I read that (#1) E=MC^2/ sqrt(1 - V^2/C^2) and (#2) E^2 = M^2C^4 +
P^2C^2 are equations which can be used for any objects

From (#1): E^2 = M^2C^4 / (1-V^2/C^2)


Therefore substituting M^2C^4 in both equations would give

E^2(1-V^2/C^2) = E^2 - P^2C^2

E^2 (C^2-V^2) = E^2C^2 - P^2C^4 (multiplying all by C^2)

E^2V^2 = P^2C^4

thus E = PC^2/V ?????

How come now the Mass is not needed at all to determine the Energy for
any object (since both equations are for any objects)....? Is it
because both equations are the same equation but written differently if
so how?

If P = momentum then how do you "measure" momentum for an object?
(static object = zero momentum?)

wrote in message oups.com...

I read that (#1) E=MC^2/ sqrt(1 - V^2/C^2) and (#2) E^2 = M^2C^4 +
P^2C^2 are equations which can be used for any objects

From (#1): E^2 = M^2C^4 / (1-V^2/C^2)


Therefore substituting M^2C^4 in both equations would give

E^2(1-V^2/C^2) = E^2 - P^2C^2
E^2 (C^2-V^2) = E^2C^2 - P^2C^4 (multiplying all by C^2)

E^2V^2 = P^2C^4

thus E = PC^2/V ?????


Yes, momentum is defined as
p = m v / sqrt(1-v^2/c^2)
and together with the definition of energy
E = m c^2 / sqrt(1-v^2/c^2) ,
this gives
E^2 = m^2 c^4 + p^2 c^2
and
v = c^2 p/E
or in a nicer form:
v/c = (c p)/E
or, like you had:
E = p c^2/v

How come now the Mass is not needed at all to determine the Energy for
any object (since both equations are for any objects)....? Is it
because both equations are the same equation but written differently if
so how?

When you have two equations with 3 'variables' p, E, m,
you can obtain a third equation with one of the unknows
eliminated. That's what algebra does for you and you
don't even have to pay for it.


If P = momentum then how do you "measure" momentum for an object?
(static object = zero momentum?)

measure v and calculate p.

Dirk Vdm

"Dirk Van de moortel" wrote
in message ...

wrote in message
oups.com...

I read that (#1) E=MC^2/ sqrt(1 - V^2/C^2) and (#2) E^2 = M^2C^4 +
P^2C^2 are equations which can be used for any objects

From (#1): E^2 = M^2C^4 / (1-V^2/C^2)


Therefore substituting M^2C^4 in both equations would give

E^2(1-V^2/C^2) = E^2 - P^2C^2
E^2 (C^2-V^2) = E^2C^2 - P^2C^4 (multiplying all by C^2)

E^2V^2 = P^2C^4

thus E = PC^2/V ?????


Yes, momentum is defined as
p = m v / sqrt(1-v^2/c^2)
and together with the definition of energy
E = m c^2 / sqrt(1-v^2/c^2) ,
this gives
E^2 = m^2 c^4 + p^2 c^2
and
v = c^2 p/E
or in a nicer form:
v/c = (c p)/E
or, like you had:
E = p c^2/v

How come now the Mass is not needed at all to determine the Energy for
any object (since both equations are for any objects)....? Is it
because both equations are the same equation but written differently if
so how?

When you have two equations with 3 'variables' p, E, m,
you can obtain a third equation with one of the unknows
eliminated. That's what algebra does for you and you
don't even have to pay for it.


If P = momentum then how do you "measure" momentum for an object?
(static object = zero momentum?)

measure v and calculate p.

Dirk Vdm


Hahahahaha!
Yeah, sure. Use radar on an electron. You should patent that, Dork.
Androcles.

:

How come now the Mass is not needed at all to determine the Energy for
any object (since both equations are for any objects)....? Is it
because both equations are the same equation but written differently if
so how?

You have three numbers. One is the mass, one is the momentum and
one is the total energy. The mass of an oject is not frame dependent.
It contributes an energy mc^2 to the total energy. The momentum is
due to the motion of the mass, which is frame dependent, since it
depends upon the relative velocity of the mass. The total energy
is the sum E^2 = (pc)^2 + (mc^2)^2. In other words, the mass is
what you have left after you subtract off the energy due to the
motion from the total energy.


If P = momentum then how do you "measure" momentum for an object?
(static object = zero momentum?)

One way to do it for a charged particle is to use a magnetic field, B.
The magnetic rigidity is defined as B times the radius of curvature
which is related to the momentum y B r = p/q. How you do measure some
momentum depends upon what methods you have available to use in a
particular circumstance.

there is one thing mathematical parrots
do not know :

IE
*any physics formula or equation
HAS ITS LIMITS OF VALIDATION!!*

now if you are a dumb mathematiciance
that has not a bit of physics touch
you can go on with those limits until
stupidity without even noticing it
but is you have some physics sense
(which most mathematicians do not have)
than once you approach a physics torpidity
a red light is lit-ten
but just mathematicians do not have
that red light in their property

so they can have momentum without mass

they can allow an em wave with
one cycle per one year or per
100 billion years etc

yet one why does have that red light
available then once he gets to a mathematical situation of 0/0
or infinite /infinite that red light is lighten
once you claim that a photon cannot move
at C 'because in that case its mass will
swell to infinity'
a red light must be lighten and
one must say
whats going on here??
is it possible that anything will have momentum but not mass
because of a formula??

so
may be that formula does not apply to that case???!!
and then he recalls the iron rule that
any formula or equation in physics
has its limit of validation-
so may be
that in that photon case we have that limit ??!!
(in which that formula does not apply anymore ??)
did anyone of those parrots ever just
thought (for a while )
about such possibility??

ATB
Y.Porat
---------------------

wrote:
I read that (#1) E=MC^2/ sqrt(1 - V^2/C^2) and (#2) E^2 = M^2C^4 +
P^2C^2 are equations which can be used for any objects

#1 is only valid for nonzero M; #2 remains valid even if M=0.
And all quantities must be measured relative to a given inertial frame.


thus E = PC^2/V ?????

Sure. Because in your notation, P=MV/sqrt(1-V^2/C^2). This is just algebra.


How come now the Mass is not needed at all to determine the Energy for
any object (since both equations are for any objects)....?

Because it is "hidden" in P.


If P = momentum then how do you "measure" momentum for an object?

Measure M and V. Or more directly, realize that the radius of a charged
particle in a constant magnetic field is inversely proportional to the
norm of its 3-momentum, so for particles of known charge you can measure
P directly using a bending magnet and detectors to measure its
trajectory's radius of curvature. This is quite common in particle
experiments.


Tom Roberts