Gravity: push, pull, or does not exist?

[Alkatran]

terrabyte: neutrinos DO affect us... u've never heard of them because sadly, u haven't studied physics as much as me.. (the rebellious teen) and the only thing i get my jollies off of is listening to u try to put up a reasonable argument with ur inferior concepts of physics. i thought u were going to be leaving when u said "good luck with life". why the heck did u even come back?? HAHAHA... and u say u don't care anymore?? i guess it WOULD be devastating to find out that everything u thought u knew about the universe is wrong...

entropy: why do u keep saying that neutrinos only come in contact with us once a month? u don't believe that site i gave u? real life is that trillions and trillions of neutrinos pass through us every second. since u helped mathematically prove they exert a pressure u just helped support my theory! wow, that was really tight! thanx. I'm not going to speak for jessep because that would be like me taking his ideas. he's the one with all the cash rewards so why no step up to his challenge and stop being weanie and just ask him urself?? and years from now, when ppl start refusing to just accept being spoon-fed einstein's fictional ideas, they will mock ur beliefs and turn to the superiority that is the push theory. i think that will get me very far in life. enjoy being wrong about the universe, entropy. when u're ready to know the truth... u know where to find me.. hehehe

-He proved that IF all the neutrinos were absorbed by the Earth that they would give off .007 joules/m^2*s. That is an INCREDIBLY small amount. Along the lines of you "crushing" someone with a hair.

-If we're being pushed by the sun into the earth, why isn't the Earth being pushed as well? Wouldn't this nullify the effect of us being pushed into the earth?

-I've read about people who "challenge" people to prove them wrong. They argue somewhere along your lines, they take every logical argument and throw hundreds of random, unsupported facts at it. And that's if they even play by the rules, I've read of people saying they get to censor the evidence before it is judged!




But really, Entropy, Terrabyte, why did the average force go down when I compressed the points (see program on page 9 or 10)? It seems that density would be a significant factor when the object is very large in comparison to the distance between the center and the observer. (Note to 15 year olds: The force appears to go DOWN and not UP)

 

[beatrix kiddo]

i should have been more clear: 500 trillion neutrinos is just an estimate. there are far more neutrinos than that.. they are just unaccounted for. trillions and trillions (more) travel through us with "a velocity nearly equal to the speed of light" -foundations of astronomy pg. 167. entropy u forgot to put that in ur equation.. hmmm

 

[Alkatran]

i should have been more clear: 500 trillion neutrinos is just an estimate. there are far more neutrinos than that.. they are just unaccounted for. trillions and trillions (more) travel through us with "a velocity nearly equal to the speed of light" -foundations of astronomy pg. 167. entropy u forgot to put that in ur equation.. hmmm

What are you talking about? He put the neutrino's energy, which would include it's kinetic energy, in his equation.

E = mv^2/2, basic physics.

If 500 trillion is an estimate, why would there definitely be far more? Does it really matter if there are 500 trillion or 600 trillion if the VAST majority only pass through us without affecting us in any way?

 

[terrabyte]

perhaps your observation is worth looking into, Alkatran. it could very well be true that compression reduces gravity, making the theory on how black holes are formed WRONG.

or perhaps the formula for gravity is indeed wrong.

or perhaps doing particle-location calculations for an object is wrong (though logically it's supposed to work)

i still say we need an expert's opinion on this one

 

[urtalkinstupid]

Entropy, thanks for proving me wrong...or so you think! Let me explain what you did and how it does not work. First, you used the wrong number of neutrinos. You should have used 5x10^{12} for the amount of neutrino flux on the earth.. Second, you only calculated the amount of energy a neutrino when it is at rest. You forgot to multiply that rest energy by the velocity at which the neutrino travels to get the momentum. Neutrinos travel at nearly the speed of light. So, 3x10^8m/s would be sufficient.

Here is your math revised:
\nu_f=5x10^{10}m^2/s
E_\nu=.81MeV
\vec{v}_\nu=3x10^8m/s
E_\nu=.81MeV \cdot 1.602^{-13}=1.29762^{-13} MeV
E_\nu=\nu_f*E_\nu=0.0064881 J \cdot m^2 \cdot s
Now for their momentum.
\vec{p}=E_\nu \cdot \vec{v}_\nu=1946430 J \cdot m^3 \cdot s

That's just what i got. Hey, I know that there is a possibility that I'm wrong. If that be the case, I'll...I dunno. Yes, I know about emission and absorption. The neutrino emitts its energy only to be absorbed by another particle. It does not "disappear," rather it passes on its energy.

Entropy: I see where you got 5x10^{10}m^2/s; you converted it form cm^2/s hehe

 

[Alkatran]

Entropy, thanks for proving me wrong...or so you think! Let me explain what you did and how it does not work. First, you used the wrong number of neutrinos. You should have used 5x10^{12} for the amount of neutrino flux on the earth.. Second, you only calculated the amount of energy a neutrino when it is at rest. You forgot to multiply that rest energy by the velocity at which the neutrino travels to get the momentum. Neutrinos travel at nearly the speed of light. So, 3x10^8m/s would be sufficient.

Here is your math revised:
\nu_f=5x10^{12}
E_\nu=.81MeV
\vec{v}_\nu=3x10^8m/s
E_\nu=.81MeV \cdot 1.602^{-13}=1.29762^{-13} MeV
\Sigma_E_\nu=\nu_f*E_\nu=.64881J
You arrived at \Sigma_E_\nu=0.007J, because you had the wrong amount of neutrinos. This was as far as you got, because this is only the rest energy that the neutrinos emitt. Now, for their momentum.
\vec{p}=\Sigma_E_\nu \cdot \vec{v}_\nu=194643000J \cdot m \cdot s

That's just what i got. Hey, I know that there is a possibility that I'm wrong. If that be the case, I'll...I dunno. Yes, I know about emission and absorption. The neutrino emitts its energy only to be absorbed by another particle. It does not "disappear," rather it passes on its energy.

That'd be like being shot in the face with a cannon 5000 times a second. Of course, that's a massive underestimation.

 

[Alkatran]

perhaps your observation is worth looking into, Alkatran. it could very well be true that compression reduces gravity, making the theory on how black holes are formed WRONG.

or perhaps the formula for gravity is indeed wrong.

or perhaps doing particle-location calculations for an object is wrong (though logically it's supposed to work)

i still say we need an expert's opinion on this one

I have a hard time believing no one has looked at this before... let's see if I can word this right:

Compressing an object reduces the gravity at a fixed point relative to its center because the increase in the square of the distance of the mass on the near side is greater than the decrease in the square of the distance of the mass on the far side.

I think that sentence has enough "of"s in there, haha.

Now that I think about it, it does make sense (oh no! assumption!) because, since the total curvature in one area is greater, for the total to stay the same the curvature in another area should be reduced, correct?

 

[terrabyte]

according to your calculations, Mass is not the sum of an objects component masses

or gravity formula is somehow wrong :O

 

[urtalkinstupid]

Are we talking about when he made the program condensing the distance between the particles?

 

[urtalkinstupid]

\textcolor{blue}{\nu_f=5x10^{10} m^2s}
\textcolor{red}{E_\nu_e=.81 MeV\approx1.29762x10^{-13} J}
\textcolor{green}{\vec{v}_\nu_e=3x10^8 \frac{m}{s}}
\textcolor{purple}{\vec{p}=}\textcolor{red}{E_\nu_e} \cdot \textcolor{green}{\vec{v_\nu_e}}=\textcolor{red}{1.29762x10^{-13} J} \cdot \textcolor{green}{3.0x10^8 \frac{m}{s}}=3.89286x10^{-5} \frac{J \cdot m}{s}}
\vec{p} \cdot \textcolor{blue}{\nu_e}=3.89286x10^{-5} \frac{J \cdot m}{s} \cdot \textcolor{blue}{5x10^{10}m^2s}=1946430 J \cdot m^3

Neutrinos are able to emitt upto 1946430 J per ever cubic centimeter of earth. This is different from my first attempt. At first, I tried to calculate the momentum by multiplying the flux rate by the energy...That wasn't the proper way. So, this time I calculated the momentum and then determined how much energy a neutrino could emitt over a given area. Maybe this is right, maybe it is not. One things for sure, it's different from Entropy's calculation, which is not momentum at all. Alkatran, this is not as much energy as you think, but a constant flow of this calculation should be enough. Remember, this is only for solar neutrinos. They are not the only source. Our body is a source of neutrinos as well, but of a lesser magnitude.

Goodnight!

 

[Alkatran]

\textcolor{blue}{\nu_f=5x10^{10} m^2s}
\textcolor{red}{E_\nu_e=.81 MeV\approx1.29762x10^{-13} J}
\textcolor{green}{\vec{v}_\nu_e=3x10^8 \frac{m}{s}}
\textcolor{purple}{\vec{p}=}\textcolor{red}{E_\nu_e} \cdot \textcolor{green}{\vec{v_\nu_e}}=\textcolor{red}{1.29762x10^{-13} J} \cdot \textcolor{green}{3.0x10^8 \frac{m}{s}}=3.89286x10^{-5} \frac{J \cdot m}{s}}
\vec{p} \cdot \textcolor{blue}{\nu_e}=3.89286x10^{-5} \frac{J \cdot m}{s} \cdot \textcolor{blue}{5x10^{10}m^2s}=1946430 J \cdot m^3

Neutrinos are able to emitt upto 1946430 J per ever cubic centimeter of earth. This is different from my first attempt. At first, I tried to calculate the momentum by multiplying the flux rate by the energy...That wasn't the proper way. So, this time I calculated the momentum and then determined how much energy a neutrino could emitt over a given area. Maybe this is right, maybe it is not. One things for sure, it's different from Entropy's calculation, which is not momentum at all. Alkatran, this is not as much energy as you think, but a constant flow of this calculation should be enough. Remember, this is only for solar neutrinos. They are not the only source. Our body is a source of neutrinos as well, but of a lesser magnitude.

Goodnight!

First of all, that much energy (the amount per cubic METER, not centimeter) is what it would need to launch you to a height of 40 kilometers (assuming you weighed a bit less than 50 kilos). 40 kilometers. per cubic meter. per second.

This number is so obviously impossible that it can't be right.


Terrabyte:
The program does the following:
1 - Pick 2 random angles
2 - Calculate the inverse of the distance of a point at distance 1 along a line project along those angles.
3 - Calculate the inverse of the distance of a point at distance 0.5 along a line project along those angles. (compressed object)
*The inverse of the distance is the force if G*m1*m2 = 1
4 - Repeat about 10000 times
5 - Sum all of these results (all the 2s and all the 3s)
6 - Compare
Does this look correct?

Of course a computer program is NOT reality and any findings would need to be confirmed experimentally.

 

[Alkatran]

P = E*v

(the red and green equation)
P -> m/s*kg
E -> J -> m^2/s^2*kg
v -> m/s

E*v -> (m^2/s^2*kg)*(m/s) -> m^3/s^3*kg

Your units on both sides of the equation don't match and therefore the equation is wrong. Momentum is mass * speed, not energy * speed. If you want to use energy in momentum you need to divide it by c^2.

1 946 430/c^2 = 1 946 430/(9*10^14) = 2.16270 * 10^-9 or about 0.000000002 Joules.

Your last equation doesn't make any sense at all either. Your multiplying momentum by speed and getting energy*m^3 and THEN saying it's PER cubic meter?? P*v = m/s*kg*m/s = m^2/s^2*kg = J NOT J*m^3.

Wait, what's Vf?

 

[Entropy]

Vf is really \nu_f, the greek letter nu which means neutrino and "f" for flux.

Entropy, thanks for proving me wrong...or so you think! Let me explain what you did and how it does not work. First, you used the wrong number of neutrinos. You should have used for the amount of neutrino flux on the earth.. Second, you only calculated the amount of energy a neutrino when it is at rest. You forgot to multiply that rest energy by the velocity at which the neutrino travels to get the momentum. Neutrinos travel at nearly the speed of light. So, would be sufficient.

First I did use the neutrion flux on Earth per meter squared.

Do you even know what neutrinos are?

Neutrions have zero rest mass (therefore they don't exist at rest), therefore zero rest energy! All there energy/mass is kinetic energy just like photons! Where do you get this "multiple by the speed of light" BS?

 

[terrabyte]

1 - Pick 2 random angles
2 - Calculate the inverse of the distance of a point at distance 1 along a line project along those angles.
3 - Calculate the inverse of the distance of a point at distance 0.5 along a line project along those angles. (compressed object)
*The inverse of the distance is the force if G*m1*m2 = 1
4 - Repeat about 10000 times
5 - Sum all of these results (all the 2s and all the 3s)
6 - Compare

looks sound to me. try plotting more than 2 points. say like 50 at once all summed together. then do comparisons on their full distance versus 1/2 their distance. and maybe twice their distance for good measure.

i loathe to even think of the possibility that they've overlooked something. this is crazy :D

 

[Alkatran]

looks sound to me. try plotting more than 2 points. say like 50 at once all summed together. then do comparisons on their full distance versus 1/2 their distance. and maybe twice their distance for good measure.

i loathe to even think of the possibility that they've overlooked something. this is crazy :D

See step 4. "Repeat 10000 times"

I'll put in a random distance when I get home. (with the compressed taking half of the randomly chosen distance)

 

[Alkatran]

Vf is really \nu_f, the greek letter nu which means neutrino and "f" for flux.




First I did use the neutrion flux on Earth per meter squared.

Do you even know what neutrinos are?

Neutrions have zero rest mass (therefore they don't exist at rest), therefore zero rest energy! All there energy/mass is kinetic energy just like photons! Where do you get this "multiple by the speed of light" BS?

Alright, thank you for the flux explanation, I assumed v was speed. But that still doesn't explain why you're result is *m^3 and not /m^3 or why there's no time involved (you need energy/time/area or energy/time/volume).

I got the "BS" from E = mc^2. But it doesn't look like you're looking for iniertia anyways, since inertia isn't in J*m/s.

 

[terrabyte]

well 10000 is certainly more than 50 :D

my reading comprehension has taken a severe nose-dive as of late :O

 

[ArmoSkater87]

This whole theory couldn't have ever been right because a neutrino is practically massless, and doesn't have a charge. Which means that it can't interact with the strong forces or the electromagnetic forces. The neutrino was created for a certain purpose, that pupose being the conservation of momentum. In certain decay reaction, it was noted that momentum was not conserved since the secondary particles that are emitted didnt go in opposite directions. This meant that there was another secondary particle involved in the decay that we could not detect. In other words neutrinos can't interact at all using any of the fundamental forces, meaning that they can't apply momentum to us and thus cannot cause gravity.

 

[Alkatran]

WAIT I think I know what I might have done wrong! I've been looking for it long enough too!

 

[Alkatran]

Here we go, I wasn't adding up the vectors. I was adding down forces to up forces instead of subtracting. The new program looks like so:

Option Explicit
Const NumPoints As Long = 100000
Const ObserverDistance As Double = 30
Const ObjectSize As Double = 20
Const CompressionFactor As Double = 0.5
Const NumTests As Long = 50
Const BaseForce as Double = 1000
Const Pi As Double = 3.14159
Private Type Point
    X As Double
    Y As Double
    Z As Double
End Type
Private Sub Form_Load()
Dim A As Long
Dim B As Long
Dim Total As Point
Dim TotalComp As Point
Dim Angle1 As Double
Dim Angle2 As Double
Dim Distance As Double
    Randomize Timer
    For B = 1 To NumTests
        Total.X = 0
        Total.Y = 0
        Total.Z = 0
        TotalComp.X = 0
        TotalComp.Y = 0
        TotalComp.Z = 0
        For A = 1 To NumPoints
            Angle1 = Rnd * 2 * Pi
            Angle2 = Rnd * 2 * Pi
            Distance = Rnd * ObjectSize + 1
            CalculateForce GeneratePoint(Angle1, Angle2, Distance), Angle1, Angle2, Total
            CalculateForce GeneratePoint(Angle1, Angle2, Distance * CompressionFactor), Angle1, Angle2, TotalComp
        Next A
        Debug.Print Format(Sqr(Total.X ^ 2 + Total.Y ^ 2 + Total.Z ^ 2), "0000.000") & ", " & Format(Sqr(TotalComp.X ^ 2 + TotalComp.Y ^ 2 + TotalComp.Z ^ 2), "0000.000")
    Next B
    Unload Me
End Sub
Private Sub CalculateForce(ByRef It As Point, ByVal Angle1 As Double, ByVal Angle2 As Double, ByRef Total As Point)
Dim Force As Double
    Force = BaseForce / Sqr(It.X ^ 2 + It.Y ^ 2 + (It.Z + ObserverDistance) ^ 2)
    Total.X = Total.X + Force * Cos(Angle1) * Sin(Angle2)
    Total.Y = Total.Y + Force * Sin(Angle1) * Sin(Angle2)
    Total.Z = Total.Z + Force * Cos(Rad(FindAngle(ObserverDistance + It.Z, Sqr(It.X ^ 2 + It.Y ^ 2))))
End Sub
Private Function GeneratePoint(ByVal Angle1 As Double, ByVal Angle2 As Double, ByVal Distance As Double) As Point
    GeneratePoint.X = Distance * Cos(Angle1) * Cos(Angle2)
    GeneratePoint.Y = Distance * Sin(Angle1) * Cos(Angle2)
    GeneratePoint.Z = Distance * Sin(Angle2)
End Function
Private Function Rad(ByVal X As Double)
    Rad = X * Pi / 180
End Function
Private Function Deg(ByVal X As Double)
    Deg = X * 180 / Pi
End Function
Private Function FindAngle(ByVal Adjacent As Double, ByVal Opposite As Double) As Double
    If Adjacent Then
        FindAngle = Deg(Atn(Opposite / Adjacent))
        If Adjacent < 0 Then
            FindAngle = FindAngle + 180
        ElseIf Opposite < 0 Then
            FindAngle = FindAngle + 360
        End If
    Else
        If Opposite > 0 Then
            FindAngle = 90
        Else
            FindAngle = 270
        End If
    End If
End Function

It now returns results that make much more sense, the first 3 digits almost always match and after that it's generally chaos, due to the accumulation of rounding errors involved in using floating point variables (double).

 

[Alkatran]

I don't know why, but it feels really good to see proof that I was wrong and to actually ACCEPT IT: HINT HINT HINT. bah.

 

[urtalkinstupid]

Well, my equation is debunked, but Entropy's is also. His equation doesn't take into account the impact force due to their velocity. I know what neutrinos are, and they have rest mass. I used the speed of light, because they are said to travel at approximately the speed of light, ergo they have rest mass. Although, this rest mass is a small number.

I haven't lost faith; I'm at my college course, and I just got done talking to a physicists. He said it is very well possible that neutrinos can provide a push force. He also said he would love to see someone prove Newton and Einstein wrong. Seeing as the sun isn't the most abundant source of neutrinos, there are others out there taht emitt many more. So, the small number that is derived from the equation is only a small fraction of the amount of the force applied by neutrinos.

Everything relies on emission and absorption. That is how the universe works. Neutrinos expresses that concept as well, making it a perfect candidate for gravity.

ArmoSkater87...emission and absorption...

 

[Alkatran]

Well, my equation is debunked, but Entropy's is also. His equation doesn't take into account the impact force due to their velocity. I know what neutrinos are, and they have rest mass. I used the speed of light, because they are said to travel at approximately the speed of light, ergo they have rest mass. Although, this rest mass is a small number.

I haven't lost faith; I'm at my college course, and I just got done talking to a physicists. He said it is very well possible that neutrinos can provide a push force. He also said he would love to see someone prove Newton and Einstein wrong. Seeing as the sun isn't the most abundant source of neutrinos, there are others out there taht emitt many more. So, the small number that is derived from the equation is only a small fraction of the amount of the force applied by neutrinos.

Everything relies on emission and absorption. That is how the universe works. Neutrinos expresses that concept as well, making it a perfect candidate for gravity.

ArmoSkater87...emission and absorption...

Where was hsi debunked? The "impact force" is technicly the energy (only technicly because force != energy) transferred to the planet, which is what he calculated.

 

[Rampantbaboon]

I have to agree with Armo. A miniscule percent of them have interaction with the Earth as they travel through the entire diamater of it. Using the term interact loosely. They would have a negligible effect on the motion of things in the universe. The shower of protons, electrons, pions, muons and other cosmic ray events (my current project) would have a much greater effect on anybody.

 

[Entropy]

This whole theory couldn't have ever been right because a neutrino is practically massless, and doesn't have a charge. Which means that it can't interact with the strong forces or the electromagnetic forces. The neutrino was created for a certain purpose, that pupose being the conservation of momentum. In certain decay reaction, it was noted that momentum was not conserved since the secondary particles that are emitted didnt go in opposite directions. This meant that there was another secondary particle involved in the decay that we could not detect. In other words neutrinos can't interact at all using any of the fundamental forces, meaning that they can't apply momentum to us and thus cannot cause gravity.

Exactly. Thank you.

 

[terrabyte]

Consider 3 particles making up an object. Particle A is 1 meter away, B is 2, C is 3, the gravity from each (they are the same mass) is m/1 + m/4 + m/9 = 36m/36 + 9m/36 + 4m/36 = 49m/36

Now if they are condensed to the same position with the same center: gravity = m/4+m/4+m/4 = 3m/4
49/36m-3/4m = 0.61 m
(3m/4 == 49m/36) = false

so what was wrong with that again?

 

[Alkatran]

so what was wrong with that again?

Like I said a few posts afterwards, the object isn't remotely spherical, which is the shape that the formula for gravity applies to.

 

[terrabyte]

weird. i thought gravity formula would apply to all shapes.

 

[JohnTownsendUCF]

I always just assumed that gravity was really just ,mass warping time-space.

When you place a bowling ball on some stretched out rubber sheet, it warps and droops. When you place a marble on that rubber, it rolls to the largest dip (being the one the bowling ball made) in the rubber sheet.

 

[Rampantbaboon]

yeah, that's what einstein said