mass under and over the limit

[AntonioLao]

The relativistic mass limit could be expressed as v/c = (1+m/M)(1-m/M). The lower case v is the absolute velocity, c is the speed of light, m is the rest mass, and M is the upper bound mass equivalent to Planck mass. For all practical purposes c is a constant. Likewise, the mass upper bound must also be a constant as the quantum of mass. Hence, only and v and m are variables. The product is physically defined as linear momentum. Furthermore, they are inversely proportional. One goes under the other goes over, vice versa. When v ® 0 then m ® M and when m ® 0 then v ® c.

It can be noted that the right-hand side and the left-hand side of the expression are both dimensionless. However, when both sides were multiplied by c the result is Dirac’s relativistic energy. How is it possible that by simple algebraic manipulations the product of dimensionless factor and c becomes square of energy?

[mkirkpatrick]

The relativistic mass limit could be expressed as v/c = (1+m/M)(1-m/M). The lower case v is the absolute velocity, c is the speed of light, m is the rest mass, and M is the upper bound mass equivalent to Planck mass. For all practical purposes c is a constant. Likewise, the mass upper bound must also be a constant as the quantum of mass. Hence, only and v and m are variables. The product is physically defined as linear momentum. Furthermore, they are inversely proportional. One goes under the other goes over, vice versa. When v ® 0 then m ® M and when m ® 0 then v ® c.

It can be noted that the right-hand side and the left-hand side of the expression are both dimensionless. However, when both sides were multiplied by c the result is Dirac’s relativistic energy. How is it possible that by simple algebraic manipulations the product of dimensionless factor and c becomes square of energy?

Maybe because you are manipulating figures that are not based in reality,or real time
events?

regards michael.

[theunify]

The relativistic mass limit could be expressed as v/c = (1+m/M)(1-m/M). The lower case v is the absolute velocity, c is the speed of light, m is the rest mass, and M is the upper bound mass equivalent to Planck mass. For all practical purposes c is a constant. Likewise, the mass upper bound must also be a constant as the quantum of mass. Hence, only and v and m are variables. The product is physically defined as linear momentum. Furthermore, they are inversely proportional. One goes under the other goes over, vice versa. When v ® 0 then m ® M and when m ® 0 then v ® c.

It can be noted that the right-hand side and the left-hand side of the expression are both dimensionless. However, when both sides were multiplied by c the result is Dirac’s relativistic energy. How is it possible that by simple algebraic manipulations the product of dimensionless factor and c becomes square of energy?

It is because energy equations take for granted a Universal particle just as a Mortician takes for granted that someone will die that night they are to train their next of kin.)

[AntonioLao]

Maybe because you are manipulating figures that are not based in reality,or real time events?

The ratios of real figures do give dimensionless figures.

It is because energy equations take for granted a Universal particle

Is this the God particle that many are searching?