New Relativity

[cwes99_03]

I just wonder because you always have 4times the energy of half your velocity. Although you can never stop by going half way slower everytime. Eventually it comes down to Reference frame A and B in contact and the first motion that is made is the same for each and gives you 4times the energy of half the speeds motion.

 

I take it back, you obviously have no math background either.

[arkain101]

I know my way around math better than the average guy.

I am refering to reality about my qestion, I am not talking about the math. The math was written to make sense of reality from observing reality.

 

You double your velocity your energy is 4 times greater correct? (I am wondering why the reason for this is).

 

If you compare your energy to what it would be at half your velocity it would be 4times less correct?

 

Velocity of the object you are trying to find the energy of is squared. That means the there is two velocities in action here at the same velocity. The mass of the object doing the work is multiplied by the total system velocity (V^2). Then it is divided in half to explain the fact that you have to make a choice as to which object will do the work. Work can only be done if one objects energy is sacrificed (conservation of momentum). It is derived from other equations but it is what it is because of reality right? lets say the equations were all derived from Ke. It would sound the same, that such and such equation was derived from ke and so on.

The equation is what it is because it works, even without including the fact it is derived from other equations. If it was the only equation we knew in physics. If say we were alive 2000 years ago. Why woudl you explain Ke is what it is. (without using any other math).

[cwes99_03]

You know math? Then why can't you understand that the equations you are confused by are derived (note the pun) from calculus.

[arkain101]

I do understand how the equations are derived from one another.

 

In the big picture they were all derived from an experiment in reality. Math works out reality for you accuratly. You can tell me about how they were written down over and over. Although it doesnt explain to me how reality functions like the math says.

 

Imagine we were alive 2000 years ago. I am wondering how you would explain Ke is what it is. (without using any other math). But explaining the reality where Ke was fabbed from.

[cwes99_03]

Although it doesnt explain to me how reality functions like the math says.

 

That means you know how to physically derive numbers, but you don't understand what a derivative or a integral means. Go back to your calculus teacher, or better yet a physics teacher and ask them what a derivative of a function means about that function.

[arkain101]

Im just asking a simple question if anyone is interested in answering it.

 

Using how reality functions why would you write the Ke equation as it is after doing an experiment to test it. Would you say it is because mass increases? Or velocity is combined between the two objects causing each reference to double when one doubles its velocity, increasing a total of 4x? Equal and oppostite reaction saying each interacting object is considered to travel the same velocity. But only 1 in the system is able to be the worker, unless you want both to do work.

[Erasmus00]

Im just asking a simple question if anyone is interested in answering it.

 

Using how reality functions why would you write the Ke equation as it is after doing an experiment to test it.

 

F=ma

 

F = m *dv/dt

 

F*dt = m *dv

 

F*v*dt = m *v *dv

 

F* dx= m *v*dv Here I have used v = dx/dt.

 

Integral F*dx = Integral m*v*dv

 

Work = 1/2 m*vfinal^2 - 1/2 m*vinitial^2. Here I have used the deffinition of work.

 

Kinetic energy and work follow directly from Newton's second. So does potential energy, if you write F as the gradient of the potential.

-Will

[arkain101]

^

I appreciate the time, effort and information.

 

None the less those are details. I want to quote Einstien for a moment here (for lack of any other famous quotes). He once said something along the lines of, I want to know how god thinks, (how the grand scheme operates) all the rest is just details. (of course they are important details, they allow you to manipulate matter in the way you want.) I too am more interested in the reasoning than the details. Maybe that helps you understand my question.

 

My question is about kenetic energy and in words I was wondering the cause for the fact that when an object accelerates to twice its velocity it has the capability of doing 4 times more work, even though it has only doubled its momentum.

 

Is it not a science fact as of now that the reason for this increase of energy is from the possible increase of the mass of the moving body?

[Erasmus00]

Is it not a science fact as of now that the reason for this increase of energy is from the possible increase of the mass of the moving body?

 

No, as has been shown above, the fact that energy is related to p^2 is a direct consequence of Newton's laws. In fact, in a relativistic treatment, Kinetic energy is no longer proportional to p^2, as E = sqrt(p^2c^2-(mc^2)^2). If p>> m, then E = pc.

 

Kinetic Energy = p^2/2m has nothing to do with a relativistic mass increase.

-Will

[arkain101]

Okay thankyou all..

 

Although my question is still un answered... I guess I will leave it alone for now.

[IDMclean]

I am once again ashamed of all you people.

 

Rather than listening to what Arkain has to say, and discussing with him you people have forced upon him your interpetations.

 

I find math completely useless unless I know why I am using it and where to use it.

 

You have done nothing to address Arkain's Question. Not why is the equation this way. That has been hammered here, without any real understanding of the Concept.

 

What you are saying, Arkain, as I understand it currently is that we have observer dependent measurement of bodies in motion. That the equation of kinetic energy is describing both the distance the observed body is moving towards, and the distance the observed body are moving away from point A to point B. Correct?

 

so essentially [math]|v_{q\rightarrow B}| * |-v_{q \leftarrow A}| = v^2 [/math].

 

The issue here, is language I think. Failure to articulate, not failure of understanding. I would hear more of what you have to say, Arkain.

[arkain101]

I may have made a bad first impression on this board, but it was not my intentions to make a single claim of fact. I did make alot of claims with ideas but those are not fact they are theory.

 

I finally came to the answer to my question without the use of math as I was trying to ask.

 

The reason reality works the way it does that requires us to square ones velocity in order to measure its moving energy is because kenetic energy is the measurement of body interaction. The inertia of the rest body and the inertia of the motion body creates a measurement that can be valued to each object in the interaction.

 

It all comes down to inertia. A moving body has only energy when it has somewhere it can expend that energy.

 

When it can and does interact with another mass (body) a measurement can be made and it can be said the object has kinetic energy, and the rest body will always return the same mass to velocity, FORCE as the moving object.

 

Hence the equal and opposite reactions of each and every thing.

[arkain101]

I expanded the equation to show this is right.

Using mass of desired KE measurement only and its velocity.

Ke= (M*V) (M*V) / (2*M)

 

KE= (M*V) (M*V) / (M+M)

[InfiniteNow]

I am once again ashamed of all you people.

As if you've never posted a comment which was inaccurate, difficult for others to understand, or poorly articulated? Please avoid comments like this KAClown.

[IDMclean]

Responce to InfiniteNow.

 

Now, back to Arkain.

 

Question, are you aware of the relativistic Kinetic Energy equation, which replaces the classic Newtonian equation?

 

it's form is:

[math]K = m_0 c^2 - \gamma m_0 c^2[/math]

Where [math]\gamma[/math] is:

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

 

I believe the mass in this equation is that of the body in motion relative to the observing mass.

 

I am curious to know your take on it.

[Erasmus00]

it's form is:

[math]K = m_0 c^2 - \gamma m_0 c^2[/math]

Where [math]\gamma[/math] is:

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

 

I believe the mass in this equation is that of the body in motion relative to the observing mass.

 

It is, but you must remember isn't the "relativistic" mass but the rest mass: i.e., it is the mass the object would have if it weren't moving.

-Will

[IDMclean]

What I ment was:

The gamma of it is calculated off of the body relative to the observing body. The m_0 was denoted, as the ground state of the mass. As per standards.