

    It looks like this entire thread to date has been obscured by this misunderstanding. Perhaps you need to start all over again!
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    Seems to me that this:

    Aer:

    I'll rewrite the equation in terms of unit quantities:

    a = v / sqrt(1 - v^2) / t

    becomes:

    (ly/yr^2) = (ly/yr) / sqrt(1 - (unitless)^2) / yr

    ly/yr^2 = ly/yr^2

    So as you can see, the first v is v, while the second v is v/c.
    Is a big problem.
    Huh?
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    Quote Originally Posted by Pete
    superluminal / Aer,
    Are you able to work through zanket's paper using the geometrized unit system? If not, why not?
    Yes. You need only refer to my analysis.

    Quote Originally Posted by Pete
    It looks like this entire thread to date has been obscured by this misunderstanding. Perhaps you need to start all over again!
    I admit that I did not realized zanket's intention at first (back on sciforums), but this had all been resolved quite awhile ago.
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    Quote Originally Posted by Aer
    ...they are just replacing "light-year" with "year" and placing some significance to it. Then miniscule brains like zanket read it and think that velocity is unitless.
    In such a system, velocity is unitless! It's a shame some miniscule brains can't find a way out of their comfort zone, or back down from their posture of intellectual arrogance.
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    Pete:

    Are you able to work through zanket's paper using the geometrized unit system? If not, why not?
    I am an engineer. Without my familiar units, I am lost, hence the horrific misunderstandigs with this whole thing. I could struggle through it, but I'm not going to.
    Huh?
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    Quote Originally Posted by Pete
    Quote Originally Posted by Aer
    ...they are just replacing "light-year" with "year" and placing some significance to it. Then miniscule brains like zanket read it and think that velocity is unitless.
    In such a system, velocity is unitless! It's a shame some miniscule brains can't find a way out of their comfort zone, or back down from their posture of intellectual arrogance.
    You are forgetting that unit of velocity, as defined as "unitless", is dependent on c = 1 unit-length/unit-time and that c is still in the equation for v (i.e. v = .9c).
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    Quote Originally Posted by superluminal
    Pete:

    Are you able to work through zanket's paper using the geometrized unit system? If not, why not?
    I am an engineer. Without my familiar units, I am lost, hence the horrific misunderstandigs with this whole thing. I could struggle through it, but I'm not going to.
    Ahh, yes - a lot of engineers never in fact do work with non-dimensional entities.
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    Quote Originally Posted by superluminal
    I am an engineer. Without my familiar units, I am lost
    Fair enough.
    I suspect that zanket might not have the patience (or the understanding?) to rework in a standard system... but it would be good exercise for him if he does.
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    My new vehicle design will have a top speed of 48.6 I told my boss, and could accelerate fron 0 to 60 in 320.

    He fired me.
    Huh?
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    Quote Originally Posted by Pete
    I suspect that zanket might not have the patience (or the understanding?) to rework in a standard system... but it would be good exercise for him if he does.
    Hopefully he does, then it will be easier for everybody to see he is a fool.
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    Quote Originally Posted by Aer
    You are forgetting that unit of velocity is dependent on c = 1 unit-length/unit-time and that c is still in the equation for v (i.e. v = .9c).
    The basis of the system is that length and time are geometrically equivalent.
    In this system:
    • The units of velocity are Length/Length.
    • c = 1 light-second / 1 light-second
      = 1
    • v = 0.9c is an identical expression to v=0.9


    You seem to imply that a geometrized unit system is not a valid system?
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    Quote Originally Posted by Aer
    Hopefully he does, then it will be easier for everybody to see he is a fool.
    See, this is why you piss me off, Aer - you're an arrogant arsehole.

    You've blundered right through the thread (both threads!) lambasting zanket for something that you just didn't understand, always working under the assumption that you know everything and that anyone who disagrees with you is a fool.

    In any intellectual discussion, I really think that one should always keep ones own fallibility in mind.

    Whether zanket is right or not is beside the point - for you to continually treat him in this insulting manner is not appropriate.
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    Quote Originally Posted by Pete
    You seem to imply that a geometrized unit system is not a valid system?
    Nope, just that it depends on the consistency of the speed of light. Then all velocities must be expressed in terms of this "speed of light". A velocity of 4380 m/s has no meaning until you convert it to 4380/299,792,458 * c.
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    Quote Originally Posted by superluminal
    My new vehicle design will have a top speed of 48.6 I told my boss, and could accelerate fron 0 to 60 in 320.

    He fired me.
    Beautiful!
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    Ha! We engineers are nothing if not practical...
    Huh?
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    Quote Originally Posted by Pete
    See, this is why you piss me off, Aer - you're an arrogant arsehole.
    Perhaps if the both of you wouldn't go off on non-important discussions and only discuss the disproof I brought up - perhaps I wouldn't be so inclined to think very little of you.

    Quote Originally Posted by Pete
    You've blundered right through the thread (both threads!) lambasting zanket for something that you just didn't understand, always working under the assumption that you know everything and that anyone who disagrees with you is a fool.
    There is nothing I didn't understand - there was only confusion because zanket wasn't expressing what he was doing properly, nevertheless - all those issues have been resolved. The only remaining issues have to do with his derivation procedure.

    Quote Originally Posted by Pete
    In any intellectual discussion, I really think that one should always keep ones own fallibility in mind.
    I started out discussing intellectually with zanket but he has proven to not be able to discuss on the same level.

    Quote Originally Posted by Pete
    Whether zanket is right or not is beside the point - for you to continually treat him in this insulting manner is not appropriate.
    I was insulted by zanket's total dismissal of my arguments without showing where my arguments were false.
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    Quote Originally Posted by Aer
    Nope, just that it depends on the consistency of the speed of light.
    I don't know if it does, really... perhaps indirectly.
    It does depend on Lorentz transforms, or some similar metric that contains some inherently defined speed.

    But accepting that it does depend on the frame independence of light speed, is it a problem? Could you not accept it as a valid premise for zanket's paper?
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    Quote Originally Posted by Pete
    I don't know if it does, really... perhaps indirectly.
    What is a "light year"?

    Quote Originally Posted by Pete
    But accepting that it does depend on the frame independence of light speed, is it a problem? Could you not accept it as a valid premise for zanket's paper?
    I already told you this is not a problem, for me - however, it makes understanding the equations a lot more difficult for other people.
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    Quote Originally Posted by Aer
    Perhaps if the both of you wouldn't go off on non-important discussions and only discuss the disproof I brought up - perhaps I wouldn't be so inclined to think very little of you.
    That might be the most absurdly narcissistic way of judging a person's worth I've heard.
    People discuss what they want, threads go where they will. Sometimes, people don't want to discuss what you want to. Get over it!

    Quote Originally Posted by Pete
    There is nothing I didn't understand
    Except the apparent validity of a geometrized system?

    Quote Originally Posted by Pete
    I started out discussing intellectually with zanket but he has proven to not be able to discuss on the same level.
    That happens, and it's OK - that's when you leave the thread, and find a worthy opponent. Hanging around beating your head against a wall and throwing insults around is... kind of immature?

    Quote Originally Posted by Aer
    I was insulted by zanket's total dismissal of my arguments without showing where my arguments were false.
    There are two possible reasons...
    1) zanket saw that you misunderstood his premises, and therefore (rightly or wrongly) dissmissed your argument. Unfortunate (if he's wrong), but understandable. The second rule of discussion is to establish common ground, rather than simply assume that everyone is working with the same set of definitions (and unit systems) as yourself.
    2) zanket is blockheaded. If that's the case, ignore him. The first rule of discussion is "never argue with a fool - they'll drag you down to their level and beat you with experience".

    I think I've said enough... probably too much.
    I apologize for lambasting you like this, and I'll allow you to close this sub-debate.
    I particular apologize for calling you an arrogant arsehole. That was uncalled for, and I withdraw it.

    Pete
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    Quote Originally Posted by Aer
    I already told you this is not a problem, for me - however, it makes understanding the equations a lot more difficult for other people.
    So it's correct but obscure?
    Is that what you're saying?
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