Science > Physics > Special relativity and velocity for given total energy
| Topic: |
Science > Physics |
| User: |
"Max" |
| Date: |
18 Feb 2006 05:48:24 PM |
| Object: |
Special relativity and velocity for given total energy |
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.
The total relativistic energy formula I know of is
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).
If it is the later, is the only way to solve for the velocity 'v' by
expanding the relativistic momentum in a Taylor series?
I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.
Thanks,
Max.
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| User: "Dirk Van de moortel" |
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| Title: Re: Special relativity and velocity for given total energy |
18 Feb 2006 06:01:29 PM |
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"Max" <mondaymystery@hotmail.com> wrote in message news:1140306504.944030.173220@o13g2000cwo.googlegroups.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/physics/Relativity/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
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| User: "Max" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 12:42:39 AM |
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Dirk Van de moortel wrote:
"Max" <mondaymystery@hotmail.com> wrote in message news:1140306504.944030.173220@o13g2000cwo.googlegroups.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/physics/Relativity/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
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| User: "Dirk Van de moortel" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 05:08:48 AM |
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"Max" <mondaymystery@hotmail.com> wrote in message news:1140331359.774823.290690@g43g2000cwa.googlegroups.com...
Dirk Van de moortel wrote:
"Max" <mondaymystery@hotmail.com> wrote in message news:1140306504.944030.173220@o13g2000cwo.googlegroups.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/physics/Relativity/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
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| User: "srp" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 03:31:16 PM |
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Max a écrit :
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.
The total relativistic energy formula I know of is
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 relativistic version is
mv/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?
I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.
Thanks,
Max.
The relativistic equation can be converted to one for direct
calculation of relativistic velocity
v = c * sqrt(1 - ((m c^2)/E)^2)
mc^2 is the energy of the rest mass of the particle.
E is the total energy of the moving particle, (E - mc^2)
c is the velocity of light
and
v is the relativistic velocity you are looking for.
André Michaud
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| User: "Max" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 06:41:02 PM |
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The relativistic equation can be converted to one for direct
calculation of relativistic velocity
v = c * sqrt(1 - ((m c^2)/E)^2)
Hi,
To derive a formula for the relativistic velocity of a particle I began
with
E = m(gamma - 1)
= m(1/sqrt(1-v^2/c^2) - 1)
The end result was that
v = c*sqrt(1-(1+E/m)^(-2))
which is not the same as the one that you posted. After some
rearrangements of our equations, an inner term of mine is
1 + E/m
and yours is
1+ E/(mc^2)
I assume the difference between your result and mine is due to units.
In the forumla E = m(gamma - 1), are c=1 units being used? and if so,
how can I convert it to SI units (MKS)?
Thanks for the help!
P.S
I'm not sure why, but for various values of E, my calculator gives
the same result, which is equal to 3x10^8. Might this be due to
precision of my calculator?
Thanks again!
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| User: "Hexenmeister" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 07:42:55 PM |
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"Max" <mondaymystery@hotmail.com> wrote in message
news:1140396062.378376.209110@f14g2000cwb.googlegroups.com...
The relativistic equation can be converted to one for direct
calculation of relativistic velocity
v = c * sqrt(1 - ((m c^2)/E)^2)
Hi,
To derive a formula for the relativistic velocity of a particle I began
with
E = m(gamma - 1)
= m(1/sqrt(1-v^2/c^2) - 1)
The end result was that
v = c*sqrt(1-(1+E/m)^(-2))
which is not the same as the one that you posted. After some
rearrangements of our equations, an inner term of mine is
1 + E/m
and yours is
1+ E/(mc^2)
I assume the difference between your result and mine is due to units.
In the forumla E = m(gamma - 1), are c=1 units being used? and if so,
how can I convert it to SI units (MKS)?
Thanks for the help!
P.S
I'm not sure why, but for various values of E, my calculator gives
the same result, which is equal to 3x10^8. Might this be due to
precision of my calculator?
Thanks again!
Not exactly... it's due to the precision of the operator.
The problem you have is choosing the wrong value for c.
You see, the time for light to go from A to B is equal to the time
it takes to go from B to A.
So, for example, if it takes 1 unit of distance to go from A to B
then it takes -1 unit of distance to go from B to A, right?
So you should have used -1 instead of 1 for c. Everyone knows
-1 = 1, and since time doesn't go backwards... well... 1/1 = 1,
but you should have used c = -1/1 = -1.
Androcles.
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| User: "Max" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 10:26:45 PM |
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I have _no_ idea what you mean!
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| User: "Bilge" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 11:07:40 PM |
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Max:
I have _no_ idea what you mean!
Consider that a plus. If you ever get the idea that you do know
what he means, it will proably be too late for psychiatry to be of
any help.
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| User: "Hexenmeister" |
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| Title: Re: Special relativity and velocity for given total energy |
20 Feb 2006 09:39:11 AM |
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Repetitive areshole-up-in-the-air-five-times-a-day "Bilge the Al Qaeda spy
and
dubious islamic *****" <dubious@radioactivex.lebesque-al.net> blared in
message
news:slrndviqnd.6g.dubious@radioactivex.lebesque-al.net...
[snip crap]
You are still a snipping, arrogant, illiterate, innumerate, illogical,
***** without a scrap of logic in you, you whining little toad.
You don't have an inkling about mathematics or physics
and live in the vain hope some moron will think you are clever,
Mr SmartArse who pretends he understands relativity and
doesn't have a clue how to synchronize his watch to Cassini
time.
Modern physics:
http://www.androcles01.pwp.blueyonder.co.uk/Synchronize/Synchronize.htm
Hey dumbfuck! Do you know how to move sideways or up?
http://www.androcles01.pwp.blueyonder.co.uk/how_to3.jpg
tau = (t-vx/c²)/sqrt(1-v²/c²)
tau = (t-uy/c²)/sqrt(1-u²/c²)
tau = (t-wz/c²)/sqrt(1-w²/c²)
xi = (x-vt)/sqrt(1-v²/c²)
eta = (y-ut)/sqrt(1-u²/c²)
zeta= (z-wt)/sqrt(1-w²/c²)
Right or wrong, dumbfuck?
If one is right they all are, if one is wrong they all are,
pathetic *****.
For v = 0.866c, u = 0.866c, w = 0.866c the resultant velocity is
sqrt( 3/4 + 3/4 +3/4) = 1.5c
Right or wrong, shitforbrains?
Einstein said
eta = y,
zeta = z
because he did not know how to move sideways or up,
anencephalous cretin.
[quote]
we establish by definition that the "time" required by a crab to travel
from A to B equals the "time" it requires to travel from B to A.
[end quote]
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/
Einstein can prove nothing can go faster than a crab.
Oops!... Did I say 'a crab'? Sorry...'light'.
"In agreement with experience we further assume the quantity
2AB/(t'A-tA) = c,
to be a universal constant--the velocity of light in empty space." --
Einstein.
In agreement with experience and without any assumption,
BA = -AB,
2AB = AC,
[AB +BA]/(t'A-tA) = 0
Hence c = 0 in Einstein's math.
Observation:
http://www.britastro.org/vss/gifc/00918-ck.gif
Explanation:
http://www.ebicom.net/~rsf1/sekerin.htm (fig 3)
(Or stars explode twice in three months, which is stupid).
In agreement with experience and without any assumption,
you remain a snipping, arrogant, illiterate, innumerate, illogical,
incompetent ***** without a scrap of logic in you,
you whining little toad.
You don't have an inkling about mathematics or physics
and live in the vain hope some moron will think you are clever,
Mr SmartArse who pretends he understands physics and
doesn't have a clue how to synchronize his watch to Cassini
time.
Modern physics:
http://www.androcles01.pwp.blueyonder.co.uk/Synchronize/Synchronize.htm
*****, useless tord!
tau = (t-vx/c²)/sqrt(1-v²/c²)
tau = (t-uy/c²)/sqrt(1-u²/c²)
tau = (t-wz/c²)/sqrt(1-w²/c²)
xi = (x-vt)/sqrt(1-v²/c²)
eta = (y-ut)/sqrt(1-u²/c²)
zeta= (z-wt)/sqrt(1-w²/c²)
If one is right they all are, if one is wrong they all are.
Carry three watches or do not move sideways or ride an elevator.
Personally I prefer three witches:
Double double, toil and trouble,
Fire burn and Einstein bubble. --- Pop!
Hexenmeister.
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| User: "srp" |
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| Title: Re: Special relativity and velocity for given total energy |
19 Feb 2006 10:33:26 PM |
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Max a écrit :
The relativistic equation can be converted to one for direct
calculation of relativistic velocity
v = c * sqrt(1 - ((m c^2)/E)^2)
Hi,
To derive a formula for the relativistic velocity of a particle I began
with
E = m(gamma - 1)
= m(1/sqrt(1-v^2/c^2) - 1)
The end result was that
v = c*sqrt(1-(1+E/m)^(-2))
which is not the same as the one that you posted. After some
rearrangements of our equations, an inner term of mine is
1 + E/m
Maybe it would help if you started with the correct formula for
kinetic energy
E = mc^2(gamma - 1)
But you can derive v from gamma itself if you wish
gamma = 1/sqrt(1-(v^2/c^2))
1/gamma^2 =1-(v^2/c^2)
v^2/c^2 = 1-1/gamma^2
v^2 = c^2 (1-1/gamma^2)
v = c sqrt(1-1/gamma^2)
and since E=gamma mc^2, gamma=E/mc^2
so,
v = c sqrt(1-1/(E/mc^2))
or finally
v = c sqrt(1-(mc^2/E)^2)
and yours is
1+ E/(mc^2)
I think mine was v = c * sqrt(1 - ((m c^2)/E)^2)
I assume the difference between your result and mine is due to units.
In the forumla E = m(gamma - 1), are c=1 units being used?
No. the kinetic energy formula is as I mentionned above, and
the speed of light is not 1, but 2.99792458 E8 m/s
Don't listen to desk corner theoreticians who play with the units.
You will only end up getting mixed up. Stick to MKS values from any
trustable source (NIST for example) for all fundamental values
and you will be ok.
and if so, how can I convert it to SI units (MKS)?
It is in MKS. Energy is in joules mass is in kg, velocity is
in m/s
Thanks for the help!
P.S
I'm not sure why, but for various values of E, my calculator gives
the same result, which is equal to 3x10^8. Might this be due to
precision of my calculator?
Thanks again!
You are welcome.
André Michaud
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| User: "Max" |
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| Title: Re: Special relativity and velocity for given total energy |
20 Feb 2006 10:10:03 AM |
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srp wrote:
Maybe it would help if you started with the correct formula for
kinetic energy
E =3D mc^2(gamma - 1)
But you can derive v from gamma itself if you wish
gamma =3D 1/sqrt(1-(v^2/c^2))
1/gamma^2 =3D1-(v^2/c^2)
v^2/c^2 =3D 1-1/gamma^2
v^2 =3D c^2 (1-1/gamma^2)
v =3D c sqrt(1-1/gamma^2)
and since E=3Dgamma mc^2, gamma=3DE/mc^2
so,
v =3D c sqrt(1-1/(E/mc^2))
or finally
v =3D c sqrt(1-(mc^2/E)^2)
Thank you for writing a derivation and explaining, as it helped me
greatly. My ambition is to some day be able to even derive the
equations you used to begin with based on Einstein's theory, but using
basic equations is fine for me now I think.
No. the kinetic energy formula is as I mentionned above, and
the speed of light is not 1, but 2.99792458 E8 m/s
Don't listen to desk corner theoreticians who play with the units.
You will only end up getting mixed up. Stick to MKS values from any
trustable source (NIST for example) for all fundamental values
and you will be ok.
The different units make learning relativity for me more difficult. The
equations are simpler, but I understand less and have to memorize
conversions from c=3D1 to c=3D2.99...E8 units. I'll follow your advice
about the units. At the moment I've been learning some general
relativity where they measure mass in length! which is even more
confusing when it comes time to solve actual problems and obtain
numerical answers.
Thank you very much Andr=E9 Michaud! I truly appreciate your help and it
has done me a lot of good.
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| User: "srp" |
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| Title: Re: Special relativity and velocity for given total energy |
20 Feb 2006 11:47:56 AM |
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Max a écrit :
srp wrote:
Maybe it would help if you started with the correct formula for
kinetic energy
E = mc^2(gamma - 1)
But you can derive v from gamma itself if you wish
gamma = 1/sqrt(1-(v^2/c^2))
1/gamma^2 =1-(v^2/c^2)
v^2/c^2 = 1-1/gamma^2
v^2 = c^2 (1-1/gamma^2)
v = c sqrt(1-1/gamma^2)
and since E=gamma mc^2, gamma=E/mc^2
so,
v = c sqrt(1-1/(E/mc^2))
or finally
v = c sqrt(1-(mc^2/E)^2)
Thank you for writing a derivation and explaining, as it helped me
greatly. My ambition is to some day be able to even derive the
equations you used to begin with based on Einstein's theory, but
using basic equations is fine for me now I think.
You are proceeding the right way. But I doubt that you will get
much of anywhere if you base your research on Einstein's theories
(both of them) because both SR and GR are based on velocity
(inertial motion) and not on acceleration, which is the real
cause of the presence of energy, a conclusion easily drawn from
Maxwell.
They are closed and can lead to nothing but themselves.
GR moreover, is based on G, which is a pseudo constant in that
it is made up of variables that have constant values only within
the Solar System system while the particulars of the underlying
third law of Kepler go unnoticed when G is applied to other
circumstances.
Both of these defects explain why no progress has been made built
on these theories since they were popularized.
The real foundation is Maxwell, and through him, Coulomb, Gauss,
Ampere, Biot-Savart. The Lorentz force equation is also valuable.
Of course, Newton is a must at the general mechanics level.
Not meaning that Einstein's theories are worthless. They must
be understood. But they are dead ends.
No. the kinetic energy formula is as I mentionned above, and
the speed of light is not 1, but 2.99792458 E8 m/s
Don't listen to desk corner theoreticians who play with the units.
You will only end up getting mixed up. Stick to MKS values from any
trustable source (NIST for example) for all fundamental values
and you will be ok.
The different units make learning relativity for me more difficult. The
equations are simpler, but I understand less and have to memorize
conversions from c=1 to c=2.99...E8 units.
You'll get used to converting. You will have to.
Just keep in mind that in equations where c is made to equal 1,
it means that the real MKS value of c has been integrated into
one of the other constants or variables of the equation (if the
equation is valid, of course). It is a very constructive exercise
to try to discover where it is hiding. The same for all other
constants set to 1.
I'll follow your advice about the units. At the moment I've been
learning some general relativity where they measure mass in length!
which is even more confusing when it comes time to solve actual
problems and obtain numerical answers.
I agree. Needless hardship. Maybe one day you will be instrumental
in bringing sense back into this mess.
Thank you very much André Michaud! I truly appreciate your help and
it has done me a lot of good.
Good luck.
André Michaud
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| User: "Jim Heckman" |
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| Title: Re: Special relativity and velocity for given total energy |
22 Feb 2006 03:29:40 PM |
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On 18-Feb-2006, "Max" <mondaymystery@hotmail.com>
wrote in message <1140306504.944030.173220@o13g2000cwo.googlegroups.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.
"Mass increase" is an outdated way to describe this, and is avoided
in newer textbooks precisely to prevent this kind of misleading
analogy. It's better to think of the kinetic energy as the total
energy minus the (rest) mass energy. But kinetic energy usually
isn't a very useful concept in relativity; in most cases it's
easier to just use the total energy.
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.
The total relativistic energy formula I know of is
E^2 = p^2*c^2 - m^2c^4,
Sign error: It's
E^2 = p^2*c^2 + m^2*c^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).
Actually, the relativistic momentum is
p = m*v / sqrt(1 - (v^2/c^2))
And it is indeed the relativistic momentum that's to be taken in
E^2 = p^2*c^2 + m^2*c^4.
If it is the later, is the only way to solve for the velocity 'v' by
expanding the relativistic momentum in a Taylor series?
Solving p = m*v / sqrt(1 - (v^2/c^2)) for v gives
v = c / sqrt(1 + (m^2*c^2/p^2))
For small p, you can, if you want to, expand this in a Taylor
series to get
v = (p/m) * (1 - (1/2)*p^2/(m^2*c^2) ...)
where the first term in the expansion, p/m, is of course the Newtonian
approximation to the relativistic solution.
I will appreciate any help in regards to improving my understanding of
this problem so that I can determine the velocity.
--
Jim Heckman
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