Realistic Relativy (Corrected)

[arkain101]

Theory was devloped from october 2nd to october 13th 2005 by Nicholis Justin Hill as the correction of relativity and physics.

 

email adress: [email protected] or/and [email protected]

 

Mailing address

1655 Aveling Coalmine Rd

Telkwa, British Columbia, Canada

V0J 2X3

 

Phone number

(250)846-5968

 

Theory was posted here at 4:00pm pacific time October 13th 2005.

 

 

 

 

 

 

 

Realistic Relativity

 

 

Introduction:

 

Einstien was on the right track his logic was flawless, but his calculations on E=mc^squared were derived from in incorrect interpretation of energy. The equations are correct but the logic looking at the equation are not. This theory describes how Velocity must always be considered doubled when in reference to how much Kinetic energy is being determined. Which is already proven but not sure if understood.

 

 

 

Velocity:

 

I want you to imagine a chalk board which represents space for the moment. There is a few atoms floating around in space. Now, let's erase all of the other atoms except 1, a hydrogen atom to keep things simple. Now all that exists in this empty vacume is 1 atom. Hypotheically lets say this atom wants to travel around and find another atom to interact with. It floats around here, than over there, than over here, but then it realizes somthing. It has no idea how fast it is going and infact now matter how much effort it puts into going fast, it still has no velocity. It has nothing to compare its velocity with. There is nothing to run into and interact with. Kinetic Energy (Moving energry) is Energy= 1/2 mass times velocity squared currently as of today. That equation can not exist with only one object in the universe. As soon as you add another atom in the path of a moving object then the universe can now have motion. Each atom would both run into eachother at the same velocity according to eachother. It is impossible to tell which object is doing the motion simply because it depends if you are looking from object A coming at B or looking from object B coming at A. They both see the same velocity coming at them. When you travel in space the universe infront of you moves at you and you move at the universe. When you are in a car we measure our speed or velocity by how much distance you cover in a cirtain time. This is not used if you want to talk how fast you are traveling in relation to a possible accident and find out kinetic energy. When you travel 100km/h in your car you are in reality traveling at a potential 200km/h if you were to hit a wall or any stationary object in your path. This is exactly why it takes 4 times longer to stop if you double the speed on your spedometer because you have actually quadrupaled your speed in reference to the road.

 

 

Application of theory on E=mc squared:

 

Energy is equal to mass times the speed of light squared in todays physics to dertime how mucn energy is contained within a chunk of matter.

remember every action has an equal and opposite reaction as verified above. Light has the ability to apply force to all mass and can be affected by gravity so lets assume it is mass. The speed of light is 300,000m/s and is considered a constant as we know it today, the universal constant of speed.

Please think of what causes this "light" to achieve the speed C. Ask where is the source?

Could there NOT be a universal constant that inforces a speed limit on which light travels and objects that travels like a speed restriction?. I think so.

It is just as possible that the force in which emmits the light is constant. This says that whatever causes the emmition of light must always have a constant mass. The force involved to cause light to travel at 300,000km/sec will cause the object emitting this force to experience the same force upon it, causing a force faster than C.

Cirtain physics forumlas show that when an object reaches the speed of light as, its mass increases towards infinate mass and in turn need an infinate force to propell it.

Or is it that, what really is happening is the point in which you would aproach the speed of light you are running out of a faster force to propell you forwards because nothing else exists to push you any faster? Yes, makes sense.. So far everything we make has a top speed pushing force. Which must mean the same for that of which pushes light from the source.

 

 

If you created a ship that had a light beam engine, and shot light out at C and propelled a ship. You would eventually reach close to speed C. Although your mass would NOT increase. Your Velocity in relation where you were headed would increase. In fact if you went from 0 to 100m/s you would actually be going 200m/s towards the source and if you doubled your speed to 200m/s the source would also come at you at 200m/s with a combined speed of 400m/s.

Eventually when you reach speed C and the light coming out of the rocket reaches speed C you no longer have any accelerating force to make you go any faster. This is why you end up with a zero in a cirtain physics equation when at speed C.

When you reach speed of C. It means you run out of a faster source to push the mass. You wouldnt run into a infinite mass and force scenario. This in turn means.. you are traveling towards an object at C and the object is traveling at you at C. 9x10^10th m/s.

 

 

Why is it that the nuclear bomb does not release the same energy that E=mc^2 would tell you? Because the same force that comes out goes in.

 

Conclusion of Realistic Relativity by : Nick Hill

[CraigD]

I've the same critique as for the previous thread 4191:

Before putting additional work into your paper, Nicholis, let me point out a basic, critical error in Math terminology.

 

You correctly state the equation for kinetic energy as “E=m*V-squared”, but appear to misunderstand how one calculates it. In you first example, you state that “10 m/s squared” is equal to “20 m/s” – that is, you are interpreting the “squared” operation as meaning “multiplied by 2”, a synonym for “doubled.” This is incorrect. The correct interpretation of “squared” is “raised to the power of 2”, or “multiplied by itself.”. “10 m/s squared”, then, is (10 m/s)^2, which evaluates to 100 m^2/s^2.

 

Note that the equation Energykinetic = Mass * Velocity^2 derives straightforwardly from the fundamental definitions of mass, distance, time, velocity, force, work and energy. Although this derivation involves some very simple Calculus, it can be easily understood by calculus-free examples. Almost any introductory Physics textbook, printed or online, will provide good explanations of this.

 

I recommend you study Math and Physics to gain a strong understanding of known theory before working on original theory. Such study is hard work, but cannot be avoided if one seeks to be taken seriously as a scientist. You can count on people here at scienceforums for advice and help in such study, and the discussions here as an informal source of study material.

[arkain101]

You must check again..

I said velocity is double of that for what I call Potential Kenetic energy.

You must re-read. I did not misinterpret squared and double anymore

that was a mistake.

[DietAnthrax]

Theory was devloped from october 2nd to october 13th 2005 by Nicholis Justin Hill as the correction of relativity and physics.

 

Here I would like to demonstrate this theory in examples.

 

Example 1a: Potential Kentic energy contained in a moving body.

Two objects weighing 10kg each are on path at one another. Object A is stationary wall (that will not move) object B is a 10kg rolling ball. The ball is rolling towards the wall at 10m/s on the radar gun. The energy involved in this system is: Object A is being expected to be hit by object B at 10m/s. Object B sees it about to be hit by object A at 10m/s . Object B rolls into and hits the solid wall. The energy contained in the ball is not transfred into motion to the solid wall. The solid wall can not move or absorb energy so it must return the energy that has been acted upon it back into object B.

(These are the forces involved when objects like cars hit solid structures attatced to ground or somthing in such a way that they do not move that

 

have a mass not much greater than the car.)

 

Example 1a needs an equation like so.

E=2(m x V^squared) for the potential kenetic energy

 

This corrects the equation of Kinetic energy.

["from wikipedia"

-The kinetic energy of a body = 1/2(mv^2)

-where m is the mass and v is the velocity of the body.

-Note that the kinetic energy increases with the square of the velocity. This means for example that if you are travelling twice as fast, you need to lose four times as much energy to stop. The braking distance of a car at 100km/hr is therefore four times as far as the braking distance at 50km/hr.]

(but really this is because the ground underneath of the car is also traveling the opposite direction under the car as you are driving on top of the road doubling your velocity.)

 

Conclusion of Realistic Relativity by : Nick Hill

 

You haven't "corrected" anything. You've just waffled and talked a lot of nonsense, and made loads of mistakes, intersperced with stunning revelations e.g. about kinetic energy being proportional to velocity squared. Wow.

 

"Potential kenetic" energy? Haha.

 

"but really this is because the ground underneath of the car is also traveling the opposite direction under the car as you are driving on top of the road doubling your velocity". No. This is totally and laughably wrong.

 

At this point I got bored and stopped reading.

[arkain101]

well you missed the point of the theory then.

[arkain101]

test

[CraigD]

You must check again..

I said velocity is double of that for what I call Potential Kenetic energy.

You must re-read. I did not misinterpret squared and double anymore

that was a mistake.

I beg your pardon, Nicholis. I read your revised paper too hastily, and failed to see the changes you made. Also, I gave the incorrect equation for kinetic energy, neglecting to include the ½ term.

 

However, I still see confusion on your part about some fundamental mechanical concepts. Although your interpretation of the kinetic energy equation (Ekinetic = (1/2)*M*V^2) now appears correct (eg: you correctly state that the same mass moving at 2 times the velocity has 2^2=4 times the kinetic energy), the following examples from the paper are incorrect:

When you travel 100km/h in your car you are in reality traveling at a potential 200km/h if you were to hit a wall or any stationary object in your path

The kinetic energy of the car in this example is calculated using the velocity of 100 km/h, not 200. For example, for a car weighing 1500 kg traveling 25 m/s (90 km/hr), the kinetic energy is (1/2)* 1500 kg * (25 m/s)^2 = 468750 joules. If its speed is reduced to 0, either by colliding with an immovable wall, colliding with an identical car traveling the same velocity in the opposite direction, or rolling up a gentle hill, the work (change in energy) done is the same. This can be proven experimentally. For collisions, the change the car can be insulated against energy loss and the change in temperature following the collision measured (using actual cars would be messy, though possible – modeling clay is more convenient and traditional). For the hill, by measuring the height (Recall that the definition of work is Work = Force * Distance, for gravitational force, Mass * Accelerationgravity , so 468750 j = 1500 kg * 9.8 m/s^2 * Height -> Height = ~32 m).

 

I find similar errors stemming from misapplied fundamental mechanics in the remaining examples. Consistently, you seek to multiply velocities by 2, but this is not supported by the classical theory of mechanics (AKA Newtonian mechanics), which has been experimentally verified to great precision for velocities not an appreciable fraction of the speed of light c, or classical mechanics and Special Relativity, which has been verified to great precision for all velocities.

 

There certainly is a connection between the equation Etotal = M*c^2 and Ekinetic = (1/2)*M*V^2. That is given by the Lorenz transformation Mmoving = Mrest/(1-V^2/c^2)^.5, which, for velocities not an appreciable fraction of c, causes Mmoving*c^2 - Mrest to be about equal to (1/2)* Mrest*V^2.

 

My recommendation to you remains the same: master fundamental Physics before attempting to write new theory.

 

An additional suggestion: when writing scientific theory, strive to make specific, exact predictions that disagree with theory you are attempting to overturn. As is constantly pointed out in this forum, writing that does not make specific, testable predictions is not scientific theory.

[arkain101]

Thank you for that. I hear exactly what you are saying.

Here is an experiment that I can seem to use to describe it best. and I thank your patience.

 

I call it Potential Kenetic energy by meaning this;

An object that interacts with another object square on the same velocity must be considered with each object regardless if one appears stationary or not.

 

 

All this time, this is the equation that explains the theory in my mind. potential kenetic energy = Mass x velocity squared. ( i left out 1/2 on purpose)

 

Let me demonstrate. This.

Using normal 1/2(m*v squared)

 

Step 1.

Ekenetic=1/2(m*v squared)

E=1/2(10*25squared)= 3125joules in ball A 0 joules in ball B.

 

Step 2.

Point of impact. Ball B is resistant to motion/transfer of energy. At this point ball A has stopped and lost all of its velocity and containing 0 kenetic energy.

 

Step 3.

We assume absolutly no energy is lost through sound, heat, light, nothing. In this experiment

Ball A rebounds carrying with it the same energy it had origionally Because energy must not be destroyed.

E=1/2(10*25squared)= 3125joules

 

In this experiment. Ball A was carrying 3125jouls of energy. It hit a stationary ball that will not move because it is perfectly held in place by some kind of hypetheical force. The energy of ball A can not be transfered so it goes back into ball A and sends it back at the same velocity as it hit ball B with.

Say What?

The problem here is at one point ball A was traveling at 0m/s and at 0m/s an object has zero kenetic energy so how does one object reverse direction at the same velocity? This goes to my theory. That of which you travel towards likewise travel towards you. But since this ball was held by hypethetical force energy must not be lost so it must return back to object A. from 10m/s to 0 back to 10m/s in the opposite direction.

 

We add up the total energy in this experimentr and we get 3125+3125= 6250joules contained in ball A to accomplish this task. it just doubled its energy? hmm

 

That is why E=M*V squared. (without the 1/2)

E=10*25 squared = 6250joules that appears to be contained in one 10kg object traveling at 25m/s.

 

This is because, the ball that looked like it wasnt moving (ball :) was actually in reality also traveling towards ball A at the same velocity Ball A was traveling towards B.

Every impact of two bodies that hit square on, the velocity is double of which one assumes! Theory of realistic relativity.

 

I am very curious on how that is otherwise explained.

[arkain101]

If you can put this theory into your mind for a moment it can undoubtable logically explain alot of problems. Even with calculations.

Using Current.

As mass speeds up it states that mass must increase. Time is said to slow down around you to explain why you seem to arrive at a destination quicker.

 

Use this theory.

As mass(a ship) speeds up, the mass in front of it reflectivly comes at it. If there was no other mass to see, or exist, it could not have velocity, it would get no where and interact with nothing. Therefore your velocity is double to what you think.

Double your velocity and the objects in front of you double at you, quadroupling your speed! Double the speed in your car takes 4 times longer to stop because you actually increased your speed from origanol 4times.

As mass speeds up, it doesnt increase in mass that of which is behind it doubles in speed away from it. Lessening and lessening the force which accels you untill you run into a point in which C and you are traveling opposite directions and equalizes out making accelertion 0.

 

 

 

The Twin Paradox is excluded.

 

Speed C is attainable for mass.

 

 

 

Gravity nearly seems explainable.

We are thousands of miles away from the center of the earth.

[Boerseun]

If I drive my van down the road at, say, 50 kph, one of two things is apparent to an objective observer:

 

1) The tree in the road is standing still, and I'm barelling towards it at 50kph.

2) I'm standing still, and the tree is coming to me at 50kph, together with the rest of the scenery.

 

The end effect of me uprooting the tree at 50kph will be the same, regardless of 1 or 2.

 

Now, say I double my speed:

 

1) The tree in the road is standing still, and I'm barelling towards it at 100kph, OR

2) I'm standing still, and the tree is coming to me at 100kph, together with the rest of the scenery.

 

Nothing has quadrupled, as in your example above.

 

Regardless of in what frame the observer stands, I fail to see your point.

[Bo]

I havent read everything written here (lack of time. etc..), but just wanted to make the following remark on your story about velocity:

You're absolutely right that velocity cannot be defined without a reference frame. however I don't agree with you that you need a particle or whatever to define your reference frame; you just have to keep n mind that any reference frame is arbitrary.

The funny thing with your next point that you actually would move at double speed is wrong, by exactly this argument: You have a certain speed with respect to the reference frame of say the road, or whatever. and now you extrapolate from this the knowledge that we would have double speed in some absolute reference frame. however, as I stated above, such a frame does not exist. (in more technical terms: you have made a coordinate transformation to a new frame where one of the objects moves at double speed while the other is standing still. while such a transformation can make your life easier (since calculations often get easier with the right choice of the coordinate system), there is no physical meaning to a coordinate transformation.

 

Bo

[cwes99_03]

I believe that the originator of this post is confusing conservation of momentum and energy as being one law.

 

When A hits B it's KE is transfered into some sort of potential energy, which results in the deforming of either A & B or just A, and we'll leave out the effects of energy loss to the two ball system by saying that this is a perfectly elastic collision.

 

The deformed ball then releases that elastic potential back into Kinetic energy. The total amount of energy is there for 1/2 m*v^2 (where v is the velocity that the ball is traveling at before or after the collision). There is no other energy in the system. The object that is anchored does not have any kinetic energy, nor does it lose any potential. The change in direction of the velocity vector is something that you'll have to learn about when you do some reading on Thermodynamics and pressure.

 

Cheers, and here's to more learning.

 

Cwes99_03

[arkain101]

1) The tree in the road is standing still, and I'm barelling towards it at 100kph, OR

2) I'm standing still, and the tree is coming to me at 100kph, together with the rest of the scenery.

 

Nothing has quadrupled, as in your example above.

 

Regardless of in what frame the observer stands, I fail to see your point.

 

I mean this.

 

Lets erase all other molecules in the universe.

 

Okay now We have two objects. Object B appears to be hitting object A , then bounced off 10meters away and eventually turns around and hits object A again repededly. Object A appears to bit sitting still. Suddenly The reference frame stops following object A and puts its frame relevant to object B. Now we see object B sitting still, and object A is doing the hitting. Both are experience force which one is really moving? you have excluded that the observer is a visible entity. Rather a hypothetical observer where only these two objects exist in a never ending blackness of space.

[arkain101]

Here is a test that someone can maybe answer me.

 

On Earth a car travels 25m/s. it takes it 100m to stop. Now the car accelerates up to 50m/s , it takes 400m to stop. simple. right.

 

In space, a ship travels 25m/s it takes 100m to stop, it accelerates up to 50m/s and now only takes 200m to stop.

 

I understand that in space it takes as long to stop as it took to accelerate but here on earth it takes a car 4 times longer to stop once it doubles its speed.

This may be incorrect but I dont mind the critiquing.

[ixa]

Here is a test that someone can maybe answer me.

 

On Earth a car travels 25m/s. it takes it 100m to stop. Now the car accelerates up to 50m/s , it takes 400m to stop. simple. right.

 

In space, a ship travels 25m/s it takes 100m to stop, it accelerates up to 50m/s and now only takes 200m to stop.

 

The (non-relativistic) kinetic energy of an object is simply Ek=(mv^2)/2. If the velocity of the object is doubled (and the mass is not changed), the kinetic energy increases by a factor of four (2^2=4). This is just the same in space as it is on earth, because the laws of physics are the same everywhere.

 

Now, if the force that slows down the object is the same in the case of 25km/h as it is in the case of 50 km/h (and why shouldn't it be?), the distance it takes for the object to stop is also increased by four, both on earth and in space. This is because the force is doing work against the kinetic energy of the object, and the kinetic energy has been quadrupled. In a simplified way the work can be calculated by W=-s F, where F is the force and s is the distance the object travels while under the effect of the force.

 

Work has the unit of energy (joules, for example), and in a sense the force converts an amount of kinetic energy into some other form of energy. The amount of kinetic energy lost is W, and for example in the case of the car the kinetic energy is converted to mostly to heat of the brakes and the tyres.

 

So, to conclude, if the distance s = -W/F, and the force remains the same while energy is quadrupled, the distance it takes for the object to stop is also increased by a factor of four. No difference between a car and a space probe.

[arkain101]

The (non-relativistic) kinetic energy of an object is simply Ek=(mv^2)/2. If the velocity of the object is doubled (and the mass is not changed), the kinetic energy increases by a factor of four (2^2=4). This is just the same in space as it is on earth, because the laws of physics are the same everywhere.

 

thank you, yes I made a mistake there,

Right! exactly, now I can apply it properly. It does take 4 times the distance to stop with the doubling of d/t velocity on land and also space right!. Read this.

In space, as you try to stop the reference frame in front of you is coming at you just as fast as you go at it, causing you to cover twice the intended distance if you had doubled your velocity with d/t which makes it take 4 times the distance to stop.

Then if you use the reference frame behind you to measure your stopping distance. The behind reference frame travels away from you just as fast causing you to cover twice the intended distance making it 4 times the distance to stop if you double your velocity. I am visioning a space ship half way between the moon and earth, using the moon as front (direction traveling towards) reference frame and earth behind it but only one reference frame at a time when measure the distance to stop compared to the distance it takes to stop if you double you d/t velocity. Yes!

 

Here is a example of 2 objects which contain there respective own reference frames. Because there is no absolute rest frame. It is undetectable to which body is doing the motion. Each object has the same mass, and the people/observers can be assumed to have the same mass and be on or in the object if you want to make things easier that way. The rectangle (the reference frame) moves with the objects as if it is all one peice.

Notice the data involved below.

 

 

[ixa]

thank you, yes I made a mistake there,

In space, as you try to stop the reference frame in front of you is coming at you just as fast as you go at it, causing you to cover twice the intended distance if you had doubled your velocity with d/t which makes it take 4 times the distance to stop.

 

Let me ask you a question: If instead of doubling the initial velocity you _tripled_ it, how many times greater would the stopping/braking distance be? For example, if at the velocity 20 km/h the distance was 30 meters, how much would it be at 60 km/h?