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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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 Posted: Mon Jul 03, 2006 1:40 pm Post subject:
Total energy: E = gamma*m*c^2 (including potential energy)
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I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct. The additional energy must be included
in the mass: m = m_0 + E_pot/c^2. The problem is, that in many cases this
will make the mass depend on other parameters, like for example time and
position.
Is this correct?
PC
__________________________________
Worth to remember: (
In your own frame things are greatest (Length
Contraction) and fastest (Time Dilation)... |
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PD
science forum Guru
Joined: 03 May 2005
Posts: 4363
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| Posted: Mon Jul 03, 2006 6:12 pm Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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Peter Christensen wrote:
| Quote: | I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct. The additional energy must be included
in the mass: m = m_0 + E_pot/c^2. The problem is, that in many cases this
will make the mass depend on other parameters, like for example time and
position.
Is this correct?
|
No. For example potential energy may have nothing to do with mass.
(Electrostatic potential energy, for example.) So there is no reason to
assume that gamma should apply to it.
There are a *bunch* of contributions to energy:
E = rest energy + linear kinetic energy + rotational kinetic energy +
thermal energy + potential energy + ....
PD |
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koobee.wublee@gmail.com
science forum Guru Wannabe
Joined: 01 Feb 2006
Posts: 141
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| Posted: Tue Jul 04, 2006 4:55 am Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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"Peter Christensen" <PeCh@MailAPS.org> wrote in message
news:44a91e44$0$15783$14726298@news.sunsite.dk...
| Quote: | I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct. The additional energy must be
included in the mass: m = m_0 + E_pot/c^2. The problem is, that in many
cases this will make the mass depend on other parameters, like for example
time and position.
Is this correct?
|
No, this is not correct under the concept of GR. In the concept of
spacetime which GR is just a special case, we have the following
spacetime
ds^2 = g_ij dq^i dq^j
Where q^0 = c dt, q^1 = x, q^2 = y, q^3 = z
Dividing both sides by ds^2, the Lagrangian dealing with the geodesics
is
L = g_ij (dq^i/ds) (dq^j/ds) = 1
For the state variable q^0 (time), the associated Euler-Lagrange
Equation is
d[2 g_0i (dq^i/ds)]/ds = 0
Perform the very simple integration, you get
g_0i (dq^i/ds) = k
Where k = the integration constant
By multiplying both sides with (ds/dq^0), the above equation becomes
ds/dq^0 = g_0i (dq^i/dq^0) / k
Now, the spacetime in general of the very first equation can also be
arranged as follows after dividing both sides by (g_00 (dq^0)^2). In
which,
we have
(ds/dq^0)^2 / g_00 = (g_ij / g_00) (dq^i/dq^0) (dq^j/dq^0)
Merging the two immediate equations above, we get
(g_0i (dq^i/dq^0))^2 / g_00 = k^2 (g_ij / g_00) (dq^i/dq^0) (dq^j/dq^0)
After simple re-arranging the equation above, we get
k^2 = ((g_0i (dq^i/dq^0))^2 / g_00) / ((g_ij / g_00) (dq^i/dq^0)
(dq^j/dq^0))
The equation above has i, j go from 0 to 3. Writing them as from 1 to
3, we
get
k = (g_0i (dq^i/dq^0)) sqrt(g_00 / (1 + 2 (g_0i / g_00) (dq^i/dq^0) +
(g_ij
/ g_00) (dq^i/dq^0) (dq^j/dq^0)))
If we have
** k = E / (m c^2)
** g_ij = diagonal metric where g_ij = 1 if i = j, g_ij = 0 if i <> j
We get the following energy equation in general.
E = m c^2 sqrt(g_00 / (1 + (g_ij / g_00) (dq^i/dq^0)^2))
Using Schwarzschild metric which is a special case to GR and GR being a
special case to the concept of spacetime in general, we have
** g_00 = 1 - 2 U
** g_11 = 1 / (1 - 2 U)
** g_22 = r^2 cos(F)
** g_33 = r^2
** q^1 = r = altitude from center of M
** q^2 = H = longitude
** q^3 = F = lattitude
** U = G M / r / c^2
The energy equation becomes
E = m c^2 sqrt((1 - 2 U) / (1 - B^2)), where
** B^2 c^2 = (dr/dt)^2 / (1 - 2 U)^2 + r^2 cos^2(F) (dH/dt)^2 / (1 - 2
U) + r^2 (dF/dt)^2 / (1 - 2 U)
Under very weak graviational field and very low speed, we have
** 1 >> 2 U
** 1 >> B^2
The energy equation becomes
E = m c^2 - m U c^2 + m B^2 c^2 / 2
You understand
** m c^2 = energy of rest mass
** m U c^2 = G M m / r = - potential energy
** m B^2 c^2 / 2 = m v^2 / 2 = kinetic energy
So, under the concept of GR, there is no such thing as kinetic energy
or potential energy. The general case of spacetime also demands the
conservation of energy. Since GR is just a special case to the general
concept of spacetime, energy must also be conserved. GR has been
misinterpreted for almost 100 years. If it can be misinterpreted, how
about the foundation that builds up GR? It is faulty right from the
very basis. |
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Bilge
science forum Guru
Joined: 30 Apr 2005
Posts: 2816
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| Posted: Wed Jul 05, 2006 8:00 am Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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Peter Christensen:
| Quote: | I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct.
|
Neither is correct. Potential energy is inherently a non-relativistic
concept, introduced out of neccessity in newtonian mechanics in order
to ``explain'' forces. In relativity, one must deal directly with the
fundamental interactions. In order to utilize a potential energy in
relativistic phenomena, one must choose a frame and make a non-relativistic
reduction in that frame. The problem is that a potential energy function
only contain part of the interaction. For example, the coulomb potential
energy is given by e\phi = e^2/r, whereas the fully relativistic coulomb
interaction contains the vector potential and electromagnetic current
density as well: e\phi + J.A. One can _reduce_ this to a non-relativistic
coulomb potential energy provided one accepts certain limitations on
the validity of the result. However, simply transforming the potential
energy does not give the correct interaction in a different frame. |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
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| Posted: Wed Jul 05, 2006 4:20 pm Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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"Bilge" <dubious@radioactivex.lebesque-al.net> skrev i en meddelelse
news:slrnean0eh.7r.dubious@radioactivex.lebesque-al.net...
| Quote: | Peter Christensen:
I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct.
Neither is correct. Potential energy is inherently a non-relativistic
concept, introduced out of neccessity in newtonian mechanics in order
to ``explain'' forces. In relativity, one must deal directly with the
fundamental interactions. In order to utilize a potential energy in
relativistic phenomena, one must choose a frame and make a
non-relativistic
reduction in that frame. The problem is that a potential energy function
only contain part of the interaction. For example, the coulomb potential
energy is given by e\phi = e^2/r, whereas the fully relativistic coulomb
interaction contains the vector potential and electromagnetic current
density as well: e\phi + J.A. One can _reduce_ this to a non-relativistic
coulomb potential energy provided one accepts certain limitations on
the validity of the result. However, simply transforming the potential
energy does not give the correct interaction in a different frame.
|
Thanks for the reply.
Can it be done within SR, or do I have to use GR? I mean, is it possible to
describe the force acting on the object with a force 4-vector or something
like this? I'm just thinking about a static electric potential, for example.
PC |
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Bilge
science forum Guru
Joined: 30 Apr 2005
Posts: 2816
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| Posted: Thu Jul 06, 2006 5:54 am Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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Peter Christensen:
| Quote: | Can it be done within SR, or do I have to use GR? I mean, is it possible to
describe the force acting on the object with a force 4-vector or something
like this? I'm just thinking about a static electric potential, for example.
|
Yes, you can write a four-vector version of the lorentz-force,
F = (qE + v x B). It's given in terms of the faraday tensor:
dp^u / d\tau = q F^uv U_v
where, p^u is the four-momentum, \tau is the proper time, U_v is
the four-velocity, and F^uv = d^u A^v - d^v A^u is the faraday
tensor. I didn't know that you were familir with four-vectors,
so I left out another detail in my previous reply. The electro-
magnetic field, A^u, has only two degrees of freedom and so
two of the four components are redundant and correspond to the
gauge freedom of the field. In order to transform the field
properly, you must perform both a lorentz transform _and_
a gauge transform to remove the additional degrees of freedom.
Hence, a lorentz transform is not enough. |
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Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
|
| Posted: Fri Jul 07, 2006 9:51 am Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
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Thanks for the long and thorough reply. I understand, that both the concepts
kinetic energy and potential energy are not very usefull when dealing with
the relativictic model. -The correct one.
I was just trying to study an experiment with a pendulum, in either a
gravitational or an electric field. I assume, that I must just use an
applied force, and not try to define potential and kinetic energies. (I
knew, that 'kinetic energy' isn't used in relativity.)
It turned out to be much more complicated, than I thought.
Rgds,
PC
"Koobee Wublee" <koobee.wublee@gmail.com> skrev i en meddelelse
news:1151988903.917000.82450@b68g2000cwa.googlegroups.com...
| Quote: | "Peter Christensen" <PeCh@MailAPS.org> wrote in message
news:44a91e44$0$15783$14726298@news.sunsite.dk...
I tried two different ways to include a potential energy (E_pot) in the
total energy:
1) E = gamma*(m_0+E_pot/c^2)*c^2 = gamma*m_0*c^2 + gamma*E_pot
2) E = gamma*m_0*c^2 + E_pot
I believe that number one is correct. The additional energy must be
included in the mass: m = m_0 + E_pot/c^2. The problem is, that in many
cases this will make the mass depend on other parameters, like for
example
time and position.
Is this correct?
No, this is not correct under the concept of GR. In the concept of
spacetime which GR is just a special case, we have the following
spacetime
ds^2 = g_ij dq^i dq^j
Where q^0 = c dt, q^1 = x, q^2 = y, q^3 = z
Dividing both sides by ds^2, the Lagrangian dealing with the geodesics
is
L = g_ij (dq^i/ds) (dq^j/ds) = 1
For the state variable q^0 (time), the associated Euler-Lagrange
Equation is
d[2 g_0i (dq^i/ds)]/ds = 0
Perform the very simple integration, you get
g_0i (dq^i/ds) = k
Where k = the integration constant
By multiplying both sides with (ds/dq^0), the above equation becomes
ds/dq^0 = g_0i (dq^i/dq^0) / k
Now, the spacetime in general of the very first equation can also be
arranged as follows after dividing both sides by (g_00 (dq^0)^2). In
which,
we have
(ds/dq^0)^2 / g_00 = (g_ij / g_00) (dq^i/dq^0) (dq^j/dq^0)
Merging the two immediate equations above, we get
(g_0i (dq^i/dq^0))^2 / g_00 = k^2 (g_ij / g_00) (dq^i/dq^0) (dq^j/dq^0)
After simple re-arranging the equation above, we get
k^2 = ((g_0i (dq^i/dq^0))^2 / g_00) / ((g_ij / g_00) (dq^i/dq^0)
(dq^j/dq^0))
The equation above has i, j go from 0 to 3. Writing them as from 1 to
3, we
get
k = (g_0i (dq^i/dq^0)) sqrt(g_00 / (1 + 2 (g_0i / g_00) (dq^i/dq^0) +
(g_ij
/ g_00) (dq^i/dq^0) (dq^j/dq^0)))
If we have
** k = E / (m c^2)
** g_ij = diagonal metric where g_ij = 1 if i = j, g_ij = 0 if i <> j
We get the following energy equation in general.
E = m c^2 sqrt(g_00 / (1 + (g_ij / g_00) (dq^i/dq^0)^2))
Using Schwarzschild metric which is a special case to GR and GR being a
special case to the concept of spacetime in general, we have
** g_00 = 1 - 2 U
** g_11 = 1 / (1 - 2 U)
** g_22 = r^2 cos(F)
** g_33 = r^2
** q^1 = r = altitude from center of M
** q^2 = H = longitude
** q^3 = F = lattitude
** U = G M / r / c^2
The energy equation becomes
E = m c^2 sqrt((1 - 2 U) / (1 - B^2)), where
** B^2 c^2 = (dr/dt)^2 / (1 - 2 U)^2 + r^2 cos^2(F) (dH/dt)^2 / (1 - 2
U) + r^2 (dF/dt)^2 / (1 - 2 U)
Under very weak graviational field and very low speed, we have
** 1 >> 2 U
** 1 >> B^2
The energy equation becomes
E = m c^2 - m U c^2 + m B^2 c^2 / 2
You understand
** m c^2 = energy of rest mass
** m U c^2 = G M m / r = - potential energy
** m B^2 c^2 / 2 = m v^2 / 2 = kinetic energy
So, under the concept of GR, there is no such thing as kinetic energy
or potential energy. The general case of spacetime also demands the
conservation of energy. Since GR is just a special case to the general
concept of spacetime, energy must also be conserved. GR has been
misinterpreted for almost 100 years. If it can be misinterpreted, how
about the foundation that builds up GR? It is faulty right from the
very basis. |
|
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| Back to top |
|
|
Peter Christensen
science forum Guru Wannabe
Joined: 02 Jan 2006
Posts: 130
|
| Posted: Sun Jul 16, 2006 1:08 pm Post subject:
Re: Total energy: E = gamma*m*c^2 (including potential energy)
|
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|
Bilge skrev:
| Quote: | Peter Christensen:
Can it be done within SR, or do I have to use GR? I mean, is it possible to
describe the force acting on the object with a force 4-vector or something
like this? I'm just thinking about a static electric potential, for example.
Yes, you can write a four-vector version of the lorentz-force,
F = (qE + v x B). It's given in terms of the faraday tensor:
dp^u / d\tau = q F^uv U_v
where, p^u is the four-momentum, \tau is the proper time, U_v is
the four-velocity, and F^uv = d^u A^v - d^v A^u is the faraday
tensor. I didn't know that you were familir with four-vectors,
so I left out another detail in my previous reply. The electro-
magnetic field, A^u, has only two degrees of freedom and so
two of the four components are redundant and correspond to the
gauge freedom of the field. In order to transform the field
properly, you must perform both a lorentz transform _and_
a gauge transform to remove the additional degrees of freedom.
Hence, a lorentz transform is not enough.
|
It's a bit strange (maybe), that I can't use the Lagrangian or the
Hamiltonian, because they are defined by kinetic energy (T) and
potential energy (V):
H = T + V
L = T - V
But I understand, that that's just how it is...
PC |
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