Ref:
http://math.ucr.edu/home/baez/physic...y/SR/mass.html

It is commonly stated that as a particle is accelerated, it's inertia
in the direction of motion, and energy increase to infinity. It seems
to me more accurate to say that the energy of the particle increases
not to infinity, but to the energy of the Universe. If correct, is this
difference significant?

In the above website, Philip Gibbs lays out a basic introduction to
mass in GR and explains that as the energy increases, it becomes easier
to accelerate a particle in a direction other than it's accumulated
inertia. To push a particle faster, the force has to be applied in an
increasingly sharper direction. Like pushing on the point of a pin that
is getting sharper and sharper. No wonder particle acceleration is so
difficult to achieve.

It is as if the Universe is set up in such a way as to not only make it
hard to go fast, but also harder to faster.

To approach the speed of light would require that *almost* all of the
energy of the Universe would have to be invested in accelerating a
particle. The remainder would be the particle itself. The energy of the
Universe would be included in the particle which seems to step over
the boundry of what is reasonalby allowed as at the point the particle
reaches the speed of light, the particle *is* the Universe.

Interestingly, if your gathering of the energy of the Universe is less
than perfect, and you forget one particle, the situation is reversed.


James

jbaloun wrote:
Ref:
http://math.ucr.edu/home/baez/physic...y/SR/mass.html

It is commonly stated that as a particle is accelerated, it's inertia
in the direction of motion, and energy increase to infinity. It seems
to me more accurate to say that the energy of the particle increases
not to infinity, but to the energy of the Universe. If correct, is this
difference significant?

We have no hope of distinguishing a particle that has "all the energy in
the universe" from one with more energy than that.


In the above website, Philip Gibbs lays out a basic introduction to
mass in GR and explains that as the energy increases, it becomes easier
to accelerate a particle in a direction other than it's accumulated
inertia.

No. It just requires less force to accelerate a moving particle
transverse to its motion compared to the force needed for the same
acceleration along its motion (here all quantities are referred to a
single inertial frame). In any case it requires more force to attain a
given acceleration of a moving particle than of an identical particle at
rest.


To push a particle faster, the force has to be applied in an
increasingly sharper direction. Like pushing on the point of a pin that
is getting sharper and sharper.

That is a very poor description. In particular, the forward component of
a force will increase the speed of the particle, even if there are
larger components of force in other directions -- this is not a "sharper
direction".


No wonder particle acceleration is so
difficult to achieve.

Well, it is, but not for the reasons you cite.


It is as if the Universe is set up in such a way as to not only make it
hard to go fast, but also harder to go faster.

Yes, the world we inhabit does indeed behave that way.


To approach the speed of light would require that *almost* all of the
energy of the Universe would have to be invested in accelerating a
particle.

Not true. Not even close. At Fermilab, SLAC and CERN (the 3
highest-energy particle accelerators in the world), particles routinely
travel at 0.999999 c relative to the lab. And they do 10^13 or more
particles in one bunch.


Tom Roberts

"Tom Roberts" wrote in message
...
| jbaloun wrote:
| Ref:
| http://math.ucr.edu/home/baez/physic...y/SR/mass.html
|
| It is commonly stated that as a particle is accelerated, it's
inertia
| in the direction of motion, and energy increase to infinity. It
seems
| to me more accurate to say that the energy of the particle increases
| not to infinity, but to the energy of the Universe. If correct, is
this
| difference significant?
|
| We have no hope of distinguishing a particle that has "all the energy
in
| the universe" from one with more energy than that.
|
|
| In the above website, Philip Gibbs lays out a basic introduction to
| mass in GR and explains that as the energy increases, it becomes
easier
| to accelerate a particle in a direction other than it's accumulated
| inertia.
|
| No. It just requires less force to accelerate a moving particle
| transverse to its motion compared to the force needed for the same
| acceleration along its motion (here all quantities are referred to a
| single inertial frame). In any case it requires more force to attain a
| given acceleration of a moving particle than of an identical particle
at
| rest.
|
|
| To push a particle faster, the force has to be applied in an
| increasingly sharper direction. Like pushing on the point of a pin
that
| is getting sharper and sharper.
|
| That is a very poor description. In particular, the forward component
of
| a force will increase the speed of the particle, even if there are
| larger components of force in other directions -- this is not a
"sharper
| direction".
|
|
| No wonder particle acceleration is so
| difficult to achieve.
|
| Well, it is, but not for the reasons you cite.
|
|
| It is as if the Universe is set up in such a way as to not only make
it
| hard to go fast, but also harder to go faster.
|
| Yes, the world we inhabit does indeed behave that way.
|
|
| To approach the speed of light would require that *almost* all of
the
| energy of the Universe would have to be invested in accelerating a
| particle.
|
| Not true. Not even close. At Fermilab, SLAC and CERN (the 3
| highest-energy particle accelerators in the world), particles
routinely
| travel at 0.999999 c relative to the lab. And they do 10^13 or more
| particles in one bunch.
|
|
| Tom Roberts

Relativist's look-up table, or how to cook the books.

gamma Desired velocity
1 0.000000000000000
10 0.994987437106620
100 0.999949998749938
1000 0.999999499999875
10000 0.999999995000000
100000 0.999999999950000
1000000 0.999999999999500
10000000 0.999999999999995

Androcles

it's - its
it's - its

Autymn D. C. wrote:
it's - its
it's - its

My pet peeve too :-)

--
Jan Bielawski

Dear jbaloun:

"jbaloun" wrote in message
ps.com...

Ref:
http://math.ucr.edu/home/baez/physic...y/SR/mass.html
....
To approach the speed of light would require that *almost* all
of the
energy of the Universe would have to be invested in
accelerating a
particle. The remainder would be the particle itself.

....a good response by Tom Roberts... Just to add a little, for
"spice":
http://www.fourmilab.ch/documents/ohmygodpart.html

David A. Smith

Tom Roberts wrote:

No. It just requires less force to accelerate a moving particle
transverse to its motion compared to the force needed for the same
acceleration along its motion (here all quantities are referred to a
single inertial frame).

That's pontification galore, care to underwrite
that statement with real dynamics?

"Ken S. Tucker" wrote in message
oups.com...

Tom Roberts wrote:

No. It just requires less force to accelerate a moving particle
transverse to its motion compared to the force needed for the same
acceleration along its motion (here all quantities are referred to a
single inertial frame).

That's pontification galore, care to underwrite
that statement with real dynamics?

Piece of cake, it's known for over a century (real dynamics as discussed
around 1900) that it's harder with an electron going for example at 0.8c to
accelerate it to 0.9c in the same direction (the next step even impossible!)
than it is to accelerate the same electron transverse from 0 to 0.1c
transverse speed.

With equations it's straightforward too. Using F = d(mv)/dt , whereby m=
gamma*m_0 :
dm/dt is higher with longitudinal dv/dt than with transverse dv/dt, because
d(gamma)/dt is higher.

Harald

Ken S. Tucker wrote:
Tom Roberts wrote:
No. It just requires less force to accelerate a moving particle
transverse to its motion compared to the force needed for the same
acceleration along its motion (here all quantities are referred to a
single inertial frame).

That's pontification galore, care to underwrite
that statement with real dynamics?

The equation of motion in Minkowski spacetime for a pointlike particle
of fixed mass m under the influence of 4-force F is:
F = dP/d\tau
P = m U
where P is the particle's 4-momentum, U is its 4-velocity, and \tau is
its proper time.

Project this equation onto inertial coordinates {x,y,z,t} in which the
particle is moving along the z axis with velocity v, and the above
equation becomes:
F^x = \gamma m d^2x(t)/dt^2
F^y = \gamma m d^2y(t)/dt^2
F^z = \gamma^3 m d^2z(t)/dt^2
\gamma = 1/sqrt(1-v^2)
here {x(t),y(t),z(t),t} is the trajectory of the particle.

As \gamma 1, it clearly requires more force for a given acceleration
along the z axis than it does along either the x or y axes.

Exercise for the reader: in those coordinates there is a 4th
equation which I omitted. What is it? Why did I omit it?


This is all just elementary SR.


Tom Roberts

Tom Roberts wrote:
Ken S. Tucker wrote:
Tom Roberts wrote:
No. It just requires less force to accelerate a moving particle
transverse to its motion compared to the force needed for the same
acceleration along its motion (here all quantities are referred to a
single inertial frame).

That's pontification galore, care to underwrite
that statement with real dynamics?

The equation of motion in Minkowski spacetime for a pointlike particle
of fixed mass m under the influence of 4-force F is:
F = dP/d\tau
P = m U
where P is the particle's 4-momentum, U is its 4-velocity, and \tau is
its proper time.

Project this equation onto inertial coordinates {x,y,z,t} in which the
particle is moving along the z axis with velocity v, and the above
equation becomes:
F^x = \gamma m d^2x(t)/dt^2
F^y = \gamma m d^2y(t)/dt^2
F^z = \gamma^3 m d^2z(t)/dt^2
\gamma = 1/sqrt(1-v^2)
here {x(t),y(t),z(t),t} is the trajectory of the particle.

As \gamma 1, it clearly requires more force for a given acceleration
along the z axis than it does along either the x or y axes.

Exercise for the reader: in those coordinates there is a 4th
equation which I omitted. What is it? Why did I omit it?



Thanks for the exercise Tom!

The other equation is

F^t = \gamma/c * dE/dt = \gamma/c * (F dot u)

Where E is energy = m_0 \gamma c^2.

I think you omitted it because this component gives information about
the energy rather than the trajectory. Energy is conserved provided
the force is always normal to the motion, as in the case of a charged
particle in a pure magnetic field.

This exercise can be done perhaps more simply without defining a proper
time \tau by defining momentum in terms of the rest mass:

p = \gamma * m_0 * u

and then applying the product rule when taking the derivative:

F = dp/dt
F = \gamma * m_0 * du/dt + m_0 * u * d\gamma/dt

The second term above is in the direction of the velocity, and so thus
only contributes to the force component parallel to the velocity.

Cheers - shevek