Since: Mar 12
Dubai, UAE

humble brother wrote: The gravity model F=(M+m)²/r² has a nonlinear relationship between force and calculable mass. It works perfectly for modeling gravity but the concept of mass is just different than the one you are used to. Your model is nonsense before we even try to test it against empirical reality because its units do not stack up. The left hand side of your equation is force. Force is: F = mass * distance * time^2 Your right hand side = mass^2 * distance^2 m * d * t^2 =/= m^2 * d^2 Your right hand side lacks any time units for starters. Your equation would only make sense...if you add a constant that balances up the units being used. You need to add: d^3 * m^1 * time^2 We will call this constant a Humble (H) in your honor. Or will we? Its the same units as G! And funnily enough, your equation will finally make sense as a coherent hypothesis but still won't match reality. Not only do you fail in your quest to demolish G, the only real difference left is your prediction that force will vary according to (M + m)^2, all else being equal, whereas Newton predicts it will vary by Mm, all else being equal. Since Newton's predictions have held up as close to reality under intense scrutiny and your prediction is significantly different even after I fix your equation (and made no sense at all before)... Your hypothesis (even after I fix your schoolboy unit errors), is falsified.

humble brother
Vanda, Finland

Chimney1 wrote: <quoted text> Your model is nonsense before we even try to test it against empirical reality because its units do not stack up. The left hand side of your equation is force. Force is: F = mass * distance * time^2 Your right hand side = mass^2 * distance^2 m * d * t^2 =/= m^2 * d^2 Your right hand side lacks any time units for starters. Your equation would only make sense...if you add a constant that balances up the units being used. You need to add: d^3 * m^1 * time^2 We will call this constant a Humble (H) in your honor. Or will we? Its the same units as G! And funnily enough, your equation will finally make sense as a coherent hypothesis but still won't match reality. Not only do you fail in your quest to demolish G, the only real difference left is your prediction that force will vary according to (M + m)^2, all else being equal, whereas Newton predicts it will vary by Mm, all else being equal. Since Newton's predictions have held up as close to reality under intense scrutiny and your prediction is significantly different even after I fix your equation (and made no sense at all before)... Your hypothesis (even after I fix your schoolboy unit errors), is falsified. Oh you came back to show your faith. This was already covered a long time ago but I can see that you understood none of it. From the new model you can directly derive the relation between mass and and rotational speed: m = M²/(v²r  M) From here you can already see what the concept of mass is within the new model. You poor little thing didn't understand the consequences of (v²r  M). Within the new model relative mass is also a function of relative rotational speed. You should go back to school to try and understand that you don't falsify mathematical models with units. Thank you for the laughs though.

Since: Mar 12
Dubai, UAE

humble brother wrote: <quoted text> Oh you came back to show your faith. This was already covered a long time ago but I can see that you understood none of it. From the new model you can directly derive the relation between mass and and rotational speed: m = M²/(v²r  M) From here you can already see what the concept of mass is within the new model. You poor little thing didn't understand the consequences of (v²r  M). Within the new model relative mass is also a function of relative rotational speed. You should go back to school to try and understand that you don't falsify mathematical models with units. Thank you for the laughs though. Yeah, I do not spend seven days a week on topix. Go figure. However, in spite of your feeble attempt at condescension, anybody can see that yet again, you have failed to align the units, which must be consistent on both sides of the equation. This is elementary stuff. You think its unnecessary? In pure maths, 5 + 3 = 8. But in pure maths, 5x + 3y =/= 8 anything The algebraic statement 5x + 3y is NOT reducible to a single variable. In physics, every variable is a quantity of a specific unit. 3 kilograms, 3 seconds, 3 somethings. Thus 3d, 3m, 3t. These are not purely numerical statements, they are algebraic of the form number, unit. 5x, not just "5". You cannot add 5 seconds to 3 kilograms. Any more than, mathematically, you can add 5x to 3y and get "8" of anything. You cannot do maths, and you certainly cannot do physics, with any competence. And its laughable that you try to redefine what "mass" is to suit your very silly nonunitaligned equations. You already have a mislignment in your m = M²/(v²r  M) nonsense, as mass on one side includes a velocity and distance component, and lacks it on the other. So you have two entirely different qualities you are trying to call mass.

Since: Mar 12
Dubai, UAE

Velocity, of course, being a time & distance component when broken down into fundamentals.
So its time, distance and distance again that you are misaligning.

humble brother
Vanda, Finland

Chimney1 wrote: Yeah, I do not spend seven days a week on topix. Go figure. However, in spite of your feeble attempt at condescension, anybody can see that yet again, you have failed to align the units, which must be consistent on both sides of the equation. This is elementary stuff. You think its unnecessary? In pure maths, 5 + 3 = 8. But in pure maths, 5x + 3y =/= 8 anything The algebraic statement 5x + 3y is NOT reducible to a single variable. In physics, every variable is a quantity of a specific unit. 3 kilograms, 3 seconds, 3 somethings. Thus 3d, 3m, 3t. These are not purely numerical statements, they are algebraic of the form number, unit. 5x, not just "5". You cannot add 5 seconds to 3 kilograms. Any more than, mathematically, you can add 5x to 3y and get "8" of anything. You cannot do maths, and you certainly cannot do physics, with any competence. And its laughable that you try to redefine what "mass" is to suit your very silly nonunitaligned equations. You already have a mislignment in your m = M²/(v²r  M) nonsense, as mass on one side includes a velocity and distance component, and lacks it on the other. So you have two entirely different qualities you are trying to call mass. It is obvious that you have absolutely no understanding of the nonlinear nature of relative mass in the new model. One kilogram is a name given to a lump of a chosen size. Then mathematically it is possible to call two of those lumps two kilograms. You can as well stop using the name kilogram and choose another name, you can start using cowabungaa as the name and choose some lump to be the fixed measure. Then you can perhaps say that one cowabungaa is 0.346333 times the lump called "kilogram". Above you can see that the unit of mass is not a kilogram or anything else that is used your your magic model. The unit of mass in the new model is m³/s². Mass in the new model should be called "mass effect" rather than just "mass".

Since: Mar 12
Dubai, UAE

humble brother wrote: It is obvious that you have absolutely no understanding of the nonlinear nature of relative mass in the new model. Actually its perfectly clear that your incoherent drivel is tripping you up. Whatever it is you are trying to define, its certainly not mass. One kilogram is a name given to a lump of a chosen size. Then mathematically it is possible to call two of those lumps two kilograms. Yes, and if we are adding 5kg to 3kg, we get 8kg. 5x + 3x = 8x WOW! However, you still CANNOT add 5 kg to 3 seconds. Like 5x + 3y, they are irreducible beyond that composite statement. And in a statement of equalities  an equation  the units must equate, as I said. If there is an expression of time on one side, there must be an expression of time on the other. The unit of mass in the new model is m³/s². Mass in the new model should be called "mass effect" rather than just "mass". Now you accept that its NOT mass you are talking about but a new derivative quality, a function of mass and time. And now "mass effect" = mass cubed over time squared. Mass is still mass, then, the same old mass, an actual, non relative quantity just like it always was, over there on the right hand side of your "mass effect" equation, the actual quantity you have decided to cube in inventing your new, utterly useless thingamejig called "mass effect". No way around it. To fix your equation, you have to add back in exactly the same units that comprise G: d^3 * m^1 * t^2 Then we just change your (M + m)^2 back to Mm, and hey presto, its Newton again and it actually works. You doofus!

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> The confusion is all in your mind. <quoted text> You are contradicting yourself still. You say here that the photons from the beacon can move faster than the speed c relative to the observer ship. So the relative speed c can then be exceeded by photons after all? So it is not a constant? No, I did NOT say that. The c+v is in the beacon's frame, not the observer's frame. The distance between the wavefront of light and the observer moving away in the beacon's frame increases at a rate of c+v, but that is not a velocity of any object nor a relative velocity in any frame. You are saying that the front beacon sees light move faster than c relative to the observer ship and therefore the number of encountered wavefronts is increased. The beacon sees the distance between the wavefront and the observer increase at a rate of c+v. But that is NOT a velocity in any frame. In particular, it is NOT the relative velocity between ligth and the observer *in the observer's frame*. Does the observer ship see light coming from front faster than the constant c? Answer: NO they do not. Right. In the observer's frame, the speed of light is always c. Once more: we know that within the model observers in different rest frames disagree on measurements. So why do you bring measurements that the two frames disagree with into one calculation? They disagree on distances, velocities, time durations, energies, momenta, and a great number of other things. But they agree on things like rest mass, proper time of an event (time experienced by someone in the frame of the event), and phase of a wave. So a count of the number of wavefronts passing the observer is independent of the reference frame. The Doppler effect compares two frames: that of the beacon and that of the observer. So of course we have to look at the system in both frames and compare them! You have drowned yourself in fallacies in your paradox. You know that the two frames disagree with the measurements and still you bring them together to calculate. They agree on number of wavefronts passing the observer. You only can say within the model that from the beacon it seems that light seemingly travels faster than the constant c relative to the ship No, the c+v is NOT the velocity of anything in any frame. And yet you have counted the encountered wavefronts for a faster than light relative speed. That's exactly what you are using in your calculations to count the wavefronts:  faster than constant light speed for the front: c+v  slower than constant light speed for the back: cv Because you change the speed like that you get different counts of wavefronts. The speed doesn't change. We are merely calculating the distance between the wavefront and the observer in the beacon's frame. The count is the same in both frames; the speed of light is the same in both frames; so the frequency differs between the frames. I must ask you again: Are you lying through your teeth to try and confuse other people here??? No, I am attempting to show you where you are making your mistakes so, perhaps, you will learn something. Clearly that isn't happening. Let's focus on the c+v. Is it the speed of light in any reference frame? Is it the speed of anything in any reference frame? What, exactly, does it represent?

humble brother
Vanda, Finland

Judged:
1
1
Chimney1 wrote: <quoted text> Actually its perfectly clear that your incoherent drivel is tripping you up. Whatever it is you are trying to define, its certainly not mass. <quoted text> Yes, and if we are adding 5kg to 3kg, we get 8kg. 5x + 3x = 8x WOW! However, you still CANNOT add 5 kg to 3 seconds. Like 5x + 3y, they are irreducible beyond that composite statement. And in a statement of equalities  an equation  the units must equate, as I said. If there is an expression of time on one side, there must be an expression of time on the other. <quoted text> Now you accept that its NOT mass you are talking about but a new derivative quality, a function of mass and time. And now "mass effect" = mass cubed over time squared. Mass is still mass, then, the same old mass, an actual, non relative quantity just like it always was, over there on the right hand side of your "mass effect" equation, the actual quantity you have decided to cube in inventing your new, utterly useless thingamejig called "mass effect". No way around it. To fix your equation, you have to add back in exactly the same units that comprise G: d^3 * m^1 * t^2 Then we just change your (M + m)^2 back to Mm, and hey presto, its Newton again and it actually works. You doofus! I have lost count of how many times I've said to you that the mass in the current gravity models is not compatible with the mass in the new model. I clearly said to you that the new concept is nonlinear. Do you at all read what you respond to??? You are constantly trying to put your primitive linear concept of mass over the new model and then start whining like a little baby when it does not work. OF COURSE THE NEWTONIAN GRAVITY MASS IS NOT COMPATIBLE WITH THE NEW CONCEPT. Suddenly you act all surprised now? You should have understood this a long time ago but I can see that you were too busy with your fingers in your ears going "lalalaa". You don't understand a word you read if its outside your existing belief system, do you.. I won't explain this to you anymore, if you still keep whining about the models not being compatible then I'll just leave you whining like a baby.


humble brother
Vanda, Finland

Judged:
1
1
polymath257 wrote: No, the c+v is NOT the velocity of anything in any frame. c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t. All of you babbling above is pure nonsense. This is what you are doing: 1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed. 2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame. You just pick and choose measurements from different frames and mix them together in a soup. polymath257 wrote: Let's focus on the c+v. Is it the speed of light in any reference frame? Is it the speed of anything in any reference frame? What, exactly, does it represent? c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits. In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it.

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t. All of you babbling above is pure nonsense. This is what you are doing: 1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed. 2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame. You just pick and choose measurements from different frames and mix them together in a soup. <quoted text> c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits. In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it. OK, let's try it another way. Suppose I send a light signal off to the right and the observer goes off to the left at .5*c in the beacon's frame. After one second, how far apart are the light signal and the observer in the beacon's frame?

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> c+v is the relative speed from which you calculate how many wavefronts the observer supposedly encounters during a specific time period t. All of you babbling above is pure nonsense. This is what you are doing: 1. From the beacon's frame you pick the relative speed v=(Vo + c) between the observer and the wavefronts. You pick this speed to calculate in the observer's frame even though the observer disagrees with this speed. No, I am calculating in the beacon's frame. I am finding the number of wavefronts that pass the observer. 2. You then use this speed inside the observer's frame and using the observer's time you magically calculate a new frequency for the light in the observer's frame. No, I use the number of wavefronts that pass the observer to calculate the frequency. You just pick and choose measurements from different frames and mix them together in a soup. You clearly don't understand the calculation and are not even trying to do so. c+v is exactly what would be used to calculate standard doppler effect for light dismissing the relative speed limits. Why is this used? Because it tells the distance between the wavefront and the observer so we can find how many wavefronts have passed the observer. In your case it is a combination of measured values that the observers do not agree on. Your case is a logical fallacy and you choose to ignore it. The number of wavefronts that pass the observer is not dependent on the frame. At this point, we have two methods for finding the frequency in the observer's frame given the frequency in the beacon's frame: 1) Use the equation describing the light wave as a function of time and distance in the beacon's frame and use a Lorenz transform to find its description in the observer's frame. 2) Count wavefronts passing the observer in the beacons frame and take into account time dilation when figuring the frequency in the observer's frame. Both methods give the same result. Both methods give results that agree with experiments. Both results keep the two frames separated and are based on a relativistic model. YOU seem to be the only one that doesn't understand what is being said. Whether this is purposeful or not I cannot say, but clearly you do not want to learn what is wrong with your understanding of this argument.

humble brother
Vanda, Finland

polymath257 wrote: OK, let's try it another way. Suppose I send a light signal off to the right and the observer goes off to the left at .5*c in the beacon's frame. After one second, how far apart are the light signal and the observer in the beacon's frame? This is where the model says they disagree. In the beacon's frame the beam has made it one light second away and the ship a half light second to the other way. In the ship's frame a lot less time has passed and that beam hasn't got nearly as far.

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> This is where the model says they disagree. In the beacon's frame the beam has made it one light second away and the ship a half light second to the other way. So how far apart are they in the beacon's frame? In the ship's frame a lot less time has passed and that beam hasn't got nearly as far. To be dealt with later...answer the above question first.

humble brother
Vanda, Finland

polymath257 wrote: So how far apart are they in the beacon's frame? 1.5 light seconds.

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> 1.5 light seconds. Good. Now, suppose that, in the beacon's frame, the observer starts 2 light seconds away and is moving towards the beacon at .5*c. One wavefront of the light signal from the beacon is passing at this point. The observer is moving towards the beacon. After one second in the beacon's frame, how far apart are the observer and that wavefront in the beacon's frame?

humble brother
Vanda, Finland

polymath257 wrote: Good. Now, suppose that, in the beacon's frame, the observer starts 2 light seconds away and is moving towards the beacon at .5*c. One wavefront of the light signal from the beacon is passing at this point. The observer is moving towards the beacon. After one second in the beacon's frame, how far apart are the observer and that wavefront in the beacon's frame? So the ship is two light seconds away from the beacon and moving at relative 0.5*c towards the beacon when the beacon calculates wavefront #1 to hit the ship. One second after this for the observer the first wave "seems" to be 1.5 light second away past the ship.

humble brother
Vanda, Finland

humble brother wrote: One second after this for the observer the first wave "seems" to be 1.5 light second away past the ship. And by observer here I mean the observer in the beacon :) So in the beacon's frame the ship appears to be 1.5 light seconds away from the first wavefront.

Since: Apr 11
Location hidden

31Jul12.....
.....a word to da wise for True members of TOPIX Forums.
Members whose monikers are in BLUE are Real members.
Wannabeeee members whose monikers are in BLACK are NOT Real members.
Ps:....Real members should NOT waste dime responding to BLACK monikers.
Forever and Ever BobLoblah

“Think&Care”
Since: Oct 07
Location hidden

humble brother wrote: <quoted text> So the ship is two light seconds away from the beacon and moving at relative 0.5*c towards the beacon when the beacon calculates wavefront #1 to hit the ship. One second after this for the observer the first wave "seems" to be 1.5 light second away past the ship. You are hedging. Why do you use the word 'seems'? In the frame of the beacon, how far away are the observer and the wavefront?

humble brother
Vanda, Finland

polymath257 wrote: You are hedging. Why do you use the word 'seems'? In the frame of the beacon, how far away are the observer and the wavefront? No I'm not hedging :) The beacon does not observe the light that the ship observes from the beacon. The light emitted by the beacon is not observed by the beacon. So the beacon can only calculate where the emitted wavefronts are traveling. As long as you understand this philosophy we can say that in the beacon's frame the first wavefront and the ship have departed 1.5 light seconds in 1.0 seconds beacon time. Carry on.

