E=mc^2 wrong?

[Carl Fischbach]

Iv'e done some calculations and found a potential error in E=mc^2.

 

 

Lets take an object with the mass of 1 kg and accelerate it to .866% of

c where gamma equals 2, the total energy or mass of this object would now be 2 kg in accordance with relativity.

 

Now assuming that the 1kg mass is moving in a striaght line at .866% of c in the +x direction,at a constant velocity, a constant force in the +y direction

is applied to the 1 kg moving mass.

 

According to E=mc^2 any small change in the velocity in any direction

of the 1 kg mass moving at .866% of c will appear like a mass of 8 kg's to a force acting on the moving mass. Note this force is acting on the the moving mass from a stationary position relative to the moving mass. Also the 1kg moving mass acting like a 8kg mass, is derived from the acceleration rate of the 1 kg mass acted on by a constant force, measured from a stationary position.

 

Iv'e looked up 2 different beam deflection equations at relativistic velocities and rearranged the equations and in both equations, Iv'e found that if the

equations are to work the 1kg mass, moving at .866% of c, would have to

appear like a mass of 2 kg's to a stationary deflecting force, not the 8 kg's that is required by E=mc^2. This would mean that E=mc^2 is wrong, or the beam deflection equations are wrong and could also bring into question the

valitity of shrinkage of space at relativistic velocities. Could someone

double check this work.

[ydoaPs]

E=mc² uses rest mass, and the full equation is E²=m²c⁴+p²c². E=mc² is a simplified case which assumes you are at rest with respect to the object you are studying.

[Carl Fischbach]

So your saying that if I am at a rest studying 1kg mass at rest and I increase the velocity of the object to a velocity .866% of c, and the object, which I will put a charge on for the sake argument, is meseared ,at a constant velocity of .866 of c ,to have a mass of 2 kg's,when passed through a magnetic field,then your saying that the 1 kg increase in mass of the object does not require me to do the work to the 1 kg mass of 1kg*c^2 to gain a mass of 1 kg.

[ydoaPs]

So your saying that if I am at a rest studying 1kg mass at rest and I increase the velocity of the object to a velocity .866% of c, and the object, which I will put a charge on for the sake argument, is meseared ,at a constant velocity of .866 of c ,to have a mass of 2 kg's,when passed through a magnetic field,then your saying that the 1 kg increase in mass of the object does not require me to do the work to the 1 kg mass of 1kg*c^2 to gain a mass of 1 kg.

 

No. You still use 1kg.

[Carl Fischbach]

One thing I would like to add is the beam deflection equations are in a

electric field not a magnetic since you are doing work to the object.

[swansont]

Iv'e looked up 2 different beam deflection equations at relativistic velocities and rearranged the equations and in both equations, Iv'e found that if the

equations are to work the 1kg mass, moving at .866% of c, would have to

appear like a mass of 2 kg's to a stationary deflecting force, not the 8 kg's that is required by E=mc^2. This would mean that E=mc^2 is wrong, or the beam deflection equations are wrong and could also bring into question the

valitity of shrinkage of space at relativistic velocities. Could someone

double check this work.

 

What equations did you use? Chances are good that they use the rest mass.

[darkenlighten]

Why is it that no one understands it's E=γmc^2 where γ= (1 - (v^2/c^2))^1/2. Therefore at rest v = 0 and γ = 1 and you get your famous E=mc^2

[Klaynos]

Why is it that no one understands it's E=γmc^2 where γ= (1 - (v^2/c^2))^1/2. Therefore at rest v = 0 and γ = 1 and you get your famous E=mc^2

 

Because it tends to be easier to use, this also allows for massless particles without the requirement of relativistic mass:

 

E²=m²c⁴+p²c²

 

Which again simplifies at p=0

 

Momentum, of cause, is not a Lorenz invariant.

[timo]

Why is it that no one understands it's E=γmc^2 where γ= (1 - (v^2/c^2))^1/2. Therefore at rest v = 0 and γ = 1 and you get your famous E=mc^2

Because neither of the two sets of words divided by the period form a sentence . EDIT: At 2nd though: if you wanted to say that at rest v=0 and v=1 then the 2nd one might be - but that probably wasn't your message.

[darkenlighten]

timo, not really sure what you are trying to say there. my point seemed pretty clear, except for one correction that gamma (γ) is 1/(1-(v^2/c^2))^1/2

 

And Klaynos, the poster's example was specifically about moving objects, where E=mc^2 does not give accurate results, only E=γmc^2 and p would not equal 0.

 

Edit: nvm, was wondering where the poster got 2 for gamma, but he states it wrong, he should have 86.6% not .866%

Edited November 4, 2009 by darkenlighten

[timo]

timo, not really sure what you are trying to say there. my point seemed pretty clear, except for one correction that gamma (γ) is 1/(1-(v^2/c^2))^1/2

The Gamov factor did not show up on my screen which made your statement word-salad. I guess you used some non-standard character that did not show up properly in my browser or on my computer (now is magically does). Hint: I've not seen TeX to cause any trouble, yet.

So what I meant: I found it funny that someone said "why is it that no one understands <something that does not make any sense at all>".

[Klaynos]

E²=m²c⁴+p²c²

 

Is valid for moving particles. Please reread my post you'll notice that I mentioned when v=0 to show that it reduces to the simplified case.

[Carl Fischbach]

For some reason my posts wouldn't go through so I'm testing this post here before I

write a page on E=mc^2.