Is our Science Wrong?

[Sciencegod]

Hello Everybody,

 

I would start by asking a simpe question, Is our Physics wrong? if So, then the whole of science is wrong too? Does any one of you think the same, then please share your views? I seriously believe that we have missed something very crucial in our understanding of nature, and its in particular related to our study of energy which is incomplete according to me. Our current science is built on the foundations of classical physics. Is it going to crumble any time soon? There are many inconsistencies and anomalies in current physics. I have discovered it myself. Has any one of you discovered the same, if so, kindly advise. To give you a simple idea,

 

Is Kinetic Energy relation = 1/2 mv square or (mv square). What do you think is the right answer? The obvious answer is very well known, but still few people have got doubts regarding it including myself.

 

Kindly express your sincere views on the subject,

 

Thanks,

 

 

Edited July 27, 2012 by Sciencegod

[imatfaal]

!

Moderator Note

As this topic is speculating that mainstream physics is/may be incorrect or poorly founded I have moved it to its more natural home.

[swansont]

Hello Everybody,

 

I would start by asking a simpe question, Is our Physics wrong? if So, then the whole of science is wrong too? Does any one of you think the same, then please share your views? I seriously believe that we have missed something very crucial in our understanding of nature, and its in particular related to our study of energy which is incomplete according to me. Our current science is built on the foundations of classical physics. Is it going to crumble any time soon? There are many inconsistencies and anomalies in current physics. I have discovered it myself. Has any one of you discovered the same, if so, kindly advise. To give you a simple idea,

 

Is Kinetic Energy relation = 1/2 mv square or (mv square). What do you think is the right answer? The obvious answer is very well known, but still few people have got doubts regarding it including myself.

 

Kindly express your sincere views on the subject,

 

Thanks,

 

 

 

What do you think is wrong or inconsistent? The work done by force through a distance gives the result of mv^2/2 for the classical kinetic energy.

[Greg H.]

Physics works. The evidence is all around you, from the car you drive (assuming you have one) to the computer you're posting these messages from. Our knowledge of physics is certainly not complete, but that's a far cry from saying it's wrong.

[Phi for All]

Is our Physics wrong?

No, it's just hard to grasp sometimes, and many people find what they think is an inconsistency, throw their hands up and use that as a good excuse to stop studying it and start claiming it's wrong.

[Bignose]

Is Kinetic Energy relation = 1/2 mv square or (mv square). What do you think is the right answer? The obvious answer is very well known, but still few people have got doubts regarding it including myself. [/size]

 

well, one of [math] E_k = \frac{1}{2}mv^2 [/math] and [math] E_k = \left(mv\right)^2 [/math] is dimensionally sound, and the other one isn't. I know that I have no doubts that the one that isn't dimensionally sound isn't right.

[newts]

Is our Physics wrong?

 

Is Kinetic Energy relation = 1/2 mv square or (mv square). What do you think is the right answer? The obvious answer is very well known, but still few people have got doubts regarding it including myself.

Physics is always partly right and partly wrong. Aristotle was right about the earth being spherical, and about space having to be full in order for objects to move; but wrong about the earth being stationary, and things being a combination of earth, air, fire and water. The situation today is not that different, with brilliant theories of atoms and the mechanics of the solar system, but absurdly bad theories about particle physics and the large scale structure of the universe. As was the case when Aristotelian philosophy ruled the roost, almost everybody accepts that the current interpretation is completely correct; whilst those who do challenge have a tendency to either criticise everything, or else criticise the correct theories and accept the nonsense.

 

The standard kinetic energy formula certainly works, and I am not clear what alternative you are considering; but since you disagree, you might find my unsuccessful attempt to persuade Bob Berenz interesting (click 'responses' at the bottom of this page http://squishtheory..../uncategorized/). It seems the real problem was not that he was too eager to challenge existing ideas; but rather that he believed the nonsense that motion is not absolute but relative between objects, and was engaged in the impossible task of trying to make sense of the universe on that basis.

[Sciencegod]

I am here with attaching the argument of my case, that the present day kinetic energy relation is incorrect. If i am wrong in my analysis, let some one please correct me. I am very happy to be corrected.

 

We know,

 

Energy = Force * distance;

 

E = F * d

 

E = ma * d (Substituting for force)

 

E = m{(v-u)/t) * d (Substituting for 'a')

 

E = m{v/t} * d ( assuming that u = 0)

 

E = mv {d/t} (re- arranging) --> simple commutative property

 

E = mv {v}

 

E = mv² (Kinetic energy relation is disproved)

 

NOTE: The above derivation seems to be alright from a mathematician's point of view but mainstream researchers call it pseudo derivation? Why? I cannot make any sense of it all. Please kindly share your sincere views on the same. Also, please kindly look at the attachment provided in favour of it.

 

KE.doc

[uncool]

I am here with attaching the argument of my case, that the present day kinetic energy relation is incorrect. If i am wrong in my analysis, let some one please correct me. I am very happy to be corrected.

 

We know,

 

Energy = Force * distance;

 

E = F * d

 

E = ma * d (Substituting for force)

 

E = m{(v-u)/t) * d (Substituting for 'a')

So here, your assumption is constant acceleration, correct? That is the only way that a = (v - u)/t always. The correct thing to say is that a = dv/dt.

E = m{v/t} * d ( assuming that u = 0)

 

E = mv {d/t} (re- arranging) --> simple commutative property

 

E = mv {v}

Now wait. The assumption behind v = d/t is that of constant velocity. The correct thing to say is that v = dd/dt - that velocity is the derivative of distance over time. You are making the assumption of no acceleration at all - since if you have a constant velocity, the acceleration must be 0. Combined with your earlier assumption that the velocity at t = 0 is 0, you are making the assumption that the velocity is constantly 0.

E = mv² (Kinetic energy relation is disproved)[/QUOTe]

Note that this is true, as long as v = 0 - which your derivation assumes.

=Uncool-

Edited July 28, 2012 by uncool

[Sciencegod]

Thanks for taking the time to correct me on that,

 

I would be very happy to know your views on the attachment which i have produced in support of my argument, If you could please kindly read and express your sincere thoughts on that, it would definitely be of great help.

 

I would be very happy, if you could point out where i have gone wrong in my analysis for the same!

 

 

 

So here, your assumption is constant acceleration, correct? That is the only way that a = (v - u)/t always. The correct thing to say is that a = dv/dt.

 

Now wait. The assumption behind $v = d/t$ is that of constant velocity. The correct thing to say is that v = dd/dt - that velocity is the derivative of distance over time. You are making the assumption of no acceleration at all - since if you have a constant velocity, the acceleration must be 0. Combined with your earlier assumption that the velocity at t = 0 is 0, you are making the assumption that the velocity is constantly 0.

 

Note that this is true, as long as v = 0 - which your derivation assumes.

=Uncool-

[studiot]

A more productive mindset from you point of view would be

 

"Why am I wrong and Science correct?" , rather than the other way round.

 

You could then move forward from something as basic and uncontroversial as this to something where there is a genuine possibility of spotting something new.

 

The problem with your pendulum analysis is quite simple.

 

The velocity of the weight varies throughout its cycle.

 

At A and B (top of the swing) the velocity is zero, as is the KE.

the energy is all potential energy.

 

At C the velocity is a maximum and the energy is all KE. The potential energy is zero.

 

As the pendulum swings from the top of the swing towards the bottom,

 

PE decreases and KE increases with the total remaining constant.

 

As the pendulum swings from the bottom back towards the top

 

KE decreases and PE increases with the total remaining constant.

Edited July 28, 2012 by studiot

[Sciencegod]

HI Friend,

 

You basically didn't answer my question?

what you have expressed is a standard answer which is known to every body including me?

 

what i am specifically interested to ask you is that, when the pendulum is released from the point A, it accelerates and thereby the velocity increases and is maximum at point C, well taken, but what happens there after is of particular interest to me and for all,

 

From the point C the pendulum is still in motion and we know that motion is associated with kinetic energy, therefore, some energy is consumed to take the pendulum from the point C to point B, so my basic question is, why then it is not added to the Kinetic energy term?

 

To give you another example, lets take a tennis ball in our hand, throw it straight up into the air, What do you think really happens?

The ball which was at rest in your hand slowly accelerates and then reaches the maximum velocity and thereafter it decelerates and reaches the point of rest, please note, all this while the ball is still in motion and therefore the energy associated should be Kinetic in nature. the point at which it attains rest is the point of PE? Therefore, didn't we account a two way process in this action? if we did, then the KE term will become (mv²) and PE term would become (2mgh)?

 

The basic argument is about the energy associated with the pendulum at the point 'A' and in this context i have already given the example of a running man in my document?

 

Energy is consumed for accelerating as well as decelerating? it's a two way process? Don't you think so?

Therefore, one has to account for this? if we account this energy, then the energy relation doubles as described in my document?

This is what is wrong in our science, very very elementary, and as a result has huge consequences for science? I hope you can imagine it too!

 

Thanks!

 

 

A more productive mindset from you point of view would be

 

"Why am I wrong and Science correct?" , rather than the other way round.

 

You could then move forward from something as basic and uncontroversial as this to something where there is a genuine possibility of spotting something new.

 

The problem with your pendulum analysis is quite simple.

 

The velocity of the weight varies throughout its cycle.

 

At A and B (top of the swing) the velocity is zero, as is the KE.

the energy is all potential energy.

 

At C the velocity is a maximum and the energy is all KE. The potential energy is zero.

 

As the pendulum swings from the top of the swing towards the bottom,

 

PE decreases and KE increases with the total remaining constant.

 

As the pendulum swings from the bottom back towards the top

 

KE decreases and PE increases with the total remaining constant.

Edited July 29, 2012 by Sciencegod

[Klaynos]

The ball at all times has both gravitational potential energy AND kinetic energy associated with it. It is not a one or the other situation. As the ball moves upwards the KE is converted into PE, the reverse happens on the way down.

[swansont]

To give you another example, lets take a tennis ball in our hand, throw it straight up into the air, What do you think really happens?

The ball which was at rest in your hand slowly accelerates and then reaches the maximum velocity and thereafter it decelerates and reaches the point of rest, please note, all this while the ball is still in motion and therefore the energy associated should be Kinetic in nature. the point at which it attains rest is the point of PE? Therefore, didn't we account a two way process in this action? if we did, then the KE term will become (mv²) and PE term would become (2mgh)?

 

We know the PE is not 2mgh, because the work done by gravity is mgh. There is PE associated with the ball at any point that h ≠ 0, and as Klaynos has pointed out, KE + PE will be a constant over the entire trajectory.

 

KE ≠ mv^2

You had a math error in your derivation. The physics is not wrong.