How many ways there are to conclude that Eo=mc^2 ?

[Herman Trivilino]

Yes. So it's a little puzzling why SR forces a tighter connection between m and E. Or actually, since "mass" is really nothing more than a parameter in the expressions for energy and momentum, it could be phrased this way: SR forces a tighter connection between E and p than exists classically. Classically, you can change the energy of a moving object (by heating it, for example) without affecting its momentum, but not relativistically.

Isn't it because the connection between space and time is "tighter" in SR than it is in Newtonian physics?

 

[DrStupid]

but if it is assumed that A(v) = 0

Than you don't conclude but only assume that Eo=mc². You could also assume that Eo is zero or infinit or something between.

 

[Herman Trivilino]

Than you don't conclude but only assume that Eo=mc². You could also assume that Eo is zero or infinit or something between.

The problem with anything other than $mc^2$ is that it doesn't match what's observed in nature.

 

[sami_m]

Okay, but I would put it the other way around: If the derivation is valid, then g(0) \neq 0
Why would you want to insist that?

i think, this to be complete derivation, the case g(0) = 0 must be included there also.

For example if g(v)m = B(v) \frac{1}{2} mv^2 (g(v)m is a product of classical kinetic energy and B(v), that is some function of v)

,then g(0) = 0 but if B(v) \neq 0, then g(v) \neq 0 , when v \neq 0
, and there may be B(v) that is nonzero.

and now the derivation for g(v) does not work. Something may be wrong in the setup or in the equations in this case, instead of that there exist no nonzero B(v).

One possibility that comes to mind is that the observation that there is only one particle M after the collision is wrong observation.

Example: if you look how electron and positron form positronium, the result is not a single positronium, it is i think positronium + 2 photons. (electron and positron both emit photon in the process.)

 

[DrStupid]

The problem with anything other than $mc^2$ is that it doesn't match what's observed in nature.

It would be OK, if sami_m would consider these observations in his derivation even though the result would be an empirical formula. Einstein derived it theoretically and predicted corresponding experimental observations.

 

[sami_m]

It would be OK, if sami_m would consider these observations in his derivation even though the result would be an empirical formula. Einstein derived it theoretically and predicted corresponding experimental observations.

I list here how i can evaluate the term A(v) based on my knowledge:

I think A(v) is nonzero is interesting possibility to think about.

1. Observations tell that the momentum has equation p = \gamma mv and kinetic energy E_{kin} = mc^2(\gamma - 1)

1.b But i think this does not have effect on that there may be also nonzero A(v) -term in the equation
W = E_{kin}(v) = [\gamma mc^2 + A(v)] - [mc^2 + A(v)]

2. In high energy particle collisions, it is possible that A(v) approaches zero when v approaches c (And this may have been verified by the observations), but A(v) may be nonzero in low velocities.

3. both terms in the kinetic energy/work equation should be positive or zero.

from 1,2,3, what i can say about A(v) is that if it is nonzero, then it must be at least that A(c) = 0. And the minimum of A(v) is -mc^2 if both energy terms in the equation are positive or zero. the last term disappears when A has minimum. It is likely that A is more or less well behaving increasing function of v. IF the minimum when v=0, then the last term in work equation totally disappears when v=0.

 

[stevendaryl]

i think, this to be complete derivation, the case g(0) = 0 must be included there also.

I don't think so. The assumption that E = m g(v) implies that g(0) = c^2. The other possibility, which is realized classically, is that E is not a function of m and v, that mass and velocity don't uniquely determine the energy.

 

[DrStupid]

2. In high energy particle collisions, it is possible that A(v) approaches zero when v approaches c (And this may have been verified by the observations), but A(v) may be nonzero in low velocities.

That would be a convincing argument. Can you give me a corresponding example and show that the experimental observations require A(v)=0?

 

[sami_m]

I don't think so. The assumption that E = m g(v) implies that g(0) = c^2. The other possibility, which is realized classically, is that E is not a function of m and v, that mass and velocity don't uniquely determine the energy.

thinking about it, the possibility that if g(0) is allowed to be 0, it may be that mg(v) then is not anymore inner property of the particle -it doesn't describe the inner energy of the particle anymore.

But then mg(v) stil might describe total energy that is associated to kinetic energy and mass energy:

Ekin + Emass = mg(v)

now Emass is not constant anymore, it is also function of velocity. Also it would be zero when v=0. but Emass may approach still mc^2 in higher velocities, but it does not in low velocities.

 

[sami_m]

That would be a convincing argument. Can you give me a corresponding example and show that the experimental observations require A(v)=0?

for seeking if A(c) = 0, or A(v) = 0 i guess it can be done by seeking the maximum of the released total energy per one collision ?

 

[sami_m]

so the relativistic form of the energy would be E= \gamma mc^2 +A(v) = \sqrt{p^2c^2 + m^2c^4 }+A(v) , but nobody talks about possible term A(v).
I am not sure about this , but E may have at least observed to have maximum:
E \to \sqrt{p^2c^2 + m^2c^4} , when v \to c
but there may be room to speculate that A(v) is not zero in low velocities.(?)

 

[Herman Trivilino]

I am not sure about this , but E may have at least observed to have maximum:
E \to \sqrt{p^2c^2 + m^2c^4} , when v \to c

As $v \to c$ the total energy increases beyond all bounds. There is no maximum.

 

[sami_m]

As $v \to c$ the total energy increases beyond all bounds. There is no maximum.

yes,that is true i said wrongly - i mean in high energy collisions it may be possible to see at least:
-whether E_{tot} \to \sqrt { p^2c^2 + m^2c^4} when v \to c ( E_{tot} approach asymptotically this Energy formula) and
-whether E_{mass} = E_{tot} - E_{kin} \to mc^2 when v \to c ( whether E_{mass} has maximum mc^2, when v \to c )

but it may be less clear whether this A(v) \neq 0 when v << c ,what would mean also that E_{mass} = mc^2 + A(v) \neq mc^2 when v &lt;&lt; c

The equation for work or kinetic energy from the work integral was:
W = [ m\gamma c^2 + A(v) ] - [mc^2 + A(v) ]

where this A(v) is usually thought to be zero. But i don't know how to prove it.

 

[Herman Trivilino]

i mean in high energy collisions it may be possible to see at least:
-whether E_{tot} \to \sqrt { p^2c^2 + m^2c^4} when v \to c ( E_{tot} approach asymptotically this Energy formula)

There's no asymptote, either. The total energy of an object increases beyond all bounds as its speed relative to the observer approaches $c$.

The equation for work or kinetic energy from the work integral was:
W = [ m\gamma c^2 + A(v) ] - [mc^2 + A(v) ]

The work-energy theorem is a dynamical relation, valid for objects that cannot posses internal energy, for example, a point particle. If you see it introduced as anything more than that, it presents work as something that cannot be generalized to thermodynamic work. This is why you see introductory physics textbooks changing the way the work-energy theorem is treated in dynamics. They want a definition of work that can be generalized to include not just dynamical considerations, but also thermodynamics.

where this A(v) is usually thought to be zero. But i don't know how to prove it.

I don't think it can be "proven". A proof is something that follows from premises, so in that sense there is no information in the end result of a proof that wasn't already present in the premises. To me, the more important consideration is evidence to support it. Can you think of any experiment that could be done to detect the presence of a nonzero value for $A(v)$?