"Max" wrote in message oups.com...
Hi,

I'm learning special relativity and am not understanding an aspect of
it relating to energy. Roughly, how some work done on a particle goes
into a mass increase and not kinetic energy, and so a final velocity
after doing work on a particle can not be found simply by using the
kinetic energy formula (1/2mv^2) of Newtonian mechanics.

In Newtonian mechanics a particle's total energy would be equal to the
sum of its kinetic and potential energies. So if a particle was
accelerated from zero to a velecity 'v' then it would have gained
energy equal to m*v*v/2. If the gain in energy 'E' was known, then the
final velocity could be found by using E=m*v*v/2 and then solving for
'v'. However, I'm not sure how to determine the final velocity using
special relativity since I know that some energy goes into a
relativistic mass increase.

It doesn't.
Have a look at
http://math.ucr.edu/home/baez/physic...y/SR/mass.html


The total relativistic energy formula I know of is
E^2 = p^2*c^2 - m^2c^4,

Make that
E^2 = p^2*c^2 + m^2c^4,


but I'm not certain if the momentum p is to be taken as

mv

or as in Newtonian mechanics or the relativistic version

mv/(1-v^2/c^2).

The momentum is defined as
p = m v / sqrt(1-v^2/c^2)



If it is the later, is the only way to solve for the velocity 'v' by
expanding the relativistic momentum in a Taylor series?

You have
E = m c^2 / sqrt(1-v^2/c^2)
and
p = m v / sqrt(1-v^2/c^2)
so if E and p are known, then
v = c^2 p/E
or in a nicer form:
v/c = (p c) / E.

As an exercise, verify that with the two expressions for
E and p, you get indeed
E^2 = p^2*c^2 + m^2c^4


I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.

Here's another couple of truly fine chapters you can have
a close look at:
http://www.physics.drexel.edu/~vogeley/SpecialRel/
specially lectures 8 and 9.


Thanks,
Max.

Enjoy :-)
[copy and followup set to sci.physics.relativity]

Dirk Vdm

Hope this helps.




AN ANALYSIS OF THE
ENERGY-MOMENTUM 4 VECTOR EQUATION
AND THE MASSLESS PARTICLE



A wise man once said, regarding theoretical physicists, "They are often
'algebraically' correct in their 'proofs', while fumbling on the
interpretation."

Einstein gives the mass of radiation as m = E/c^2, that of the photon
being
h nu/c^2.

The energy of the photon is given as E = pc = h nu.

Then there is the oft misinterpretation of the energy-momentum 4 vector
equation.

One interpretation of the said equation, E^2 = (mc^2)^2 + (pc)^2, is as
follows:

"If we set the m in the right hand first term to zero, then we get E =
pc
which we know is true. This shows that the mass of the photon is zero."

THE FALLACY:

Following is a brilliant analysis by Jim Redgewell:




EINSTEIN'S ENERGY FORMULA

Page created 9 February 2003 www.daf****.com

Many people are familiar with the formula E = mc² , but this is a
simplified version of the following formula:

E^2 = m^2 c^4 + p^2 c^2


This formula takes into account the momentum 'p'. When p = 0, then E =
mc².
-------------------------------------------------------------------------------
INTERJECTION by vergon:
It is imperative to note here that first and foremost, mathematical
expressions are required to reflect the physical conditions.
In this case there are two ways to set p to zero. One is to have the
object at rest - then as a result mc^2 is the rest energy at rest.
The other is to set p to 0 by setting the mass to 0 then both m and p
would be 0 and E would equal 0. Redgewell has obviously chosen the
former.
We must exercise caution and practice astuteness when manipulating
mathematics.
-------------------------------------------------------------------------------

PYTHAGORAS
The formula can be written differently to show that it is in fact based
upon the Pythagoras theorem:

E^2= (mc^2)^2 + (pc)^2


The next observation to be made, is that the above formula is in fact
the following two formulas applied at right angles to each other.


E = hf E = mc^2

Obviously, the next step is to prove that pc = hf. So let's start with
the formula for wavelength:


lambda = h/p


THE MATHS

Dirk Van de moortel wrote:
"Max" wrote in message oups.com...
Hi,

I'm learning special relativity and am not understanding an aspect of
it relating to energy. Roughly, how some work done on a particle goes
into a mass increase and not kinetic energy, and so a final velocity
after doing work on a particle can not be found simply by using the
kinetic energy formula (1/2mv^2) of Newtonian mechanics.

In Newtonian mechanics a particle's total energy would be equal to the
sum of its kinetic and potential energies. So if a particle was
accelerated from zero to a velecity 'v' then it would have gained
energy equal to m*v*v/2. If the gain in energy 'E' was known, then the
final velocity could be found by using E=m*v*v/2 and then solving for
'v'. However, I'm not sure how to determine the final velocity using
special relativity since I know that some energy goes into a
relativistic mass increase.

It doesn't.
Have a look at
http://math.ucr.edu/home/baez/physic...y/SR/mass.html


The total relativistic energy formula I know of is
E^2 = p^2*c^2 - m^2c^4,

Make that
E^2 = p^2*c^2 + m^2c^4,


but I'm not certain if the momentum p is to be taken as

mv

or as in Newtonian mechanics or the relativistic version

mv/(1-v^2/c^2).

The momentum is defined as
p = m v / sqrt(1-v^2/c^2)



If it is the later, is the only way to solve for the velocity 'v' by
expanding the relativistic momentum in a Taylor series?

You have
E = m c^2 / sqrt(1-v^2/c^2)
and
p = m v / sqrt(1-v^2/c^2)
so if E and p are known, then
v = c^2 p/E
or in a nicer form:
v/c = (p c) / E.

As an exercise, verify that with the two expressions for
E and p, you get indeed
E^2 = p^2*c^2 + m^2c^4


I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.

Here's another couple of truly fine chapters you can have
a close look at:
http://www.physics.drexel.edu/~vogeley/SpecialRel/
specially lectures 8 and 9.


Hi Dirk Van de moortel,

I read a lot of lecture 8 from the link you kindly gave. Am I correct
in understanding that I should use K = m(gamma-1) as the relativistic
kinetic energy and then solve for the velocity term of the gamma in
terms of K if I know what K is?

Roughly. I seek to know the velocity of a particle after it has been
subjected to a constant acceleration for a given amount of time by
using the special theory of relavitiy. If a particle initially has no
velocity and then an amount of work W is done on the particle to
accelerate it, what will be the resulting velocity in the theory of
relativity? In the past Newtonian mechanics would be used and W =
1/2mv^2 would then be solved for v, but from my understanding it won't
work using the ideas of special relativity since as the velocity of the
particle approaches the speed of light it takes greater and greater
amounts of work for a similar increase in acceleration. Is using W =
m(gamma - 1) the correct expression to use to solve for the resulting
velocity? Is it the relativistic equivalent of W = 1/2mv^2?

Thank you for your help. I truly appreciate it and my understanding has
indeed improved.


Thanks,
Max.

Enjoy :-)
[copy and followup set to sci.physics.relativity]

Dirk Vdm

wrote in message
oups.com...
Hope this helps.




AN ANALYSIS OF THE
ENERGY-MOMENTUM 4 VECTOR EQUATION
AND THE MASSLESS PARTICLE



A wise man once said, regarding theoretical physicists, "They are often
'algebraically' correct in their 'proofs', while fumbling on the
interpretation."

Yeah... pity they can't complete a square correctly.
http://www.androcles01.pwp.blueyonder.co.uk/E^2/EnergySquare.htm

Androcles.

"Max" wrote in message oups.com...

Dirk Van de moortel wrote:
"Max" wrote in message oups.com...
Hi,

I'm learning special relativity and am not understanding an aspect of
it relating to energy. Roughly, how some work done on a particle goes
into a mass increase and not kinetic energy, and so a final velocity
after doing work on a particle can not be found simply by using the
kinetic energy formula (1/2mv^2) of Newtonian mechanics.

In Newtonian mechanics a particle's total energy would be equal to the
sum of its kinetic and potential energies. So if a particle was
accelerated from zero to a velecity 'v' then it would have gained
energy equal to m*v*v/2. If the gain in energy 'E' was known, then the
final velocity could be found by using E=m*v*v/2 and then solving for
'v'. However, I'm not sure how to determine the final velocity using
special relativity since I know that some energy goes into a
relativistic mass increase.

It doesn't.
Have a look at
http://math.ucr.edu/home/baez/physic...y/SR/mass.html


The total relativistic energy formula I know of is
E^2 = p^2*c^2 - m^2c^4,

Make that
E^2 = p^2*c^2 + m^2c^4,


but I'm not certain if the momentum p is to be taken as

mv

or as in Newtonian mechanics or the relativistic version

mv/(1-v^2/c^2).

The momentum is defined as
p = m v / sqrt(1-v^2/c^2)



If it is the later, is the only way to solve for the velocity 'v' by
expanding the relativistic momentum in a Taylor series?

You have
E = m c^2 / sqrt(1-v^2/c^2)
and
p = m v / sqrt(1-v^2/c^2)
so if E and p are known, then
v = c^2 p/E
or in a nicer form:
v/c = (p c) / E.

As an exercise, verify that with the two expressions for
E and p, you get indeed
E^2 = p^2*c^2 + m^2c^4


I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.

Here's another couple of truly fine chapters you can have
a close look at:
http://www.physics.drexel.edu/~vogeley/SpecialRel/
specially lectures 8 and 9.


Hi Dirk Van de moortel,

I read a lot of lecture 8 from the link you kindly gave. Am I correct
in understanding that I should use K = m(gamma-1) as the relativistic
kinetic energy and then solve for the velocity term of the gamma in
terms of K if I know what K is?

Roughly. I seek to know the velocity of a particle after it has been
subjected to a constant acceleration for a given amount of time by
using the special theory of relavitiy. If a particle initially has no
velocity and then an amount of work W is done on the particle to
accelerate it, what will be the resulting velocity in the theory of
relativity? In the past Newtonian mechanics would be used and W =
1/2mv^2 would then be solved for v, but from my understanding it won't
work using the ideas of special relativity since as the velocity of the
particle approaches the speed of light it takes greater and greater
amounts of work for a similar increase in acceleration. Is using W =
m(gamma - 1) the correct expression to use to solve for the resulting
velocity? Is it the relativistic equivalent of W = 1/2mv^2?

Yes, replace it with
W = m ( 1/sqrt(1-v^2/c^2) - 1 )
you get
v/c = +/- sqrt( (W/m)^2 + 2 W/m ) / ( W/m + 1 )
with appropriate limits -1 and +1 as W/m -- infinity


Thank you for your help. I truly appreciate it and my understanding has
indeed improved.

Cheers,
Dirk Vdm

"Hexenmeister" wrote in message k...

wrote in message
oups.com...
Hope this helps.




AN ANALYSIS OF THE
ENERGY-MOMENTUM 4 VECTOR EQUATION
AND THE MASSLESS PARTICLE



A wise man once said, regarding theoretical physicists, "They are often
'algebraically' correct in their 'proofs', while fumbling on the
interpretation."

Yeah... pity they can't complete a square correctly.
http://www.androcles01.pwp.blueyonder.co.uk/E^2/EnergySquare.htm

Androcles.

Vergon and Androcles - the Clash of the Morons.
Fun ahaed :-)

Dirk Vdm

Max wrote:

Roughly. I seek to know the velocity of a particle after it has been
subjected to a constant acceleration for a given amount of time by
using the special theory of relavitiy.

This is known as hyberbolic motion and not surprisingly, is most
straightforwardly described by using hyberbolic trig functions.
I like the approach taken in the last section of this web page.

http://casa.colorado.edu/~ajsh/sr/wheel.html

The more general solution is given in the post below.
http://groups.google.com/group/sci.p...a?dmode=source
(note minor typo in post, v.v=-1 is correct)


---Tim Shuba---